Solutions to exercises in Morandi’s Field and Galois Theory

July 16, 2020

I. Galois Theory
Section 1. Field Extensions
Problem 1.1. Let K be a field extension of F . By defining scalar multiplication for α ∈ F and a ∈ K by
α · a = αa, the multiplication in K, show that K is an F -vector space.
This is a routine check of the vector space axioms, which all follow from the field axioms for K.

Problem 1.2. If K is a field extension of F , prove that [K : F ] = 1 is and only if K = F .

(⇒) Suppose K ⊃ F . Then there exists α ∈ K \ F . I claim that {1, α} is linearly independent. To see
this, let 0 = c1 · 1 + c2 α. If c1 = 0, then c2 = 0 as well, since α 6= 0 (because 0 ∈ F ). If c1 6= 0, then
α = −c1 /c2 ∈ F , a contradiction. We conclude that [K : F ] ≥ 2.
(⇐) If K = F , then {1} is a basis for K/F , implying that [K : F ] = 1.

Problem 1.3. Let K be a field extension of F , and let a ∈ K. Show that the evaluation map eva : F [x] → K
given by eva (f (x)) = f (a) is a ring and an F -vector space homomorphism. (Such a map is called an F -
algebra homomorphism.) Pn Pn
First we check that eva is a ring homomorphism. To see this, let f (x) = i=0 bi xi and g(x) = i
Pn i=1 ciix ,
where
Pn wei do not assume Pn that the coefficients
Pn are non-zero.
Pn Then evP a (f (x) + g(x)) = eva ( i=0 bi x +
n
i=0 ci x ) = eva ( i=0 (bi + ci )xi ) = i=0 (bi + ci )ai = i=0 bi ai + i=0 ci ai = eva (f (x)) + eva (g(x)).
P2n Pi P2n Pi
Next, eva (f (x)g(x)) = eva ( i=0 k=0 ak bi−k xi ) = i=0 k=0 ak bi−k ai = eva (f (x))eva (g(x)).
Finally, we check that eva is an F -vector space homomorphism. For this, assume that α ∈ F . Then it is
clear that eva (αf (x)) = α · eva (f (x)).

Problem 1.4. Prove Proposition 1.9: Let K be a field extension of F and let a1 , . . . , an ∈ K. Then

F [a1 , . . . , an ] = {f (a1 , . . . , an ) : f ∈ F [x1 , dots, xn ]}

and  
f (a1 , . . . , an )
F (a1 , . . . , an ) = : f, g ∈ F [x1 , . . . , xn ], g(a1 , . . . , an ) 6= 0 ,
g(a1 , . . . , an )
so F (a1 , . . . , an ) is the quotient field of F [a1 , . . . , an ].
By closure under ring operations, {f (a1 , . . . , an ) : f ∈ F [x1 , . . . , xn ]} ⊆ F [a1 , . . . , an ]. To prove the
other inclusion, it is enough to show that {f (a1 , . . . , an ) : f ∈ F [x1 , . . . , xn ]} is a ring, since F [a1 , . . . , an ] is
by definition the smallest ring containing F and {a1 , . . . , an }. This is clear, since the sum and product of
polynomials is again a polynomial.
By closure under field operations,
 
f (a1 , . . . , an )
: f, g ∈ F [x1 , . . . , xn ], g(a1 , . . . , an ) 6= 0 ⊆ F (a1 , . . . , an ).
g(a1 , . . . , an )
n o
Moreover, fg(a (a1 ,...,an )
1 ,...,an ) is a field since the sum and product of rational functions is again a rational function.
Moreover, every non-zero element is invertible, since in this case the numerator would be non-zero.

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 √ √ √ √
Problem
√ 1.5.√ Show that
√ √ Q( 5, 7)√
= Q(√5 + 7).√ √
(⊇) 5 + 7√∈ Q(√ 5, 7), so Q( 5 +√ 7) ⊆ Q( 5,√ 7).
(⊆) Let a = 5 + 7. Then 7 = (a − 5)2 = a2 − 2 5a + 5. Rearranging gives
√ a2 − 2 √ √
5= ∈ Q( 5 + 7).
2a
√ √
Similarly, 5 = (a − 7)2 = a2 − 2a 7 + 7. Rearranging, we have
√ a2 + 7 √ √
7= ∈ Q( 5 + 7).
2a
√ √ √ √
Hence, Q( 5, 7) ⊆ Q( 5 + 7).

Problem 1.6. Verify the following universal mapping property for polynomial rings:
(a) Let A be a commutative ring containing a field F . If a1 , . . . , an ∈ A, show that there is a unique F -ring
homomorphism φ : F [x1 , . . . , xn ] → A with φ(xi ) = ai for each i.
The map eva1 ,...,an : F [x1 , . . . , xn ] → A, f (x1 , . . . , xn ) 7→ f (a1 , . . . , an ) is the desired ring homomor-
phism. Properties of F -ring homomorphisms show that the map is uniquely determined by its action
on the set {x1 , . . . , xn }.
(b) Moreover, suppose that B is a commutative ring containing F , together with a function f : {x1 , . . . , xn } →
B, satisfying the following property: For any commutative ring A containing F and elements a1 , . . . , an ∈
A there is a unique F -ring homomorphism φ : B → A with φ(f (xi )) = ai . Show that B is isomorphic
to F [x1 , . . . , xn ].
Take A = B, and ai = f (xi ). Part (a) says that there is a homomorphism ψ : F [x1 , . . . , xn ] → B with
ψ(xi ) = f (xi ). On the other hand, if we take A = F [x1 , . . . , xn ] and ai = xi , then part (b) says there
is a map φ : B → F [x1 , . . . , xn ] such that φ(f (xi )) = xi . We will show that ψ and φ are inverse maps.
First, note that ψ ◦ φ(f (xi )) = ψ(xi ) = f (xi ). The uniqueness part of part (b) guarantees that ψ ◦ φ is
then the identity map. Next, φ ◦ ψ(xi ) = φ(f (xi )) = xi . The uniqueness part of part (a) then implies
that φ ◦ ψ is the identity. Hence, B ∼ = F [x1 , . . . , xn ].
Problem 1.7. Let A be a ring. If A is also an F -vector space and α(ab) = (αa)b = a(αb) for all α ∈ F and
a, b ∈ A, then A is said to be an F -algebra. If A is an F -algebra, show that A contains an isomorphic copy
of F . Also show that if K is a field extension of F , then K is an F -algebra.
Consider the map i : F → A, α → α · 1. This is clearly an embedding of F into A. If K is an extension
of F , then K is an F -algebra by endowing it with field multiplication.

Problem 1.8. Let K = F (a) be a finite extension of F . For α ∈ K, let Lα be the map from K to K
defined by Lα (x) = αx. Show that Lα is an F -linear transformation. Also show that det(xI − La ) is the
minimal polynomial min(F, a) of a. For which α ∈ K is det(xI − Lα ) = min(F, α)?
Lα is clearly F -linear. Now, suppose that [K : F ] = n. Proposition 1.14 tells us that {1, a, a2 , . . . , an−1 }
is a basis for K/F and that the degree of min(F, a) is n. In this basis, La is represented by
 
0 0 0 ··· 0 −a0
1 0 0 · · · 0 −a1 
 
0 1 0 · · · 0 −a2 
 ,
 .. .. .. . . .. .. 
. . . . . . 
0 0 0 ··· 1 −an−1

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where min(F, a) = xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 . Then
x 0 0 ··· 0 a0
−1 x 0 ··· 0 a1
det(xI − La ) = 0 −1 x ··· 0 a2 = xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 = min(F, a)
.. .. .. .. .. ..
. . . . . .
0 0 0 ··· −1 x + an−1
as desired.
Note that for α ∈ K, the degree of det(xI − Lα ) is n. So det(xI − Lα ) = min(F, α) is and only if the
degree of min(F, α) is n, which occurs if and only if K = F (α).

Problem 1.9. If K is an extension of F such that [K : F ] is prime, show that there are no interme-
diate fields between K and F .
Let F ⊆ L ⊆ K. Then [K : F ] = [K : L][L : F ] by Proposition 1.20. Since [K : F ] is prime, either
[K : L] or [L : F ] equals 1. By problem 1.2, either L = K or L = F .

Problem 1.10. If K is a field extension of F and if a ∈ K such that [F (a) : F ] is odd, show that
F (a) = F (a2 ). Give an example to show that this can be false if the degree of F (a) over F is even.
Note that [F (a) : F ] = [F (a) : F (a2 )][F (a2 ) : F ]. Moreover F (a) = F (a2 )(a) and a satisfies x2 − a2
over F (a2 ). Hence [F (a) : F (a2 )] equals 1 or 2. Since [F (a) : F ] is odd, we must have [F (a) : F (a2 )] = 1,
implying F (a) =√F (a2 ). √ √
Consider √Q( 4 2)/Q.√ Note that [Q( 4 2) : Q] = 4, since √ min(Q, 4√2) = x4 − 2 by Eisenstein’s
√ criterion.
√
However, Q( 4 2) 6= Q( 2). To see this, we will show √ that 4
2 ∈
/
√ Q( 2). Assume not,
√ then 4
2 = a+b 2
for some a, b ∈ Q. Squaring both sides, we have √ 2 =√ a2 + 2ab 2 + 2b2 ⇒ (1 − 2ab) 2 = a2 + 2b2 . Then
2 2 4
a + 2b = 0, implying a = b = 0. But then 2 = a + b 2 = 0, a contradiction.

Problem 1.11. If K is an algebraic extension of F and if R is a subring of K with F ⊆ R ⊆ K,

show that R is a field.
Let r ∈ R; we want to show that r is invertible. Since K/F is algebraic, r satisfies a minimal, monic
polynomial with coefficients in F , say
rn + an−1 rn−1 + · · · + a0 = 0.
Note that a0 6= 0, otherwise the minimal polynomial would not be irreducible. Then
 
1
r − (an−1 rn−2 + · · · + a1 ) = 1.
a0
Hence, r is invertible, and R is a field.
√ √
Problem 1.12. Show that Q( 2) and Q( 3) are not isomorphic as fields but are isomorphic as vector
spaces over Q. √ √ √
Suppose that φ : Q( 2)√→ Q( 3) is a field √ isomorphism,
√ and let φ(√ 2) = a. √Then a2 = φ(2) =
√ a = ± 2, implying
φ(1) + φ(1) = 2. Then that√ 2 ∈ Q( 3). Suppose that 2 = a + b 3 forp some a, b ∈ Q.
Then 2 = a2 +2ab 3+3b2 ⇒ √ 2−a2 −3b2 = 2ab 3. Then either a = 0 or b = 0. If a = 0,
√ then √ = b ∈ Q,
2/3
which is false. If b = 0, then 2 = a ∈ Q, which is again false. We conclude that Q( 2) → Q( 3) are not
isomorphic as fields.
√ √ √ √
However, [Q( 2) : Q] = [Q( 3) : Q] = 2, since min(Q, 2)√= x2 − 2 and 2
√ min(Q, 3) = x − 3. Since
vector spaces of the same (finite) dimension are isomorphic, Q( 2) and Q( 3) are isomorphic as Q-vector
spaces.

Problem 1.13. If L1 = F (a1 , . . . , an ) and L2 = F (b1 , . . . , bm ), show that the composite L1 L2 is equal
to F (a1 , . . . , an , b1 , . . . , bm ).
Note that F ⊆ L1 L2 , and a1 , . . . , an , b1 , . . . , bm ∈ L1 L2 , so F (a1 , . . . , an , b1 , . . . , bm ) ⊆ L1 L2 . Moreover,
L1 , L2 ⊆ F (a1 , . . . , an , b1 , . . . , bm ). Then L1 L2 ⊆ F (a1 , . . . , an , b1 , . . . , bm ).

3
Problem 1.14. If L1 and L2 are field extensions of F that are contained in a common field, show that L1 L2
is a finite extension of F if and only if both L1 and L2 are finite extensions of F .
(⇒) Assume L1 L2 /F is finite. Then L1 /F and L2 /F are finite since L1 , L2 ⊆ L1 L2 .
(⇐) Let L1 and L2 be finite extensions of F . Then, by Lemma 1.19, L1 = F (a1 , . . . , an ) and L2 =
F (b1 , . . . , bm ) for some a1 , . . . , an , b1 , . . . , bm in the common field. Moreover, by Lemma 1.19 again, the ai
and bj are algebraic over F . By the previous problem, L1 L2 = F (a1 , . . . , an , b1 , . . . , bm ). By Proposition
1.21,

[L1 L2 : F ] = [F (a1 , . . . , an , b1 , . . . , bm ) : F ] ≤ [F (a1 ) : F ] · · · [F (an ) : F ][F (b1 ) : F ] · · · [F (bn ) : F ] < ∞.

Problem 1.15. If L1 and L2 are field extensions of F that are contained in a common field, show that L1 L2
is algebraic over F if and only if both L1 and L2 are algebraic over F .
(⇒) Let L1 L2 /F be algebraic. Then, since F ⊆ L1 , L2 ⊆ L1 L2 , L1 /F and L2 /F are algebraic.
(⇐) Let L1 /F and L2 /F be algebraic. We will show that L1 L2 = F (L1 ∪ L2 ). Note that L1 L2 is the
smallest field that contains L1 and L2 . Therefore L1 L2 ⊆ F (L1 ∪ L2 ). On the other hand F (L1 ∪ L2 ) is the
smallest field containing F , L1 , and L2 . Hence, F (L1 ∪ L2 ) ⊆ L1 L2 , proving the claim. Since every element
of L1 ∪ L2 is algebraic over F , L1 L2 = F (L1 ∪ L2 ) is algebraic over F by Proposition 1.23.

Problem 1.16.√ Prove that [A : Q] = ∞.

Let A be the algebraic closure of Q in C. √
Note
√ that n
2 ∈ A for every n ∈ N. Moreover,
√ min(Q, n
2) = xn − 2 by Eisenstein’s criterion. Therefore
n n
[Q( 2) : Q] = n. Moreover, A : Q] ≥ [Q( 2) : Q] = n for every n ∈ N. Hence, [A : Q] = ∞.

Problem 1.17. Let K be a finite extension of F . If L1 and L2 are subfields of K containing F , show
that [L1 L2 : F ] ≤ [L1 : F ][L2 : F ]. If gcd([L1 : F ], [L2 : F ]) = 1, prove that [L1 L2 : F ] = [L1 : F ][L2 : F ].
We begin by proving the following claim: if F ⊆ L ⊆ K is an inclusion of fields, and a1 , . . . , an ∈ K are
algebraic over F , then [L(a1 , . . . , an ) : L] ≤ [F (a1 , . . . , an ) : F ]. We will do this by induction. For the base
case, note that min(L, a1 ) divides min(F, a1 ). Therefore, [L(a1 ) : L] ≤ [F (a1 ) : F ].
Now assume the claim holds for n. By induction,

[L(a1 , . . . , an+1 ) : L] = [L(a1 , . . . , an )(an+1 ) : L(a1 , . . . , an )][L(a1 , . . . , an ) : L]

≤ [F (a1 , . . . , an )(an+1 ) : F (a1 , . . . , an )][f (a1 , . . . , an ) : F ] = [F (a1 , . . . , an+1 ) : F ],

proving the claim.

Let’s now turn back to the problem. Since K is a finite extension of F , so are L1 and L2 . Then
L1 = F (a1 , . . . , an ) and L2 = F (b1 , . . . , bm ) for some ai , bj ∈ K. By the claim,

[L1 L2 : F ] ≤ [F (a1 , . . . , an , b1 , . . . , bm ) : F ]
= [F (a1 , . . . , an )(b1 , . . . , bm ) : F (a1 , . . . , an )][F (a1 , . . . , an ) : F ]
≤ [F (b1 , . . . , bm ) : F ][F (a1 , . . . , an ) : F ] = [L1 : F ][L2 : F ].

Now suppose the [L1 : F ] and [L2 : F ] are coprime. Note that

[L1 L2 : F ] = [L1 L2 : L1 ][L1 : F ] = [L1 L2 : L2 ][L2 : F ].

Then [L2 : F ] divides [L1 L2 : L1 ], implying that [L1 L2 : L1 ] ≥ [L2 : F ], and therefore [L1 L2 : F ] ≤ [L1 :
F ][L2 : F ]. Having shown both inequalities, we have the desired result.
√ √
Problem 1.18. Show
√ √that [Q( 4 2, 3)√: Q]√= 8. √ √ √
Note that [Q( 4 2, 3) √: Q] = [Q(
√
4
2, 3) : Q( 4 2)][Q( 4 2) : Q] and [Q( 4 2) :√Q]√= √4. It is there-
fore enough to show that 3 ∈ / Q( 4 2). To show
√ this,
√ we will
√ first prove that {1, 2, 3, 6} is linearly
independent over Q. To this end, let 0 = a + b 2 + c 3 + d 6 with a, b, c, d ∈ Q. On one hand, we have
√ √ √ √ √
(a + c 3)2 = (b 2 + 6)2 ⇒ (a2 + 3c2 ) + 2ac 3 = (2b2 + 6d2 ) + 4bd 3.

4
 √
Since 3 is irrational, we must have that a2 + 3c2 = 2b2 + 6d2 . On the other hand,
√ √ √ √ √
(a + d 6)2 = (b 2 + c 3)2 ⇒ (a2 + 6d2 ) + 2ad 6 = (2b2 + 3c2 ) + 2bc 6

that a2 + 6d2 = 2b2 + 3c2 . We then have 6d2 − 3c2 = 3c2 − 6d2 ⇒ 2d2 = c2 , which is
similarly implies √
impossible since 2 is √irrational.
√ √ √ √ √
Now, suppose that 3 ∈ Q( 4 2), then 3 = α + β 4 2 + γ 2 + δ 4 8 for α, β, γ, δ ∈ Q. Then,
√ √ √ √ √ √ √ √
3 − γ 2 − α = β 2 + δ 8 ⇒ 3 + 2γ 2 + α2 − 2γ 6 − 2α 3 + 2αγ 2 = (β 2 + 2δ 2 ) 2 + 4βδ.
4 4

From the above√equation and our linear independence proof, we immediately have that α = γ = 0. Then
3 = (β 2 + 2δ 2 ) 2 + 4βδ, which implies√that β 2 √ + 2δ 2 = 0 ⇒
√β √ = δ = 0.√ But then the previous equation
becomes 3 = 0, a contradiction. Hence, 3 ∈ / Q( 2) and [Q( 4 2, 3) : Q( 4 2)] = 2, concluding the solution.
4

Note that this exercise shows that the converse to the last part of the previous problem does not hold,
i.e. [L1 L2 : F ] = [L1 : F ][L2 : F ] does not imply that gcd([L1 : F ], [L2 : F ]) = 1.

Problem 1.19. Give an example of field extensions √ L1 , L2 of F for which [L1 L2 : F ] < [L1 : F ][L2 : F ].
Very √trivially, we could
√ choose L 1 = L 2 = Q( 2) and F = Q. Slightly less trivially, we could choose
L1 = Q( 4 2), L2 = Q( 2), and F = Q.
Let’s insist that √
L1 is√not a subfield of L2 , and vice versa. An example is given immediately after Propo-
sition 1.21: L = Q( 4 2, 4 18), and it uses the previous problem.
√
Problem 1.20. Give an example of a field extension K/F with [K : F ] = 3, but with K 6= F ( 3 b) for
any b ∈ F .
Let F = F2 and let α be a root of x3 + x + 1 (which is irreducible over F2 since it has no roots in F2 ).
Let K =√F2 (α). Proposition 1.15 tells us that [K : F ] = 3, and {1, α, α2 } is an F -basis for K. Suppose that
K = F ( 3 b) for some b ∈ F . Then b = (c0 + c1 α + c2 α2 )3 for some c0 , c1 , c2 ∈ F . It is a straightforward
computation to check that (c0 + c1 α + c2 α2 )3 is never in F for all
√ triplets (c0 , c1 , c2 ) other than (0, 0, 0) and
(1, 0, 0). √However, in these two exceptional cases we would have 3 b ∈ F , which rules out the possibility that
K = F ( 3 b).
Qn−1
Problem 1.21. Let a ∈ C be a root of xn − b, where b ∈ C. Show that xn − b factors as i=0 (x − ω i a),
where ω = e2πi/n .
The ω i a are all roots of xn − b, and they are all distinct for i ∈ {0, . . . , n − 1}. Since xn − b has n roots
Qn−1
in C, this is the entire collection of roots. Therefore xn − b = i=0 (x − ω i a).

Problem 1.22.
(a) Let F be a field, and let f (x) ∈ F [x]. If f (x) = i ai xi and α ∈ F , let f (x + α) = i ai (xi + α)i .
P P
Prove that f is irreducible over F if and only if f (x + α) is irreducible over F for any α ∈ F .
(⇒) Let f (x) be irreducible, and suppose there is an α for which f (x + α) is reducible, say f (x + α) =
g(x)h(x). Then f (x) = g(x − α)h(x − α), contradicting the fact that f is irreducible.
(⇐) Take α = 0.
(b) Show that xp−1 + xp−2 + · · · + x + 1 is irreducible over Q if p is a prime.
xp −1
Let φ(x) = xp−1 + xp−2 + · · · + x + 1. Note that φ(x) = x−1 . Then
Pp p

p  
(x + 1)p − 1 i=1 xi X p
φ(x + 1) = = i
= xi−1 .
x x i=1
i

Since p is prime, coefficient of φ(x

+ 1) is divisible by p, except the leading coefficient which is 1.
Moreover, the constant term is p1 = p, which is not divisible by p. Then, by Eisenstein’s criterion,
φ(x + 1) is irreducible. By part (a), φ(x) is irreducible.

5
Problem 1.23. Recall that the characteristic of a ring R with identity is the smallest positive integer n for
which n · 1 = 0, if such an n exists, or else the characteristic is 0. Let R be a ring with identity. Define
φ : Z → R by φ(n) = n · 1, where 1 is the identity of R.Show that φ is a ring homomorphism and that
ker(φ) = mZ for a unique nonnegative integer m, and show that m is the characteristic of R.
Note that φ(n1 + n2 ) = (n1 + n2 ) · 1 = n1 · 1 + n2 · 1 = φ(n1 ) + φ(n2 ) and φ(n1 n2 ) = (n1 n2 ) · 1 =
(n1 · 1) · (n2 · 1) = φ(n1 ) · φ(n2 ). Therefore, φ is a ring homomorphism. Since φ is a ring homomorphism,
ker(φ) is an ideal of Z. Hence, ker(φ) = mZ for a unique non-negative integer Z. Suppose m is non-zero.
Let n · 1 = 0 for some non-negative integer n. Then n ∈ mZ, which means that m divides n and thus m ≤ n.
Therefore, m is the characteristic of R. If m = 0, then ker(φ) = {0}, implying that n · 1 6= 0 for all n ∈ N.
Hence, the characteristic of R is 0 in this case.

Problem 1.24. For any positive integer n, give an example of a ring of characteristic n.
Z/nZ.

Problem 1.25. If R is an integral domain, show that either char(R) = 0 or char(R) is prime.
Suppose char(R)n > 0, and suppose n is composite. Say n = pq, with p, q > 1. Then 0 = n·1 = (p·1)·(q·1).
Since R is an integral domain, either p · 1 = 0 or q · 1 = 0, contradicting the assumption that char(R) = n.

Problem 1.26. Let R be a commutative ring with identity. The prime subring of R is the intersection
of all subrings of R. Show that this intersection is a subring of R that is contained inside all subrings of R.
Moreover, show that the prime subring of R is equal to {n · 1 : n ∈ Z}, where 1 is the multiplicative identity
of R.
It is a standard fact that the intersection of subrings is a gain a subring. By its construction, the prime
subring is contained in all subrings of R. Since 1 is an element all subrings, it is an element of the prime
subring; moreover, all the multiples of 1 are in the prime subring, so {n · 1 : n ∈ Z} is contained in the prime
subring. On the other hand, {n · 1 : n ∈ Z} is a subring, so it contains the prime subring.

Problem 1.27. Let F be a field. If char(F ) = p > 0, show that the prime subring of R is isomorphic
to the field Fp , and if char(F ) = 0, then the prime subring is isomorphic to Z.
This follows easily from the previous problem.

Problem 1.28. Let F be a field. The prime subfield of F is the intersection of all subfields of F . Show that
this subfield is the quotient field of the prime subring of F , that it is contained inside all subfields of F , and
that it is isomorphic to Fp or Q depending on whether the characteristic of F is p > 0 or 0.
The prime subfield is indeed contained in all subfields by its construction. The prime subfield is larger
than the prime subring, so it contains its fraction field. On the other hand, the fraction field of the prime
subring is a field, and thus it contains the prime subfield. The fraction fields of Fp and Z are Fp and Q
respectively, so the last part of the problem follows from the previous problem.

Section 2. Automorphisms
Problem 2.1. Show that the only automorphism of Q is the identity.
If φ : Q → Q is an automorphism, note that φ(1) = 1 forces φ(a/b) = φ(a)/φ(b) = φ(a · 1)/φ(b · 1) = a/b.
Hence, the only automorphism of Q is the identity.

Problem 2.2. Show that the only automorphism of R is the identity. It is an interesting fact that there are
infinitely many automorphisms of C, even though [C : R] = 2. Why is this fact not a contradiction to this
problem?
Let σ : R → R be an automorphism. By the same proof as in the previous √problem σ|Q = id. We will
now show that σ is increasing. √Let a < b. Then σ(b) − σ(a) = σ(b − a) = σ( b − a)2 > 0, since σ is an
automorphism and therefore σ( b − a) 6= 0. Hence σ(a) < σ(b), proving that σ is increasing.
Now let c ∈ R be arbitrary. By the density of Q in R, we can find rational numbers p and q such that
p < c < q and q − p <  for any  > 0. Applying σ, we have p < σ(c) < q. Therefore, σ(c) = c, proving that
the only automorphism of of R is the identity.

6
 Since C/R is Galois, [C : R] = 2 implies that there are only 2 automorphisms of C that fix R. The rest
of the infinitely many automorphisms of C then do not fix R. This is interesting, since they do fix Q; the
automorphisms other than the two in Gal(C/R) must be highly discontinuous.
√
Problem 2.3. Show that the six functions given in Example 2.21 extend to Q-automorphisms of Q( 3 2, ω).
√ Some √ details will be omitted from this problem. I will also √ only show that the function ρ mapping
3
2 7→ ω 3 2 and ω 7→ ω 2 extends to an automorphism of Q( 3 2, ω). The other functions are dealt with
similarly. √ √ √ √
First, note that {1, 3 2, 3 4, ω, ω 3 2, ω 3 4}. This follows from the discussion in Example 2.21 and the
proof of Proposition 1.20. Define
√ √ √ √ √ √ √ √
ρ(c1 + c2 2 + c3 4 + c4 ω + c5 ω 2 + c6 ω 4) = c1 + c2 ω 2 + c3 ω 2 4 + c4 ω 2 + c5 2 + c6 ω 4
3 3 3 3 3 3 3 3

√
3
√
3
√
3
√
3
= (c1 − c4 ) + c5 2 − c3 4 − c4 ω + c2 ω 2 + (c6 − c3 )ω 4.

Checking that ρ, defined as above, is a field homomorphism is busywork. Field homomorphisms are always
injective, so we just need to check surjectivity. For an arbitrary element
√3
√3
√
3
√
3
α = d1 + d2 2 + d3 4 + d4 ω + d5 ω 2 + d6 ω 4,
√
in Q( 3 2, ω), let √ √ √ √
3 3 3 3
β = (d1 − d4 ) + d5 2 − d3 4 − d4 ω + d2 ω 2 + (d6 − d3 )ω 4.
Then σ(α) = β, proving that ρ is surjective and thus an automorphism.

Problem 2.4. Let B be an integral domain with quotient field F . If σ : B → B is a ring automorphism,
show that σ induces a field automorphism σ 0 : F → F defined by σ 0 (a/b) = σ(a)/σ(b).
First, let’s check that σ 0 is well defined. Suppose that a/b = c/d, that is, suppose that ad − bc = 0. Then
σ (a/b) = σ(a)/σ(b) = σ(c)/σ(d) = σ 0 (c/d), since σ(a)σ(d) − σ(b)σ(c) = σ(ad − bc) = 0. Moreover, since σ
0

is an automorphism, b 6= 0 ⇒ σ(b) 6= 0, so the image of a fraction will not have a zero in the denominator.
Now, let’s check that σ 0 is a field homomorphism. First,
a  
c ad + bc σ(ad + bc) σ(a)σ(d) + σ(b)σ(c) σ(a) σ(c)
σ + =σ = = = + = σ 0 (a/b) + σ 0 (b/c).
b d bd σ(bd) σ(b)σ(d) σ(b) σ(d)

Second,
a c  ac  σ(ac) σ(a)σ(c) σ(a) σ(c)
σ0 · = σ0 = = = · = σ 0 (a/b) · σ 0 (c/d).
b d bd σ(bd) σ(b)σ(d) σ(b) σ(d)
Thus, σ 0 is a field homomorphism. To see that σ 0 is surjective, let a/b ∈ F be arbitrary. Since σ is surjective,
there are elements a0 , b0 ∈ R such that σ(a0 ) = a and σ(b0 ) = b. Then σ 0 (a0 /b0 ) = a/b.

Problem 2.5. Let K = k(x1 , . . . , xn ) be the field of rational functions in n variables over a field k. Show
that the definition  
f (x1 , . . . , xn ) f (xσ(1) , . . . , xσ(n) )
σ =
g(x1 , . . . , xn ) g(xσ(1) , . . . , xσ(n) )
makes a permutation σ ∈ Sn into a field automorphism of K.
If we can show that σ(f (x1 , . . . , xn )) = f (xσ(1) , . . . , xσ(n) ) makes σ into a ring automorphism of k[x1 , . . . , xn ],
then the previous problem will ensure that the extended definition of σ is a field automorphism of K. By
Problem 1.6, there is a unique k-homomorphism φ : k[x1 , . . . , xn ] → k[x1 , . . . , xn ] such that xi 7→ xσ(i) . By
the properties of ring homomorphisms, this homomorphism must be σ. Since σ is a permutation, we also
have that the induced map σ on polynomials is surjective. Hence, σ induces an automorphism of K.

Problem 2.6. Let√F be a field of characteristic not 2, and let K be an extension of F with [K : F ] = 2.
Show that K = F ( a) for some a ∈ F ; that is, show that K = F (α) with α2 = a ∈ F . Moreover, show that
K is Galois over F .

7
 There exists β ∈ K such that {1, β} is an F -basis for K. Then K = F (β). Then β is the root of a monic
polynomial x2 + bx + c. Since the characteristic of F is not 2,
√
−b ± b2 − 4c
β= .
2
Therefore, K = F (α), where α2 = b2 − 4c.
We now show that K/F is Galois. Let φ : K → K be an F -automorphism. Then φ is completely
determined by φ(α). We know that φ(α)2 = φ(α2 ) = α2 , so φ(α) = ±α (that these are distinct depends on
the fact that F does not have characteristic 2). If φ(α) = α, then φ is the identity. We will now show that
φ(α) = −α gives rise to a well-defined automorphism. Linearity and surjectivity are obvious, we just need
to show that φ is compatible with multiplication. Indeed,

φ((a+bα)(c+dα)) = φ((ac+bdα2 )+(ad+bc)α) = ac+bdα2 −(ad+bc)α = (a−bα)(c−dα) = φ((a−bα)(c−dα)).

Therefore, |Gal(K/F )| = 2 = [K : F ], which implies that K/F is Galois by Corollary 2.16.

Problem 2.7.Let F = F2 and K = F (α), where α is a root of 1 + x + x2 . Show that the function
σ : K → K given by σ(a + bα) = a + b + bα for a, b ∈ F is an F -automorphism of K.
It is clear that σ is additive and that it fixes F . Let’s now show that σ is multiplicative. Indeed,

σ((a + bα)(c + dα)) = σ(ac + (ad + bc)α + bdα2 ) = σ(ac + (ad + bc)α + bd(1 + α))
= σ((ac + bd) + (ad + bc + bd)α) = (ac + bd + ad + bc + bd) + (ad + bc + bd)α
= (ac + ad + bc) + (ad + bc + bd)α = (a + b + bα)(c + d + dα) = σ(a + bα)σ(c + dα).

Therefore, σ is a field automorphism. For a, b ∈ F2 , note that

σ(a − b + bα) = a − b + b + bα = a + bα.

Therefore σ is an F -automorphism of K.
Note that√this problem gives an example of a field F of characteristic 2 and extension K/F of degree 2
with K 6= F ( a) for any a ∈ F .

Problem 2.8. Suppose that a is algebraic over Q with p(x) = min(Q, a), and let b be any root in C of
p. Show that the map σ : Q(a) → C given by σ(f (a)) = f (b) is a well defined Q-homomorphism.
Suppose f (a) = g(a). We want to show that f (b) = g(b). Then a is a root of the polynomial (f − g)(x),
which means that p divides f − g. Since b is a root of p, b is also a root of f − g, so f (b) = g(b) as desired.
Moreover, σ fixes Q, and the field homomorphism properties follow directly from the definition.
√ √
Problem 2.9. Show that the complex numbers i 3 and 1 + i 3 are roots of f (x) = x4 − 2x3 + 2
√7x − 6x +√12.
Let K be the field generated by Q and the roots of f . Is there an automorphism σ of K with σ(i 3) = 1+i 3?
The first part of the problem is a direct computation, and should be straightforward given some familiarity
with multiplication of complex numbers. √ √
Let σ be an automorphism of K satisfying σ(i 3) = 1 + i 3. Note that σ fixes Q, since it fixes 1. Then
√ √ √ √
−3 = σ(−3) = σ((i 3)2 ) = σ(i 3)2 = (1 + i 3)2 = −2 + 2i 3,

a contradiction.
Combining this with the result
√ of the previous
√ problem, we have a proof that f (x) = x4 −2x3 +7x2 −6x+12
is reducible over Q, and that i 3 and 1 + i 3 are roots of distinct irreducible components of f .

Problem 2.10. Determine whether the following fields are Galois over Q.

(a) Q(ω), where ω = exp(2πi/3).

Note that min(Q, ω) = x2 + x + 1, so [Q(ω) : Q] = 2. The other root of min(Q, ω) is ω 2 ∈ Q(ω).
Therefore, by Corollary 2.17, Q(ω) is Galois over Q.

8
 √
(b) Q( 4 2).
√ √ √ √
Note that min(Q, 4 2) = x4 −√ 2. Its roots are ± 4 2, ±i 2 4. Only two of these are in Q( 4 2 ⊆ R, so
Corollary 2.17 implies that Q( 4 2)/Q is not a Galois extension.
√ √
(c) Q( 5, 7).
√ √ √ √ √ √ √
From Problem
√ 1.5, we have that Q( 5, 7) = Q( 5+ 7). Let a = 5+ 7. Note that a2 = 12+ 35.
Isolating 35 and squaring, we obtain

a4 − 24a2 + 109 = 0.
4 2
p √
Now,√by the
√ quadratic formula, the roots of x − 24x + 109 are ± 12 ± 35 (which can be rearranged
to ± 5 ± 7), which are pairwise distinct. We have not proven that x4 − 24x2 + 109 is the minimal
polynomial of a over Q (which it is), but we don’t need that since min(Q, a) will divide x4 − 24x2 +
109,
√ implying
√ that
√ it will
√ have as many distinct roots as its √degree. √ Moreover, all these roots lie in
Q( 5, 7) = Q( 5 + 7). Corollary 2.17 then gives that Q( 5, 7) is Galois over Q.

Problem 2.11. Prove or disprove the following assertion and its converse: If F ⊆ L ⊆ K are fields with
K/L and L/F Galois, then K/F is Galois. √ √ √ √ √
(⇒) This direction is not true. √ Consider
√ Q ⊆ Q( 2)√ ⊆ Q( 4 2). Then [Q( 2) : Q] = [Q( 4 2) : Q(
√ 2)] =
2, so Problem 2.6 implies√that Q( 4 2)/Q( 2) and Q( 2)/Q are √ Galois extensions. However, Q( 4
2)/Q is
not Galois, since min(Q, 4 2) = x4 − 2 only has two roots
√ in Q( 4
2).
(⇐) This direction is also false.
√ For example,
√ Q( 3 2, ω)
√ is Galois, where ω = e2πi/3 (shown in Example
3 3 3
2.21). However, we have Q ⊆ Q( 2) ⊆ Q( 2, ω), but Q( 2)/Q is not Galois (shown in Example 2.18).

Problem 2.12. Galois connections. The relationship given in Corollary 2.10 between the set of inter-
mediate fields of a Galois extension and the set of subgroups of its Galois group appears in other situations,
so we study it here. We first need a definition. If S is a set, a relation ≤ on S is called a partial order on
S provided that a ≤ a for all a ∈ S; if a ≤ b and b ≤ a, then a = b; and if a ≤ b and b ≤ c, then a ≤ c. Let
S and T be sets with partial orders ≤S and ≤T , respectively. Suppose that there are functions f : S → T
and g : T → S such that (i) if s1 ≤S s2 , then f (s2 ) ≤T f (s1 ), (ii) if t1 ≤T t2 , then f (t2 ) ≤S f (t1 ), and
(iii) s ≤S g(f (s)) and t ≤T f (g(t)) for all s ∈ S and t ∈ T . Prove that there is a 1-1 order reversing
correspondence between the image of g and the image of f , given by s 7→ f (s), whose inverse is t 7→ g(t).
Let s ∈ g(T ). We want to show that g(f (s)) = s. We already know that s ≤S g(f (s)). On the other
hand, we have that s = g(t) for some t ∈ T , so f (s) = f (g(t)) ≥T t. Hence, g(f (s)) ≤S g(t) = s, implying
g(f (s)) = s. Now, let t ∈ f (S). A similar proof as above will show that t = f (g(t)). Therefore, f induces a
1-1 correspondence between g(T ) and f (S), whose inverse is g.

Problem 2.13. Let k be a field, and let K = k(x) be the rational function field in one variable over
k. Let σ and τ be the automorphisms of K defined by σ(f (x)/g(x)) = f (1/x)/g(1/x) and τ (f (x)/g(x)) =
f (1 − x)/g(1 − x), respectively. Determine the fixed field F of {σ, τ }, and determine Gal(K/F ). Find an
h ∈ F so that F = k(h).
The images of x under all possible compositions of σ and τ are
1 1 x−1 x
s1 = x, s2 = , s3 = 1 − x, s4 = , s5 = , s6 = .
x 1−x x x−1
Therefore, the order of the group generated by {σ, τ } is 6. Since (x − 1)/x = στ (x) 6= τ σ(x) = 1/(1 − x),
hσ, τ i is not Abelian. Hence, hσ, τ i ∼ = S3 , the only non-Abelian group of order 6. By Proposition 2.14,
Gal(K/F ) ∼ = S6 , and [K : F ] = 6.
Let’s now determine F . A direct computation shows that the rational function h := s1 s2 +s1 s3 +· · ·+s5 s6
can be written as a quotient f /g, where f is a monic polynomial of degree 6, and g(x) = x2 (x − 1)2 .
Moreover, the computation will show that neither 1 nor 0 are roots of f , so gcd(deg(f ), deg(g)) = 1. Hence,
Example 1.17 shows that [K : k(h)] = 6. But clearly h ∈ F , so k(h) ∈ F . Therefore k(h) = F , since
6 = [K : k(h)] = [K : F ][F : k(h)] = 6[F : k(h)] ⇒ [F : k(h)] = 1.
A (perhaps) more satisfying description of F is also possible. Let t1 = s1 + s2 + · · · + s6 , t2 =
s1 s2 + s1 s3 + · · · + s5 s6 , . . . , t6 = s1 s2 · · · s6 . Then, trivially, k(t1 , . . . , tn ) ⊆ F . Bute we also have that

9
h = t2 , so F = k(h) ⊆ k(t1 , . . . , t6 ). Hence, F = k(t1 , . . . , t6 ).

Problem 2.14. Let k be a field, and let K = k(x) be the rational function field in one variable  over
 k. If
a b
u ∈ k, show that K = k(u) if and only if u = (ax + b)/(cx + d) for some a, b, c, d ∈ k with det 6= 0.
c d
(⇒) Let K = k(u). From Example 1.17, we know that we can write u = f /g, where f, g ∈ k[x], and
max{deg f, deg
 g} = 1. Then u = (ax + b)/(cx + d). Without loss of generality, we may assume that a 6= 0.
a b
Suppose det = 0. Then ad − bc = 0, which implies c 6= 0, and
c d

ax + b a(ax + b) a(ax + b) a
u= = = = ∈ K.
cx + d acx + ad acx + bc c
But then k(u) = k 6= k(x). We conclude that ad − bc 6= 0.
(⇐) Let ad − bc 6= 0. I claim that gcd(ax + b, cx + d) = 1. Suppose not; then there is a linear polynomial
px+q such that ax+b = k1 (px+q) and cx+d = k2 (px+q). This implies ad−bc = 0, so gcd(ax+b, cx+d) = 1.
Example 1.17 implies [K : k(u)] = 1, so k(u) = K.
 
a b
Problem 2.15. Use the previous problem to show that any invertible 2 × 2 matrix determines
c d
an element of Gal(k(x)/k) with x 7→ (ax + b)/(cx + d). Moreover, show that every element of Gal(k(x)/k) is
such aformula. Show that the map from the set of invertible 2 × 2 matrices over k to Gal(k(x)/k)
given by 
a b
given by 7→ φ, where φ(x) = (ax + b)/(cx + d), is a group homomorphism. Determine the kernel to
c d
show that Gal(k(x)/k) ∼ = PGL2 (k), the group of invertible 2 × 2 matrices over k modulo the scalar matrices.
(This group is the projective general linear group over k of 2 × 2 matrices.)
Let φ : k(x) → k(x) be given by φ(x) = (ax + b)/(cx + d), where ad − bc 6= 0, and extended to all of
k(x) naturally. By its construction, φ fixes k. Moreover, by the previous problem, φ is surjective. Hence,
φ ∈ Gal(k(x)/k).
Let τ ∈ Gal(k(x)/k). Then τ (k(x)) = k(u), where u = τ (x). Since τ is an automorphism, k(u) = k(x),
so by the previous problem τ (x) = (ax + b)/(cx + d) for some a, b, c, d ∈ k with ad − bc 6= 0.
Next, let  0
a b0
  
a b
7→ φ and 7→ ψ,
c d c0 d0
where the matrices are invertible. We want to show that
 0
a b0
 
a b
7→ φ ◦ ψ.
c d c0 d0

Note that  0
b0 aa0 + bc0 ab0 + bd0
   
a b a
= .
c d c0 d0 ca0 + dc0 cb0 + dd0
We also have
a0 x + b0
 
a 0 +b
c x + d0 (aa0 + bc0 )x + (ab0 + bd0 )
φ ◦ ψ(x) =  0 = ,
a x + b0 (ca0 + dc0 )x + (cb0 + dd0 )

c 0 + d
c x + d0
 
a
b
proving that 7→ φ is a group homomorphism.
c
d  
a b
Suppose that A = 7→ id. Then x = (ax + b)/(cx + d) ⇒ cx2 + dx = ax + b, implying c = 0 = b
c d
and a = d. Therefore A is a scalar matrix. It is also clear that scalar matrices map to the identity. Hence
Gal(k(x)/k) ∼
= PGL2 (k).

10
Problem 2.16. Let k = R, and let A be the matrix
 √ 
−1/2 − 3/2
√
3/2 −1/2
given by rotating the plane around the origin by 120◦ . Using the previous problem, show that A determines
a subgroup of Gal(k(x)/k) of order 3. Let F be the fixed field. Show that k(x)/F is Galois, find a u so that
F = k(u), find the minimal polynomial min(F, x), and find all the roots of this polynomial.
Since A is invertible, it determines an element of Gal(k(x)/k) (by the previous problem). Since it has
order 3, it determines a cyclic subgroup G of Gal(k(x)/k) of order 3. By Proposition 2.14, k(x)/F is Galois,
and [k(x) : F ] = 3.
The images of x under G are √ √
−x − 3 −x + 3
x, √ , √ .
3x − 1 − 3x − 1
Therefore
√ √ √ √ √ √ √ √
−x − 3 −x + 3 x( 3x − 1)(− 3x − 1) + (−x + 3)(− 3x − 1) + (−x + 3)( 3x − 1)
u=x+ √ + √ = √ √
3x − 1 − 3x − 1 ( 3x − 1)(− 3x − 1)
is in the fixed field. Moreover, the numerator and denominator of u have no factor in common. Since the
numerator has the larger degree, and it has degree 3, Example 1.17 implies that [k(x) : k(u)] = 3. Hence
F = k(u). √
By foiling the numerator and denominator
√ of u, we find that x satisfies x3 − 2ux2 − x + 2 3/3 = 0.
Hence, min(F, x) = t3 − 3ut2√− t + 2 3/3.√ Since the coefficients are invariant under applications of A, the
roots of min(F, x) are x, −x−
√ 3
3x−1
, and −−x+
√ 3
3x−1
.

Problem 2.17. Let k = Fp , and let k(x) be the rational function field in one variable over k. Define
φ : k(x) → k(x) by φ(x) = x + 1. Show that φ has finite order in Gal(k(x)/k). Determine this order, find a
u so that k(u) is the fixed field of φ, determine the minimal polynomial over k(u) of x, and find all the roots
of this minimal polynomial.
Since φk (x) = x + k, φ has order p. The polynomial u = x(x + 1)(x + 2) · · · (x + p − 1) is contained in
the fixed field, so k(u) ⊆ F(φ). Since φ has order p, p = |hφi| = [k(x) : F(φ)] by Proposition 2.14. But by
Example 1.17, [k(x) : k(u)] = p, so k(u) = F(φ).
The minimal polynomial must be of degree p, so it is simply t(t + 1) · · · (t + p − 1) − u, and the roots are
x, x + 1, . . . , x + p − 1.

Problem 2.18. Let k be a field of characteristic p > 0, and let a ∈ k. Let f (x) = xp − ap−1 x. Show that f
is fixed by the automorphism φ of k(x) defined by φ(f (x)/g(x)) = f (x + a)/g(x + a) for any f (x), g(x) ∈ k[x].
Show that k(f ) is the fixed field of φ.
I will assume a 6= 0, otherwise the last part of the problem is false. Note that (x + a)p − ap−1 (x + a) =
xp + ap − ap−1 x − ap = xp − ap−1 x, so xp − ap−1 x is fixed by φ. Since the order of φ is p, Example 1.17 and
Proposition 2.14 imply that k(f ) is the fixed field of φ.

Problem 2.19. Prove that (t − x1 ) · · · (t − xn ) = tn − s1 tn−1 + · · · + (−1)n sn , as we claimed in Example

2.22.
We will prove this by induction on n. For the base case, s1 = x1 , so t − x1 = t − s1 .
Now assume the result holds for some n ∈ N. Let s0i denote the i-th elementary symmetric in n variables,
and si the i-th elementary symmetric function in n + 1 variables. By induction,

(t − x1 ) · · · (t − xn )(t − xn+1 ) = (tn − s01 tn−1 + · · · + (−1)n s0n )(t − xn+1 )

= tn+1 − (s01 + xn+1 )tn + (s02 + s01 xn+1 )tn−1 + · · · + (−1)n+1 s0n xn+1
= tn+1 − s1 tn + s2 tn−1 + · · · + (−1)n+1 sn+1 ,

as desired.

11
Section 3. Normal Extensions
Problem 3.1. Show that K is a splitting field over F for a set {f1 , . . . , fn } of polynomials in F [x] if and
only if K is a splitting field over F for the single polynomial f1 · · · fn .
(⇒) Assume that K is a splitting field for {f1 , . . . , fn }. Then K = F (X), where X ⊆ K is the set of
roots of the fi . It is easy to see that X is also the set of roots of f1 · · · fn , so K is a splitting field for f1 · · · fn .
(⇐) Let K be a splitting field for f1 · · · fn . Then K = F (X) where X is the set of roots of f1 · · · fn . But
X is also the set of roots of {f1 , . . . , fn }, so K is a splitting field for {f1 , . . . , fn }.

Problem 3.2. Let K be a splitting field of a set S of polynomials over F . If L is a subfield of K containing
L for which each f ∈ S splits over L, show that L = K.
Let X be the set of all roots of all polynomials in S. Then K = F (X). Since each f ∈ S splits over L,
X ⊆ L. Moreover, since F ⊆ L, K = F (X) ⊆ L. Hence L = K.

Problem 3.3. If F ⊆ L ⊆ K are fields, and if K is a splitting field of S ⊆ F [x] over F , show that
K is also a splitting field for S over L.
Let X ⊆ K be the set of all roots of all the polynomials in S. Then F (X) = K. Moreover, F (X) ⊆
L(X) ⊆ K, so L(X) = K. Therefore K is a splitting field for S over L.

Problem 3.4.
(a) Let K be an algebraically closed field extension of F . Show that the algebraic closure of F in K is an
algebraic closure of F .
Recall that the algebraic closure of F in K is

F = {a ∈ K : a is algebraic over F }.

We want to show that F is algebraically closed, that is, we want to show that every non-constant
polynomial in F [x] has a root in F . To this end, let f be a non-constant polynomial over F . Then f
has a root a ∈ K, since K is algebraically closed. Since a is algebraic over F , and F is algebraic over
F , a is algebraic over F , i.e. a ∈ F .
(b) If A = {a ∈ C : a is algebraic over Q}, then, assuming that C is algebraically closed, show that A is an
algebraic closure of Q.
By definition, A is the algebraic closure of Q in C. By part (a), A is an algebraic closure of Q.

Problem 3.5. Give an example of fields F ⊆ K ⊆ L where L/K and K/F are normal but L/F is not
normal. √ √ √ √
4
Consider the tower Q ⊆ Q( √2) ⊆ (Q √ 2). Then Q( 2)/Q is normal,
√ since Q( 2) is the splitting√ field
x2 − 2 over Q. Similarly,
of √ √ Q( 4
2)/Q( 2) is normal because Q( 4
2 is the splitting field of x
√
2
− 2 over
Q( 2). However,√Q( 4 2)/Q is not normal since x4 − 2 is irreducible over Q, has a root in Q( 4 2), but does
not split over Q( 4 2).

Problem 3.6. Let f (x) be an irreducible polynomial over F of degree n, and let K be a field extension
of F with [K : F ] = m. If gcd(n, m) = 1, show that f is irreducible over K.
Let α be a root of f in some extension of F . Then [F (α) : F ] = n. Note that the composite
KF (α) = K(α). By Problem 1.17, [K(α) : F ] = mn. Then [K(α) : K][K : F ] = mn, so [K(α) : K] = m,
so deg(K, α) = m, which implies that f is the minimal polynomial of α over K, which means that f is
irreducible.
√ √
Problem 3.7. Show that x5 − 9x3 + 15x + 6 is irreducible over Q( 2, 3). √ √
By Eisenstein’s criterion, x5 − 9x3 + 15x + 6 is irreducible over Q, using p√= 3.
√ Since [Q( 2, 3) : Q] = 4,
the previous problem shows that x5 − 9x3 + 15x + 6 is irreducible over Q( 2, 3).

Problem 3.8. Find the degree of the splitting field of x6 + 1 over

12
 (a) Q,
q √
Note that x6 + 1 = (x2 + 1)(x4 − x2 + 1). The roots of this polynomial (in C) are ±i and ± 1±2 3i
q √ 
1+ 3i
by the quadratic formula. I claim that the splitting field is L = Q 2 . Clearly L is contained
in the splitting field, so it is a matter of showing that all the roots of x6 + 1 lie in L.
To this end, note that
s s √ s √
2 2(1 − 3i) 1 − 3i
√ = = .
1 + 3i 4 2
q √
1± 3i
Hence, ± 2 ∈ L. Furthermore, a simple computation shows that
2
√ √
s s
1 + 3i 1 − 3i 
 − = −1.
2 2

q q
Therefore, 1+2√3i − 1−2√3i = i ∈ L. This proves that L contains the splitting field of x6 + 1, and
therefore that L is the splitting field.
To find the degree of the splitting field, we need to show that x4 − x2 + 1 is irreducible over Q. We
already know that x4 − x2 + 1 has no roots in Q, so if it is reducible, then it factors as the product of
two quadratics over Q. Suppose x4 − x2 + 1 = (x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (ac + b +
d)x2 + (ad + bc)x + bd. Hence, c = −a, and ad + bc = a(d − b) = 0. Say a = c = 0. Then b + d = −1
and bd = 1, this implies b2 + d2 = −1, a contradiction. On the other hand, suppose b = d. Then bd = 1
implies b = d = ±1. Then −a2 ± 2 = −1, which has no real and rational solutions for a (regardless of
the sign). We conclude that x4 − x2 + 1 is irreducible, and that the degree of L over Q is 4.
(b) F2 .
Note that x6 + 1 = (x + 1)2 (x2 + x + 1)2 over F2 . Hence, the splitting field of x6 + 1 is formed by
appending a root of x2 + x + 1 to F2 . Therefore, the degree of the splitting field is 2.
Problem 3.9. Determine the splitting field of x4 − 7 over
(a) Q,
√ √ √ √ √
Since x4 − 7 splits as (x + i 4 7)(x − i 4 7)(x + 4 7)(x − 4 7), the splitting field is Q(i, 4 7).
(b) F5 ,
A simple calculation shows that x4 −7 = x4 +3 has no roots in F5 . Suppose x4 +3 factors as the product
of two quadratics, say x4 +3 = x2 +ax+b)(x2 +cx+d) = x4 +(a+c)x3 +(ac+b+d)x2 +(ad+bc)x+bd.
We immediately have that a = −c. Moreover, either b = 1 and d = 3, or b = 2 and d = 4. Suppose
b = 1 and d = 3. Then ad + bc = 2a = 0, implying a = 0. But then ac + b + d = 4, a contradiction.
Then suppose that b = 2 and d = 4. The ad + bc = 2a = 0, so a = 0. Then ac + b + d = 1, a
contradiction. We conclude that x4 + 3 is irreducible over F5 , and therefore that F5 [α]/(α4 + 3) is a
field.
I claim that L := F5 [α]/(α4 + 3) is the splitting field of x4 + 3 over F5 . Note that L is contained
in the splitting field, since L = F ([α]), and [α] is a root of x4 + 3. We just need to show that L
contains the three other roots of x4 + 3. By Fermat’s Little Theorem, (α)4 = (2α)4 = (3α)4 = (4α)4 ,
so α, 2α, 3α, 4α ∈ L are all roots of x4 + 3. Thus, L is the splitting field of x4 + 3 over F5 .
(c) F11 .
Again, x4 − 7 = x4 + 4 has no roots in F11 . However, it does factor as

x4 + 4 = (x2 + 2x + 2)(x2 − 2x + 2).

13
 I claim that the field L := F11 [α]/(α2 + 2α + 2) is the splitting field of x4 + 4. By its construction, L
contains a root of x2 + 2x + 2, namely α, and therefore it contains −α, which is a root of x2 − 2x + 2.
Hence, x4 + 4 splits over L. Moreover, L = F11 ([α]), so L is the splitting field of x4 + 4.
Problem 3.10. Let F be a field, and let f (x) be a polynomial of prime degree. Suppose for every field
extension K of F that if f has a root in K, then f splits over K. Prove that either f is irreducible over F
or f has a root (and hence splits) in F .
Let f1 and f2 be two irreducible factors of f , and let α1 be a root of f1 in some extension K of F . Then
f splits in F (α1 ); in particular, F (α1 ) contains a root of f2 , which we will call α2 . Then f also splits over
F (α2 ), so α1 ⊆ F (α2 ). Therefore, F (α1 ) = F (α2 ), which in turn implies that deg(f1 ) = deg(F2 ). Since f1
and f2 were arbitrarily chosen, we conclude that every irreducible factor of f has the same degree. Since
deg(f ) is prime, either every irreducible factor of f is linear, or f is irreducible.

Problem 3.11. Show that the hypotheses of the previous problem hold for
(a) f (x) = xp − a, where char(F ) = p and a ∈ F .
Let K be an extension of F containing α, a root of f . Then f splits as (x − α)p in K.

(b) f (x) = xp − x − a, where char(F ) = p and a ∈ F .

Let K be an extension of F containing α, a root of f . Then α + n is a root of f for any n ∈ N. Indeed,
(α + n)p − (α + n) + a = αp + np − α − n + a = αp + n − α − n + a = αp − α − a = 0, where we have
used Fermat’s Little Theorem. Hence, f splits as (x − α)(x − (α + 1)) · · · (x − (α + p − 1)) in K.
(c) f (x) = xp − a, where char(F ) 6= p and F contains an element ω with ω p = 1 and ω 6= 1.
If a = 0, then f splits in F . Suppose a 6= 0. Note that p is the order of ω in F ∗ since p is prime (if
ω n = 1 with n ≤ p, then n must divide p, so n = 1 or n = p). Say f has a root α in an extension K
of F . Then, αω i is a root of f for each i ∈ N, and αω i 6= αω j for i 6= j and i, j ∈ {0, 1, . . . , p − 1}.
Therefore, f splits as (x − α)(x − αω) · · · (x − αω p−1 ) in K.
Problem 3.12. Let K be a field, and suppose that σ ∈ Aut(K) has infinite order. Let F be the fixed field
of σ. If K/F is algebraic, show that K is normal over F .
Let f (x) ∈ F [x] be an irreducible polynomial with a root α ∈ K. Note that σ n (α) is a root of f
for every n ∈ Z. Since f only has finitely many roots, there exists and n ∈ N such that the elements of
{α, σ(α), · · · , σ n−1 (α)} are pairwise distinct and they form the complete set of elements of the form σ k (α).
Let σ : K[x] → K[x] be the ring homomorphism induced by σ. Then σ((x−α)(x−σ(α)) · · · (x−σ n−1 (α))) =
(x − α)(x − σ(α)) · · · (x − σ n−1 (α)), implying that p(x) := (x − α)(x − σ(α)) · · · (x − σ n−1 (α)) ∈ F [x]. But
p divides f , so p = f since f is irreducible. Hence, f splits in K, implying that K/F is normal.

Problem 3.13. Let K be a normal extension of F , and let f (x) ∈ F [x] be an irreducible polynomial over
F . Let g1 (x) and g2 (x) be monic irreducible factors of f (x) in K[x]. Prove that there is a σ ∈ Gal(K/F )
with σ(g1 ) = g2 .
Let M be an algebraic closure of F , and let α1 and α2 be roots of g1 and g2 respectively. By the Isomor-
phism Extension Theorem, there is an F -isomorphism σ of M such that σ(α1 ) = α2 . I claim that σ(g1 ) = g2 .
Indeed, by part 2 of Proposition 3.28, σ(K) = K, so σ(g1 ) ∈ K[x]. Then, since σ(α1 ) = α2 is a root of
σ(g1 ), we have that g2 divides σ(g1 ). Since σ is an isomorphism of K, and g1 is irreducible, σ(g1 ) = g2 is
irreducible as well. Hence, σ(g1 ) = g2 .

Problem 3.14. Let K be a normal extension of F , and let p(x) be an irreducible polynomial in F [x].
If p is not irreducible over K, show that p factors over K into a product of irreducible polynomials of the
same degree. In particular, if p has a root in K, then p splits over K.
From the previous problem, for any pair of irreducible factors of p (in K[x]), there is an element of
Gal(K/F ) one to the other. Thus, any pair of irreducible factors over K[x] must have the same degree. If p
has a root in K, then it has a factor of degree one in K[x]. Thus, p splits over K.

14
Problem 3.15. Let K and L be extensions of F . Show that KL is normal over F if both K and L
are normal over F . Is the converse true?
Let K and L be the splitting fields of SK ⊆ F [x] and SL ⊆ F [x], respectively. Let XK and XL be the set
of roots of SK and SL , respectively. Then K = F (XK ) and L = F (XL ). I claim that KL = F (XK ∪ XL ).
On one hand, KL is the smallest field containing K and L. Since F (XK ∪ XL ) contains F (XK ) = K and
F (XL ) = L, it contains KL. On the other hand, F (XK ∪ XL ) is the smallest field containing F , XK , and
XL . Since KL contains each of these, it contains F (XK ∪ XL ). Therefore, KL = F (XK ∪ XL ) as claimed.
This implies that KL is the splitting field of SK ∪ SL , so KL/F is normal. √
The converse√is not true in general. For example, note that although Q(e2πi/3 )Q( 3 2)/Q is normal (Ex-
ample
√ 3.26), Q( 3 2)/Q is not normal.√ To see this, note that x3 − 2 is irreducible over Q, it has a root in
3 3
Q( 2), but it does not split over Q( 2).

Problem 3.16. Let M be a normal extension of F . Suppose that a, a0 ∈ M are roots of min(F, a), and
that b, b0 are roots of min(F, b). Determine whether or not there is an automorphism σ ∈ Gal(M/F ) with
σ(a) = a0 and σ(b) = b0 . √
I think there is an issue with the way the question is√worded. For√instance, let M√= Q( 3 2, e2πi/3 ) and
F = Q. Then M/F is normal. Now choose a = a0 = 3 2 and b = 3 2, b0 = e2πi/3 3 2. Then there is no
automorphism σ satisfying the desired properties.
To avoid trivial examples such as the √ one above, lets insist that min(F, a)√6= min(F, b), and moreover
0 0
that
√ a 6
= a and b 6
= b . Now take M = Q( 6
2, eπi/3 ) and F =√Q. Note0 that√Q( 2, e
6 πi/3
)/Q is normal, since
6 πi/3
Q( 2, e√ ) is the splitting field √ of x 6
− 2 over Q.
√ Let a = 2 and a = − 2, so that a and
√ a0 are roots of
2 6 0 2πi/3 6 0 6 6
min(Q, 2) = x − 2. Let b = 2 and b = e 2, so that b and
√ b are roots of min(Q, √ 2) = √x − 2. If
0 3 0 3
there exists σ ∈ Gal(M/F ) such that σ(b) = b , then σ(a) = σ( 2) = σ(b ) = (b ) = 2 6= − 2. So the
answer is again negative; in general, there is not an F -automorphism of M satisfying the desired properties.

Problem 3.17. This problem will prove that any symmetric polynomial is a polynomial in the elementary
symmetric functions. This problem requires some knowledge of integral ring extensions along with theorems
about algebraic independence from Section 19. Let K = k(x1 , . . . , xn ) be the field of rational functions in the
xi over a field k. Then the group Sn acts as automorphisms on K as in Example 2.22. Let f ∈ k[x1 , . . . , xn ]
be a symmetric polynomial; that is, σ(f ) = f for all σ ∈ Sn . Show that f ∈ k[s1 , . . . , sn ].
Let s1 , . . . , sn be the elementary symmetric polynomials, defined as in Example 2.22. First, we show that
k[x1 , . . . , xn ] is an integral ring extension of k[s1 , . . . sn ]. Indeed, every xi is integral over k[s1 , . . . , sn ], since
xi is a root of
(t − x1 ) · · · (t − xn ) = tn − s1 tn−1 + · · · + (−1)n sn ,
where the above equality can be derived by induction on n.
Next, we show that the elementary symmetric polynomials are algebraically independent over k. To this
end, let f ∈ k[x1 , . . . , xn ] be non-zero. Define an order of the monomials as follows: xa1 1 · · · xann > xb11 · · · xbnn
if and only aj + · · · an > bj + · · · + bn for the first value of j for which aj + · · · + an 6= bj + · · · + bn . Call
this the S-order. It is easy to see that the S-order is well-defined, and that every polynomial has a unique
maximal term with respect to it.
Let cxa1 1 · · · xann be the maximal term of f (x1 , . . . , xn ) with respect to the S-order (assume c 6= 0). Then
the maximal term of ksa1 1 · · · sann with respect to the lexicographic order (call it the L-order) is

cx1a1 +···+an xa2 2 +···+an · · · xann .

Suppose xb11 · · · xbnn appears in f (s1 , · · · sn ). Then it must appear in some term sc11 · · · scnn , where xc11 · · · xcnn
appears in f (x1 , . . . , xn ). Then by definition of the S-order

xa1 1 +···+an xa2 2 +···+an · · · xann ≥ x1c1 +···+cn x2c2 +···+cn · · · xcnn ≥ xb11 · · · xbnn ,

with respect to the L-order. Therefore, cx1a1 +···+an xa2 2 +···+an · · · xann is the maximal element of f (s1 , . . . , sn ).
Since c 6= 0, we conclude that f (s1 , . . . , sn ) 6= 0, and therefore that the elementary symmetric polynomials
are algebraically independent over k.
Finally, we show that if f is symmetric, then f ∈ k[s1 , . . . , sn ]. Since k[x1 , . . . , xn ]/k[s1 , . . . , sn ] is an
integral ring extension, f satisfies a monic polynomial with coefficients in k[s1 , . . . , sn ]. We note that since f

15
is symmetric, f ∈ k(s1 , . . . , sn ), the field of rational functions in the elementary symmetric polynomials. This
is shown in Example 3.9. By Lemma 19.5, k[x1 , . . . , xn ] is k-isomorphic to k[s1 , . . . , sn ]. Since k[x1 , . . . , xn ]
is integrally closed in its fraction field, so is k[s1 , . . . , sn ]. Therefore f ∈ k[s1 , . . . , sn ].

Problem 3.18. Give an example of fields k ⊆ K ⊆ L and l ⊆ L for which l/k and L/K are algebraic, k is
algebraically closed in K, and lK = L, but l is not algebraically closed in L.
I don’t know how to do this problem, and I think there is possibly an error in the statement. In any
case, I asked about it on Stack Exchange and I was given a partial answer, which can be found here. Using
the primitive element theorem (more specifically, Corollary 5.8), the user gives a proof that the claim cannot
hold in characteristic 0.

Problem 3.19. This problem gives a construction of an algebraic closure of a field, due to E. Artin.
Let F be a field, and let S be the set of all monic irreducible polynomials in F [x]. Let A = F [xf : f ∈ S]
be a polynomial ring in one variable for each polynomial in S. Let I be the ideal generated by all f (xf ) for
f ∈ S Show that I 6= A. Let M ⊇ I be a maximal ideal of A, and let F1 = A/M . Then F1 is an extension
of F in which each f ∈ S has a root. Given the field Fi , construct
S∞ the field Fi+1 by repeating this procedure
starting with Fi as the base field in place of F . Let L = n=1 Fn . Show that each f ∈ S splits into linear
factors over L, and show that the algebraic closure of F in L is an algebraic closure of F .
Suppose that 1 ∈ I. Then there is a finite collection {f1 , . . . , fk } ⊆ S such that

1 = a1 (x1 , . . . , xk )f1 (x1 ) + · · · + ak (x1 , . . . , xk )fk (xk )

for some ai ∈ A. We can actually assume that the ai lie in F [x1 , . . . , xn ], since if they did, we could set the
additional variables to zero in the above equation. Moreover, we assume that k is the minimal integer such
that such an equation holds. Therefore, the ideal J = (f1 , . . . , fn−1 ) is not the entire ring, so in the quotient
A/J, the equation becomes
1 = ak fk ,
where the bar notation denotes the images of polynomials under A 7→ A/J. But fk cannot be a unit, as it
is monic and of degree at least 1. We conclude, from this contradiction, that I is a proper ideal of A.
We now show that the natural map F → F1 is injective. Suppose that α ∈ F , and α 7→ 0. Then α ∈ M ,
which implies α = 0, because otherwise M = A. Thus, we can view F1 as an extension of F . Moreover,
every f ∈ S has a root in F1 , namely xf .
Now let f ∈ S; we will show that f splits over L. Suppose that f has degree n. We know that f has a
root in F1 , so it factors over F1 as a product of a degree n − 1 polynomial f1 and a degree 1 polynomial.
By the same reasoning, f1 factors over F2 as a product of a degree n − 2 polynomial f2 and a degree 1
polynomial. By induction, we find that f factors as a product of linear factors over Fn ⊆ L.
Let K be the algebraic closure of F in L, and let f be polynomial in K[x]. We can construct an
extension of K in which f has a root β. Then K(β) is algebraic over F , so there exists a minimal polynomial
p = min(F, β). Then f divides p. But then, since p splits into linear factors over K, f must splits over K as
well, so f has a root in K. Hence, K is algebraically closed, and therefore an algebraic closure of F .

Section 4. Separable and Inseparable Extensions

Problem 4.1. Prove the sum, product, and chain
Pn rules for formal polynomial
Pn differentiation in F [x].
Let f, g ∈ F [x]. We can then let f (x) = i=0 ai xi and g(x) = i=0 bi xi (some of the coefficients may
be zero). Then,
n
!0 n n n
X X X X
(f + g)0 (x) = (ai + bi )xi = i(ai + bi )xi−1 = iai xi−1 + ibi xi−1 = f 0 (x) + g 0 (x),
i=0 i=1 i=1 i=1

which proves the sum rule.

16
 For the product rule, we will induct on n. The base case n = 0 is trivial. Now suppose the product rule
holds for polynomials of degree smaller than n. We will use the sum rule freely. Then,
 0  !  0
Xn n
X n−1
X n−1
X
(f g)0 (x) =  ai xi · bj xj  =  an xn + ai xi · bn xn + bj xj 
i=0 j=0 i=0 j=0
 0
n−1
X n−1
X n−1
X n−1
X
= an bn x2n + an bj xn+j + bn ai xn+i + ai xi · bj x j 
j=0 i=0 i=0 j=0
n−1
X n−1
X
= 2nan bn x2n−1 + an (n + j)bj xn+j−1 + bn (n + i)ai xn+i−1
j=0 i=0
!0 n−1  ! n−1 0
n−1
X X n−1
X X
+ ai xi  bj xj  + ai xi  bj xj 
i=0 j=0 i=0 j=0
n−1
X n−1
X n−1
X
2n−1 n−1 j n j−1 n−1
= 2nan bn x + nan x bj x + an x jbj x + nbn x ai xi
j=0 j=0 i=0
n−1
X n−1
X n−1
X n−1
X n−1
X
+ bn xn iai xi−1 + iai xi−1 · bj x j + ai xi · jbj xj−1
i=0 i=0 j=0 i=0 j=0
n
X n
X n
X n−1
X n−1
X n
X
n j−1 n i−1 i−1 j i
= an x jbj x + bn x iai x + iai x · bj x + ai x · jbj xj−1
j=0 i=0 i=0 j=0 i=0 j=0
n
X n
X n
X n
X
= iai xi−1 · bj xj + a i xi · jbj xj−1
i=0 j=0 i=0 j=0
0 0
= f (x)g(x) + f (x)g (x),

which proves the product rule.

Finally, we prove the chain rule. From the product rule, it follows that
m
!0
Y
fi (x) = f10 (x)f2 (x) · · · fm (x) + f1 (x)f20 (x)f3 (x) · · · fn (x) + · · · + f1 (x) · · · fn−1 (x)fn (x),
i=1

where each fi is a polynomial. From this we have that the formal derivative of (f (x))m is m(f (x))m−1 f 0 (x).
It is also clear that the product rule implies that (af (x))0 = af 0 (x) for any a ∈ F . From the above, and an
application of the sum rule, we have
n
!0 n
X X
0
(g(f (x))) = bn (f (x)) i
= bn i(f (x))i−1 f 0 (x) = g 0 (f (x))f 0 (x).
i=0 i=0

Problem 4.2. If F ⊆ L ⊆ K are fields such that K/F is separable, show that L/F and K/L are separable.
Since every element in K is separable over F , so is every element in L. Therefore, L/F is a separable
extension. Next, let α and K. Note that min(L, α) divides min(F, α), which has no repeated roots in any
splitting field by separability of K/F . Hence, min(L, α) has no repeated roots in any splitting field, implying
that K/L is separable.
m
Problem 4.3. If K is a field extension of F and if α ∈ K is not separable over F , show that αp is
separable over F for some m ≥ 0, where p = char(F ).
m
Let f (x) = min(F, α). By Proposition 4.6, f (x) = g(xp ) for some m ≥ 0, where g ∈ F [x] is a separable
m m
irreducible polynomial over F . Since αp is a root of g and g is separable, we have that min(F, αp ) = g.
m
Since g is separable over F , αp is separable over F .

17
Problem 4.4. Let F ⊆ L ⊆ K be fields such that K/L is normal and L/F is purely inseparable. Show that
K/F is normal.
First assume that the fields have characteristic 0. The only way that L/F can be purely inseparable is
if L = F . Then K/F is normal by assumption.
Now assume that the fields have characteristic p > 0. Let α ∈ K and f (x) := min(F, α). Our goal is to
show that f (x) splits over K. By normality of K/L, l(x) := min(L, α) splits over K. Since L/F is purely
n
inseparable, Lemma 4.16 implies that for each β ∈ L, there is an n ∈ N such that β p ∈ F . Then, we can find
n n n
an n ∈ N such that l(x)p ∈ F [x]. Then f divides lp . Since l splits over K, so does lp , and consequently
so does f .

Problem 4.5. Let F be a field of characteristic p > 0, and let a ∈ F − F p . Show that xp − a is irre-
ducible over F .
Let K be an algebraic closure of F and let α ∈ K be a root of xp − a. Then αp ∈ F . By Lemma 4.16,
min(F, α) = (x − α)p = xp − a. Therefore, xp − a is irreducible.

Problem 4.6. Let F be a field of characteristic p > 0, and let K be a purely inseparable extension of
n
F with [K : F ] = pn . Prove that ap ∈ F for all a ∈ K.
m m
By Lemma 4.6, we know that min(F, a) = (x − a)p for some m. Then ap ∈ F . Moreover, [F (a) : F ] =
n
pm ≤ pn = [K : F ], so ap ∈ F .

Problem 4.7. Let K and L be extensions of F . Show that KL is separable over F if both K and L
are separable over F . Is the converse true?
Note that KL = F (K ∪ L), i.e. KL is the smallest field containing K and L. By Proposition 1.10,
[
KL = {F (a1 , . . . , an ) : a1 , . . . , an ∈ K ∪ L}.

Now, let α ∈ KL. Then α ∈ F (a1 , . . . , an ) for some a1 , . . . , an ∈ K ∪ L. By Corollary 4.10, F (a1 , . . . , an )
is separable over F , since each ai is separable over F . Hence, α is separable over F . Since α ∈ KL was
arbitrary, KL/F is separable.
The converse is true by Problem 4.2.

Problem 4.8. Let K and L be extensions of F . Show that KL is purely inseparable over F if both K
and L are purely inseparable over F . Is the converse true?
Note that KL = F (K ∪ L) (as explained in the previous problem). Since every element of K ∪ L is purely
inseparable over F , we have that KL is purely inseparable over F (by Lemma 4.17).
Yes, the converse is true. If KL is purely inseparable over F , then every element of KL is purely insep-
arable over F . Since K, L ⊆ KL, every element of K and every element of L is purely inseparable over F .
Therefore, K and L are each purely inseparable over F .

Problem 4.9. Let K and L be extensions of F . Show that KL is Galois over F if both K and L are
Galois over F . Is the converse true?
By Theorem 4.9, being a Galois is equivalent to being normal and separable. Thus, we will show that
KL is normal and separable over F . By Problem 4.7, KL is separable over F . By Problem 3.15, KL is also
normal over F , so KL is Galois over F . √
The converse is not true. In our solution of Problem 3.15, we showed that Q(e2πi/3 )Q( 3 2)/Q √ is normal.
Moreover, this extension is also separable since the fields are of√characteristic 0, so Q(e2πi/3 )Q( 3 2)/Q is Ga-
lois. However, as discussed in the solution of Problem 3.15, Q( 3 2)/Q is not normal and therefore not Galois.

Problem 4.10. Let K and L be subfields of a common field, both of which contain a field F . Prove
the following statements.
(a) If K = F (X) for some set X ⊆ K, then KL = L(X).
Since L, X ⊆ KL, we have that L(X) ⊆ KL. On the other hand, since K is the smallest field containing

18
 F and X, and L(X) contains both L and X, we have that K ⊆ L(X). Moreover, L ⊆ L(X). Therefore,
KL ⊆ L(X), so KL = L(X).
(b) [KL : F ] ≤ [K : F ] · [L : F ].
From Problem 1.17, we know that this inequality holds if L and K are finite extensions. The inequality
is obvious if either extension is infinite, so there is nothing to show.
(c) If K and L are algebraic over F , then KL is algebraic over F .
Again, note that KL = F (K ∪ L). Since K/F and L/F are algebraic, Proposition 1.23 implies that
KL/F is algebraic.
(d) Prove that the previous statement remains true when “algebraic” is replaced by “normal,” “separable,”
“purely inseparable,” or “Galois.”
This is the result of Problems 4.7, 4.8, and 4.9.
Problem 4.11. Let K be the rational function field k(x) over a perfect k of characteristic p > 0. Let
F = k(u) for some u ∈ K, and write u = f (x)/g(x) with f and g relatively prime. Show that K/F is a
separable extension if and only if u ∈ / K p.
p
(⇒) Suppose u ∈ K . From Example 1.17, we have that the minimal polynomial of x is m(t) =
ug(t) − f (t). Since u ∈ K p , we have that f, g ∈ k[x]p . Let u = ũp , f = f˜p , and g = g̃ p (this actually requires
a little bit of work!). Then the minimal polynomial factors in K as m(t) = (ũg̃(t) − f˜(t))p . Hence, x is a
repeated root of its minimal polynomial, so K/F is not separable.
(⇐) Let u ∈ / K p . Again, let m(t) = ug(t) − f (t) be the minimal polynomial of x over F . By Proposition
n
4.6, m(t) = h(tp ) for some n ≥ 0 and some h(t) ∈ F [t] that is irreducible and separable. Now, since u ∈ / K p,
p p s
either f (t) ∈
/ K or g(t) ∈ / K . Then, since k is perfect, either f or g must contain a t where s is not a
multiple of p. Then ug(t) − f (t) contains a ts where s is not a multiple of p. Therefore, n must be zero, so
that m(t) = h(t). Hence, m is separable, which makes K/F separable.

Problem 4.12. Let K be a finite extension of F with charf = p > 0 and K p ⊆ F . Thus, K/F is
purely inseparable. A set {a1 , . . . , an } ⊆ K is said to be a p-basis for K/F provided that there is a chain of
proper extensions
F ⊂ F (a1 ) ⊂ F (a1 , a2 ) ⊂ · · · ⊂ F (a1 , . . . , an ) = K.
Show that if {a1 , . . . , an } is a p-basis K/F , then [K : F ] = pn , and conclude that the number of elements
in a p-basis is uniquely determined by K/F . Then number n is called the p-dimension of K/F . Also, show
that any finite purely inseparable extension has a p-basis.
Note the K/F is purely inseparable as a direct consequence of Lemma 4.16. Next we show that

[F (a1 , . . . , ai+1 ) : F (a1 , . . . , ai )] = p.

First, F (a1 , . . . , ai ) ⊂ F (a1 , . . . , ai+1 ), so

[F (a1 , . . . , ai+1 ) : F (a1 , . . . , ai )] > 1.

Next, api+1 ∈ F ⊂ F (a1 , . . . , ai ), so min(F (a1 , . . . , ai ), ai+1 ) = (x − ai+1 )p by Lemma 4.16. Then,

[F (a1 , . . . , ai+1 ) : F (a1 , . . . , ai )] = p.

This in turn implies that [K : F ] = pn . The number of elements in a p-basis is therefore determined by the
degree of the extension, which means that every p-basis has the same number of elements.
Let K/F be an arbitrary finite purely inseparable extension. If the fields are of characteristic 0, then
K n= F and there is nothing to show. Let charF = p. By Lemma 4.16, there is a minimal n1 ∈ N such that
1
α1p ∈ F . We will construct a p-basis with n elements. Let α1 ∈ K − F . By Lemma 4.17, [K : F ] = pn for
n1 −1
some n. Then, let a1 = α1p . Then a1 ∈ K − F and ap1 ∈ F . Hence, F ⊂ F (a1 ) and [F (a1 ) : F ] = p. Now
suppose that a1 , . . . , ai have been chosen such that

F ⊂ F (a1 ) ⊂ F (a1 , a2 ) ⊂ · · · ⊂ F (a1 , . . . , ai )

19
and
[F (a1 , . . . , aj+1 ) : F (a1 , . . . , aj )] = p
for each j ∈ {1, . . . , i − 1}. Now choose αi+1 ∈ K − F (a1 , . . . , ai ). By Lemma 4.16, there is a minimal
pni+1 pni+1 −1
ni+1 ∈ N such that αi+1 ∈ F (a1 , . . . , ai ). Then, let ai+1 = αi+1 , so that

F ⊂ F (a1 ) ⊂ F (a1 , a2 ) ⊂ · · · ⊂ F (a1 , . . . , ai ) ⊂ F (a1 , . . . , ai+1 )

and
[F (a1 , . . . , ai+1 ) : F (a1 , . . . , ai )] = p.
By inductively continuing this procedure, we obtain a p-basis {a1 , . . . , an } for K/F . Note that the assump-
tion K p ⊆ F is unnecessary for the existence of a p-basis.

Problem 4.13. Give three examples of a field extension K/F which is neither normal nor separable.
Note that two such examples are given in the section.
The field extension
√ F2 (x1/6 )/F2 (x) given in Example 4.18 is neither normal√nor separable. As discussed
in the example, x is purely√ inseparable over F2 (x). We will now show that 3 x is not normal √ over F2 by
showing that min(F2 (x), x) does not split over F2 (x1/6 ). First of all, we claim that min(F2 (x), 3 x) = t3 +x.
3

We can use Eisenstein’s criterion here, since F2 [x] is a UFD. It then follows directly that t3 + x is irreducible
by using Eisenstein’s criterion with x as our prime element. Now we must show that t3 + x does not split
over F2 (x1/6 ), which is equivalent to showing that t2 + x1/3 t + x2/3 does not split over F2 (x1/6 ). If it did,
then it would have a root in F2 (x1/6 ). By the rational root test, the root would be of the form α/β where α
divides x2/3 and β divides 1. Then we can let β = 1. We then have that α equals 1, x1/6 , x1/3 , x1/2 , or x2/3 .
Simply plugging in each of the options shows that none of them can be roots of t2 + x1/3 t + x2/3 , which
concludes that the extension
The field extension in Example 4.24 is neither normal nor separable. More precisely, the extension is
√
K/F where F = k(x, y), k is a field of characteristic 2, and K = k(x, y, u, uy) where u is a root of t2 + t + x.
The example explains why K/F is not separable. Theorem 4.23 states that if K/F is normal, then K = SI.
Since Example 4.24 shows that K 6= SI, we have that K/F is not normal.
The last example will be a generalization of the first one. We claim that the extension Fp (x1/pq )/Fp (x)
(where p is a prime, and q is any integer greater than 1 and coprime to p − 1 and p) is neither normal
nor separable. To show that the extension is not separable, simply note that min(Fp (x), x1/p ) = tp − x
splits as (t − x1/p )p over Fp (x1/pq ). We will now show that x1/q is not normal over Fp (x). Note that
min(Fp (x), x1/q ) = tq − x, since tq − x is irreducible over Fp (x) by Eisenstein’s criterion. So we want to show
that tq − x does not split over Fp (x1/pq ). Note that the roots of tq − x in Fp (x1/pq ) are of the form ax1/q ,
where a ∈ Fp and aq = 1. Since gcd (q, p − 1) = 1, it follows that a = 1 and that x1/q is the only root of
tq − x. However, tq − x does not split as (t − x1/q )q , since gcd (p, q) = 1. Therefore, tq − x does not split in
Fp (x1/pq ), and hence x1/q is not normal over Fp (x). Note that it might be possible to relax the conditions
on p and q listed above.

Problem 4.14. Let k be a field of characteristic p > 0, let K = k(x, y) be the rational function field
over k in two variables, and let F = k(xp , y p ). Show that K/F is a purely inseparable extension of degree
p2 . Show that K 6= F (a) for any a ∈ K.
Before beginning, note that this problem provides an example where Corollary 5.8 fails for fields of char-
acteristic p > 0. Since x and y satisfy the polynomials tp − xp = (t − x)p and tp − y p = (t − y)p , it follows that
x and y are purely inseparable over F . Since K = F (x, y), it follows that K/F is a purely inseparable exten-
sion. Since k is a field of characteristic p, we have that K p ⊆ F . Next, note that F ⊂ F (x) ⊂ F (x, y) = K,
so {x, y} is a p-basis for K/F . Therefore, [K : F ] = p2 . Suppose that K = F (a) for some a ∈ K. Since
K p ⊂ F , a satisfies the polynomial tp − ap ∈ F [t], implying that p2 = [K : F ] = [F (a) : F ] ≤ p, a contradic-
tion. Therefore, K 6= F (a) for any a ∈ K.

Problem 4.15. Prove the following product formulas for separability and inseparability degree: If F ⊆
L ⊆ K are fields, then show that [K : F ]s = [K : L]s [L : F ]s and [K : L]i [L : F ]i .
I will assume that each field extension is finite, since separability and inseparability degree are only
defined for finite extensions. The inseparability product formula is exactly the statement of Lemma 8.11.

20
From this, we have
[K : L][L : F ] [K : F ]
[K : L]s [L : F ]s = = = [K : F ]s .
[K : L]i [L : F ]i [K : F ]i

Section 5. The Fundamental Theorem of Galois Theory

Problem 5.1. A transitive subgroup of Sn is a subgroup G such that for each i, j ∈ {1, . . . , n}, there is a
σ ∈ G with σ(i) = j. If K is the splitting field over F of a separable irreducible polynomial f (x) ∈ F [x] of
degree n, show that |Gal(K/F )| is divisible by n and that Gal(K/F ) is isomorphic to a transitive subgroup
of Sn . Conclude that [K : F ] divides n!.
The splitting field of an irreducible separable polynomial is normal and separable over F . Therefore, K/F
is Galois, and | Gal(K/F )| = [K : F ] = [K : F (α)][F (α) : F ], where α is a root of f . Since [F (α) : F ] = n, n
divides | Gal(K/F )|.
Since every element of Gal(K/F ) maps the set of roots of f to itself, and the n roots are pairwise distinct,
every element of the Galois group corresponds to a permutation in Sn . Hence, Gal(K/F ) is isomorphic to
a subgroup of Sn . By the Isomorphism Extension Theorem, for each pair of roots {α, β} of f , there exists
τ ∈ Gal(K/F ) such that τ (α) = β. Hence, Gal(K/F ) is isomorphic to a transitive subgroup of Sn . By
Lagrange’s theorem, | Gal(K/F )| = [K : F ] divides n!.

Problem 5.2. Write down all the transitive subgroups of S3 and S4 .

We begin with S3 . Clearly, S3 is a transitive subgroup of S3 . Moreover, h(123)i ∼ = Z3 is a transitive
subgroup. It is easy to check that every other subgroup is not transitive.
Now for S4 . I appealed to a classification of the subgroups of S4 to complete this list. Again, S4 is clearly
a transitive subgroup. The subgroups generated by 4-cycles are also transitive; namely, h(1234)i, h(1243)i,
and h(1324)i, which are all isomorphic to Z4 , are transitive. We also have three copies of D4 , the dihedral
group of order 8, which are all transitive: h(1234)(13)i, h(1243)(14)i, and h(1324)(12)i. Next we have a copy
of the Klein group Z2 × Z2 , given by h(12)(34), (13)(24)i. Finally, there is the alternating group A4 given by
h(123), (12)(34)i.

Problem 5.3. Determine all the transitive subgroups G of S5 for which |G| is a multiple of 5. For each
transitive subgroup, find a field F and an irreducible polynomial of degree 5 over F such that if K is the
splitting field of f over F , then Gal(K/F ) is isomorphic to the given subgroup.
First of all, all transitive subgroups of S5 are multiples of 5. To see this, let G ≤ S5 act on {1, 2, 3, 4, 5}
by permutation. Since G is transitive, the Orbit-Stabilizer Theorem states that, for any x ∈ {1, 2, 3, 4, 5},
|G| = |Gx| · |Stab(x)| = 5|Stab(x)|. Therefore, the question is simply asking to find all transitive subgroups
of S5 .
We begin with the most obvious subgroup: S5 itself. Consider the polynomial f (t) = (t − x1 )(t − x2 )(t −
x3 )(t − x4 )(t − x5 ) ∈ Q(s1 , s2 , s3 , s4 , s5 )[t]. The splitting field of f is K := Q(x1 , x2 , x3 , x4 , x5 ), as shown
in Example 3.9. Moreover, Gal(K/Q(s1 , s2 , s3 , s4 , s5 )) ∼ = S5 , which is also shown in Example 3.9. We know
that f is irreducible because the Galois group acts transitively on its roots.
Now, for each transitive subgroup G ⊆ Gal(K/Q(s1 , s2 , s3 , s4 , s5 )), we have that Gal(K/F(G)) ∼ = G, by
the Fundamental Theorem of Galois Theory. Moreover, K is clearly the splitting field of f (as defined above)
over G. Moreover, f is irreducible over F(G), since G acts transitively on the roots.
What remains to do is to find the transitive subgroups of S5 . The possible orders are 5, 10, 15, 20, 30,
40, and 60.
For order 5, we have Z5 , generated by any 5-cycle.
There are two groups of order 10: D5 and Z10 . It is easy to see that S5 contains a copy of D5 . Since
there is no element of order 10 in S5 , Z10 is not isomorphic to a subgroup of S5 .
There is only one group of order 15: Z15 . Since S5 does not contain an element of order 15, Z15 is not
isomorphic to a subgroup of S5 .
There are five groups of order 20, three of which we can do away with immediately; namely, Z20 , Z2 ×Z10 ,
and D10 are not isomorphic to subgroups of S5 since they all contain elements of order 10. The next group
is F20 , the Frobenius group of order 20. This is the group of all affine maps ax + b, where a ∈ F∗5 and b ∈ F5 .
F20 can be seen to be a subgroup of S5 by observing how the affine maps act on the elements of F5 . It is

21
also clear that F20 is a transitive subgroup of S5 . The last group of order 20 is the dicyclic group Dic10 ; it
contains an element of order 10, and is therefore not isomorphic to a subgroup of S5 .
We can quickly show that S5 contains no group of order 30 as follows. Suppose G ≤ S5 and |G| = 30. If
G ≤ A5 , then G is a normal subgroup of A5 , as [A5 : G] = 2. However, A5 is simple so this cannot be the
case. Then |A5 ∩ G| = 15. However, we have already shown that S5 does not contain a subgroup of order
15.
Next, S5 cannot contain a group G of order 40. If it did, G cannot be a subgroup of A5 , since 40 - 60.
Therefore G ∩ A5 = 20, implying that G ∩ A5 ∼ = F2 0. However, F20 contains odd elements, such as 2x, so
this cannot be the case. hence S5 contains no subgroups of order 40.
Finally, it is well known that the only subgroup of order 60 is A5 , and it is transitive.

Problem 5.4. In the following problems, let K be the splitting field of f (x) over F . Determine Gal(K/F )
and find all the intermediate subfields of K/F .

(a) F = Q and f (x) = x4 − 7.

√ √ √ √
In C, we √ have x4 − 7 = (x − 4 7)(x + 4 7)(x − i 4 7)(x + i 4 7). Therefore, √the splitting √ field
√ is
K = Q(i, 4 7). Let φ ∈ Gal (K/F ). Then it is easy to check that φ(i) = ±i and φ( 4 7) ∈ {± 4 7, ±i 4 7}.
We see that there are eight possible F -automorphisms. Moreover, these eight possibilities are indeed
F -automorphisms, since they can be generated by two known F √-automorphisms;
√ namely complex con-
jugation, denoted by τ , and the map σ which fixes i and maps 4 7 to i 4 7. Therefore, | Gal (K/F )| = 8,
which means Gal(K/F ) ∼ = D4 by Problem 5.2. The subgroups of D4 and the corresponding interme-
diate subfields are shown below.
D4

hσ 2 , τ i hσi hσ 2 , τ σi

hτ σ 2 i hτ i hσ 2 i hτ σi hτ σ 3 i

hidi

Subgroup lattice for D4 .

√
4
Q(i, 7)

√ √ √ √ √
Q(i 4 7) Q( 4 7) Q(i, 7) Q((i − 1) 4 7) Q((i + 1) 4 7)

√ √
Q( 7) Q(i) Q(i 7)

Q
√
4
Intermediate subfields of Q(i, 7)/Q.

22
(b) F = F5 and f (x) = x4 − 7.
In Problem 3.9, we found the splitting field to be K = F5 [α]/(α4 − 7) and that x4 − 7 = (x − α)(x −
2α)(x − 3α)(x − 4α). Since the splitting field is K = F ([α]) and min(F, [α]) = x4 − 7, we have that
[K : F ] = 4, and therefore, Gal(K/F ) must be a transitive subgroup of S4 of order 4. Let σ be the
F -automorphism that maps α to 2α. Then Gal(K/F ) = hσi ∼ = Z4 . The subgroups of Z4 and the
corresponding intermediate subfields are shown below.

Z4 F5 ([α])

hσ 2 i F5 ([α2 ])

hidi F5

Subgroup lattice for Z4 . Intermediate subfields of F5 ([α])/F5 .

(c) F = Q and f (x) = x5 − 2.

√ √ √ √ √
Let ω = e2πi/5 . In C, we have x5 − 2 = (x −√5 2)(x − 5 2ω)(x − 5 2ω 2 )(x − 5 2ω 3 )(x − 5 2ω 4 ). Hence,
the splitting field of f over F is K√= Q(ω, 5 2). Therefore, the elements of Gal(K/F ) are completely
2
determined by their action on {ω, 5 2}.√ Let τ√be the automorphism determined by ω 7→ ω , and let σ
be the automorphism determined by 5 2 7→ 5 2ω. Then Gal(K/F ) = hσ, τ i = {τ j σ i : 1 ≤ i ≤ 5, 1 ≤
j ≤ 4}. Since the Gal(K/F ) has order 20 and it is isomorphic to a transitive subgroup of S5 , we have
that Gal(K/F ) ∼ = F20 by the previous problem. The subgroup lattice of F20 and the corresponding
intermediate subfields are shown below.
F20

hτ i hτ σ 4 i hτ σ 3 i hτ σ 2 i hτ σi hσ, τ 2 i

hτ 2 i hτ 2 σi hτ 2 σ 2 i hτ 2 σ 3 i hτ 2 σ 4 i hσi

hidi

Subgroup lattice for F20 .

√
5
Q(ω, 2)
√ √ √ √ √
Q(ω 2 + ω 3 , 5
2) Q(ω 2 + ω 3 , 5
2) Q(ω 2 + ω 3 , 5
2) Q(ω 2 + ω 3 , 5
2) Q(ω 2 + ω 3 , 5
2) Q(ω)

√ √ √ √ √
Q( 5 2) Q(ω 2 5 2) Q(ω 4 5 2) Q(ω 5 2) Q(ω 3 5 2) Q(ω 2 + ω 3 )

Q
√
5
Intermediate subfields of Q(ω, 2)/Q.

23
 (d) F = F2 and f (x) = x6 + 1.
Note that x6 +1 = (x3 +1)2 = (x+1)2 (x2 +x+1)2 , so really we are after the splitting field of x2 +x+1.
Since x2 +x+1 has no roots, it is irreducible. Hence the splitting field is K = F2 [α]/(α2 +α+1) = F ([α]).
Hence, Gal(K/F ) ∼= Z2 . The subgroup lattice of Z2 is trivial, and therefore, there are no intermediate
subfields.
Z2 F2 ([α])

hidi F2

Subgroup lattice for Z2 . Intermediate subfields of F2 ([α])/F2 .

(e) F = Q and f (x) = x8 − 1.

The splitting field is K = Q(ω) where ω = eπi/4 . Note that x8 − 1 has an irreducible factor of x4 + 1, so
Gal(K/F ) will permute its roots, namely ω, ω 3 , ω 5 , and ω 7 . Note that none of the F -automorphisms
defined by ω 7→ ω k (with k = 1, 3, 5, 7) generates all four automorphisms. Hence, Gal(K/F ) ∼ = Z2 × Z2 .
Let σ be defined by σ(ω) = ω 3 and τ be defined by τ (ω) = ω 5 . Then, we have the following subgroup
lattice and intermediate subfields.
Z2 × Z2 Q(ω)

hτ i hστ i hσi Q(ω + ω 5 ) Q(ω + ω 7 ) Q(ω + ω 3 )

hidi Q

Subgroup lattice for Z2 × Z2 . Subgroup lattice for Z2 × Z2 .

Problem 5.5. Let K be a Galois extension of F with [K : F ] = n. If p is a prime divisor of n, show that
there is a subfield L of K with [K : L] = p.
By the FTGT, we have that | Gal(K/F )| = n. Since n is divisible by p, Gal(K/F ) contains an element
of order p, and thus a subgroup H of order p. Then, choosing L = F(H), by the FTGT we have that
[K : L] = |H| = p.

Problem 5.6. Let N be a Galois extension of F with Gal(N/F ) = A4 . Show that there is no intermediate
field L of N/F with [L : F ] = 2.
By the FTGT, if there was a subfield L such that [L : F ] = 2, then A4 would have a subgroup H such
that [A4 : H] = 2. Hence, |H| = 6, which means that either H ∼ = Z6 or H ∼ = S3 . We know that H ∼ = Z6 is
not possible, because Z6 contains an element of order 6, while A4 does not. Suppose that H ∼ = S3 . Then
H must contain three elements of order 2, so H must contain a product of transpositions. But H must
also contain an element of order 3, i.e. a 3-cycle. But A4 is generated by a product of transpositions and a
3-cycle, which implies that H = A4 , a contradiction. We conclude that there is no intermediate subfield L
with [L : F ] = 2.

Problem 5.7. Give examples of field extensions K/F with

(a) K/F normal but not Galois,
Let M be the algebraic closure of F = Fp (x), and let α ∈ K be a root of f (t) = tp − x ∈ F [t]. By
Problem 4.5, tp − x is irreducible. Moreover, f (t) splits as (t − α)p over M . Then let K = F (α). Then
K/F is normal since K is the splitting field of f over F . However, K/F is not separable, because
α∈/ F is purely inseparable over F . Hence, K/F is not Galois.

24
 (b) K/F separable but not Galois.
√
Let K = Q( 3 2) and F = Q. Then K/F
√ is separable because the fields are characteristic 0. However,
K/F is not normal because min(Q, 3 2) = x3 − 2 does not split over K. Hence, K/F is not Galois.
Problem 5.8. Let K/F be Galois with G = Gal(K/F ), and let L be an T intermediate field. Let N ⊆ K be
−1
the normal closure of L/F . If H = Gal(K/L), show that Gal(K/N ) = σ∈G σHσ .
that Gal(K/N ) ⊆ σ∈G σHσ −1 . Let τ ∈ Gal(K/N ). Let τ ∈ Gal(K/N ). We want
T
We begin by showing
to show that τ ∈ σ∈G σHσ −1 , which is equivalent to showing τ ∈ σHσ −1 for all σ ∈ G, which is equivalent
T
to showing that σ −1 τ σ ∈ H for all σ ∈ G. To this end, let α ∈ L. We will first show that σ(α) ∈ N . Note
that α is the root of the minimal polynomial f (t) = min(F, α), and therefore so is σ(α). Since N is the
splitting field of min(F, a) : a ∈ L, σ(α) ∈ N . Therefore, σ −1 τ σ(α) = σ −1 σ(α) = α. Hence, σ −1 τ σ fixes L,
implying σ −1 τ σ ∈ H, as desired.
Now we will show that show that Gal(K/N ) = σ∈G σHσ −1 . I am going to assume that K/F is finite
T
so that
T we can use the FTGT. Otherwise, I’m not sure how to prove this inclusion.

−1 We T begin by −1showing
−1 −1
τ −1 =
T
that σ∈G σHσ is normal in G. To see normality, we have τ σ∈G σHσ τ = σ∈G τ σHσ
−1 −1
T T
σ∈G (τ σ)H(τ σ) = σ∈G σHσT after re-indexing.
We have that Gal(K/N ) ⊆ σ∈G σHσ −1 ⊆ Gal(K/L). Then, let σ∈G σHσ −1 correspond
T
to the in-
termediate field M under the Galois correspondence, we have that N ⊇ M ⊇ L. Since σ∈G σHσ −1 is a
T
normal subgroup of G, M/F is Galois and therefore normal. T Hence, M/L is normal as well. Hence, either
M = N or M = L by Proposition 5.9. If M = N , then σ∈G σHσ −1 = Gal(K/N ) by the FTGT, and we
are done. If M = L, then L/F is normal, implying that L = N = M , which brings us back to the previous
case.

Problem 5.9. Let K be a Galois extension of F and let a ∈ K. Let n = [K : F ], r = [F (a) : F ],

and H = Gal(K/F (a)). Let τ1 , . . . , τr be left coset representatives of H in G. Show that min(F, a) =
Q r
i=1 (x − τi (a)). Conclude that
Y
(x − σ(a)) = min(F, a)n/r .
σ∈Gal(K/F )

First, note that τi (a) does not depend on the choice of coset representative, since two representatives
for the same coset differ on the right by an element of Gal(K/F (a)), i.e. by an element that fixes a. Since
[F (a) : F ] = r, we know that deg(min(F, a)) = r. Moreover, we can let the τi representing
Qr the coset eH = H
be the identity (since theQchoice doesn’t matter). Then it is clear that a is a root of i=1 (x − τi (a)), which
r
proves that min(F, a) = i=1 (x − τi (a)). By the FTGT, [G : H] = [F (a) : F ] = r, so |H| = n/r. Since τi (a)
is independent of the choice of coset representative τi , and there are n/r representatives for each coset, it
follows immediately that Y
(x − σ(a)) = min(F, a)n/r .
σ∈Gal(K/F )

Problem 5.10.Let K be a finite Galois extension of F , and let a ∈ K. Let La : K → K be the F -

linear
Q transformation defined by La (b) = ab. Show that the characteristic polynomial of La is equal to
σ∈Gal(K/F ) (x − σ(a)) and the minimal polynomial of La is min(F, a).
Let [K : F ] = n and [F (a) : F ] = r. Then {1, a, a2 , . . . , ar−1 } is a basis for F (a)/F and we let
{b1 , . . . , bn/r } be a basis for K/F (a). Hence, {ai bj : 0 ≤ i ≤ r − 1, 1 ≤ j ≤ n/r} is a basis for K/F . Then
 
  x 0 0 ··· 0 a0
A 0 ··· 0 −1 x 0 · · · 0
0 A · · · 0  a1  
 0 −1 x · · · 0 a2 
det(xI − La ) =  . . . , where A =  ,
 
.
 .. .. . . .. 

 .. .. .. . . .. .. 
 . . . . . . 
0 0 ··· A
0 0 0 · · · −1 x + ar−1

and min(F, a) = xr +ar−1 xr−1 +· · ·+a2 x2 +a1 x+a0 (see Problem 1.9). Then det(xI − La ) = (min(F, a))n/r =
Q
σ∈Gal(K/F ) (x − σ(a)) by the result of the previous problem. Since matrices satisfy their characteristic poly-
nomial, it follows that La satisfies min(F, a), and therefore that min(F, a) is the minimal polynomial of La .

25
Problem 5.11. Let K be a finite Galois extension of F with Galois group G. Let L be an intermedi-
ate extension, and let H be the corresponding subgroup of G. If N (H) is the normalizer of H in G, let L0
be the fixed field of N (H). Show that L/L0 is Galois and that if M is any subfield of L containing F for
which L/M is Galois, then M contains L0 .
Since K/F is a finite Galois extension, so is K/L0 . Since H is normal in N (H), L/L0 is Galois by the
FTGT. Now, let M correspond to the subgroup S. Then F ⊆ M ⊆ L and G ⊇ S ⊇ H. By the FTGT, L/M
being Galois implies that H / S. Then S ≤ N (H), so L0 ⊆ M .

Problem 5.12. Let F be a field of characteristic not √ 2, and

√ let K be a Galois extension with [K : F ] = 4.
Prove that if Gal(K/F ) ∼ = Z/2Z × Z/2Z, then K = F ( a, b) for some a, b ∈ F .
Gal(K/F ) contains two distinct copies of Z2 , so K/F has two distinct intermediate subfields,√M and
N , such √that [M : F ] = [N : F ] = 2. The result of Problem √ √ 2.6 implies √ that M = √ F ( a) and
N = F ( b)√ for some a, b ∈ F . We will
√ √ now √show that [F ( a,√ b) : F √( a)] = 2. √Since b√satisfies
x2 − b ∈ F√( a)[x], we have
√ that [F ( a, b) : F ( a)] ≤
√ 2. Now, if
√ b ∈ F ( a), then F ( b)
√ ⊆ F ( a), √and
since [F ( a) : F ] = [F ( b) : F ] = 2, we have
√ √ that F ( √a) = F ( b). But we know that
√ √ F ( a) and
√ F ( b)
are distinct,
√ √ so this can’t be. Hence, [F
√ √ ( a, b) : F ( a)] ≥ 2, and therefore [F ( a, b) : F ( a)] = 2.
Then [F ( a, b) : F ] = 4, so K = F ( a, b).

Problem 5.13. If K is the splitting field of x4 − 2 over Q, find Gal(K/Q) and find all intermediate
subfields. To what group is Gal(K/Q) abstractly isomorphic?
The solution is the same as that of Problem 5.4(a), except all occurences of 7 should be replaced with 2.

Problem 5.14. If K is the splitting field of x5 − 11 over Q, find Gal(K/Q) and find all intermediate
subfields.
The solution is the same as that of Problem 5.4(c), except all occurences of 2 should be replaced with 11.

Problem 5.15. Let K be a finite normal extension of F such that there are no proper intermediate extensions
of K/F . Show that [K : F ] is prime. Give a counterexample if K is not normal over F .
We assume the extension is proper. Assume the extension is separable. Then K/F is Galois and finite.
By the FTGT, Gal(K/F ) has no non-trivial subgroups. Therefore | Gal(K/F )| is prime. Hence [K : F ] is
prime.
Now assume K/F is not separable. Let S be the separable closure of F in K. Then S = F , since K/F
is not separable and there are no non-trivial intermediate subfields. By Theorem 4.23, K = SI, where I
is the purely inseparable closure of F in K. Then K = SI = F I = I, so K/F is purely inseparable and
[K : F ] = pn for some n ∈ N. Choose some a ∈ K − F . Then, since there are no non-trivial intermediate
n n n−1 n
subfields, K = F (a) and min(F, a) = xp − ap . Suppose n > 1. Then ap satisfies xp − ap , implying that
n−1 n−1
[F (ap ) : F ] = p, and therefore that F (ap ) is a non-trivial intermediate subfield of K/F . We conclude
that n = 1, and that [K : F ] = p.
Let’s now produce the counterexample. Let L/F be any Galois extension such that Gal(L/F ) = S4
(Example 3.9 gives such an example). Let K = F(S3 ), where S3 is viewed as a subgroup of S4 . Since S3
is maximal in S4 , it follows from the FTGT that K/F has no strictly intermediate subfields. Moreover,
[K : F ] = [S4 : S3 ] = 4, which is not prime. Note that this in fact proves that K/F is not normal.

Problem 5.16. Let K be a Galois extension of Q. View K as a subfield of C. If σ is complex con-

jugation, show that σ(K) = K, so σ|K ∈ Gal(K/Q). Show that F(σ|K ) = K ∩ R, and conclude that
[K : K ∩ R] ≤ 2. Give examples to show that both [K : K ∩ R] = 1 and [K : K ∩ R] = 2 can occur.
To show that σ(K) = K, we will show that for any a ∈ K, we have that a ∈ K, where a denotes the
conjugate of a. Let f (x) = min(Q, a), and note that a is also a root of f (x), since 0 = f (a) = f (a). Since
K/Q is Galois, K contains all the roots of min(Q, a), including a. Hence, σ(K) = K. Since σ fixes Q,
σ|K ∈ Gal(K/Q). It is clear that a complex number is fixed under conjugation if and only if it is real,
therefore F(σ|K ) = K ∩ R. Since σ has order 2, σ|K has order either 1 or 2. Proposition 2.14 then implies
that [K : K ∩ R] ≤ 2. If we let K = Q, then K/Q is Galois and [K : K ∩ R] = 1. If we let K = Q(i), then
K/Q is Galois, K ∩ R = Q, and [K : K ∩ R] = 2.

26
Problem 5.17. Prove the normal basis theorem: If K is a finite Galois extension of F , then there is
an a ∈ K such that {σ(a) : σ ∈ Gal(K/F )} is a basis for K as an F -vector space.
We first prove the theorem in the case where F is infinite. We follow Artin’s proof. Let [K : F ] = n.
By the Primitive Element Theorem, K = F (α) for some α ∈ K. Let f (x) = min(F, α), Gal(K/F ) =
{σ1 , . . . , σn }, and αi := σi (α). Moreover, define
f (x) f (x)
g(x) = 0
and gi (x) := σi (g(x)) = .
(x − α)f (α) (x − αi )f 0 (αi )
Note that gi (x)gj (x) ≡ 0 (mod f (x)) for i 6= j, since every linear factor of f (x) appears in the product
gi (x)gj (x). Next, we remark that the αi are pairwise distinct, since K/F is Galois and thus separable.
Furthermore, gi (αi ) = 1 and gi (αj ) = 0 when i 6= j. These facts imply that g1 (x) + · · · + gn (x) − 1 = 0
identically, since the LHS of the previous equation is a polynomial of degree at most n−1, and it has n distinct
roots (namely, α1 , . . . , αn are roots). Multiplying the equation by gi (x) yields gi (x)2 ≡ gi (x) (mod f (x)).
Now, let A(x) := (σi (σj (g(x))))ij be a matrix and let D(x) := det(A(x)). We will prove that D(x)
2
is not identically

zero by showing that D(x)2 is not identically zero. Note that D(x)2 = det(A(x)) =
det A(x)2 = det(A(x)t A(x)). The (i, j) entry of A(x)t A(x) is
n
X n
X
σk (σi (g(x)))σk (σj (g(x))) = σk (gi (x)gj (x)).
k=1 k=1
Pn Pn Pn
If i = j, then the above expression reduces to k=1 σk (gi (x)) = k=1 σk (σi (g(x))) = k=1 σk (g(x)) =
P
k=1 gk (x) = 1, where everything has been taken modulo f (x). If i 6= j, then gi (x)gj (x) ≡ 0, and therefore
the (i, j) entry is zero modulo f (x). Hence, At (x)A(x) ≡ Id (modf (x)), and hence D(x)2 ≡ 1. Therefore,
D(x) is not identically zero, and therefore has finitely many roots as it is a polynomial in x. Since F , and
thus K, is infinite, we can choose some θ such that D(θ) 6= 0. Letting a = g(θ), we have that the determinant
of B := (σi (σj (a)))ij is non-zero. We then claim that σ1 (a), . . . , σn (a) are linearly independent. Indeed,
let c1 , . . . , cn ∈ F be constants such that c1 σ1 (a) + · · · + cn σn (a) = 0. We then have n equations of the
form c1 σi (σ1 (a)) + · · · + cn σi (σn (a)) = 0, which can be translated into the matrix equation Bc = 0, where
c = (c1 , . . . , cn )t . Since B is invertible, c = 0, proving the theorem in the infinite field case.
We now assume that F is a finite field. For this case, we will need the structure theorem for finitely
generated modules over a PID. The theorem states that if M is a finitely generated module over a PID R,
then
M∼ = R/(d1 ) ⊕ · · · ⊕ R/(d1 ),
where di divides di+1 . Since F is a finite field, and K/F is finite, say [K : F ] = n, we have that K ∼ = Fn
as a vector space, and therefore K is finite as well. Corollary 6.7 then implies that Gal(K/F ) is cyclic, so
let Gal(K/F ) = {1, σ, . . . , σ n−1 }. Note that σ i can be viewed as a group homomorphism from K ∗ → K ∗ .
Hence we can apply Dedekind’s lemma (Lemma 2.12) to conclude that the σ i are linearly independent over
K (we will use this later).
Now, viewing K as an F -vector space and σ as an F -linear transformation of K, we can view K as an
F [x]-module, where the action x · α = σ(α) is extended in the obvious way to all of F [x]. Since K is a finite
dimensional vector space, K is finitely generated as a F [x]-module; indeed, the basis for K/F will do as a
generating set for K viewed as a module. Moreover, since F is a field, F [x] is a PID. Hence, by the structure
theorem,
K∼ = F [x]/(f1 ) ⊕ · · · ⊕ F [x]/(fk ),
where f1 is σ’s minimal polynomial, and f1 · · · fk is σ’s characteristic polynomial.
We will now show that the minimal polynomial and the characteristic polynomial of σ is xn − 1. First of
all, the σ i are linearly independent, so σ cannot satisfy a polynomial of degree less than n. Since σ satisfies
xn − 1, it follows that xn − 1 is σ’s minimal polynomial. Since the degree of the characteristic polynomial of
a linear transformation is equal to the dimension of the space on which it acts, σ’s characteristic polynomial
has degree n. Moreover, the fact the minimal polynomial divides the characteristic polynomial implies that
xn − 1 is also σ’s characteristic polynomial. Therefore,
K∼
= F [x]/(xn − 1)

27
as an F [x]-module. Then, let a ∈ K be the element corresponding to 1 under the above isomorphism. Since
every element of F [x]/(xn − 1) is of the form (c0 + c1 x + · · · + cn−1 xn−1 ) · 1, every element of K can be
written as a linear combination of a, σ(a), . . . , σ n−1 (a). This concludes the finite field case.

Problem 5.18. Let Q8 be the quaternion group {±1, ±i, ±j, ±k}, where multiplication is determined by the
relations i2 = j 2 = −1 and ij = k = −ji. Show that Q8 is not isomorphic to a subgroup of S4 . Conclude
that Q8 is not the Galois group of the splitting field of a degree 4 polynomial over a field.
Note that Q8 an element of order 4 (take i for example) and an element of order 2 (−1 is the only one).
So if Q8 was isomorphic to a subgroup of S4 , such a subgroup would need to contain a transposition and
a 4-cycle. But these two elements generate all of S4 , which would imply that Q8 is isomorphic to S4 . But
|Q8 | = 8 and |S4 | = 24, so clearly this cannot be. Hence, Q8 does not occur as a subgroup of S4 .
Now, if f (x) is a degree 4 irreducible polynomial, then the Galois group of any splitting field would
necessarily be a subgroup of S4 , so clearly Q8 is not such a Galois group. However, f could have degree 2
or 3 irreducible factors. The Galois group of the splitting field would then be a subgroup of either S2 or
S3 . But Q8 is larger than both of these groups, so clearly it cannot occur as a subgroup of either of them.
Hence, Q8 is not the Galois group of the splitting field of a degree 4 polynomial.

Problem 5.19.
(a) Let K ⊆ N be Galois extensions of a field F . Show that the map φ : Gal(N/F ) → Gal(K/F ) given
by φ(σ) = σ|K is a surjective group homomorphism. Therefore, Gal(K/F ) = {σ|K : σ ∈ Gal(N/F )}.
Show that ker(φ) = Gal(N/K).
That φ is a group homomorphism is clear. Now, let σ ∈ Gal(K/F ) be arbitrary. Since N/F is Galois,
it is normal. Now apply Proposition 3.28, using the chain of inclusions F ⊆ K ⊆ N ⊆ N . The
proposition tells us that there is a τ ∈ Gal(N/F ) such that τ |K = σ. This proves surjectivity.
Now, let σ ∈ ker(φ). Then σ|K = id, which implies that σ ∈ Gal(N/K). Conversely, suppose
σ ∈ Gal(N/K). Then σ|K = id, implying σ ∈ ker(φ). The result of all this is that Gal(K/F ) ∼
=
Gal(N/F )/ Gal(N/K).

(b) Let K and L be Galois extensions of F . Show that the restriction of function map defined in (a)
induces an injective group homomorphism Gal(KL/F ) → Gal(K/F ) ⊕ Gal(L/F ). Show that this map
is surjective if and only if K ∩ L = F .
The induced map is σ 7→ σ|K ⊕σ|L . That this is a homomorphism is clear. To show injectivity, suppose
that σ|K ⊕ σ|L = τ |K ⊕ τ |L . Then σ and τ agree on both K and L. Since KL = F (K ∪ L), this implies
that σ = τ by Lemma 2.2. Hence the induced homomorphism is injective.
Now suppose the map is surjective. Let α ∈ K ∩ L. Moreover, let σ ⊕ id ∈ Gal(K/F ) ⊕ Gal(L/F ).
By surjectivity, there is a τ ∈ Gal(KL/F ) such that τ |K = σ and τL = id. But then τ (α) = σ(α) =
id(α) = α. Since σ was arbitrarily chosen, it follows that α ∈ F(Gal(K/F )) = F . Hence K ∩ L = F .
The converse is trickier. Let’s first suppose that K and L are finite extensions of F . We will then
see what adjustments need to be made for the infinite case. We start by proving that the map
φ : Gal(KL/L) 7→ Gal(K/K ∩ L), σ 7→ σ|K induces a a group isomorphism. First we show injectivity.
Let σ|K = τ |K . Since KL = L(K), σ and τ agree on the generating set of the extension, and therefore
σ = τ . Now we show surjectivity of φ. Note that K/K ∩ L is Galois: every element of K is normal and
separable over F , so every element of K must also be normal and separable over K ∩ L ⊇ F . Similarly
KL/L is Galois. We will prove that the fixed field of φ(Gal(KL/L)) is K ∩ L. Let α ∈ K be fixed by
φ(Gal(KL/L)), then α ∈ L since KL/L is Galois. If α ∈ K ∩ L, then clearly α is in the fixed field.
Hence, by the FTGT, φ(Gal(KL/L)) = Gal(K/K ∩ L), which concludes the proof of surjectivity and
of our claim.
We will now prove the converse, i.e. that the map Gal(KL/F ) → Gal(K/F ) ⊕ Gal(L/F ) is surjective,
given that K ∩ L = F . Let σ1 ⊕ σ2 ∈ Gal(K/F ) ⊕ Gal(L/F ) be arbitrary. By the above claim,
Gal(K/F ) = Gal(K/K∩L) ∼ = Gal(KL/L). Hence, there exists a σ̃1 ∈ Gal(KL/L) such that σ̃1 |K = σ1 .
Similarly, there is a σ̃2 ∈ Gal(KL/K) such that σ̃2 |L = σ2 . I claim that σ := σ̃1 σ̃1 maps to σ1 ⊕ σ2 .

28
 Indeed, σ̃1 |L = id, so σ|L = σ̃2 |L = σ2 . Similarly σ|K = σ1 . Hence, σ 7→ σ1 ⊕ σ2 . This concludes the
proof of the finite case.
We now drop the assumption that all extensions are of finite degree. The only part of the proof that
doesn’t carry over is the use of the FTGT to conclude that φ(Gal(KL/L)) = Gal(K/K ∩ L). It turns
out the we need to use the Fundamental Theorem of Infinite Galois Theory (Theorem 17.8) for this.
To do so, we need to justify that φ(Gal(KL/L)) is a closed subgroup of Gal(K/K ∩ L), with respect
to the Krull topology defined in Section 17. First, we show that φ is a continuous map. For this, it is
enough to show that the preimage of an arbitrary basis set is open. Let σN be such a basis set. Here,
σ ∈ Gal(K/K ∩ L) and N = Gal(K/E),Swhere K ∩ L ⊆ E ⊆ K, [E : K ∩ L] < ∞, and E/K ∩ L is
Galois. We then claim that φ−1 (σN ) = i {σi Gal(KL/EL) : σi ∈ Gal(KL/L), σi |E = σ}.
(⊆) Let τ ∈ φ−1 (σN ). Then τ |E = σ and τ is in some coset σi Gal(KL/EL), so τ = σθ, where
θ ∈ Gal(KL/EL). Hence, σ = τ |E = σi θ|E = σi |E .
(⊇) Let τ ∈ σi Gal(KL/EL) for some σi ∈ Gal(KL/L) such that σi |E = σ. Then τ = σi θ for some
θ ∈ Gal(KL/EL), and τ |E = σi θ|E = σi |E = σ. Hence, φ(τ ) = τ |K ∈ σN .
Note that [EL : L] = [EL : (K ∩ L)L] ≤ [E : K ∩ L] < ∞. Moreover, EL is Galois over L, since
E is Galois over K ∩ L. Therefore φ−1 (σN ) is open in the Krull topology, so φ is continuous. By
Theorem 17.6, Gal(KL/L) is compact; therefore φ(Gal(KL/L) is compact. Again by Theorem 17.6,
Gal(K/K ∩ L) is Hausdorff; therefore φ(Gal(KL/L) is closed, which is what we wanted to show.
Problem 5.20. Let k be a field of characteristic p > 0, let K = k(x, y) be the rational function field in two
variables over k, and let F = k(xp , y p ).
(a) Prove that [K : F ] = p2 .
See Problem 4.14.
(b) Prove that K p ⊆ F .
See Problem 4.14.
(c) Prove that there is no α ∈ K with K = F (α).
Se Problem 4.14.
(d) Exhibit an infinite number of intermediate fields of K/F .
I claim that for every a ∈ F , F (x + ay) is an intermediate extension, and that if a, b ∈ F are distinct,
then F (x + ay) 6= F (x + by).
/ F , so F ⊂ F (x + ay). Moreover, (x + ay)p ∈ F , so [F (x + ay) : F ] = p. Hence
Clearly, x + ay ∈
F ⊂ F (x + ay) ⊂ K.
Now, suppose that a 6= b and F (x + ay) = F (x + by). Then (x + ay) − (x + by) = (a − b)y ∈ F (x + ay).
Since a 6= b, we have that y ∈ F (x + ay). But then x ∈ F (x + ay), which implies that F (x + ay) = K,
which is impossible by part (c). Hence, the F (x + ay) give an infinite class of intermediate extensions
(since F is infinite).
Problem 5.21. This problem gives an alternative proof of the primitive element theorem for infinite fields.
(a) Let V be a finite dimensional F -vector space, where F is an infinite field. Show that V is not the union
of finitely many proper subspaces.
If dim V = 1, then the statement is trivial. Let dim V = n > 1. Define the (n − 1)-dimensional
subspace        
1 0 0 0
0 1 0  0 
       
 ..   ..   ..   .. 
Wα = Span   ,   , . . . ,  .  ,  .  .
 .   .     
0 0 1  0 
       
0 0 0  1 
0 0 0 α

29
 Then Wα = Wβ if and only if α = β. Since F is infinite, V contains infinitely many distinct (n − 1)-
dimensional subspaces.
We now prove the following claim: if W, U1 , . . . , Uk is a collection of (n − 1)-dimensional subspaces of
V , then W ⊆ U1 ∪ · · · ∪ Uk implies W = Ui for some i. Let W ⊆ U1 ∪ · · · ∪ Uk , and let {e1 , . . . , en−1 }
be a basis for W . We can order the basis such that {e1 , . . . , em } ⊆ U1 and {em+1 , . . . , en−1 } ∩ U1 = ∅.
If m = n − 1, then we are done. Otherwise, note that em+1 ∈ U2 ∪ · · · ∪ Uk , which implies that
{e1 + em+1 , . . . , em + em+1 } ⊆ U2 ∪ · · · ∪ Uk . This in turn implies that {e1 , . . . , en−1 } ⊆ U2 ∪ · · · ∪ Uk .
Repeat the procedure described above with U2 in the role of U1 . Then either the entire basis is
contained in U3 and we are done, or it is contained in U3 ∪ · · · ∪ Uk . We continue this procedure until
we find that the entire basis is contained in a single subspace Ui . Then W ⊆ Ui , implying W = Ui
since dim W = dim Ui = n − 1.
Now, assume that V is the union of finitely many proper subspaces Ũi . Each Ũi is contained in some
subspace Ui of dimension n − 1, so we have V = U1 ∪ · · · ∪ Uk , i.e. V is the union of finitely many
subspaces of dimensions n − 1. Then Wα ⊆ U1 ∪ · · · ∪ Uk for every α ∈ F . By the pigeonhole principle
and our claim, Wα = Wβ = U for some α 6= β. But this is impossible, so we conclude that V cannot
be written as a union of finitely many proper subspaces.

(b) Let K/F be a finite extension of infinite fields. Show that K is not the union of the proper intermediate
S
fields of K/F . Conclude that if {Ki } is the set of proper intermediate fields and a ∈ K − Ki , then
K = F (a).
I believe that we are meant to assume that K/F has finitely many intermediate fields, given that we
are providing an alternative proof of the primitive element theorem. View K, F , and each of the
intermediate fields as F -vector spaces. By part (a), K cannot be the union of the proper intermediate
S
fields (using our assumption that there are only finitely many of them). Then there is an a ∈ K − Ki
an we must have K = F (a), since F (a) is not a member of the collection of intermediate fields.
√
Problem 5.22. Let K = Q(X), where X = { p : p is prime}. Show that K is Galois over Q. If
√ √ √
σ ∈ Gal(K/Q), let Yσ = { p : σ( p) = − p}.
√
Since min(Q, p) = x2 − p is normal and separable for each prime p, K/Q(X) is Galois.

Prove the following statements.

(a) If Yσ = Yτ , then σ = τ .
√ √ √
Note that min(Q, p) = x2 − p, so for any σ ∈ Gal(K/Q), σ( p) = ± p. Hence, if Yσ = Yτ , then
σ|X = τ |X ⇒ σ = τ .
(b) If Y ⊆ X, then there is a σ ∈ Gal(K/Q) with Yσ = Y .
√ √
Claim 1. If {p1 , . . . , pn } is a set of distinct rational numbers such that no subset of { p1 , . . . , pn }
√ √
has product in Q, then [Q( p1 , . . . , pn ) : Q] = 2n .
√ √
Proof. We proceed by induction on n. We clearly have that p1 ∈ / Q, so [Q( p1 ) : Q] = 2.
√ √ √ √
Now consider Q( p1 , . . . , pn )/Q. By induction, [Q( p3 , . . . , pn ) : Q] = 2n−2 . We will then be done
√ √ √ √ √
if we can show that {1, p1 , p2 , p1 p2 } is linearly independent over Q( p3 , . . . , pn ). By induction,
√ √ √ √ √ √ √ √ √
[Q( p1 , p3 , . . . , pn ) : Q] = [Q( p2 , p3 , . . . , pn ) : Q] = [Q( p1 p2 , p3 , . . . , pn ) : Q] = 2n−1 ,
√ √ √ √ √ √
so p1 , p2 , p1 p2 ∈/ Q( p3 , . . . , pn ). Now assume we have a linear combination a + b p1 p2 =
√ √ √ √
c p1 + d p2 , with a, b, c, d ∈ Q( p3 , . . . , pn ). Then squaring both sides and rearranging yields
√
a2 + b2 p1 p2 − c2 p1 − d2 p2 = 2(cd − ab) p1 p2 .
√ √ √
Hence, cd = ab. Rearranging our linear combination to a − d p2 = c p1 − b p1 p2 and squaring both
sides yields ad = bcp1 in similar fashion. Then a d = abcp1 , implying that a2 d = c2 dp1 . Then either
2
√ √ √
c = 0 or d = 0, because otherwise p1 ∈ Q( p3 , . . . , pn ).

30
 √ √ √
Suppose d 6= 0. Then a + b p1 p2 = d p2 , implying that a2 + 2ab p1 p2 + b2 p1 p2 = d2 p2 . Then ab = 0,
so either a = 0 or b = 0. It is easily seen that neither of these cases can occur. Similarly, we can
√ √ √
show that d = 0 cannot occur. We conclude that {1, p1 , p2 , p1 p2 } is linearly independent over
√ √ √ √
Q( p3 , . . . , pn ), and therefore that [Q( p1 , . . . , pn ) : Q] = 2n .

Note that Claim 1 holds in the particular case where the pi are primes (as the notation suggested).
√ √
This implies that that | Gal(Q( p1 , . . . , pn )/Q)| = 2n for distinct primes pi . Since every σ ∈
√ √ √
Gal(Q( p1 , . . . , pn )/Q) is determined by where it sends the pi ’s, and they can only be sent to
√ √ √ √ √
± pi , we have that for every subset S := { pi1 , . . . , pik } ⊆ T := { p1 , . . . , pn }, there is exactly
√ √
one element σST ∈ Gal(Q( p1 , . . . , pn )/Q) that maps every element in S to its negative and fixes the
rest of T .
We are now ready to give a tentative definition of σ. Let Y ⊆ X be an arbitrary subset. Let α ∈ K.
Then α ∈ Q(T ) for some finite subset T of X. We then define
σ(α) := σYT ∩T (α).
From the definition it is clear that Yσ = Y . However, what is not clear is that σ is well-defined. We
show that now.
Claim 2. Let S and T be finite subsets of X. Then Q(S) ∩ Q(T ) = Q(S ∩ T ).

Proof. It is clear that Q(S ∩ T ) ⊆ Q(S) ∩ Q(T ). Now, note that for Galois extensions K and L of
F , we have Gal(KL/L) ∼ = Gal(K/K ∩ L) (we proved this in the solution of Problem 5.19(b)). Let
|S| = s, |T | = t, and |S ∩ T | = u. We then have that Gal(Q(S ∪ T )/Q(T )) ∼ = Gal(Q(S)/Q(S ∩
T )). Since all the extensions being considered are finite and Galois, this isomorphism implies that
[Q(S ∪ T ) : Q(T )] = [Q(S) : Q(S) ∩ Q(T )]. By Claim 1, [Q(S ∪ T ) : Q(T )] = 2s+t−u /2t = 2s−u and
[Q(S) : Q(S) ∩ Q(T )] = 2s /[Q(S) ∩ Q(T ) : Q]. Hence, [Q(S ∩ T ) : Q] = [Q(S) ∩ Q(T ) : Q] = 2u . Since
Q(S ∩ T ) ⊆ Q(S) ∩ Q(T ), we have that Q(S ∩ T ) ⊆ Q(S) = Q(T ) as desired.

Claim 3. Let S ⊆ T and let α ∈ Q(S). Then σYS ∩S (α) = σYT ∩T (α).
√ √ √
Proof. Let S = { p1 , . . . , pn }. If α = pi for some i, then the claim is obvious. In general, we have
√ √ √ √
that α = f ( p1 , . . . , pn )/g( p1 , . . . , pn ) for polynomials f, g ∈ Q[x1 , . . . , xn ]. Then
√ √ √ √
f (σYS ∩S ( p1 ), . . . , σYS ∩S ( pn ))
 
f ( p1 , . . . , pn )
σYS ∩S (α) = σYS ∩S √ √ = √ √
g( p1 , . . . , pn ) g(σYS ∩S ( p1 ), . . . , σYS ∩S ( pn ))
√ √  √ √
f (σYT ∩S ( p1 ), . . . , σYT ∩S ( pn ))

T f ( p1 , . . . , pn )
= √ √ = σ Y ∩S √ √ = σYT ∩S (α).
g(σYT ∩S ( p1 ), . . . , σYT ∩S ( pn )) g( p1 , . . . , pn )

We can now show that σ is well defined. Suppose that α ∈ Q(S) and α ∈ Q(T ), where S and T are
finite subsets of X. The concern is that our definition of σ might depend on whether we view α as an
element of Q(S) or Q(T ). We now show that this is not the case. By Claim 2, α ∈ Q(S ∩ T ). By Claim
3, σYS ∩S (α) = σYS∩T T
∩S∩T (α) = σY ∩T (α). Hence, σ is well-defined. Moreover, σ ∈ Gal(K/Q) is clear.

(c) If P(X) is the power set of X, show that | Gal(K/Q)| = |P(X)| and that |X| = [K : Q], and conclude
that | Gal(K/Q)| > [K : Q].
Before starting, we remark that for a finite extension K/F , we always have | Gal(K/F )| ≤ [K : F ].
Now, consider the map Gal(K/Q) → P(X), σ 7→ Yσ . By part (a), this map is injective; by part (b),
this map is surjective. Hence, | Gal(K/Q)| = |P(X)|. By Claim 1 of part (b), [K : Q] is certainly not
finite, so we just need to show that it is
√ countable,
√ √ which we can do by showing that K is countable.
√
Now, K = ∪n∈N Kn , where Kn := Q( 2, 3, 5, . . . , pn ) and pn is the nth prime. Each Kn is a
finite extension of Q, and is therefore countable, since finite dimensional vector spaces over countable
fields are countable. Since countable unions of countable sets are countable, K is countable. Hence,
|X| = [K : F ] and therefore, | Gal(K/Q)| = |P(X)| > |X| = [K : Q].

31
Problem 5.23. Suppose that K is an extension of F with [K : F ] = 2. If char(F ) 6= 2, show that K/F is
Galois. √ √ √ √
By Problem 2.6, K = F ( a) for some a ∈ K. Then min(F, a) = x2 − a = (x + a)(x − a) is both
normal and separable over F (since −1 6= 1). Therefore, K/F is Galois.

Problem 5.24. Let F ⊆ L ⊆ K be fields such that L/F is purely inseparable. Let a ∈ K be separa-
ble over F . Prove that min(F, a) = min(L, a). Use this to prove the following statement: Suppose that
F ⊆ L ⊆ K are fields such that L/F is purely inseparable, K/L is separable, and [K : F ] < ∞. Let S be the
separable closure of F in K. Then K = SL and [K : L] = [S : F ].
In characteristic 0, F = L is forced and all of the above claims become trivial, so we will assume that
we are in characteristic p from now on. Let M be an algebraic closure of M . Then min(F, a) factors as
(x − a1 ) · · · (x − an ) over M , where the ai are pairwise distinct. On one hand, we know that min(L, a) divides
n
min(F, a). On the other hand, since L/F is purely inseparable, we have that min(L, a)p ∈ F [x] for some
n ≥ 0. Hence, min(F, a) divides min(L, a)p . In particular, every root of min(F, a) is a root of min(L, a). We
conclude that min(F, a) = min(L, a).
Now for the second statement. Let α ∈ K. Then min(SL, α) divides min(L, α), which is separable.
Hence α is separable over SL. But min(SL, α) also divides min(S, α), which is purely inseparable by Propo-
sition 4.20. Hence, α is also purely inseparable over SL. We conclude that K = SL. By the Primitive
Element Theorem, S = F (a) for some a ∈ S. By the first part of the problem, min(F, a) = min(L, a), so
[S : F ] = [L(a) : L]. But L(a) = SL = K, so [S : F ] = [K : L].

Problem 5.25. This problem outlines a proof that the separable degree [K : F ]s of a finite extension
K/F is equal to the number of F -homomorphisms from K to and algebraic closure of F .
(a) Suppose that K = F (a), and let f (x) = min(F, a). If N is an algebraic closure of F and b ∈ N is a
root of f , show that there is an F -homomorphism from K → N that sends a to b.
Let M ⊆ N be the splitting field of f , i.e. the subfield of N generated by F and the roots of f . By
the Isomorphism Extension Theorem, there is an F -homomorphism σ : M → M such that σ(a) = b.
Restricting this map to K and taking the range to be N , we have the desired F -homomorphism.
(b) If K = F (a) as above, show that all F -homomorphisms from K to N are obtained in the manner of
the previous step. Conclude that [K : F ]s is equal to the number of such F -homomorphisms.
I’m fairly certain that we need to assume that a is separable. Let’s do it, because we will still get to
a proof of the desired result. Let σ : K → N be some F -homomorphism. Then σ(a) is necessarily a
root of min(F, a). Moreover, since {1, a, a2 , . . . , an−1 } (where n = deg min(F, a)) is a basis for K/F , σ
is completely determined by where it sends a. Hence, the number of F -homomorphisms from K to N
is equal to the number of distinct roots of min(F, a), which is n, since we assumed that a is separable.
Hence n = [K : F ] = [K : F ]s is equal to the number of F -homomorphisms from K to N .
(c) Let K/F be a finite extension, and let S be the separable closure of F in K. Show that any F -
homomorphism from S to N extends uniquely to K. Use the previous step to conclude that [S : F ] =
[K : F ]s is the number of F -homomorphisms from K to N .
We have that S = F (a) for some a ∈ S. From part (b), we have that [S : F ] = [K : F ]s equals the
number of F -homomorphisms from S to N . If we can show that any such homomorphism extends
uniquely to K, then we will be done. From Proposition 4.20, we have that K/S is purely inseparable;
hence, from Lemma 4.17, we have that K/S is normal. Using S ⊆ S ⊆ K ⊆ N as our tower of
extensions, we apply the third point of Proposition 3.28 to conclude that every F -homomorphism from
S to N extends to a homomorphism from K to N . To show that the extension is unique, let σ1 and
σ2 be any homomorphisms agreeing on S. Again by the second point of Proposition 3.28, we have
that σ1 σ2−1 is an automorphism of K that fixes S. But since K/S is a purely inseparable extension,
Gal(K/S) = 1 by Lemma 4.17. Hence σ1 = σ2 , which concludes the proof.

32
Problem 5.26. Let K/F be a normal extension and let L/F be an algebraic extension. If either K/F or
L/F is separable, show that [KL : L] = [K : K ∩ L]. Give an example to show that this can be false without
the separability hypothesis.
First suppose that K/F is separable. Then K/F is Galois. We proved that Gal(KL/L) ∼ = Gal(K/K ∩ L)
in part (b) of Problem 5.19. Note that the proof only made use of the fact that KL/L and K/K ∩ L are
Galois, which is true in this case. For the moment, I do not know how to do the case where K/F is normal
and L/F is separable. Note that if F is a field of characteristic 0, or more generally if F is a perfect field, then
K/F is automatically Galois. So the interesting cases are for infinite characteristic p fields. Unfortunately,
I also do not have an example where K/F is normal and the degree formula fails.

Problem 5.27. Let F be a field. Show that the rational function field F (x) is not algebraically closed.
Consider the polynomial tn − x ∈ F (x)[t], where n > 1. We claim that this polynomial does not have a
root in F (x). If it did, then we would have an element f (x)/g(x) ∈ F (x) such that
 n
f (x)
= x ⇔ f (x)n = xg(x)n .
g(x)

But the degree of xg(x)n is not a multiple of n, while the degree of f (x)n is. We conclude that tn − x has
no roots, and therefore that F (x) is not algebraically closed.

Problem 5.28. Let F be a finite extension of Q. Show that F is not algebraically closed.
Let [F : Q] = n. Let p be any prime larger than n. In part (b) of Problem 1.22 we showed that
f (x) := xp−1 + xp−2 + · · · + x + 1 is irreducible over Q. If we assume that F is algebraically closed, then F
contains a root of f , which we will call a. Then [Q(a) : Q] = p > [F : Q], contradicting the fact that Q(a) is
a subfield of F . We conclude that F is not algebraically closed.

33
II. Some Galois Extensions
Section 6. Finite Fields
Problem 6.1. Let G be a finite Abelian group.

(a) If a, b ∈ G have orders n and m, respectively, and if gcd(n, m) = 1, show that there is an element of G
whose order is nm.
The statement is trivial if either a = 1 or b = 1, so we ignore this case. We clearly have that
(ab)nm = 1. Now let r be the order of ab. Then r divides nm. Since gcd(n, m) = 1, we have that either
r = nm, in which case we would be done, or r divides either n or m. Assume that either r divides
n or m. Without loss of generality, assume that r divides n. Then n = kr for some integer k. Then
1 = 1k = (ab)kr = (ab)n = bn . Hence, m divides n. But this is impossible since m and n are coprime.
We conclude that r = nm.
(b) If a, b ∈ G have orders n and m, respectively, show that there is an element of G whose order is
lcm(n, m).
Let d := gcd(n, m). Note that we have an element a0 ∈ G of order n0 = n/d (this is because a finite
cyclic group contains elements of all orders dividing the order of the group). Hence, gcd(n0 , m) = 1.
By part (a), the order of a0 b is n0 m = nm/d = lcm(n, m).
(c) Show that there is an element of G whose order is exp(G).
Let G = {a1 , . . . , an }. We will prove that there is an element of order lcm(a1 , . . . , aj ) by induction on
j. For j = 1, this is trivial. Now suppose that there is an element a ∈ G of order lcm(a1 , . . . , aj−1 ). By
part (b), there is an element of order lcm(lcm(a1 , . . . , aj−1 ), aj ) = lcm(a1 , . . . , aj−1 , aj ). By induction,
we have that there is an element of order exp(G).
The fact that lcm(lcm(a1 , . . . , aj−1 ), aj ) = lcm(a1 , . . . , aj ) needs some justification, which we provide
here. First of all, note that lcm(lcm(a1 , . . . , aj−1 ), aj ) is clearly a multiple of each ai . Hence, by
definition, lcm(lcm(a1 , . . . , aj−1 ), aj ) ≥ lcm(a1 , . . . , aj ). Now, lcm(a1 , . . . , aj−1 , aj ) is a multiple of aj .
We claim that it is also a multiple of lcm(a1 , . . . , aj−1 ). To see this, write

lcm(a1 , . . . , aj ) = k · lcm(a1 , . . . , aj−1 ) + r,

where 0 ≤ r < lcm(a1 , . . . , aj−1 ). From this we see that ai divides r for each i ∈ {1, . . . , j − 1}. Hence,
we must have that r = 0 in order to not violate the definition of lcm. Hence, lcm(a1 , . . . , aj ) is a multiple
of both aj and lcm(a1 , . . . , aj−1 ). We conclude that lcm(lcm(a1 , . . . , aj−1 ), aj ) ≤ lcm(a1 , . . . , aj ),
whence equality holds.
Problem 6.2. Let p be a prime, and let F be a field with |F | = p2 . Show that there is an a ∈ F with a2 = 5.
Generalize this statement, and prove the generalization.
Here is a generalized statement: Let p be a prime, and let F be a field with |F | = pk , with k ∈ {2, 3}.
Then there is an a ∈ F with ak = n for any n ∈ Fp .
To prove this, consider the polynomial f (x) = xk − n ∈ Fp [x]. If F has a root in Fp , then we are done.
Otherwise, f is irreducible, since k ∈ {2, 3}. Let a be a root of f in an algebraic closure of Fp . Then
Corollary 6.9 implies that [Fp (a) : Fp ] = k. Hence, |Fp (a)| = pk . Corollary 6.6 then gives us that F ∼ = Fp (a),
and the isomorphism fixes Fp . Hence, there is an element of a ∈ F such that ak = n.
It turns out that we don’t need to restrict ourselves to the case where k ∈ {2, 3}. The argument is
not mine, I found several versions of it on StackExchange. Let k = k1 k2 , where gcd(p − 1, k2 ) = 1. First
of all, there is an element n1 ∈ Fp such that nk12 = n. To see this, we can simply prove that the map
Fp → Fp , m 7→ mk2 is surjective (and therefore a bijection). Let j ∈ Fp − {0} be arbitrary. Then, for some
integers s and t, we have j = j sk2 +t(p−1) = (j s )k2 . Hence, j s 7→ j. If j = 0, then simply note that j 7→ j.
We would now like to find a root a of xk1 − n1 has a root in Fpk , since then ak = nk12 = n. We now prove
the following: If q is any prime factor of k1 , and β ∈ Fpm for some m, then xq − β has a root in Fpmq . Note
that since q is a prime divisor of k1 , q must divide p − 1. Then, F× p contains a subgroup of order q, implying
that Fp contains a primitive qth root of unity, which we will call ω. Let γ be a root of xq − β in the algebraic

34
closure of Fpm . Then ω i γ is also a root for each i ∈ {0, . . . , q − 1}. Hence, Fpm (γ) is the splitting field of
xq − β. By Corollary 6.6, Gal(Fpm (γ), Fpm ) is cyclic, so it has a generator σ such that σ(γ) = ω i γ for some
i ∈ {0, . . . , q − 1}. Hence, [Fpm (γ) : Fpm ] = q, implying that Fpm (γ) = Fpqm , and thus proving our claim.
Let k1 = q1 . . . ql . By the preceding paragraph, there exists γ1 ∈ Fpq1 such that γ1q1 = n1 . Similarly,
there exists γ2 ∈ Fpq1 q2 such that γ2q2 = γ1 . Inductively define γi ∈ Fpq1 ···qi such that γiqi = γi−1 . Then
q ...qk−1
γlk1 = γlq1 ...qk = γn−1
1
= · · · = γ1q1 = n1 .

Hence, we have found a := γl ∈ Fpk1 ⊆ Fpk such that ak = γlk1 k2 = nk12 = n.

Problem 6.3. Let F be a finite field. Prove that there is an irreducible polynomial of degree n over F
for any n.
Identify F with Fpk for some k. Corollary 6.4 implies that Fpnk /Fpk is a simple extension. Let Fpn = Fp (α)
and let min(Fpk , α) be of degree m. Then [Fpnk /Fpk ] = m, which implies that Fpnk contains pmk elements.
Hence m = n, showing that min(Fpk , α) is an irreducible polynomial of degree n.

Problem 6.4. Let K be a field with |K| = 4. Show that K = F2 (α), where α2 + α + 1 = 0.
By Theorem 6.5, K is a splitting field of x4 −x. By Proposition 6.14, x4 −x factors over F2 as the product
of all monic irreducible polynomials over F2 of degree a divisor of 2. x2 +x+1 is such a polynomial, so K con-
tains all of its roots. Hence, F2 (α) ⊆ K. Since [F2 (α) : F2 ] = 2, we have that |F2 (α)| = 4. Hence, K = F2 (α).

Problem 6.5. Determine the irreducible factors if x4 + 1 over F3 .

x4 +1 has no rrots in F3 , so it would need to factor as the product of two degree 2 irreducible polynomials.
Hence,
x4 + 1 = (x2 + x + 2)(x2 − x + 2)
is the factorisation.

Problem 6.6. Let F be a finite field. If f, g ∈ F [x] are irreducible polynomials of the same degree, show
that they have the same splitting field. Use this to determine the splitting field of x4 + 1 over F3 .
Let deg f = deg g = n and let K and K 0 be splitting fields of f and g, respectively, over F . Corollary
6.9 implies that [K : F ] = [K 0 : F ] = n. Hence |K| = |K 0 |. Viewing K and K 0 as subfields of the algebraic
closure of F (and hence of Fp ), Theorem 6.8 tells us that K = K 0 .
Hence, the splitting field of x4 + 1 is the splitting field of x2 + x + 2, which is isomorphic to F32 .

Problem 6.7. Let q be a power of a prime p, and let n be a positive integer not divisible by p. We
let Fq be the unique up to isomorphism finite field of q elements. If K is the splitting field of xn − 1 over Fq ,
show that K = Fqm , where m is the order of q in the group of units (Z/nZ)∗ of the ring Z/nZ.
We have that K/Fq is generated by σ, where σ(a) = aq . Note that q m ≡n 1, so q m = 1 + cn for some
m
integer c. If α is any root of xn − 1, then σ m (α) = αq = α1+cn = α. Hence, σ m = id, since it is the identity
on the roots of xn − 1, which in turn generate K over Fq .
Note that the roots of xn − 1 form a subgroup of K ∗ . Moreover xn − 1 is separable by the derivative
test. Hence, the subgroup of nth roots of unity has size n. Since any finite subgroup of K ∗ is cyclic, we
have that there exists β ∈ K of order n in K ∗ . Let 0 < l < m, then q l 6≡n 1, i.e. q l = j + cn for some
l
j ∈ {2, 3, . . . , n} and some integer c. Then τ l (β) = β q = β j+cn = β j 6= β. Hence, τ has order m, implying
that | Gal(K/Fq )| = [K : Fq ] = m, and therefore that K = Fqm .

Problem 6.8. Let F be a field of characteristic p.

(a) Let F p = {ap : a ∈ F }. Show that F p is a subfield of F .
We have 1 = 1p , 0 = 0p , ap + bp = (a + b)p , ap bp = (ab)p . Then only thing left to prove is that F p
is closed under taking inverses. Let ap ∈ F p , where a ∈ F . Then ap (a−1 )p = (aa−1 )p = 1p = 1.
Therefore, (ap )−1 = (a−1 )p ∈ F p .
(b) If F = Fp (x) is the rational function field in one variable over Fp , determine F p and [F : F p ].
Since Fp is perfect, we have that F p = Fp (xp ). Hence [F : F p ] = p.

35
Problem 6.9. Show that x4 − 7 is irreducible over F5 .
See Problem 3.9, part (b).

Problem 6.10. Show that every element of a finite field is a sum of two squares.
Let F be a finite field containing pn elements (where p is prime). If p = 2, then for every a ∈ F , we have
n n−1
that a = a2 = (a2 )2 + 02 , so this case is settled.
Now let p be an odd prime. We first count the number of squares in F . Consider the homomorphism
φ : F × → F × , x 7→ x2 . Then ker φ = {±1}. Hence, |Im(φ)| = |F × |/| ker φ| = (pn − 1)/2. Accounting for
0, which is a square, we have (pn + 1)/2 squares in F . Let a ∈ F be arbitrary. Then there are (pn + 1)/2
elements of the form a − b2 . If none of these are squares, then F contains at least pn + 1 elements, a contra-
diction. Hence a − b2 = c2 for some c, b ∈ F , i.e. a = b2 + c2 .

Problem 6.11. Let F be a field with |F | = q. Determine, with proof, the number of monic irreducible
polynomials of prime degree p over F , where p need not be the characteristic of F .
Let K be the splitting field of all degree p irreducible polynomials over F (see Problem 6.6). Then
|K| = q p . Since [K : F ] = p, there are no intermediate fields. Hence, every element of K − F is a root of
some degree p monic irreducible polynomial over F . Since K/F is Galois, the roots p roots of each monic
p
irreducible polynomial over F are distinct. Hence, there are q p−q monic irreducible polynomials of degree p
over F .

Problem 6.12. Let K and L be extensions of a finite field F of degrees n and m, respectively. Show
that KL has degree lcm(n, m) over F and that K ∩ L has degree gcd(n, m) over F .
Since KL contains both K and L, its degree must be a multiple of n and m (by Theorem 6.8). Hence
[KL : F ] ≥ lcm(n, m). Also by Theorem 6.8, we have that there exists a field whose degree is lcm(n, m)
over F , and that it contains K and L. Since KL is the smallest field containing K and L, we have that
[KL : F ] = lcm(n, m).
Since K ∩ L is contained in both K and L, Theorem 6.8 implies that [K ∩ L : F ] is a divisor of n and m.
Hence, [K ∩ L : F ] ≤ gcd(n, m). Moreover, Theorem 6.8 implies that there is a field whose degree over F
is gcd(n, m), and that it is contained in both K and L. Hence it is contained in K ∩ L, which implies that
[K ∩ L : F ] = gcd(n, m).

Problem 6.13.
(a) Show that x3 + x2 + 1 and x3 + x + 1 are irreducible over F2 .
Simply note that neither polynomial has any roots in F2 .

(b) Give an explicit isomorphism between F2 [x]/(x3 + x2 + 1) and F2 [x]/(x3 + x + 1).

I claim that φ : F2 [x]/(x3 + x2 + 1) → F2 [x]/(x3 + x + 1), [a2 x2 + a1 x + a0 ] 7→ [a2 (x + 1)2 + a1 (x + 1) + a0 ]
is an isomorphism (the square brackets are to denote equivalence classes). First, the map is well
defined: Let [a2 x2 + a1 x + a0 ] = [b2 x2 + b1 x + b0 ]. Then a2 x2 + a1 x + a0 and b2 x2 + b1 x + b0
differ by a multiple of x3 + x2 + 1, so it is enough to check that φ([x3 + x2 + 1]) = [0]. Indeed,
[(x + 1)3 + (x + 1)2 + 1] = [x3 + x + 1] = [0]. Moreover, φ is clearly additive and multiplicative. Finally,
φ is injective, since φ−1 ([0]) = [0]. We don’t need to check surjectivity, since φ is a map between finite
sets of the same cardinality.
Problem 6.14. Let k be the algebraic closure of Zp , and let φ ∈ Gal(k/Zp ) be the Frobenius map φ(a) = ap .
Show that φ has infinite order, and find a σ ∈ Gal(k/Zp ) with σ ∈ / hφi.
pk
We will use Fp to denote Zp . Suppose φ had order k. Then a = a for all a ∈ k. This implies that every
k
element of k is a root of xp − x, implying that |k| ≤ pk . However, by Theorem 6.8, k contains a subfield of
order pn for every n ∈ N. Hence |k| is not finite. We conclude that φ has infinite order.
Now consider the map · · · ◦ φn! ◦ · ·S
· ◦ φ3! ◦ φ2! ◦ φ1! =: σ. We now justify why this map is a well defined
Fp -automorphism. We claim that k = n∈N Fpn! . Indeed, every element in k is algebraic over Fp , and hence
lies in Fpn for some n large enough. The fact that Fpn ⊆ Fpn! justifies the claim. Now, the fixed field of φk!

36
is Fpk! , so if a ∈ Fpk! , then σ(a) = φ(k−1)! ◦ · · · ◦ φ1! (a). Since we have the tower Fp ⊆ Fp2! ⊆ · · · ⊆ Fpn! ⊆ · · · ,
σ is well-defined. Moreover, σ is a Fp -isomorphism since it is an isomorphism restricted to each Fpn!
We now show that τ ∈ / hφi. Suppose not, i.e. suppose that τ = φn for some n ∈ Z. Then for every k ∈ N,
we have that τ |F k! = σ|F k! , which implies that φ(k−1)! ◦ · · · ◦ φ1! = φn as elements of Gal(Fpk! /Fp ). Since
p p
Gal(Fpk! /Fp ) is cyclic, generated by φ, and contains k! elements, we have that (k − 1)! + · · · + 1! ≡k! n for
every k ∈ N. Equivalently, (k − 1)! + · · · + 1! − n is a multiple of k! for every k ∈ N, which is again equivalent
to
(k − 1)! + · · · + 1! − n
k!
being an integer for all k ∈ N. This is impossible, since

(k − 1)! + · · · + 1! − n
lim = 0.
k→∞ k!
Indeed,
   
(k − 1)! + · · · + 1! 1 1 1 1 k−2
lim = lim + + ··· + ≤ lim + = 0.
k→∞ k! k→∞ k k(k − 1) k! k→∞ k k(k − 1)

Therefore, τ ∈
/ hφi.

Problem 6.15. Let N be an algebraic closure of a finite field F . Prove that Gal(N/F ) is an Abelian
group and that any automorphism in Gal(N/F ) is of infinite order.
Note that N is also an algebraic closure of Fp , and that Gal(N/F ) ≤ Gal(N/Fp ), so we may as well assume
that F = Fp . Let σ1 , σ2 ∈ Gal(N/F ) be arbitrary. We will show that σ1 (σ2 (a)) = σ2 (σ1 (a)) for any a ∈ N . If
a ∈ N , then a ∈ Fpn for some n. Hence σ1 (σ2 (a)) = σ1 |Fpn ( σ2 |Fpn (a)). Since σ1 |Fpn , σ2 |Fpn ∈ Gal(Fpn /F ),
which is cyclic and thus Abelian, we have σ1 (σ2 (a)) = σ1 |Fpn ( σ2 |Fpn (a)) = σ2 |Fpn ( σ1 |Fpn (a)) = σ2 (σ1 (a)),
which proves that Gal(N/F ) is Abelian.
Let σ ∈ Gal(N/F ) have order m < ∞. Moreover, let x ∈ N and let [F (x) : F ] = n. Since Fpmn is Galois
over F = Fp , we have that σ|Fpmn ∈ Gal(Fpmn /Fp ) (Proposition 3.28). Let k be the order of σ|Fpmn . Then
k | m. Hence, [Fpmn : F(σ|Fpmn )] | m (Proposition 2.14), which implies that n | [F(σ|Fpmn ) : F ]. Since
F (x) = Fpn , we have that F (x) ⊆ F(σ|Fpmn ), and therefore σ(x) = x. We conclude that σ = id. Hence, the
only finite order automorphism of Gal(N/F ) is the identity.

Section 7. Cyclotomic Extensions

Problem 7.1. Determine all of the subfields of Q12 .
From Corollary 7.8, we have that Gal(Q12 /Q) ∼ = (Z12 )∗ = {1, 5, 7, 11} ∼
= Z2 × Z2 . Hence, Gal(Q12 /Q) =
{id, σ5 , σ7 , σ12 }. There are then four subfields of Q12 . The easy ones are Q and Q12 . Let ω be a primitive
12th root of unity. Then F(σ5 ) = Q(ω 7 + ω 11 ) and F(σ1 1) = Q(ω 5 + ω 7 ) are the two other intermediate
fields. Since Q has non-trivial subfields, we are done.

Problem 7.2. Show that cos(π/9) is algebraic over Q, and find [Q(cos(π/9)) : Q].
Let ω = e2πi/18 . Then α := 2 cos(π/9) = ω + ω = ω + ω −1 . Hence, α3 = ω 3 + 3ω + 3ω −1 + ω −3 =
ω + 3α + ω −3 . We have that Ψ18 (x) = x6 − x3 + 1, so 1 = ω 3 + ω −3 . Thus, α3 = 3α + 1. This implies that
3

cos(π/9) satisfies 8x3 − 6x − 1. We can check that this polynomial is irreducible by the rational root test.
Hence, cos(π/9) is algebraic over Q and [Q(cos(π/9)) : Q] = 3.

Problem 7.3. Show that cos(2π/n) and sin(2π/n) are algebraic over Q for any n ∈ N.
We have that 1 = (e2πi/n )n = (cos(2π/n) + i sin(2π/n))n . Using the binomial theorem,
n  
X n
1= ik cosn−k (2π/n) sink (2π/n).
k
k=0

37
Suppose that n is even. Then we have
n/2  
X n
1= (±1)k cosn−2k (2π/n) sin2k (2π/n).
2k
k=0

On one hand,
n/2  
X n
1= (±1)k (1 − sin2 (2π/n))n/2−k sin2k (2π/n)
2k
k=0

and on the other

n/2  
X n
1= (±1)k cosn−2k (2π/n)(1 − cos2 (2π/n))k ,
2k
k=0

which show that cos(2π/n) and sin(2π/n) are algebraic over Q when n is even. The proof for the n odd case
is similar, though requires a bit more work.

Problem 7.4. Prove that Q(cos(2π/n)) is Galois over Q for any n. Is the same true for Q(sin(2π/n))?
Note that cos(2π/n) = 21 (ω + ω −1 ), where ω = e2π/n . Hence, Q ⊆ Q(cos(2π/n)) ⊆ Q(ω). By the FTGT,
if a subgroup of a Galois group (of Galois extension) is normal, then the corresponding intermediate field
is Galois over the base field. Hence, since Q(ω)/Q is Galois and has Abelian Galois group, the subgroup
corresponding to Q(cos(2π/n)) is normal, and hence Q(cos(2π/n))/Q is Galois.
This is not true in general for Q(sin(2π/n)). For instance, the minimal polynomial of sin(2π/5) is
x4 − 21 x2 − 14 , but it only has two real roots. However, Q(sin(2π/n)) is Galois

over Q if n is even. Indeed,
in this case i ∈ Q(ω), which implies that Q(sin(2π/n)) = Q(sin 2i 1
(ω − ω −1 ) ) ⊆ Q(ω). Then we can apply
the same argument as in the first paragraph.

Problem 7.5. If p is prime, prove that φ(pn ) = pn−1 (p − 1).

Note that kp + r is coprime to p for every r ∈ {1, 2, . . . , p − 1}. Letting k run from 0 to pn−2 we obtain
that there are pn−1 (p − 1) integers between 1 and pn that are coprime to p. Hence, φ(pn ) = pn−1 (p − 1).

i i
P P
Problem 7.6. Let θ : Z[x] → Fp [x] be the map that sends i ai x to i ai x , where a is the equiva-
lence class of a modulo p. Show that θ is a ring homomorphism.
mod p
This is a direct result of the fact that Z → Fp is a ring homomorphism.

Problem 7.7. If gcd(n, m) = 1, show that φ(nm) = φ(n)φ(m).

This is a simple consequence of the Chinese Remainder Theorem. Note that Zmn ∼
= Zm × Zn since
gcd(m, n) = 1. Hence, Z× ∼ × ×
mn = Zm × Zn , implying that φ(mn) = φ(m)φ(n).

α1 αr
Q αi −1
1 · · · pr , show that φ(n) =
Problem 7.8. If the prime factorization of n is n = pQ i pi Q(pi − 1).
By the CRT, as in the preceding problem, φ(n) = i φ(pi ). By Problem 7.5, φ(n) = i piαi −1 (pi − 1).
αi

Problem 7.9. Let n, m be positive integers with d = gcd(n, m) and l = lcm(n, m). Prove that φ(n)φ(m) =
φ(d)φ(l).
Write down the prime factorizations of m and n, use lcm(m, n) = mn/ gcd(m, n), and the previous prob-
lem.

Problem 7.10. Show that (Z/nZ)∗ = {a + nZ : gcd(a, n) = 1}.

Note that a ∈ (Zn )∗ if and only if there is a b ∈ Zn such that ab ≡n 1, i.e. ab + kn = 1 for some k ∈ Z.
By Bézout’s theorem, this is equivalent to gcd(a, n, ) = 1.

Problem 7.11. If n is odd, prove that Q2n = Qn .

It is clear that Q2n ⊇ Qn . Conversely, note that if n is odd, say n = 2m + 1, then (e2πi/n )m+1 = eπi/n .
Hence, Q2n ⊆ Qn .

38
Problem 7.12. Let n, m be positive integers with d = gcd(n, m) and l = lcm(n, m).
(a) If n divides m, prove that Qn ⊆ Qm .
Let m = kn. Then (e2πi/m )k = e2πi/n .

(b) Prove that Qn Qm = Ql .

Since both n and m divide l, we have that Qn Qm ⊆ Ql . On the other hand,
1 d rn + sm r s
= = = +
l nm mn m n
for some integers r, s. Hence, e2πi/l = (e2πi/m )r (e2πi/n )s , proving that Qn Qm = Ql .
(c) Prove that Qn ∩ Qm = Qd .
Since d divides n and m, we have that Qd ⊆ Qn ∩ Qm . On the other hand, by Theorem 5.5, [Ql :
Qm ] = [Qn Qm : Qm ] = [Qn : Qn ∩ Qm ]. Hence, φ(l)/φ(m) = φ(n)/[Qn ∩ Qm : Q], so [Qn ∩ Qm : Q] =
φ(m)φ(n)/φ(l) = φ(d) = [Qd : Q] by Problem 7.9. Hence, Qn ∩ Qm = Qd .
Problem 7.13. Determine for√which n and m there is an inclusion Qn ⊆ Qm . From this, determine which
cyclotomic extensions contain −1.
Suppose that m is even, that Qn ⊆ Qm , and that n does not divide m. Note that Qm = Qn QmQ= Ql
(see the preceding problem). Hence, φ(m) = φ(l). I claim that this is a contradiction. Let m = 2a i pα
i
i

β γ
and l = 2b i pi i j qj be the prime factorizations of m and l respectively. Note that b ≥ a ≥ 1 and
Q Q j

βi ≥ αi since m divides l. Since n does not divide m, we have that l > m. By Problem 7.8, we have that
φ(l) > φ(m), where the fact that a ≥ 1 was necessary here. Hence, if m is even, n must divide m.
Now suppose that m is odd. Then Qm = Q2m , and therefore Qn ⊆ Q2m . By the previous paragraph,
n must divide 2m. So the pairs (n, m) such that Qn ⊆ Qm are (n, m) where m is even and n divides m or
where m is odd√ and n divides 2n.
Note that −1 ∈ Qm if and only if Q4 ⊆ Qm . This forces m to be a multiple of 4.

Problem 7.14. Find a positive integer n such that there is a subfield of Qn that is not a cyclotomic
extension of Q. √
8. Then Q8 contains e±2πi/8 , and therefore e2πi/8 + e−2πi/8 = 2 cos(2π/8) = 2 cos(π/4) = 2 ∈
Take n = √
Qn . Thus Q( 2) is a subfield of Q8 but it is not a cyclotomic extension of Q.
√
Problem 7.15. If d ∈ Q, show that Q( d) lies in some cyclotomic extension of Q.
√
We begin by showing that p lies in some cyclotomic extension of Q for every prime p. In the previous
problem, we did this for p = 2, so we only need to consider the case where p is an odd prime. Let ω := e2πi/p .
Pp−1 2 √ √
I claim that α := n=0 ω n is equal to either ± p or ±i p. Note that
p−1 X
p−1 p−1 X
p−1 p−1 X
p−1
2
−m2
X X X
αα = ωn = ω (n−m)(n+m) = ω st ,
n=0 m=0 n=0 m=0 s=0 t=0

since the system n − m = s, n + m = t has a unique solution for each n, m, s, t ∈ Zp . Note that we used the
Pp−1
fact that p > 2 here. For each s ∈ {1, . . . , p − 1}, we have that t=0 ω st = 0. Therefore, αα = p, which
√ √
implies |α| = p. Suppose that −1 is a square in Zp . Then α = α, implying that α2 = p ⇒ α = ± p. In
√
this case, we have shown that p ∈ Qp . Now suppose that −1 is not a square in Zp . Recall that half of the
elements of Z× × × 2
p are squares, and the squaring map Zp → Zp , x 7→ x has kernel {±1}. Hence, each non-zero
Pp−1 n
square is the square of exactly two elements. Hence, α + α = 2 n=0 ω = 0. Therefore α is pure imaginary,
√ √ √
and hence α = ±i p. Therefore, p ∈ Q4p . Note that in all cases, we can say that p ∈ Q4p .
Now let d ∈ Q and let d = a/b, √ where gcd(a, b) = 1. Let {p1 , . . . , pi } and {q1 , . . . , qj } be the primes
dividing a and b, respectively. Then d ∈ Q4p1 . . . Q4pi Q4q1 . . . Q4qj = Q4p1 ···pi q1 ···qj , where we have induc-
tively applied the result of Problem 7.12(b).

39
Problem 7.16. The group (Z/nZ)∗ is a finite Abelian group; hence, it decomposes into a direct product of
cyclic groups. This problem explicitly describes this decomposition.
(a) If n = pr11 ·Q · · prkk is the prime factorization of n, show that (Z/nZ) ∼ = i (Z/pri i Z) as rings; hence,
Q
(Z/nZ)∗ ∼ = i (Z/pri i Z)∗ .
We will show that the ring homomorphism Z/nZ → i (Z/pri i Z), [m]n 7→ ([m]pr11 , . . . , [m]prk ) is injec-
Q
k
tive, and therefore surjective, since the map is between finite sets of the same cardinality. Suppose
that ([l]pr11 , . . . , [l]prk ) = ([m]pr11 , . . . , [m]prk ). Then l = m + c1 pr11 , . . . , l = m + ck prkk for some integers
k k
c1 , . . . , ck . Note that pri i divides c1 for every i > 1. Hence, l = m + c1 pr11 implies [l]n = [m]n .
∗
The second isomorphism follows immediately from the easy fact that ( i R) = i Ri∗ , where the Ri
Q Q
are commutative rings.
t
(b) If p is an odd prime, show that (1 + p)p ≡ 1 + pt+1 (mod pt+2 ) if t ≥ 0. Use this to find an element
of large order in Z/pri i Z, and then conclude that (Z/pri i Z)∗ is cyclic if pi is an odd prime.
By the Binomial Theorem, we have
t t
p  t p
pt pt − 1 k
 
pt
X p k
X
t
(1 + p) = p =1+p + p .
k k k−1
k=0 k=2

Let k = lps , where gcd(l, p) = 1. Then,

pt
pt+k−s pt − 1
 
pt
X
t
(1 + p) =1+p + .
l k−1
k=2

By induction on s, it is easily proven that k ≥ s + 2 (the fact that p is an odd prime is needed).
t
Hence, (1 + p)p ≡ 1 + pt+1 (mod pt+2 ). We will now show that 1 + pi has order pri i −1 in (Z/pri i Z)∗ .
ri −1
First of all, (1 + pi )pi ≡ (1 + pri i −1 )pi ≡ 1 (mod pri i ). Thus the order must divide pri i . But
ri −2
(1 + pi )pi ≡ 1 + pri i −1 6≡ 1 (mod pri i ), so the order of 1 + pi is pri i −1 .
Let a be relatively prime to pi and let a have order pi − 1 in (Z/pi Z)∗ (such an element exists because
ri −1
Z/pi Z is a finite field and thus its group of units is cyclic). We will show that api has order pi − 1 in
ri −1 ri −1
Z/pri i Z. Clearly, (api )pi −1 = api (pi −1) ≡ 1 (mod pri i ) since pri i −1 (pi − 1) is the order of (Z/pri i Z)∗
ri −1 ri −1
(Problem 7.5). Let 0 < m < p − 1. Then (api )n ≡ an 6≡ 1 (mod pi ). Hence, (api )n 6≡ 1 (mod pri i ),
ri −1 ri −1
so api has order pi − 1 in (Z/pri i Z)∗ . Since gcd(pri i −1 , pi − 1) = 1, we conclude that api (1 + pi )
has order pri i −1 (pi − 1) (see the solution of Problem 6.1(a)); hence, (Z/pri i Z)∗ is cyclic.

(c) Show that 52 ≡ 1 + 2t+2 (mod 2t+3 ), and then that (Z/2ri Z)∗ ∼
t
= Z/2ri −2 Z × Z/2Z if ri ≥ 3. Note that
r ∗
(Z/2i Z) is cyclic if ri ≤ 2.
We have
t t
2  t 2
2t 2t − 1 2k
 
2t 2t
X 2 X
5 = (1 + 4) = 4k = 1 + 2t+2 = 2 .
k k k−1
k=0 k=2
s
Let k = 2 l, where l is not divisible by 2. Then
2t
2t + 2k − s 2t − 1
 
2t
X
t+2
5 =1+2 + 22k .
k k−1
k=2

t
By induction on s, it is easily shown that 2k − s ≥ 3; hence, 52 ≡ 1 + 2t+2 (mod 2t+3 ). We will
ri −2
now show that 5 has order 2ri −2 in Z/2ri Z. Indeed, 52 ≡ (1 + 2ri −1 )2 ≡ 1 (mod 2ri ). Since
ri −3
52 ≡ 1 + 2ri −1 6≡ (mod 2ri ), we conclude that 5 has order 2ri −2 in Z/2ri Z. Hence, h5i is an index
2 subgroup of Z/2ri Z. We now show that −1 ∈ / h5i. Suppose not; since −1 has order 2, we must have

40
 ri −3
−1 ≡ 52 ≡ 1 + 2ri −1 (mod 2ri ) ⇒ 2 + 2ri −1 ≡ 0 (mod 2ri ). This is a contradiction, since ri ≥ 3
implies that 2 + 2ri −1 < 2 · 2ri −1 = 2 · 2ri −1 . It then follows that (Z/2ri Z)∗ = h−1, 5i, and therefore
that (Z/2ri Z)∗ ∼
= Z/2ri −2 Z × Z/2Z.
We note that (Z/2Z)∗ = {1} and (Z/4Z)∗ = {1, 3} ∼ = Z/2Z, so (Z/2r Z)∗ is cyclic if ri ≤ 2.
i
Problem 7.17 Let G be a finite Abelian group. Show that there is a Galois extension K/Q with Gal(K/Q) ∼ =
G.
By the structure theorem for finitely generated Abelian groups, we have that G ∼ = Z/n1 Z × Z/n2 Z ×
· · · × Z/nk Z, where the ni are distinct integers greater than 1. Dirichlet’s theorem on arithmetic progressions
states that there are infinitely many primes of the form 1 + ki ni for each i ∈ {1, . . . , k}. Hence, we can choose
distinct primes p1 , . . . , pk such that pi ≡ 1 (mod ni ) for each i. Note that these primes are all odd primes.
Let n := p1 p2 . . . pk . By the previous problem, (Z/nZ)∗ ∼ = Gal(Qn /Q) contains a subgroup isomorphic to
G. Since G is necessarily normal in (Z/nZ)∗ (since (Z/nZ)∗ is Abelian), we have that K := F(G) is Galois
over Q and Gal(K/Q) ∼ = G.

Section 8. Norms and Traces

Problem 8.1. Let K/F be an extension of finite fields. Show that the norm map NK/F is surjective.
Let |K| = pn and |F | = pm . Let α ∈ K ∗ be a generator of K ∗ . Recall that Gal(K/F ) is generated by
m
σ(a) = ap . Then Y m m 2 m (n/m−1)
NK/F (α) = τ (α) = α1+p +(p ) +···+(p ) .
τ ∈Gal(K/F )
n m (n/m)
Since p − 1 = (p ) − 1 = (p − 1)(1 + pm + (pm )2 + · · · + (pm )(n/m−1) ), we conclude that NK/F (α)
m

has order p − 1, and therefore generates F ∗ . Since NK/F (0) = 0, NK/F is surjective.
m

Problem 8.2. Let p be an odd prime, let ω be a primitive pth root of unity, and let K = Q(ω). Show
that NK/Q (1 − ω) = p.
The matrix representation of Lω−1 in the basis {1, ω, ω 2 , . . . , ω p−2 } is
 
1 0 0 0 ··· 0 1
−1 1 0 0 · · · 0 1
 
 0 −1 1 0 · · · 0 1
 
0
 0 −1 1 · · · 0 1 .
 .. .. .. .. . . .. .. 
 .
 . . . . . . 

0 0 0 0 · · · 1 1
0 0 0 0 · · · −1 2
Taking the determinant of the above matrix along the (p − 2)th column and using the fact that p is odd
shows that NK/Q (1 − ω) = p.

Problem 8.3. Let n ≥ 3 be an integer, let ω be a primitive nth root of unity, and let K = Q(ω). Show that
NK/Q (ω) = 1.
First we prove the following claim: Let Ψm (x) be the mth cyclotomic polynomial, where m ≥ 2. Then
Ψm (0) = 1. We proceed by induction on m. Note that Ψ2 (x) = x Q + 1. Now suppose that the claim holds for
all integers less than m. By Lemma 7.6, we have that xm − 1 = d|m Ψd (x). By the inductive hypothesis,
and since Ψ1 (x) = x − 1, we conclude that Ψm (0) = 1.
Now, the matrix representation of ω in the basis {1, ω, ω 2 , . . . , ω φ(n)−1 } is
 
0 0 0 ··· 0 −a0
1 0 0 · · · 0 −a1 
 
0 1 0 · · · 0 −a2 
 
0 0 1 · · · 0 −a3 
 ,
 .. .. .. . . .. .. 
. . .
 . . . 

0 0 0 · · · 0 −aφ(n)−2 
0 0 0 · · · 1 −aφ(n)−1

41
where Ψn (x) = xφ(n) + aφ(n)−1 xφ(n)−1 + · · · + a1 x + a0 . Note that a0 = 1 by our claim. Taking the determi-
nant of the above matrix along the first row, we then find that NK/Q (ω) = (−1)φ(n) . By Problem 7.8, φ(n)
is even for all n ≥ 3. Hence, NK/Q (ω) = 1.

Problem 8.4. In Examples 7.9 and 7.10, generators consisting of traces were found for intermediate fields.
Let K be a Galois extension of F . If L is an intermediate field of K/F , show that L is generated over F by
traces from K to L. In other words, show that L = F ({TK/L (a) : a ∈ K}).
By the Normal Basis Theorem (see Problem 5.17), there is an a ∈ K such that {σ(a) : σ ∈ Gal(K/F )}
is a basis for K as an F -vector space. Let
X
α = TrK/L (a) = σ(a).
σ∈Gal(K/L)

Then τ (α) = α for any τ ∈ Gal(K/L). Conversely, suppose τ (α) = α for some τ ∈ Gal(K/F ). Since
{σ(a) : σ ∈ Gal(K/F )} is a basis, τ σ must be in Gal(K/L) for every σ ∈ Gal(K/L). In particular,
τ ∈ Gal(K/L). Hence,
Gal(K/L) = Gal(K/F (α)) ⇒ F (α) = F ({TK/L (a) : a ∈ K}) = L.
Problem 8.5. Let K be a Galois extension of F . Prove or disprove that any intermediate field L of K/F
is of the form L = F ({NK/L (a) : a ∈ K}).
The statement is true. First, suppose that F is a finite field (and so is K, since we are assuming K/F is
a finite extension). By Problem 8.1, NK/L : K → L is surjective. Hence, L = F ({NK/L (a) : a ∈ K}).
Now suppose that F is infinite. By the Primitive Element Theorem, there is an α ∈ K such that
K = F (α). Let
Y
f (x) = min(L, α) = (x − σ(α)) = xn + ln−1 xn−1 + · · · + l1 x + l0 .
σ∈Gal(K/L)

We will prove that L = F (l0 , . . . , ln−1 ). It is clear that F ({l0 , . . . , ln−1 }) ⊆ L, so

Gal(K/L) ⊆ Gal(K/F (l0 , . . . , ln−1 )).
Let τ ∈ Gal(K/F (l0 , . . . , ln−1 )). Then, since τ fixes f (x), it permutes the σ(α) for σ ∈ Gal(K/L). Therefore,
τ σ(α) = σ 0 (α) for some σ 0 ∈ Gal(K/L). Since K = F (α), τ σ = σ 0 ⇒ σ ∈ Gal(K/L). Hence,
Gal(K/L) = Gal(K/F (l0 , . . . , ln−1 )) ⇒ L = F (l0 , . . . , ln−1 ).
For a ∈ F , we have
Y Y Y
f (a) = (a − σ(α)) = (σ(a) − σ(α)) = σ(a − α) = NK/L (a − α),
σ∈Gal(K/L) σ∈Gal(K/L) σ∈Gal(K/L)

and the problem is reduced to showing that l0 , . . . , ln−1 ⊆ F ({f (a) : a ∈ F }). Since |F | = ∞, we can take n
distinct elements a0 , . . . , an−1 ∈ F to obtain the following equation:

a20 · · · an−1 f (a0 ) − an0

    
1 a0 0 l0
 .. .. .. .. . .
..   ..  = 
    ..
. . . .  .
2 n−1 n
1 an−1 an−1 · · · an−1 l n−1 f (an−1 ) − an−1 .
Q
The n × n matrix above is a Vandermonde matrix, which has determinant 0≤i<j≤n−1 (aj − ai ) (see
Lemma 12.5). Since the ai are distinct, the determinant is nonzero, and the matrix is invertible. Hence,
l0 , . . . , ln−1 ⊆ F ({f (a) : a ∈ F }), as desired.

Problem 8.6. Let f ⊆ K ⊆ L be fields with L/F a finite extension. Use the product theorem for the
purely inseparable degree proved in this section to prove the corresponding product formula for separable
degree; that is, prove that [L : F ]s = [L : K]s [K : F ]s .
See Problem 4.15.

42
Section 9. Cyclic Extensions
Problem 9.1. Suppose that F is a field containing a primitive nth root of unity, and let a ∈ F . Show that
xn − a is irreducible over F if and only if a is not an mth power for any m > 1 dividing n.
(⇒) Suppose that a = bm for some b ∈ F and m dividing n. Then xn − a = (xk )m − bm is divisible by
k
x − b, and is therefore not irreducible over F .
(⇐) Suppose that xn − a is reducible. Let ω ∈ F be a primitive nth root of unity. Then
n
Y √
xn − a = (x − ω i n
a) = f (x)g(x)
i=1

where√f and g√are non-constant and deg(f ) = k and deg(g) = n − k. The constant √term in f (g, resp.)
n n n
is ω b ak (ω c an−k , resp.) for some integers b and c. By the Euclidean algorithm, ad = b ∈ F , where
d = gcd(n − k, k). For some positive integers s and t, we have n − k = sd and k = td, implying n = (s + t)d.
Hence, d divides n and a = bn/d , contradicting the assumption that a is not an mth power for m dividing n.

Problem 9.2. Suppose that F is a field, and let ω be a primitive nth root of unity in an algebraic clo-
sure of F . If a ∈ F is not an mth power in F (ω) for any m > 1 that divides n, show that xn − a is
irreducible over F .
By the result of the preceding problem, xn −a is irreducible over F (ω), and is therefore irreducible over F .

Problem 9.3. This problem describes cyclic extensions of degree four of a base field that does not con-
tain a primitive
√ fourth root of unity. Let F be a field that
√ does not contain a primitive fourth root of unity.
Let L = F ( a) for some a ∈ F − F 2 , and let K = L( b) for some b ∈ L − L2 . Show that the following
statements are equivalent:
(a) a is a sum of two squares in F .
(b) −1 = NL/F (α) for some α ∈ L.
(c) a = NL/F (α) for some α ∈ L.

(d) NL/F (b) ≡ a mod F ∗ 2 for some b ∈ L.

(e) K/F is a cyclic extension (with the b in Problem 3d).
(f) L lies in a cyclic extension of F of degree 4.
We will assume char(F ) 6= 2 throughout, though it is not necessary
√ for most implications. √
((a) ⇒ (b)) Let a = b2 + c2 , where b, c ∈ F . Let α = cb + 1c b2 + c2 . With respect to the basis {1, a},
we have
b bc + c2
 
c c 
α =  ,
1 b
 
c c
and NL/F (α) = −1. √ 2 √ 2
2
((b)√⇒ (c)) Let NL/F (α) = −1. Note that N √ L/F (a) = a = NL/F ( a) , so NL/F ( a) = ±a. If
NL/F ( a)2 = a, we are done. Otherwise, NL/F (α a)2 = a.
((c) ⇒ (d)) Trivial. √
((d) ⇒ (e)) Let b = k1 + k2 a, with k1 , k2 ∈ F . Then NL/F (b) = k12 − ak22 = ak 2 for some k ∈ F . Note
√
that b satisfies the polynomial
√ √ p p 1
f (x) = (x − b)(x + a/b)(x + k a/b) = x4 − (ak 2 + b2 )x2 + ak 2 .
b)(x − k
b
√
If k2 = 0, then a = (k1 /k)2 ∈ F 2 , a contradiction.
√ Hence, k2 6= 0, and therefore K = F ( b). Since [K : F ] =
[K : L][L : F ] = 2 · 2 = 4, and K = F ( b), f is irreducible over F . Since f ’s roots are distinct and f splits

43
 √ p
over K, the extension K/F is Galois, which means there is a σ ∈ √ Gal(K/F ) such that σ( b) =p
k a/b (see
√ p √
Problem 5.1). We will show that σ has order 4. First, note that σ 2 ( b) = kσ( √a)·(k √a/b)−1 = b/aσ( a).
If σ does not have order 4, then it must have order 2, which would imply σ( a) = a by the previous line.
However,
√ q
√ p q
√ √ √
σ( b) = σ( k1 + k2 a) = k a/b = k1 − k2 a ⇒ σ( a) = − a,
so we conclude that σ has order 4, and therefore generates Gal(K/F ).
((e) ⇒ (f)) Since [K : F ] = 4, Gal(K/F ) ∼ = Z4 . Hence, L is contained in K, which is a cyclic extension
of F of degree 4.
((f) ⇒ (a)) Let K be the cyclic extension
√ of F containing L. Since [K : F ] = [K :√
L][L : F ] = 4, we have
that [K √ : L] = 2, √and therefore K = L( b) for some b ∈ L − L2 . If b ∈ F , then [F ( b) : F ] = 2, implying
that F ( b) = F ( a), since there is only√one degree 2 intermediate extension of K/F (using the fact that
Gal(K/F √) is cyclic). But then K = L( b) = L, contradicting the assumption [K : F ] = 4. Therefore,
K = F ( b). √ √ √ √
Let b = k1 + k2 a. Note that k1 6= 0 since F ( a) 6= F ( b). Then min(F, b) = f (x) = x4 − 2k1 x2 +
k12 − k22 a. Since we are assuming that char(F ) 6= 2, the roots of f are of the form ±α, ±β and the constant
term is k12 − k22 a =pα2 β 2 . Since K/F is Galois, f splits over K and therefore α, β ∈ K. Hence, k12 − k22 a is a
squarepin K, and k12 − k22 a ∈ K.
2 2 2 2 2
If k12 − kp 2
2 a ∈ F , then k1 − k2 a = k and hence a = (k/k1 ) + (k2 a/k1 ) and we are done. Otherwise,
√ 2 2
F ( a) = F ( k1 −p k2 a) since there is only one intermediate degree 2 extension of K/F . It is then not
hard to show that a(k12 − k22 a) ∈ F and therefore a(k12 − k22 a) = k 2 for some k ∈ F . Again, we have
a = (k/k1 )2 + (k2 a/k1 )2 and we are done.

Problem 9.4. This problem investigates the splitting field of the polynomial xn − a over a field F that
does not contain a primitive nth root of unity.

(a) If a ∈ F , show that the splitting field of xn − a over F is F (α, ω), where αn = a and ω is a primitive
nth root of unity. Since xn − a = (x − α)(x − ωα) · · · (x − ω n−1 α) in F (α, ω), and F (α, ω) is generated
by the roots of xn − a over F , F (α, ω) is the splitting field of xn − a over F .
(b) Let N = F (α, ω), let K = F (α), and let L = F (ω). Show that L/F is Galois and N/L is cyclic.
The minimal polynomial min(F, ω) divides xn − 1 which splits over L and has n distinct roots since ω
is primitive. Hence, L/F is normal and separable, and therefore Galois.
√
Since L is a field containing a primitive nth root of unity and K = L( n a) where a ∈ L, Proposition
9.6 implies that N/L is cyclic.
(c) Suppose that min(F, ω) = (x − ω)(x − ω −1 ) and that [N : L] = n. Show that there is an element
σ ∈ Gal(N/F ) with σ(α) = ωα and σ(ω) = ω, and a τ with τ (ω) = ω −1 and τ (α) = α. Moreover,
show that the order of σ is n, the order of τ is 2, and τ στ = σ −1 . Recall the definition of the dihedral
group Dn , and show that Dn = Gal(N/F ).
Since [N : L] = n, the minimal polynomial is min(L, α) = xn − a. Since N/L is cyclic, and in particular
Galois, there is an element σ ∈ Gal(N/L) such that σ(α) = ωα. Since σ fixes L, σ(ω) = ω. Since L/F
is Galois, there is a τ ∈ Gal(L/F ) such that τ (ω) = ω −1 . Since N is a splitting field for xn − a over
F , it is also a splitting field over L. Hence, by the Isomorphism Extension Theorem, we can extend τ
to an isomorphism of K such that τ (α) = α.
Now, σ k (α) = ω k α. Since ω has order n in N ∗ , the minimal k such that σ k (α) = α is n. Since
σ n is the identity on F ∪ {α, ω}, σ n = id. Hence, σ has order n. Similarly, τ 2 is the identity on
F ∪ {α, ω}, so τ has order 2. Note that τ στ (α) = ω −1 α = σ −1 (α) and τ στ (ω) = ω = σ −1 (ω), so
τ στ = τ −1 . Note that |hσi| = n and τ ∈ / hσi. Therefore, |hσ, τ i| ≥ 2n. On the other hand, min(F, ω)
has degree 2, so [N : F ] = [N : L][L : F ] = 2n, and Gal(N/F ) = 2n ⇒ hσ, τ i = Gal(N/F ). Since
Dn = hs, t | sn , tst = s−1 i, there is a homomorphism Dn → hσ, τ i sending s 7→ σ and t 7→ τ . The
image of Dn must have order ≥ 2n, since hsi maps onto hσi and τ does not map into hσi. Therefore,
Dn → hσ, τ i is an isomorphism, and Gal(N/F ) ∼ = Dn .

44
 (d) Let p be an odd prime, and let ω ∈ C be a primitive pth root of unity. √ let F = Q(ω) ∩ R. Let a ∈ Q
be a rational
√ number that is not a pth power in Q. Show that [F ( p
a) : F ] = p and that if L = F (ω),
then [L( p a) : L] = p. Conclude that if N is the splitting field of xp − a over F , then Gal(N/F ) = Dp .
Qp−1
We begin by showing that xp − a is irreducible over Q. Suppose not. Then xp − a = i=0 (x − ω i ) =
(xk + · · · + ω j ak/p )(xp−k + · · · + ω l a(p−k)/p ) for some k ∈ {1, . . . , p − 1}. Then ω j ak/p , ω l a(p−k)/p ∈
Q. This implies that j and l are multiples of p, so that ω j = ω l = 1. Then ak/p , a(p−k)/p ∈ Q,
implying that agcd(k,p−k)/p ∈ Q.√ But gcd(k, p − k) must divide p, and since gcd(k, p − k) < 1, we have
gcd(k, p − k) √ = 1. Therefore, p a ∈ Q, a contradiction. We conclude that xp − a is irreducible, and
therefore [Q( a) : Q] = p.
p

Suppose that xp −a is reducible over Q(ω). Then, by√the same argument as above, ω j ak/p , ω l a(p−k)/p ∈
Q(ω).
√ Hence,
√ ak/p , a(p−k)/p ∈ Q(ω),
√ and therefore p a ∈ Q(ω). However, p − 1 = [Q(ω) :√Q] = [Q(ω) :
Q( a)][Q( a) : Q] = [Q(ω) : Q( a)] · p, a contradiction. Note that L = Q(ω), so [L( √
p p p p
a) : L] = p.
p
Moreover, since x − a is irreducible over Q(ω), it is irreducible over Q(ω) ∩ R. Hence, [F ( p a) : F ] = p.
√
By part (a), the splitting field is N = F ( p a, ω). Next, min(F, ω) = (x − ω)(x − ω −1 ) = x2 − (ω +
ω −1 )x+1, since ω +ω −1 ∈ R. From the previous paragraph, [N : L] = p. By part (c), Gal(N/F ) ∼ = Dp .
Problem 9.5. In this problem, we prove the following√result: Suppose that K/F is a finite extension with
K algebraically closed. Then char(F ) = 0 and K = F ( −1). Use the following steps to prove this:
r
(a) If char(F ) = p > 0 and β ∈ F − F p , then xp − β is irreducible over F for all r > 0.
r r
Let M be an algebraic closure of F , and let α be a root of xp − β. Then β = αp ∈ F , so Proposition
m m
4.16 implies that min(F, α) = xp − αp for some m. If m < r, then β ∈ F p , a contradiction. Hence,
r r r r
min(F, α) = xp − αp = xp − β, and therefore xp − β is irreducible.
(b) If char(F ) = p > 0 and there is a cyclic extension of degree p, then there are cyclic extensions of F of
degree pr for any r ≥ 1.
This seems tough. I think it requires some Kummer theory and Witt vectors, which I know nothing
about right now.
(c) Let p be a prime, and suppose that either F contains a primitive pth root of unity for p odd, or that F
contains a primitive fourth root of unity for p = 2. If there is an a ∈ F with xp − a irreducible over F ,
2
then xp − a is irreducible over F .
2 2
Let M be an algebraic closure of F , and let α ∈ M be a root of xp − a. Then xp − a is irreducible if
[F (α) : F ] = p2 . Since (αp )p − a = 0 and xp − a is irreducible, [F (αp ) : F ] = p. Hence, it is enough to
show that [F (α) : F (αp )] = p, or equivalently that xp − αp is irreducible over F (αp ).
Suppose xp −αp is reducible over F (αp ). In both cases, F contains a primitive pth root of unity. Hence,
Qp−1
the roots of xp − αp are α, ωα, . . . , ω p−1 α, so xp − αp = i=0 (x − ω i α) = (xk + · · · + ω j αk )(xp−k +
· · · + ω l αp−k ). Then, as argued in the previous problem, α ∈ F (αp ). Hence, it suffices to show that
F (αp ) does not contain any root of xp − αp to show that xp − αp is irreducible over F (αp ).
Let β be a root of xp − αp . Then

NF (αp )/F (β)p = NF (αp )/F (β p ) = NF (αp )/F (αp ) = (−1)p+1 a,

since min(F, αp ) = xp − a (see Proposition 8.6). If p is odd, then NF (αp )/F (β)p ∈ F is a root of xp − a,
contradicting the assumption that xp − a is irreducible over F . Now suppose that p = 2, then i ∈ F ,
where i is a primitive fourth root of unity. Then

(iNF (αp )/F (β))2 = a,

which is again a contradiction.

(d) Use the previous steps to prove the result.
Assume that char(F ) = p > 0. By part (a), we must have F = F p , since otherwise there is a β ∈ F −F p
r
and xp − β is irreducible for every r > 0. But then [K : F ] ≥ pr for all r > 0, contradicting the

45
 assumption that K/F is a finite extension. Hence, F is perfect, and therefore K/F is separable. Since
K is algebraically closed, K/F is also normal, and therefore Galois.
Suppose that |Gal(K/F )| is divisible by a prime q. Then there is an intermediate field L such that
[K : L] = q and Gal(K/L) ∼ = Zq . If q = p, then by part (b), there are cyclic extensions of L of degree
pr for every r > 0, contradicting the fact that K/F is a finite extension. Hence, q 6= p. Suppose q is
odd. Then xq − 1 is separable (by the derivative test), and therefore there is a primitive qth root of
unity ω ∈ K (since K is algebraically closed). Hence, L ⊆ L(ω) ⊆ K. Since [L(ω) : L] ≤ q − 1 √ and
[K : L] = q, we have [L(ω) : L] = 1 ⇒ ω ∈ L. By Theorem 9.5, there is a b ∈ L such that K = L( q b).
2
By part (c), xq − b is irreducible over L, implying that [K : L] ≥ q 2 since K is algebraically closed, a
contradiction. We conclude that |Gal(K/F )| is not divisible by an odd prime.
Now suppose
√ [K : F ] is even and p 6= 2. Let L be an intermediate field such that [K : L] = 2. Then
K = L( b) for some b ∈ L − L2 . Since x4 − 1 is separable by the derivative test, K contains a primitive
fourth root of unity, which we call i. If i ∈ L, then, by part (c), x4 − b is irreducible, contradicting the
fact that [K : L] = 4, since K is algebraically closed. Now assume i ∈ / L, then i2 = −1, so K = L(i).
We will now show that a sum of squares in L is always a square in L. Since K is algebraically closed,
every element of K is a square in K. Let a, b ∈ L, then a + ib = (c + id)2 for some c, d ∈ L. Then
Pp−1
a = c2 − d2 and b = 2cd. Then a2 + b2 = (c2 + d2 )2 . Then, since −1 = i=1 1, −1 is a sum of squares,
and therefore a square in L. Therefore, L contains a primitive 4th root of unity, so x4 − b is irreducible,
which is a contradiction.
We conclude that char(F ) = 0. Then K/F is separable, and it is also normal since K is algebraically
closed. Note that [K : F ] cannot be divisible by an odd prime by the same argument given in
characteristic p above. Now suppose [K : F ] is divisible by 4, so that there is an intermediate field L
such that [K : L] = 2. If L contains a primitive fourth root of unity i, then x4 − b is irreducible for
some b ∈ L, contradicting [K : L] = 2. Then i ∈ / F , so [F (i) : F ] = 2 = [K : F (i)]. But [K √
: F (i)] = 2
is a contradiction, since i ∈ F (i). We conclude that [K : F ] = 2 and i ∈ / F . Hence, K = F ( −1).

Section 10. Hilbert Theorem 90 and Group Cohomology

Problem 10.1. Let M be a G-module. Show that the boundary map δn : C n (G, M ) → C n+1 (G, M ) defined
in this section is a homomorphism.
Let f and g be n-cochains. For n ≥ 1,
n
X
δn (f + g)(σ1 , . . . , σn+1 ) = σ1 (f + g)(σ2 , . . . , σn+1 ) + (−1)i (f + g)(σ1 , . . . , σi σi+1 , . . . , σn+1 )
i=1
n+1
+ (−1) (f + g)(σ1 , . . . , σn )
n
X
= σ1 f (σ2 , . . . , σn+1 ) + (−1)i f (σ1 , . . . , σi σi+1 , . . . , σn+1 )
i=1
n+1
+ (−1) f (σ1 , . . . , σn )
X n
+ σ1 g(σ2 , . . . , σn+1 ) + (−1)i g(σ1 , . . . , σi σi+1 , . . . , σn+1 )
i=1
+ (−1)n+1 g(σ1 , . . . , σn )
= δn (f )(σ1 , . . . , σn+1 ) + δn (g)(σ1 , . . . , σn+1 ).

For n = 0,

δ0 (m1 + m2 )(σ) = σ(m1 + m2 ) − (m1 + m2 ) = σm1 − m1 + σm2 − m2 = δ0 (m1 )(σ) + δ0 (m2 )(σ).

Therefore, δn is a homomorphism for all n.

46
Problem 10.2. With notation as in the previous problem, show that δn+1 ◦ δn is the zero map.
For n ≥ 1,
n+1
X
δn+1 ◦ δn (f )(σ1 , . . . , σn+2 ) = σ1 δn (f )(σ2 , . . . , σn+1 ) + (−1)i δn (f )(σ1 , . . . , σi σi+1 , . . . , σn+2 )
i=1
+ (−1)n+2 δn (f )(σ1 , . . . , σn+1 )
 n
X 
= σ1 σ2 f (σ3 , . . . , σn+2 ) + (−1)i f (σ2 , . . . , σi+1 σi+2 , . . . , σn+2 ) + (−1)n+1 f (σ2 , . . . , σn+1 )
i=1
 n
X
− σ1 σ2 f (σ3 , . . . , σn+2 ) − f (σ1 σ2 σ3 , σ4 , . . . , σn+2 ) + (−1)i f (σ1 σ2 , σ3 , . . . , σi+1 σi+2 , . . . , σn+2 )
i=2
 Xn 
n+1 i
+ (−1) f (σ1 σ2 , . . . , σn+1 ) + (−1) σ1 f (σ2 , . . . , σi σi+1 , . . . , σn+2 )
i=2
i−2
X
+ (−1)j f (σ1 , . . . , σj σj+1 , . . . , σi σi+1 , . . . , σn+2 ) + (−1)i−1 f (σ1 , . . . , σi−1 σi σi+1 , . . . , σn+2 )
j=1
n
X
+ (−1)i f (σ1 , . . . , σi σi+1 σi+2 , . . . , σn+2 ) + (−1)j f (σ1 , . . . , σi σi+1 , . . . , σj+1 σj+2 , . . . , σn+2 )
j=i+1
 
(−1)n+1 f (σ1 , . . . , σi σi+1 , . . . , σn+1 ) + (−1)n+1 σ1 f (σ2 , . . . , σn+1 σn+2 )
n−1
X
+ (−1)j f (σ1 , . . . , σj σj+1 , . . . , σn+1 σn+2 ) + (−1)n f (σ1 , . . . , σn σn+1 σn+2 )
j=1
  n
X
n+1 n+2
+ (−1) f (σ1 , . . . , σn ) + (−1) σ1 f (σ2 , . . . , σn+1 ) + (−1)i f (σ1 , . . . , σi σi+1 , . . . , σn+1 )
i=1

n+1
(−1) f (σ1 , . . . , σn )
n
X
= − (−1)i f (σ1 σ2 , σ3 , . . . , σi+1 σi+2 , . . . , σn+2 )
i=2
n X
X i−2
+ (−1)i (−1)j f (σ1 , . . . , σj σj+1 , . . . , σi σi+1 , . . . , σn+2 )
i=3 j=1
n+1
X n
X
− f (σ1 , . . . , σi−1 σi σi+1 , . . . , σn+2 ) + f (σ1 , . . . , σi σi+1 σi+2 , . . . , σn+2 )
i=2 i=1
n−1
X n
X
+ (−1)i (−1)j f (σ1 , . . . , σi σi+1 , . . . , σj+1 σj+2 , . . . , σn+2 )
i=2 j=i+1
n−1
X
+ (−1)n+1 (−1)j f (σ1 , . . . , σj σj+1 , . . . , σn+1 σn+2 )
j=1
n+1
XX i−2
= (−1)i (−1)j f (σ1 , . . . , σj σj+1 , . . . , σi σi+1 , . . . , σn+2 )
i=3 j=1
n−1
X n
X
+ (−1)i (−1)j f (σ1 , . . . , σi σi+1 , . . . , σj+1 σj+2 , . . . , σn+2 )
i=1 j=i+1

= 0,

47
as desired. For n = 0, we have

δ1 ◦ δ0 (m)(σ1 , σ2 ) = σ1 δ0 (m)(σ2 ) − δ0 (m)(σ1 σ2 ) + δ0 (m)(σ1 )

= σ1 (σ2 m − m) − (σ1 σ2 m − m) + (σ1 m − m)
= 0.

Problem 10.3. Let M be a G-module, and let f ∈ Z 2 (G, M ). Show that f (1, 1) = f (1, σ) = f (σ, 1) for all
σ ∈ G.
First, note that

δ2 (f )(σ1 , σ2 , σ3 ) = σ1 f (σ2 , σ3 ) − f (σ1 σ2 , σ3 ) + f (σ1 , σ2 σ3 ) − f (σ1 , σ2 ) = 0

for any σ1 , σ2 , σ3 ∈ G. Substituting σ1 = σ2 = 1 gives f (1, σ) = f (1, 1). Substituting σ2 = σ3 = 1, we obtain

σf (1, 1) = f (σ, 1). Note that there is an error in the question.

Problem 10.4. If E is a group with an Abelian normal subgroup M , and if G = E/M , show that the
action of G on M given by σm = eme−1 if eM = σ is well defined and makes M into a G-module.
Suppose σ = e1 M = e2 M . Then e1 = e2 m̃ for some m̃ ∈ M . Then σm = e1 me−1 1 = e2 m̃mm̃
−1 −1
e2 =
−1
e2 me2 since M is Abelian. Thus, the action of G on M is well defined.
Now, 1m = 1m1−1 = m, σ(τ m) = σ(eτ me−1 −1 −1
τ ) = eσ eτ meτ eσ = (eσ et au)m(eσ eτ )
−1
= (στ )m, and
−1 −1 −1
σ(m1 m2 ) = e(m1 m2 )e = (em1 e )(em2 e ) = σ(m1 )σ(m2 ) (we use multiplicative notation here instead
of additive in M since the group G is not necessarily Abelian). Thus, M is a G-module.

Problem 10.5. With E, M, G as in the previous problem, if eσ is a coset representative of σ, show that the
function defined by f (σ, τ ) = eσ eτ e−1
στ is a 2-cocycle.
We use multiplicative notation since G is not, in general, Abelian. We have,

δ2 (f )(σ1 , σ2 , σ3 ) = σ1 f (σ2 , σ3 )f (σ1 σ2 , σ3 )−1 f (σ1 , σ2 σ3 )f (σ1 , σ2 )−1

= σ1 (eσ2 eσ3 e−1 −1
σ2 σ3 )(eσ1 σ2 eσ3 eσ1 σ2 σ3 )
−1
(eσ1 eσ2 σ3 e−1 −1
σ1 σ2 σ3 )(eσ1 eσ2 eσ1 σ2 )
−1

= (eσ1 eσ2 eσ3 e−1 −1 −1

σ2 σ3 eσ1 )(eσ1 σ2 eσ3 eσ1 σ2 σ3 )
−1
(eσ1 eσ2 σ3 e−1 −1
σ1 σ2 σ3 )(eσ1 eσ2 eσ1 σ2 )
−1

= (eσ1 eσ2 eσ3 e−1 −1 −1 −1

σ2 σ3 eσ1 )(eσ1 eσ2 σ3 eσ1 σ2 σ3 )(eσ1 σ2 eσ3 eσ1 σ2 σ3 )
−1
(eσ1 eσ2 e−1
σ1 σ2 )
−1

= eσ1 eσ2 eσ3 e−1 −1 −1 −1 −1 −1 −1

σ2 σ3 eσ1 eσ1 eσ2 σ3 eσ1 σ2 σ3 eσ1 σ2 σ3 eσ3 eσ1 σ2 eσ1 σ2 eσ2 eσ1
= 1.

Therefore, f is a 2-cocycle. Note that we used the commutativity of M to go from the third to the fourth line.

Problem 10.6. Suppose that M is a G-module. For each σ ∈ G, let mσ ∈ M . Show that the cochain
f defined by f (σ, τ ) = mσ + σmτ − mστ is a coboundary.
We have a 1-chain f : G → M, σ 7→ mσ . By definition, δ1 (g)(σ, τ ) = σg(τ ) − g(στ ) + g(σ) =
σmτ − mστ + mσ = f (σ, τ ). Hence, f ∈ B 2 (G, M ).

Problem 10.7. If M is a G-module and f ∈ Z 2 (G, M ), show that Ef = M × G with multiplication

defined by (m, σ)(n, τ ) = (m · σn · f (σ, τ ), στ ) makes Ef into a group.
We begin with associativity. On one hand,

((m, σ)(n, τ ))(r, ρ) = ((m · σn · f (σ, τ ), στ ))(r, ρ)

= (m · σn · f (σ, τ ) · στ r · f (στ, ρ), στ ρ).

On the other,

(m, σ)((n, τ )(r, ρ)) = (m, σ)(n · τ r · f (τ, ρ), τ ρ)

= (m · σ(n · τ r · f (τ, ρ)) · f (σ, τ ρ), στ ρ)
= (m · σn · στ r · σf (τ, ρ) · f (σ, τ ρ), στ ρ).

48
Since f is a 2-cocyle, f (σ, τ )·f (στ, ρ) = σf (τ, ρ)·f (σ, τ ρ). Therefore, the product defined above is associative.
We now show that (f (1, 1)−1 , 1) is the identity element. We have

(f (1, 1)−1 , 1)(m, σ) = (f (1, 1)−1 · m · f (1, σ), σ) = (m, σ),

since f (1, 1) = f (1, σ) (see Problem 10.3). Moreover,

(m, σ)(f (1, 1)−1 , 1) = (m · σf (1, 1)−1 · f (σ, 1), σ) = (m, σ),

since σf (1, 1) = f (σ, 1) (see Problem 10.3) and σf (1, 1)−1 = (σf (1, 1))−1 (this is obscured because of the
multiplicative notation, but is obvious with additive notation). Hence, (f (1, 1)−1 , 1) is the identity element.
Finally, we show that (m, σ)−1 = (σ −1 (f (σ, σ −1 ) · m · f (1, 1))−1 , σ −1 ) (note that the formula for the
inverse given on page 99 of Morandi’s book is incorrect). On one hand,

(m, σ)(σ −1 (f (σ, σ −1 ) · m · f (1, 1))−1 , σ −1 ) = (m · (f (σ, σ −1 ) · m · f (1, 1))−1 · f (σ, σ −1 ), 1) = (f (1, 1)−1 , 1).

On the other hand,

(σ −1 (f (σ, σ −1 ) · m · f (1, 1))−1 , σ −1 )(m, σ) = (σ −1 (f (σ, σ −1 ) · m · f (1, 1))−1 · σ −1 m · f (σ −1 , σ), 1)

= (σ −1 f (σ, σ −1 )−1 · σ −1 f (1, 1)−1 · f (σ −1 , σ), 1)
= (f (1, 1)−1 , 1).

To get the last line, make the substitution σ = σ −1 , τ = σ, and ρ = σ −1 into the cocycle condition, we obtain
f (σ −1 , σ) · f (1, σ −1 ) = σ −1 f (σ, σ −1 ) · f (σ −1 , 1), implying f (1, 1) = σ −1 f (σ, σ −1 ) · σ −1 f (1, 1) · f (σ, σ −1 )−1 ,
where we have used Problem 10.3. We conclude that Ef is a group with the given product.

Problem 10.8. If M is a G-module, show that the group extensions constructed from 2-cocycles f, g ∈
Z 2 (G, M ) are isomorphic if f and g are cohomologous.
Suppose that f and g differ by the coboundary h. More precisely,

f (σ, τ )g(σ, τ )−1 = δ1 (h)(σ, τ ) = σh(τ ) · h(στ )−1 · h(σ)

for some h ∈ B 1 (G, M ). We claim that the map φ : Ef → Eg , (m, σ) 7→ (m · h(σ), σ) is an isomorphism.
First, we check that φ is bijective. Let (m · h(σ), σ) = (m0 · h(σ 0 ), σ 0 ). From the second components, we
have σ = σ 0 , which implies m = m0 from the first components. For an arbitrary (m, σ) ∈ Eg , we have
(m · h(σ)−1 , σ) 7→ (m, σ). Hence, φ is a bijection.
We now check that φ is a homomorphism. On one hand,

(m, σ)(n, τ ) = (m · σn · f (σ, τ ), στ ) 7→ (m · σn · f (σ, τ ) · h(στ ), στ ).

On the other hand,

(m · h(σ), σ)(n · h(τ ), τ ) = (m · h(σ) · σn · σh(τ ) · g(σ, τ ), στ ) = (m · σn · f (σ, τ ) · h(στ ), στ ).

Thus, φ is an isomorphism.

Problem 10.9. In the crossed product construction given in this section, show that the multiplicative
identity is f (1, 1)−1 xid .
Let id = 1 and let xσ ∈ A be an arbitrary symbol. By definition,

f (1, 1)−1 x1 xσ = f (1, 1)−1 f (1, σ)x1·σ = xσ ,

and
xσ · f (1, 1)−1 x1 = [σ(f (1, 1))]−1 xσ x1 = [σ(f (1, 1))]−1 f (σ, 1)xσ·1 = xσ
where we have used the fact that f is a 2-cocycle. Using distributivity for the general products, we see that
f (1, 1)−1 x1 is the multiplicative identity.

49
Problem 10.10. A normalized cocycle is a cocycle f that satisfies f (1, σ) = f (σ, 1) = 1 for all σ ∈ G. Let
A = (K/F, G, f ) be a crossed product algebra. Show that xid = 1 if and only if f is a normalized cocycle.
Since f is a cocycle, f (1, 1) = f (1, σ) = σ −1 f (σ, 1) = 1 for every σ ∈ G. Then f (σ, 1) = 1. Hence, the
condition of being a normalized cocycle is equivalent to f (1, 1) = 1. By Problem 10.9, x1 = 1 if and only if
x1 = f (1, 1)−1 x1 if and only if f (1, 1) = 1.

Problem 10.11. In the construction of group extensions, show that if eid is chosen to be 1, then the
resulting cocycle is a normalized cocycle.
By definition, f (1, 1) = e1 e1 e−1
1 = 1, so f is a normalized cocycle.

Problem 10.12. Show that any 2-cocycle is cohomologous to a normalized cocycle.

We return to additive notation for this problem. Let f ∈ Z 2 (G, M ) be arbitrary. Define h(σ) = f (σ, 1).
Then f is clearly cohomologous to g = f − δ1 (h). Using that f is a 2-cocycle, we have
δ1 (h)(σ, τ ) = σh(τ ) − h(στ ) + h(σ) = σf (τ, 1) − f (στ, 1) + f (σ, 1) = f (σ, 1).
Therefore, g(1, 1) = f (1, 1) − f (1, 1) = 0, so g is a normalized cocycle.

Problem 10.13. If two crossed products (K/F, G, f ) and (K/F, G, g) are isomorphic as F -algebras, show
that f and g are cohomologous.
By Problem 10.12, we may assume that f and g are normalized. Let A = (K/F, G, f ) and B =
(K/F, G, g), and let φ : A → B be an isomorphism. Note that there are natural inclusions ιA , ιB : K ,→ A, B.
By the Skolem-Noether theorem, there is a b ∈ B such that ιB (a) = β · φ(ιA (k)) · β −1 for all k ∈ K and for
some unit β ∈ B.
We now show that all the symbols xσ ∈ A and yσ ∈ B are units. Indeed,
xσ · σ −1 (f (1, 1)−1 f (σ, σ −1 )−1 )xσ−1 = f (1, 1)−1 x1
and
σ −1 (f (1, 1)−1 f (σ, σ −1 )−1 )xσ−1 · xσ = σ −1 (f (1, 1)−1 f (σ, σ −1 )−1 )f (σ −1 , σ)x1 = f (1, 1)−1 x1
where the last equality follows from the cocycle condition and Problem 10.3. Since f (1, 1)−1 x1 is the identity
(Problem 10.9), we see that the sympols xσ are all invertible.
I will now stop indicating the inclusion maps; for example, the identity ιB (k) = β · φ(ιA (k)) · β −1 will
be written k = β · φ(k) · β −1 . By definition, for every k ∈ K and σ ∈ G, we have σ(k) = xσ · k · x−1 σ and
σ(k) = yσ · k · yσ−1 . Moreover,
σ(k) = β · φ(σ(k)) · β −1 = β · φ(xσ · k · x−1
σ )·β
−1
= β · φ(xσ ) · φ(k) · φ(xσ )−1 · β −1
= β · φ(xσ ) · β −1 · k · β · φ(xσ )−1 · β −1
= yσ · k · yσ−1 .
Therefore, [yσ−1 βφ(xσ )β −1 ] · k · [yσ−1 βφ(xσ )β −1 ]−1 · k −1 =
P1 for every k ∈ K. We will use this to show that
yσ−1 βφ(xσ )β −1 ∈ K for all σ ∈ G. Let yσ−1 βφ(xσ )β −1 = σ∈G aσ yσ . Then
X X X X
aσ yσ · k = aσ σ(k)yσ = k · aσ yσ = aσ kxσ
σ∈G σ∈G σ∈G σ∈G

and therefore aσ σ(k) = aσ k for every σ ∈ G and every k ∈ K. Thus, a1 is the only non-zero coefficient,
implying that yσ−1 βφ(xσ )β −1 ∈ K since y1 = 1.
For every σ ∈ G, let yσ−1 βφ(xσ )β −1 = kσ0 ∈ K. Then βφ(xσ )β −1 = yσ kσ0 = σ(kσ0 )yσ . Letting kσ =
σ(kσ0 )−1 , we have yσ = kσ βφ(xσ )β −1 . Hence,
yσ yτ = g(σ, τ )yστ
−1
kσ βφ(xσ )β kτ βφ(xτ )β −1 = g(σ, τ )kστ βφ(xστ )β −1
βφ(kσ )φ(xσ )φ(kτ )φ(xτ )β −1 = βφ(g(σ, τ ))φ(kστ )φ(xστ )β −1
kσ xσ kτ xτ = g(σ, τ )kστ xστ
kσ σ(kτ )f (σ, τ )xστ = g(σ, τ )kστ xστ

50
for every σ, τ ∈ G. Since the symbols xσ form a basis, we have g(σ, τ )f (σ, τ )−1 = kσ kστ
−1
σ(kτ ) = δ1 (h)(σ, τ ),
∗
where h : G → K , σ 7→ kσ . Therefore, f and g are cohomologous.

Problem 10.14. Let G be a group of order n. Show that nH 2 (G, M ) = 0.

Since f is a 2-cocycle,
X X X
g(σ, τ ) = f (σ, ρ) + σf (τ, ρ) − f (στ, ρ)
ρ∈G ρ∈G ρ∈G
X X X
= f (σ, τ ρ) + σf (τ, ρ) − f (στ, ρ)
ρ∈G ρ∈G ρ∈G
X
= [σf (τ, ρ) − f (στ, ρ) + f (σ, τ ρ)]
ρ∈G
X
= f (σ, τ )
ρ∈G

= nf (σ, τ ).

So nf = g, and thus nH 2 (G, M ) = 0 since g is a coboundary.

Ln−1
Problem 10.15. A = (K/F, σ, a) be a cyclic algebra. If A = i=0 Kxσi , show that xnσ = a.
By definition, f (1, 1)−1 = 1, so 1 = x1 = xσn . If n = 2, then x2σ = xσ xσ = f (σ, σ)xσ2 = ax1 = a.
Now assume n ≥ 3. We will now show that xiσ = f (σ, σ)f (σ 2 , σ) · · · f (σ i−1 , σ)xσi . For the base case,
xσ = f (σ, σ)xσ , since f (σ, σ) = 1. By induction,

xiσ = xσi−1 xσ = f (σ, σ)f (σ 2 , σ) · · · f (σ i−2 , σ)xσi−1 xσ = f (σ, σ)f (σ 2 , σ) · · · f (σ i−2 , σ)f (σ i−1 , σ)xσi .

Hence,
xnσ = f (σ, σ)f (σ 2 , σ) · · · f (σ n−1 , σ)xσn = f (σ n−1 , σ)xσn = a,
since f (σ, σ i ) = 1 for i < n − 1.

Problem 10.16. Cohomology of a cyclic group. In this problem, we determine H 2 (G, M ) for a cyclic
group G. Suppose that G = hσi is a cyclic group of order n. If M is a G-module, let M G = {m ∈
Pn−1
M : σm = m}. Also, define the norm map N : M → M G by N (m) = i=0 σ i m. We will prove that
H 2 (G, M ) ∼
= M G / im(N ) in the following steps.

(a) If m ∈ M G , let fm be the cochain given by fm (σ i , σ j ) = 0 if i + j < n, and fm (σ i , σ j ) = m if i + j ≥ n.

Prove that fm is a cocycle.
Let i, j, k ∈ {0, 1, . . . , n − 1}. Our goal is to show that

σ i fm (σ j , σ k ) − fm (σ i+j , σ k ) + fm (σ i , σ j+k ) − fm (σ i , σ j ) = 0.

Since im(fm ) ⊆ M G , the above equation reduces to

fm (σ j , σ k ) + fm (σ i , σ j+k ) = fm (σ i , σ j ) + fm (σ i+j , σ k )

For an arbitrary integer s, let (s)n denote the unique integer in {0, 1, . . . , n − 1} congruent to s modulo
n.
Case 1 (j + k < n and i + (j + k)n < n). Then i + j + k < n, which implies that i + j, (i + j)n + k < n.
Thus, fm satisfies the cocycle condition.
Case 2 (j + k ≥ n and i + (j + k)n < n). We will show that either i + j ≥ n and (i + j)n + k < n
or i + j < n and (i + j)n + k ≥ n. By Case 1 and by symmetry, we cannot have i + j < n and
(i + j)n + k < n. Suppose i + j ≥ n and (i + j)n + k ≥ n. Then (i + j)n = i + j − n, so i + j + k ≥ 2n.
But we know that i + (j + k)n < n, which implies i + j + k < n, a contradiction. Thus, fm satisfies
the cocycle condition.

51
 Case 3 (j + k < n and i + (j + k)n ≥ n). We will show that either i + j ≥ n and (i + j)n + k < n
or i + j < n and (i + j)n + k ≥ n. By Case 1 and by symmetry, we cannot have i + j < n and
(i + j)n + k < n. Suppose i + j ≥ n and (i + j)n + k ≥ n. Then (i + j)n = i + j − n, so i + j + k ≥ 2n.
But we know that j + k < n, so we must have i > n, a contradiction. Thus, fm satisfies the cocycle
condition.
Case 4 (j + k ≥ n and i + (j + k)n ≥ n). We will show that i + j ≥ n and (i + j)n + k ≥ n. We know
that i + j + k ≥ 2n. Therefore, if i + j < n then k > n, a contradiction. If (i + j)n + k < n, then
i + j + k < 2n, since i + j ≥ n + (i + j)n . We have another contradiction, so we conclude that fm is a
cocycle.
(b) Suppose that fm1 and fm2 are cocycles that are cohomologous. Then there are ci ∈ M with fm1 (σ i , σ j ) =
fm2 (σ i , σ j ) + ci + σ i (cj ) − ci+j , where we are writing ci for cσi . Show that m1 − m2 = N (c1 ).
We have that σ i (c1 ) = fm1 (σ i , σ) − fm2 (σ i , σ) − ci + ci+1 . Therefore,
n−1
X n−1
X
N (c1 ) = σ i (c1 ) = [fm1 (σ i , σ) − fm2 (σ i , σ) − ci + ci+1 ] = m1 − m2 .
i=0 i=0

Pn−1
(c) Prove that a cocycle f ∈ Z 2 (G, M ) is cohomologous to fm , where m = i=0 f (σ i , σ). Make use of the
cocycle condition
f (σ i , σ k ) + f (σ i+k , σ) = σ i (f (σ k , σ)) + f (σ i , σ k+1 ).

Let’s first check that m ∈ M G . Indeed,

n−1
X n−1
X n−1
X
σm = σ(f (σ i , σ)) = [f (σ, σ i ) + f (σ 1+i , σ) − f (σ, σ i+1 )] = f (σ 1+i , σ) = m.
i=0 i=0 i=0

Note that

f (σ i , σ j ) = f (σ i , σ j−1 ) − σ i f (σ j−1 , σ) + f (σ i+j−1 , σ)

= f (σ i , σ j−2 ) − σ i f (σ j−2 , σ) + f (σ i+j−2 , σ) − σ i f (σ j−1 , σ) + f (σ i+j−1 , σ)
j−1
X i+j−1
X
i i l
= f (σ , 1) − σ f (σ , σ) + f (σ l , σ)
l=0 l=i
j−1
X i+j−1
X i−1
X
i i l l
= f (σ , 1) − σ f (σ , σ) + f (σ , σ) − f (σ l , σ).
l=0 l=0 l=0

By Problem 10.12, we will assume that f is normalized so that

j−1
X i+j−1
X i−1
X
f (σ i , σ j ) = − σ i f (σ l , σ) + f (σ l , σ) − f (σ l , σ).
l=0 l=0 l=0
P(i)n −1
Define h(σ i ) = l=0 f (σ l , σ). Suppose that i + j < n. Then fm (σ i , σ j ) − f (σ i , σ j ) = −f (σ i , σ j ) =
σ i h(σ j ) − h(σ i+j ) + h(σ i ). If i + j ≥ n, then

(i+j)n −1 i+j−n−1
X X
i+j
h(σ ) = f (σ l , σ) = f (σ l , σ)
l=0 l=0
i+j−1
X i+j−1
X i+j−1
X
l l
= f (σ , σ) − f (σ , σ) = f (σ l , σ) − m.
l=0 l=i+j−n l=0

Therefore, if i + j ≥ n, then fm (σ i , σ j ) − f (σ i , σ j ) = m − f (σ i , σ j ) = σ i h(σ j ) − h(σ i+j ) + h(σ i ).

Therefore, f is cohomologous to fm .

52
 (d) Conclude from these steps that the map m 7→ fm induces an isomorphism M G /im(N ) ∼
= H 2 (G, M ).
Clearly, fm1 +m2 = fm1 + fm2 , so the map φ : M G → H 2 (G, M ), m 7→ fm is a homomorphism
(the codomain is Z 2 (G, M ) by part (a)). By part (c), every cocycle is cohomologous to some fm ,
so φ is a surjection. Now suppose that m 7→ [fm ] = [f0 ]. Then m − 0 = m ∈ im(N ) by part (b).
Pn−1
Conversely, suppose that m = i=0 σ i p ∈ im(N ). An easy calculation shows that fm = δ1 (h), where
h(σ i ) = l=0 σ l p. Hence, ker(φ) = im(N ). We conclude that M G /im(N ) ∼
Pi−1
= H 2 (G, M ).

Problem 10.17. In this problem, we calculate H 1 (G, M ) for a cyclic group G. Let N be the norm map
defined in the previous problem, and let D : M → M be defined by D(m) = σm − m. We show that
H 1 (G, M ) ∼
= ker(N )/im(D).

(a) Let m ∈ M satisfy N (m) = 0. Define a 1-cochain f by f (σ i ) = m + σm + · · · + σ i−1 m. Show that f

is a 1-cocycle. For the rest of this problem, fm will denote this cocycle.
Suppose that i + j < n. Then

δ1 (f )(σ i , σ j ) = σ i f (σ j ) − f (σ i+j ) + f (σ i )
= σ i (m + · · · + σ j−1 m) − m − · · · − σ i+j−1 m + m + · · · + σ i−1 m
= σ i m + · · · + σ i+j−1 m − m − · · · − σ i+j−1 m + m + · · · + σ i−1 m
= 0.

Suppose that i + j ≥ n. Then

δ1 (f )(σ i , σ j ) = σ i f (σ j ) − f (σ i+j ) + f (σ i )
= σ i (m + · · · + σ j−1 m) − m − · · · − σ i+j−n−1 m + m + · · · + σ i−1 m
= σ i m + · · · + σ i+j−1 m − m − · · · − σ i+j−n−1 m + m + · · · + σ i−1 m
= m + σm + · · · + σ n−1 m
= N (m)
= 0.

Therefore f = fm is a 1-cocycle.

(b) If fm1 and fm2 are cohomologous, show that m1 − m2 = σp − p for some p ∈ M .
For some p ∈ M , we have fm1 (σ i ) − fm2 (σ i ) = σ i p − p. Setting i = 1, we have m1 − m2 = σp − p.
(c) Let f be a 1-cocycle. If m = f (σ), show that f is cohomologous to fm .
Since f is a cocycle, we have σ i f (σ) − f (σ i+1 ) + f (σ i ) = 0. Then

f (σ i ) = σ i−1 f (σ) + f (σ i−1 )

= σ i−1 f (σ) + σ i−2 f (σ) + f (σ i−2 )
..
.
= σ i−1 f (σ) + · · · + f (σ) + f (1)
= fm (σ i ).

Therefore, f is in fact equal to fm where m = f (σ).

(d) Conclude that H 1 (G, M ) ∼
= ker(N )/im(D).
The map φ : ker(N ) → H 2 (G, M ), m 7→ [fm ] is well-defined by part (a) and a surjective by part (c).
By part (b), if m 7→ [f0 ], then m − 0 = m ∈ im(D). Conversely, if m = σp − p ∈ im(D), then
fm (σ i ) = m + σm + · · · + σ i−1 m = σ i p − p = δ0 (p)(σ i ). Hence, [fm ] = 0, and ker(φ) = im(D). We
conclude that H 1 (G, M ) ∼= ker(N )/im(D).

53
Section 11. Kummer Extensions
√ √
Problem 11.1. Let p1 , . . . , pn be distinct primes. Show that [Q( p1 , . . . , pn ) : Q] = 2n .
See Claim 1 of Problem 5.22(b).
√
Problem 11.2. Let F = Q({ n : 1 ≤ n ≤ 28}). Determine [F : Q].
See Problem 11.3. There are 9 primes less than or equal to 28, namely 2, 3, 5, 7, 11, 13, 17, 19, and 23.
Thus [F : Q] = 512.
√
Problem 11.3. Let N be a positive integer, and let FN = Q({ n : 1 ≤ n ≤ N }). Determine [FN : Q].
√
Every integer n satisfying 1 ≤ n ≤ N is a product of primes less than or equal to N . Thus FN = Q({ p :
1 ≤ p ≤ N, p prime}). By Problem 11.1, [FN : Q] = 2π(N ) , where π(N ) is the number of primes less that or
equal to N .

Problem 11.4. Let G be a finite Abelian group whose exponent divides the order of a cyclic group C.
Show that hom(G, C) ∼ = hom(G, C∗ ).
∗
Let φ : G → C be an arbitrary homomoprhism. If n is the order of g, then φ(g)n = 1, implying that
φ(g) is an nth root of unity for every g ∈ G. Since the order of φ(g) divides the order of g, which divides the
exponent of G, which divides c = |C|, we have that im(φ) ⊆ hωi, where ω is a primitive cth root of unity.
Then hωi ∼= C, so hom(G, C) ∼ = hom(G, C∗ ).

Problem 11.5. If G is a finite Abelian group, show that hom(G, C∗ ) ∼ = G.

We begin by showing that hom(G1 × G2 , C∗ ) ∼ = hom(G1 , C∗ ) × hom(G2 , C∗ ). More specifically, we will
prove that Ψ : hom(G1 × G2 , C∗ ) → hom(G1 , C∗ ) × hom(G2 , C∗ ) : φ 7→ (φ1 , φ2 ), where φ1 : G1 → C∗ , g 7→
φ(g, e) and φ2 : G2 → C∗ , g 7→ φ(e, g), is an isomorphism.
Let φ, ψ ∈ hom(G1 × G2 , C∗ ). Then Ψ(φψ) = ((φψ)1 , (φψ)2 ) = (φ1 ψ1 , φ2 ψ2 ) = (φ1 , φ2 )(ψ1 , ψ2 ) =
Ψ(φ)Ψ(ψ), since (φψ)1 (g) = ((φψ)(g), e) = (φ(g)ψ(g), e) = (φ(g), e)(ψ(g), e) = φ1 (g)ψ1 (g), and similarly
(φψ)2 (g) = φ2 (g)ψ2 (g). Hence, Ψ is a homomorphism.
Now suppose φ ∈ ker Ψ. Then φ(g, e) = 1 = φ(e, h) for every g, h ∈ G. Then 1 = φ(g, e)φ(e, h) =
φ(ge, eh) = φ(g, h), which implies φ = id. Thus, Ψ is injective.
Let (φ, ψ) ∈ hom(G1 , C∗ ) × hom(G2 , C∗ ). Note that τ : G1 × G2 → C∗, (g1 , g2 ) 7→ φ(g1 )ψ(g2 ) is in
hom(G1 × G2 , C∗ ) (this uses the fact that C∗ is Abelian). Moreover, Ψ(τ ) = (φ, ψ). Hence Ψ is surjective,
and therefore is an isomorphism.
Using the work done above and the structure theorem for finite Abelian groups, it is now enough to show
that hom(G, C∗ ) ∼= G where G is a finite cyclic group. Indeed, let G ∼ = hσ : σ n i. Let φ ∈ hom(G, C∗ ). As
we showed in the previous problem, φ(σ) must be an nth root of unity. Let ω be any nth root of unity
(not necessarily primitive). We claim that φ(σ i ) = ω i gives rise to a well-defined group homomorphism.
Suppose that σ i = σ j . Then i − j is a multiple of n, and φ(σ i ) = ω i = ω j−i + i = ω j = φ(σ j ). Thus, φ is
a well-defined map. Next, φ(σ i σ k ) = φ(σ i+k ) = ω i+k = ω i ω k = φ(σ i )φ(σ k ), so φ is a homomorphism. We
conclude that the elements of hom(G, C∗ ) are in one-to-one correspondence with the nth roots of unity in
C. Thus, for each nth root of unity, we have a distinct homomorphism φω : G → C∗ . It is then easy to see
that hom(G, C∗ ) is isomorphic to the group of nth roots of unity in C, where the isomorphism is given by
ω 7→ ψω . Since the group of nth roots of unity is cyclic of order n, we have hom(G, C∗ ) ∼ = G as desired.

Problem 11.6. Let F be √ a field containing a primitive nth root of unity, and let G be a subgroup of
F ∗ /F ∗n . Let F (G) = F ({ n a : aF ∗n ∈ G}). Show that F (G) is an n-Kummer extension of F and that G is
the image of kum(F (G)/F ) under the map f defined in Proposition 11.10. Conclude that Gal(F (G)/F ) ∼ =G
and [F (G) : F ] = |G|.
The problem should specify that G is a finite subgroup of F ∗ /F ∗n . For example, if F = Q, n = 2, and
G = Q∗ /Q∗2 . Then [F (G) : F√ ] = ∞ by√ Problem 11.1. In this case F (G)/F is not a Kummer extension.
√ √
Now, if aF ∗n = bF ∗n , then n a = c n b for some c ∈ F . Hence, F (G) = F ( n a1 ,√ . . . , n ak ) for some
∗
ai √∈ F . By Theorem 11.4, F (G)/F is an n-Kummer extension. Let aF ∈ G. Then n a ∈ F (G). Hence,
f ( n aF ∗ ) = aF ∗n , so G ⊆ im(f ). √ √
Now, consider the pairing B : Gal(F (G)/F ) × G → µ(F ), (σ, aF ∗n ) 7→ σ( n a)/ n a. It is not hard to see

54
that B is well defined. The argument that B is bilinear is the same as the one given in Proposition 11.9.
We will now show that√ B is√non-degenerate. Suppose B(σ, aF ∗n ) = 1 for every
√ σ ∈ Gal(F (G)/F ) and some
aF ∈ G. Then σ( a) = n a for every σ ∈ Gal(F (G)/F ), implying that n a ∈ F ∗ ⇔ a ∈ F ∗n ⇔ aF ∗n = 1.
∗n n

On the
√ other hand, if B(σaF ∗n ) = 1 for every aF ∗n ∈ G and some σ ∈ Gal(F (G)/F ), then σ is the identity
on { a : aF ∗n ∈ G}, and therefore the identity on all of F (G). Hence, B is non-degenerate, and thus
n

induces an isomorphism G ∼ = Gal(F (G)/F ) by Lemma 11.8. Hence, [F (G) : F ] = |G|.

We can now retroactively prove that G = f (kum(F (G)/F )). By Proposition 11.10 and the paragraph
above, f is injective so |f (kum(F (G)/F ))| = |kum(F (G)/F )| = [F (G) : F ] = |G|. Since we have shown that
G ⊆ f (kum(F (G)/F )), we have that G = f (kum(F (G)/F )).

55
III. Applications of Galois Theory
Section 12. Discriminants
Problem 12.1. Let B : V × V → F be a bilinear form. If V = {v1 , . . . , vn } is a basis for V , another
basis W = {w1 , . . . , wn } is called a dual basis to V provided that B(vi , wi ) = 1 for all i, and B(vi , wj ) = 0
whenever i 6= j. If V and W are dual bases, show that disc(B)V · disc(B)W = 1.
LetPA = (aij ) be the change of P basis matrix from V to W, and let A−1 =P (aij ) be its inverse. Then
w
P i = k a ki v k and B(w i , wj ) = k aki B(v k , wj ) = aji . Moreover, v j = k akj wk and B(vi , vj ) =
a
k kj B(v i , w k ) = aij . By definition,

disc(B)V · disc(B)W = det(B(vi , vj )) det(B(wi , wj ))

= det(B(vi , vj )) det B(wi , wj )T

= det B(vi , vj ) · B(wi , wj )T

!
X
= det B(vi , vk )B(wj , wk )
k
!
X
= det aik akj
k
= det(In )
= 1.

Problem 12.2. If B is a nondegenerate bilinear form on V , show that any basis has a dual basis.
P
P V = {v1 , . . . , vn } be an arbitrary basis for V , and let A = (B(vi , vj )). Let v =
Let i αi vi and
w = i βi vi be arbitrary vectors. Note that
!
X X X
B(v, w) = B αi vi , βi vi = αi βj B(vi , vj ) = v T Aw.
i i i,j

By the discussion following Definition 12.17, the fact that B is nondegenerate implies that A is invertible.
For each i ∈ {1, . . . , n}, let wi = A−1 vi . Since A is invertible, W = {w1 , . . . , wn } is linearly independent,
and therefore a basis. Moreover,

B(vi , wj ) = viT Awj = viT AA−1 vj = viT vj = δij ,

so W is dual to V.

Problem 12.3. Let {ei } be a basis for F n , and choose an ai ∈ F for each i. Define B on this basis
by B(ei , ej ) = 0 if i 6= j and B(ei , ei ) = ai ∈ F . Prove that this function extends uniquely to a bilinear form
B : F n × F n → F , and determine P the discriminant P of B.
Let v, w ∈ F n , and let v = i αi ei and w = i βi ei be their basis representations. Define
X
B(v, w) = αi βi ai .
i
P
Clearly our definition of B extends the definition given in the problem statement. Let u = i γi ei . Then
X X X
B(u + v, w) = (γi + αi )βi ai = γi βi ai + αi βi ai = B(u, w) + B(v, w).
i i i

Since B(v, w) = B(w, v) for all v, w ∈ F n , we have that B is bilinear. We now show that this extension is
unique. Let C : F n × F n → F be a bilinear form that agrees with B on elements of the form (ei , ej ). Then
!
X X X X
C(v, w) = C αi ei , βi e i = αi βj C(ei , ej ) = αi βi ai = B(v, w).
i i i,j i

56
Thus, B extends uniquely to F n × F n . Note that (B(ei , ej )) is a diagonal matrix, so disc(B) =
Q
i ai .

Problem 12.4. Let A be a symmetric n×n matrix, and define a map B : F n ×F n → F by B(v, w) = vAwT ,
where v and w are viewed as row vectors. Show that B is bilinear. Using the fact that a symmetric matrix
can be diagonalized by an orthogonal transformation, use the previous problem to determine the discriminant
of B in terms of A.
Clearly B is bilinear. Let P AP T be diagonal, where P is an orthogonal matrix. Let {e1 , . . . , en } be the
standard basis for F n . We will compute the discriminant of B with respect to the basis V = {e1 P, . . . , en P }.
Then B(ei P, ej P ) = ei P AP T eTj = 0 for i 6= j and B(ei P, ei P ) = ei P AP T eTi = λi , where λi is the eigenvalue
Q
of A at the ith position on the diagonal. By the previous problem, the discriminant is disc(B)V = i λi ,
i.e. the discriminant is the product of the eigenvalues of A, counting multiplicities.

Problem 12.5. Let K be a finite separable extension of F , and let A be an integrally closed ring with
quotient field FP
. Let B be the integral closure of A in K. Show that there is an F -basis v1 , v2 , . . . , vn of K
such that B ⊆ Avi .
We begin by finding a basis of elements in B. Let {u1 , . . . , un } be a basis for K/F . Since K/F is
algebraic, each ui satisfies a minimal polynomial fi (x) = xm + αm−1 xm−1 + · · · + α1 x + α0 where αj ∈ F .
Multiplying by the denominators of the coefficients, we have that
m−1
am um
i + am−1 ui + · · · + a1 ui + a0 = 0
for aj ∈ A. Multiplying the above equation by am−1
m , we have
(am ui )m + am−1 (am ui )m−1 + am−2 am (am ui )m−2 + · · · + a1 am−2
m
m−1
(am ui ) + a0 am = 0.
Therefore, am ui is integral over A, and therefore am ui ∈ B. For each i, let ki ∈ F ∗ be such that
wi = ki ui ∈ F . Then W = {w1 , . . . , wn } is a basis for K/F that is contained in B. By Corollary 8.17, the
trace form B(v, w) = TrK/F (vw) is nondegenerate. By Problem 12.2, there is a dual basis V = {v1 , . . . , vn }.
Let b ∈ B and let b = c1 v1 + · · · + cn vn . We will show that each ci ∈ A. We have that TrK/FP (bwi ) = ci , so
it is enough to show that TrK/F (β) ∈ A for every β ∈ B. By Theorem 8.12, TrK/F (β) = i σi (β), where
σ1 , . . . , σr are the distinct F -homomorphisms from K to an algebraic closure of F . Since β is integral over
A, so is every σi (β). It is a standard fact that the sum of integral elements is also integral, so TrK/F (β) is
integral over P A. Since TrK/F (β) ∈ F and A is integrally closed, we have that TrK/F (β) ∈ A. We conclude
that B ⊆ Avi .

Problem 12.6. Let K be an algebraic number field, and let B be the integral closure of Z in K. Use
the previous problem to show that B is a finitely generated Z-module, and conclude that B is a Noetherian
ring. Moreover, show that there is a basis of K that is also a basis for B as a Z-module. Such a basis is
called an integral basis for B/Z.
P Z is integrally closed, the previous problem implies that there is an Q-basis {v1 , . . . , vn } such that
Since
B ⊆ Zvi . Therefore B is a submodule of a finitely Z-module. Since Z is a PID, B is finitely generated as
a Z-module. Since Z is Noetherian, B is a Noetherian module. Since the ideals of B viewed as a ring are
Z-submodules of B viewed as a module, B is also a Noetherian ring.
Let {w1 , . . . , wn } ⊆ B be a basis for K/Q (we constructed such a basis in the previous problem). Define
a homomorphism φ : K → Qn by φ(α) = (TrK/Q (αw1 ), . . . , TrK/Q (αwn )). Suppose that φ(α) = (0, . . . , 0).
Then α = 0, because otherwise {αw1 , . . . , αwn } is basis, which would imply that TrK/Q is the zero-map.
However, this is impossible since K/Q is separable. Thus, φ is injective. Moreover, if α ∈ B, then φ ∈ Zn .
Hence, φ restricts to an injective homomorphism φ : B → Zn . Therefore, B is a free Z module of rank at
most n. Since {w1 , . . . , wn } ⊆ B is linearly independent, B is free of rank n. Therefore, there {w1 , . . . , wn }
is a basis for B/Z and for K/Q.

Problem 12.7. With the notation of the previous problem, let d be the discriminant of K/F relative
to an integral basis {v1 , . . . , vn } of B/Z. Prove that d ∈ Z. The integer d is called the discriminant of B/Z.
Show that if we use a different integral basis, then the two discriminants are equal.
Since vi vj ∈ B for every pair i, j, we have
TrK/Q (vi vj ) ∈ Z, as discussed in Problem 12.5. There-
fore, d = disc(v1 , . . . , vn ) = det TrQ/K (vi vj ) ∈ Z. Let {w1 , . . . , wn } be another integral basis. Since

57
every vi is a Z-linear combination of the wj , if A is the transition matrix between the two bases, we have
2
det(A) ∈ Z. Reversing the roles of the bases, A−1 ∈ Z as well, so det(A) = 1. By Proposition 12.14,
disc(v1 , . . . , vn ) = disc(w1 , . . . , wn ).

Problem 12.8. Calculate the discriminant of B/Z for the following fields, where B is the integral clo-
sure of Z in that field.
√
(a) Q( −1).
√
Let i = −1. We will show that B = Z[i], the ring of Gaussian integers. First, let a + ib ∈ B, where
a, b ∈ Q. Using {1, i} as a basis for Q(i)/Q, the element a + ib has the matrix representation
 
a −b
.
b a

Then NQ(i)/Q (a + ib) = a2 + b2 and TrQ(i)/Q (a + ib) = 2a. Since the trace and norm of integral elements
are in Z, we let a2 + b2 = m and 2a = n, where m, n ∈ Z. Letting b = c/d, with c, d ∈ Z, we have
(after some rearranging) 4(md2 − c2 ) = n2 , implying that n2 is even. Therefore, n is even, and thus
a ∈ Z. Hence, b2 ∈ Z, and therefore b ∈ Z. We conclude that B ⊆ Z[i]. Now let a + ib ∈ Z[i]. If
f (x) = x2 − 2ax + a2 + b2 , then f (a + ib) = 0. Therefore, a + ib ∈ B, and B = Z[i].
This proves that {1, i} is an integral basis for B/Z. Then
   
TrQ(i)/Q (1) TrQ(i)/Q (i) 2 0
d = det = det = −4.
TrQ(i)/Q (i) TrQ(i)/Q (−1) 0 −2
√
(b) Q( d), where d > 0 is a square-free integer.
√ √ √
Let a + b d ∈ Q( d). A simple computation shows that NQ(√d)/Q (a + b d) = a2 − b2 d and
√
TrQ(√d)/Q (a + b d) = 2a. We will show that a, b ∈ 21 Z. Let a2 − b2 d = m and 2a = n, where
m, n ∈ Z. This gives a ∈ 21 Z. If n is even, then a is an integer, so clearly b must be an integer as
well. Now suppose that n is odd. Then, letting b = s/t (where s, t ∈ Z and gcd(s, t) = 1), we have
t2 n2 = 4s2 d + 4mt2 . Then t must be even; let t = 2k, which gives sd = k 2 (n2 − 4m). Since k and
s have no common factors, k 2 must divide d. However, we assumed that d is square-free, so k = 1,
implying that b ∈ 21 Z. We can then let a = x/2 and b = y/2; note that

x2 − y 2 d
m= ∈ Z.
4

Case 1. (d ≡4 1) We have the equation 4m = x2 − y 2 d, which, taken modulo 4, gives x2 ≡4 y 2 . Since

x2 ≡4 0 if x is even and√ x2 ≡4 1 if x is odd, we√have that√x and y must have the same parity. We
conclude that B ⊆ Z[ 1+2 d ]. Conversely, α + β 1+2 d ∈ Z[ 1+2 d ] satisfies the polynomial
 
d−1
f (x) = x2 − (2α + β)x + α2 + αβ − β 2 .
4
√ √
Note that f has integer coefficients since d ≡4 1. We conclude that B = Z[ 1+2 d ], and that {1, 1+2 d }
is an integral basis. Therefore, the discriminant is
  √  
TrQ(√d)/Q (1) TrQ(√d)/Q 1+2 d
 
det   √   √  = det 2 1+d 1
= d.
TrQ(√d)/Q 1+2 d TrQ(√d)/Q 1+2 4 d+d 1 2

Case 2. (d 6≡4 1) Then d ≡4 2, 3, since d is square-free. Suppose that x is odd; taking 4m = x2 − y 2 d

modulo 4 yields y 2 d ≡4 1. Then y must be odd as well, so d ≡4 1, a contradiction. We√ conclude that
x is even. Then y 2 d ≡4 0, which implies that y must be even as well. Hence, B ⊆ Z[ d]. Conversely,

58
 √ √ √ √
α + β d ∈ Z[ d] satisfies f (x) = x2 − 2α + α2 − β 2 α, so B = Z[ d] and {1, d} is an integral basis.
Therefore, the discriminant is
√ !
TrQ(√d)/Q (1) TrQ(√d)/Q ( d)
 
√ 2 0
det = det = 4d.
TrQ(√d)/Q ( d) TrQ(√d)/Q (d) 0 2d

(c) Q(ω), where ω is a primitive pth root of unity and p is prime.

We will assume that p is an odd prime. The case where p = 2 is uninteresting since Q(ω) = Q in this
case, so d = 1.
Claim 1. (1 − ω)B ∩ Z = pZ.
Let pn ∈ pZ. Note that pn = (1 − ω)(1 − ω 2 ) · · · (1 − ω p−1 )n ∈ (1 − ω)B ∩ Z. Now suppose that
pZ ⊂ (1 − ω)B ∩ Z is a strict inclusion. Then, since pZ is a maximal ideal of Z, we have that
(1 − ω)B ∩ Z = Z. Hence, there is a b ∈ B such that (1 − ω)b = 1. As shown in Example 12.10,
NQ(ω)/Q (1 − ω) = p. Taking the norm on both sides of (1 − ω)b = 1, we obtain pNQ(ω)/Q (b) = p − 1,
which is a contradiction since NQ(ω)/Q (b) ∈ Z. Thus, (1 − ω)B ∩ Z = pZ, which proves Claim 1.
Claim 2. TrQ(ω)/Q (b(1 − ω)) ∈ pZ for every b ∈ B.
Pp−1
We have that TrQ(ω)Q (b(1 − ω)) = i=1 σi (b)(1 − ω i ), where the σi are the elements of Gal(Q(ω)/Q).
Since 1 − ω i factors as (1 + ω + · · · + ω i−1 )(1 − ω), we have that TrQ(ω)Q (b(1 − ω)) ∈ (1 − ω)B. Moreover,
b(1 − ω) ∈ B, so TrQ(ω)Q (b(1 − ω)) ∈ Z. By Claim 1, TrQ(ω)Q (b(1 − ω)) ∈ pZ. This proves Claim 2.
Claim 3. B = Z[ω].
It is clear that Z[ω] ⊆ B. Let α = k0 +k1 ω+· · ·+kp−2 ω p−2 ∈ B be arbitrary. Note that TrQ(ω)/Q (ω j ) =
Pp−1 i
i=1 ω = −1 for every j ∈ {1, . . . , p − 1}. Thus,

p−2
!
X
i i+1
TrQ(ω)/Q (α(1 − ω)) = TrQ(ω)/Q ki (ω − ω ) = TrQ(ω)/Q (k0 (1 − ω)) ∈ pZ,
i=0

by Claim 2. But TrQ(ω)/Q (k0 (1 − ω)) = k0 (TrQ(ω)/Q (1) − TrQ(ω)/Q (ω) = k0 (p − 1 + 1) = k0 p ∈ pZ.
Therefore, k0 ∈ Z. Note that ω p−1 (α − k0 ) = k1 + k2 ω + · · · + kp−2 ω p−3 ∈ B. Repeating the above
argument shows that k1 ∈ Z. Continuing in this manner, we find that k0 , k1 , . . . , kp−2 ∈ Z, proving
Claim 3.
Note that {1, ω, . . . , ω p−2 } is an integral basis for B/Z. Then the discriminant is
 
p − 1 −1 −1 · · · −1 −1 −1 −1
 −1 −1 −1 · · · −1 −1 −1 −1 
 
 −1 −1 −1 · · · −1 −1 −1 p − 1
 
 −1 −1 −1 · · · −1 −1 p − 1 −1 
d = det  −1 −1 −1 · · · −1 p − 1 −1 .
 
 −1  
 −1 −1 −1 · · · p − 1 −1 −1 −1 
 
 .. .. .. .. .. .. .. .. 
 . . . . . . . . 
−1 −1 p − 1 · · · −1 −1 −1 −1

Subtracting the second row from all other rows (we could do the same with columns), we obtain
 
p 0 0 ··· 0 0 0 0
−1 −1 −1 · · · −1 −1 −1 −1
 
0
 0 0 ··· 0 0 0 p 
0 0 0 ··· 0 0 p 0
d = det  0 .
 
 0 0 ··· 0 p 0 0 
0
 0 0 · · · p 0 0 0 

 .. .. .. .. .. .. .. .. 
 . . . . . . . . 
0 0 p ··· 0 0 0 0

59
 We take the determinant along the second row; the only non-zero term will come from the row 2 -
column 2 entry. We have
 
p 0 ··· 0 0 0 0
0 0
 ··· 0 0 0 p 
0 0
 ··· 0 0 p 0 
d = − det 0 0
 ··· 0 p 0 0 p−3 p−1
 = −(−1) 2 pp−2 = (−1) 2 pp−2 .
0 0
 ··· p 0 0 0 
 .. .. .. .. .. .. .. 
. . . . . . .
0 p ··· 0 0 0 0

Section 13. Polynomials of Degree 3 and 4

Problem 13.1. Ferrari’s solution of the quartic. Here is Ferrari’s method for finding the roots of a quartic.
Let g(x) = x4 + ax3 + bx2 + cx + d. Starting with g(x) = 0, move the quadratic part of f to the right-hand
side. Show by completion of squares that the equation becomes
 2  
2 1 1 2
x + ax = a − b x2 − cx − d.
2 4

Ferrari’s idea is to add to both sides the expression y(x2 + ax/2) + y 2 /4 for some y, so that the left-hand
side is a perfect square. The equation then becomes
 2    
1 1 1 2 1 1
x2 + ax + y = a − b + y x2 + ay − c x + y 2 − d.
2 2 4 2 4

We wish to choose y so that the right-hand side becomes a square, (ex + f )2 . Writing the right-hand side
as Ax2 + Bx + C, this is possible if and only if B 2 − 4AC = 0. Show that this gives an equation in y to be
solved, and if r is the resolvent of g, then this equation is r(x) = 0. Given such a y, take the equation
 2
1 1
x2 + ax + y = (ex + f )2
2 2
and obtain two quadratic equations in x and solve them to find the general solution to g(x) = 0. Relate
Ferrari’s method to the method of the section by showing that e = 12 u and that the discriminants of the two
quadratic equations are equal to v 2 and (v 0 )2 .
One can check that  2  
1 1 2
x2 + ax = a − b x2 − cx − d
2 4
simply by foiling both sides. The equation B 2 − 4AC is then
 2   
1 1 2 1 2
ay − c − 4 a −b+y y − d = 0.
2 4 4
After foiling and multiplying by −1, we have

0 = y 3 − by 2 + (ac − 4d)y + 4bd − a2 d − c2 = r(y).

Let y be a solution of the preceding equation. Then we have

 2
1 1
x2 + ax + y = (ex + f )2 ,
2 2
which yields the two quadratic equations
   
1 1 1 1
x2 + a−e x+ y−f = 0 and x2 + a + e x + y + f = 0.
2 2 2 2

60
Thus, the roots of g(x) are
   
  s 2   s 2
1 1 1 1 1 1
− a−e ± a − e − (2y − 4f ) , − a+e ± a + e − (2y + 4f ) .
2 2 2 2 2 2

As on page 128 of Field and Galois

q Theory, let theq resolvent be r(x) = (x − β1 )(x − β2 )(x − β3 ),
1 2 1 2 1
and say we take y = β1 . Then e = 4a − b + y = 4 a − b + β1 = 2 u. The discriminants our two
quadratic equations are ( 12 a ± e)2 − 4( 12 y ± f ). Replacing e with 12 u and y with β1 , the discriminants become
1 2 1 1 1
4 (a ± u) − 4( 2 y ± f ). We also have that 2ef = 2 ay − c, implying that f = 2u (aβ1 − 2c). Then the discrimi-
nants are 4 (a±u) −2(β1 ± u (aβ1 −2c)). Thus, the discriminants of the two quadratic equations are v 2 and v 02 .
1 2 1

Problem 13.2. Solve x4 + 4x − 1 = 0 by Ferrari’s method and by the method of the section.
For both methods, we will need to find a root of the resolvent cubic, which is

r(x) = x3 + 4x − 16.

By inspection, we see that x = 2 is a root. However, I didn’t see this initially and I did a bunch of work to
find the roots, which I record here. This also gives us a chance to practice Cardano’s method. Te resolvent
is already in the form x3 + px + q, where p = 4 and q = −16. We set x = u + v, and write r(x) = 0 as

u3 + v 3 − 16 + (3uv + 4)(u + v) = 0,
4
from which we get the equations u3 + v 3 − 16 = 0 and 3uv + 4 = 0. Hence, v = − 3u . Substituting this back
6 3 64
into the first equation, we have u − 16u − 27 = 0. Thus,

√
q
64
16 ± 256 + 4 · 27 72 ± 16 21
3
u = = .
2 9
√ √
Let A = 19 (72 + 16 21) and B = 91 (72 − 16 21), and set u3 = A and v 3 = B. Since we require that
√ √
3
A 3 B = − 43 , the solutions of r(x) = 0 are
s √ s √ s √ s √ s √ s √
3 72 + 16 21 3 72 − 16 21 3 72 + 16 21 3 72 − 16 21 3 72 + 16 21 3 72 − 16 21
2 2
+ , ω +ω , ω +ω ,
9 9 9 9 9 9

where ω is a primitive third root of unity. We then get the neat identity
s √ s √
3 72 + 16 21 3 72 − 16 21
2 = + ,
9 9

so the work wasn’t completely useless.

Ferrari’s method. The equation to be solved is x4 + 4x − 1 = 0. Rearranging and adding yx2 + y 2 /4 to
both sides, we have
 2
2 1 1
x + y = yx2 − 4x + y 2 + 1.
2 4
We take y = 2, a root of the resolvent cubic, and the above equation becomes
√ √
(x2 + 1)2 = ( 2x − 2)2 ,
√ √ √ √
which yields the quadratic equations x2 − 2x + 1 + 2 = 0 and x2 + 2x + 1 − 2 = 0. The roots of
x4 + 4x − 1 are then √ p √ √ p √
2±i 4 2+2 − 2± 4 2−2
x= , .
2 2

61
 Method from the section. Let t1 , t2 , t3 , t4 be the roots of f (x) = (x−t1 )(x−t2 )(x−t3 )(x−t4 ) = x4 +4x−1.
Let u = (t1 + t2 ) − (t3 + t4 ) and v = t1 − t2 . As shown in the section,

u2 = a2 − 4b + 4β1 = 8

(since β1 = 2 is a root of the resolvent cubic),

√
 
1 1
v2 = (u − a)2 − 2 β1 + (2c − aβ1 ) = −2 + 4 2,
4 u
and
√
 
1 1
v 02 = (u + a)2 − 2 β1 − (2c − aβ1 ) = −2 − 4 2.
4 u
The roots are given in the section, and are of the same form as those we found using Ferrari’s method.

Problem 13.3. Show that 2 cos(2π/15) is a root of x4 − x3 − 4x2 + 4x + 1. What are the other roots?
Let ω be any primitive 15th root of unity in C. A computation shows that
 4  3  2  
1 1 1 1 1 1 1
ω+ − ω+ −4 ω+ +4 ω+ + 1 = ω4 + 4 − ω3 − 3 + ω + − 1
ω ω ω ω ω ω ω
1 8
= 4 (ω − ω 7 + ω 5 − ω 4 + ω 3 − ω + 1)
ω
1
= 4 Ψ15 (ω) = 0.
ω
Taking ω = ei2π/15 , ei4π/15 , ei8π/15 , ei14π/15 , we obtain that 2 cos(2π/15), 2 cos(4π/15), 2 cos(8π/15), and
2 cos(14π/15) are roots of x4 − x3 − 4x2 + 4x + 1.

Problem 13.4. Solve the equation ((x + 2)2 + x2 )3 = 8x4 (x + 2)2 by setting y = x + 1.
After making the substitution y = x + 1, we obtain the equation y 5 + 2y 4 − 2y 3 + 2y 2 + y = 0. This
equation has y = 0 as a solution, and thus x = −1 is a solution. To find the other solutions, we must solve
√
y 4 + 2y 3 − 2y 2 + 2y + 1 = 0, which can be solved as described in Problem 13.5. We have that a± = −1 ± 5,
and therefore the solutions are
√ p √ √ p √
−1 + 5 ± 2 − 2 5 −1 − 5 ± 2 + 2 5
and .
2 2
Subtract one from the solutions above to obtain the remaining solutions of the original equation.

Problem 13.5. Find the roots of x4 + px3 + qx2 + px + 1, and notice that cube and fourth roots are
not needed.
Since 0 is not a root of x4 + px3 + qx2 + px + 1, any root of this polynomial will satisfy
x4 + px3 + qx2 + px + 1
   
2 1 1
= x + + p x + + q = 0.
x2 x2 x
1
Note that x2 + x2 = (x + x1 )2 − 2, so we have y 2 + py + q − 2 = 0, where y = x + x1 . Hence,
p
1 −p ± p2 − 4(q − 2)
x+ = = a± .
x 2
Then x2 − a± x + 1 = 0, so the roots are q
a± ± a2± − 4
.
2
p
3
√ p
3
√ p
3
√
Problem 13.6. Use the ideas of this section to show that 5+2− 5 − 2 = 1 and that 7+ 50 +
p
3
√
7 − 50 = 2.

62
 I’m assuming that we are meant to be working in R. In fact, the identities are false in certain fields of
positive characteristic.
For the first identity, we want to find a cubic polynomial x3 + px + q that has 1 as a unique real root,
2 3
and such that − 2q = 2 and q4 + p27 = 5. Then q = −4 and p = 3, and indeed, 1 is a root of x3 + 3x − 4.
The derivative of this polynomial is 3x2 + 3, which is positive for all x, so 1 is the unique real root of the
polynomial (we could have also computed the roots directly).
2 3
For the second identity, we want − 2q = 7 and q4 + p27 = 50. Then q = −14 and p = 3, and indeed 2 is a
root of x3 + 3x − 14. Again, this polynomial has only on real root, so we have proven the second identity.

Problem 13.7. Find the roots of x3 − 6x − 6 and the roots of 2x3 + 6x + 3.

I won’t go through all the steps for each polynomial. For a more detailed example, see Problem 13.2.
We
√ begin
√ with
√ x3 −√6x − 6,√ which√is already depressed. We have A = 4 and B = 2. Then the roots are
3
4 + 2, ω 4 + ω 2 3 2, ω 2 3 4 + ω 3 2.
3 3

For the second cubic, we divide by 2 to apply Cardano’s method to the polynomial x3 + 3x + 32 . Then
q √ q √ q √
A = 12 and B = −2. The roots are then 3 12 − 3 2, ω 3 12 − ω 2 3 2, ω 2 3 12 − ω 3 2.

Problem 13.8. If the specific gravity of cork is 0.25, to what depth will a sphere of radius r made of
cork sink in water? Archimedes’ principle is that the weight of water displaced is equal to the weight of the
cork.
Let ρ be the density of water. Then the density of cork is 14 ρ, so the weight of the sphere is 14 ρ · 43 πr3 =
1
ρ · 3 πr3 . Let x be the depth that the cork sinks into the water. Then the volume of the displaced water is the
volume of the spherical cap of height x of the sphere of radius r, which is 13 πx2 (3r − x). The weight of the
displaced water is then ρ · 13 πh2 (3r − x). By Archimedes’ principle, we then have ρ · 13 πr3 = ρ · 13 πx2 (3r − x),
which reduces to the cubic equation
x3 − 3rx2 + r3 = 0.
Making the substitution x = y + r, we obtain

y 3 − 3r2 y − r3 = 0.

The A = eiπ/3 r3 and B = e−iπ/3 r3 . Then, the solutions for x are (1 + 2 cos(π/9))r, (1 + 2 cos(7π/9))r, and
(1 + 2 cos(13π/9))r. We must have 0 ≤ x ≤ 2r, so the cork sinks to a depth of (1 + 2 cos(13π/9))r ≈ 0.65r.

Problem 13.9. Let f (x) = x4 + ax2 + b ∈ Q[x]. Determine the Galois group of f .
Let G be the Galois group of f . Note that in what follows, Q can always be replaced by an arbitrary
field of characteristic not 2. The resolvent cubic of f is r(x) = x3 − ax2 − 4bx + 4ab = (x − a)(x2 − 4b). Since
r(x) is reducible, we are in case 3, 4, or 5 of Theorem 13.4. If b ∈ Q2 , then r(x) splits over Q, and therefore
we are in case 3, so G ∼ = Z2 × Z2 .
Now suppose that b ∈ / Q2 . If a2 − 4b ∈ Q2 , then f is reducible over Q, and therefore
√ we are in case 4, so
G∼ = Z4 . Suppose that a2 − 4b ∈ / Q2 , we will show that f is irreducible over L = Q( b), and therefore we
are in case 5, so G ∼= D4 . Suppose for a contradiction that f is reducible over L. Since the roots of f are
s √
−a ± a2 − 4b
± ,
2

an even number of roots lie in L. Therefore, if f has a factorization of the form

f (x) = x4 + ax2 + b = (x2 + α0 x + β0 )(x2 + α1 x + β1 )

= x4 + (α0 + α1 )x3 + (α0 α1 + β0 + β1 )x2 + (α0 β1 + α1 β0 )x + β0 β1 .

If α0 = α1 = 0, then x2 + ax + b is reducible, so a2 − 4b ∈ Q2 , contradicting

√ √our initial assumption. Hence
√
α0 and α1 , and both are non-zero. We then obtain that β0 = β1 = b, and 2 b − α02 = a. Let α0 = c + d b,
where c, d ∈ Q. Then √ √
−a + 2 b = (c2 + d2 b) + 2cd b.

63
Then cd = 1, so −a = c2 + cb2 ⇒ 0 = c4 + ac2 + b. But then c is a root of f in Q; by observing the form of
the roots given above, we conclude that a2 − 4b ∈ Q2 , a contradiction.
To recap, if b ∈ Q2 , then G ∼ / Q2 , then there are two cases. If a2 − 4b ∈ Q2 , then G ∼
= Z2 × Z2 . If b ∈ = Z4 .
2
If a − 4b ∈ 2 ∼
/ Q , then G = D4 .

Problem 13.10. Let K be a field extension of F with [K : F ] = 4. Show that K contains an intermediate
4 2
subfield L with [L : F ] = 2 if and √ where c satisfies a polynomial x + ax + b ∈ F√[x].√
√ only if K = F (c),
(⇒) We have √ that L = F ( α) and K√= L( β) for some α ∈ F and β ∈ L. Hence, K = F ( α, β).
√
Suppose that [F ( β) √ : F ]√= 2. Then √ F ( β) √ = F ( γ) for some
√ γ√∈ F . Therefore,
√ we√may assume √ that
β ∈ F . Note that√ α + β ∈
/ F ( α), F ( β). Indeed, if α + β = F ( α),
√ then
√ β ∈ √ α) and
F (
[K √: F] =√ [F ( α) : F ] = 2, a contradiction;
√ √ a similar
√ argument
√ applies if α + β ∈ F ( β). Then
4 2 2
[F ( α + β) : F ] = 4, so √ K = F ( α + β). Note that √ α + β is a root
√ of x − 2(α + β)x + (α −
√ β) = 0.
√ Now suppose that [F ( β) : F ] = 4, so K = F ( β). Since β ∈ F ( α), we have β = k0 + k1 α. Then
β satisfies x4 − 2k0 x2 + k02 − k12 a = 0.
(⇐) Suppose c is a root of x4 + ax2 + b (we must assume that the polynomial is irreducible). Then a2 − 4b
is not a square in F . Since c is of the form
s √
−a ± a2 − 4b
± ,
2
√ √
we know that a2 − 4b ∈ K. Hence, L = F ( a2 − b) is an intermediate subfield with [L : F ] = 2.

Problem 13.11. Given the splitting field k(t1 , t2 , t3 , t4 ) of the general quartic (x − t1 )(x − t2 )(x − t3 )(x − t4 )
over k(s1 , s2 , s3 , s4 ), for each pair L2 /L1 of intermediate subfields for which there is no proper intermediate
subfield, find a single element that generates L2 over L1 , and find this element’s minimal polynomial over
L1 .

64
 Let K = k(t1 , t2 , t3 , t4 ) and let F = k(s1 , s2 , s3 , s4 ).
Above is the subgroup lattice of S4 , courtesy Q of Wikipedia. The unique subgroup of index 2 is A4 . The
corresponding subfield is F (∆), where ∆ = i<j (ti − tj ). The minimal polynomial of ∆ is x2 − ∆2 , since
∆2 ∈ F is the discriminant.
A4 contains one subgroup of order 4, namely Z2 × Z2 ∼ = h(12)(34), (13)(24)i. The corresponding subfield
is L = F (β1 , β2 , β3 ), where the βi are the roots of the resolvent cubic. Since (for each i) [F (βi ) : F ] = 3 and
[F (∆) : F ] = 2, we know that βi ∈ L \ F (∆). Since L/F (∆) has no intermediate subfields, we have that
L = F (∆)(βi ), for any i. The minimal polynomial of βi over F (∆) is then the resolvent cubic.
h(12)(34), (13)(24)i contains three subgroups of order 2, each generated by an element of the form (ab)(cd).
Note that u = (ta + tb ) − (tc + td ) ∈ / L since it is not fixed by h(12)(34), (13)(24)i, however u is contained
in the fixed field F(h(ab)(cd)i). Hence, F(h(ab)(cd)i) = L(u), and the minimal polynomial of u over L is
x2 − u2 (it is shown on page 128 of Morandi’s book that u2 ∈ L).
K is generated by v = ta − tb over L(u). To see this, first note the form of the roots of the general
quartic given on page 129 of Morandi’s book. From this we have that ta , tb ∈ L(u, v). This implies that
tc + td ∈ L(u, v); then ta (tc + td ) − (ta tc + tb td ) = td (ta − tb ) ∈ L(u, v), so td ∈ L(u, v) and tc ∈ L(u, v).
Hence, K = L(u, v). The minimal polynomial of v over L(u) is x2 − v 2 (page 129 shows that v 2 ∈ L(u)).
S4 contains four copies of S3 . Consider the copy of S3 that permutes tb , tc , td and fixes ta . Then F (ta ) is
contained in the fixed field of S3 . Moreover, ta is a root of the general irreducible quartic, so [F (ta ) : F ] = 4
and F (ta ) is the subfield corresponding to S3 .
Every copy of S 3 contains a subgroup of order 3. Note that ∆ is not contained in F (ta ), since these fields
correspond to copies of S3 , which do not fix ∆. Since ∆2 ∈ F , we then have that [F (ta , ∆) : F (ta )] = 2 and
F (ta , ∆) corresponds to the subgroup h(bcd)i.
A4 also contains the four subgroups of order 3, each having index 4 in A4 . The previous paragraph shows
that corresponding extensions are F (ta , ∆)/F (∆).
S4 contains three copies of the dihedral group, each of the form D4 ∼ = h(abcd), (ac)i. Each subgroup fixes
one of the roots of the resolvent cubic, so the correpsonding field extension is F (βi )/F for some i, and the
minimal polynomial of βi is of course the resolvent cubic.
The dihedral group D4 ∼ = h(abcd), (ac)i contains Z2 × Z2 ∼ = h(12)(34), (13)(24)i as an index 2 subgroup,
and the corresponding field extension is L/F (βi ). Note that βj ∈ / F (βi ) for i 6= j, since βj is not fixed by the
dihedral group corresponding to F (βi ). Moreover, [F (βi , βj ) : F (βi )] = 2, since βj satisfies r(x)/(x − βi ) ∈
F (βi )[x], where r is the resolvent cubic. Thus, L = F (βi , βj ).
The dihedral group D4 ∼ = h(abcd), (ac)i contains Z2 × Z2 ∼ = h(ac), (bd)i as an index 2 subgroup. Note
0
that u = (ta + tc ) − (tb + td ) ∈ / F (βi ) since u0 is not fixed by D4 . However, u is in the subfield corresponding
to h(ac), (bd)i, and u02 ∈ F (βi ). Thus, the corresponding field extension is F (βi , u)/F (βi ).
D4 ∼ = h(abcd), (ac)i contains Z4 ∼ = h(abcd)i as an index 2 subgroup. Note that u0 ∆ is not fixed by D4 , but
it is fixed by Z4 . Moreover, (u ∆) ∈ F (βi ). Hence, F (βi , u0 ∆)/F (βi ) is the corresponding field extension.
0 2

Z4 ∼ = h(abcd)i contains Z2 ∼ = h(ac)(bd)i; the corresponding extension is L(u0 )/F (βi , u0 ∆). Note that
u , ∆, βi ∈ F (βi , u ∆, ∆). Since L = F (βi , ∆), we have that L(u0 ) = F (βi , u0 ∆)(∆). The minimal polynomial
0 0

of ∆ is x2 − ∆2 .
For the extension L(u)/F (βi , u) corresponding to the subgroup Z2 ∼ = h(ab)(cd)i contained in Z2 × Z2 ∼ =
h(ab), (cd)i, we note that L(u) = F (βi , βj , u) for j 6= i, and the minimal polynomial of βj over F (βi , u) is
r(x)/(x − βi ).
Let S3 be the subgroup permuting the roots tb , tc , td , and let Z3 ∼ = h(bcd)i ≤ S3 . Note that ∆ ∈ / F(S3 ) =
F (ta ). Hence, F (ta , ∆)/F (ta ) is the corresponding extension.
Z2 ∼= h(cd)i is contained in Z2 × Z2 ∼ = h(ab), (cd)i. Note that v is not contained in F(Z2 × Z2 ), but it is
fixed by Z2 . Then the corresponding field extension is then F (βi , u, v)/F (βi , u).
We then want an element that generates F (βi , u, v) over F (ta ). Since there are no intermediate subgroups
between Z2 ∼ = h(cd)i and S3 , and βi is not fixed by S3 , we have that βi generated the extension. Moreover,
since the extension is of degree 3, we have that the minimal polynomial is the resolvent cubic.
We know K = L(u, v) = F (βi , ∆, u, v). This immediately tells us how to generate the extension
K/F (βi , u, v) corresponding to the subgroup h(cd)i containing {1}.
We just need to an element to generate K/F (t1 , ∆). Since there are no intermediate extensions and
βi ∈ / F (t1 , ∆), which corresponds to h(bcd)i, we conclude that βi generates the extension. It’s minimal
polynomial is the resolvent cubic, since the extension is degree 3.

65