W10 Lec 1

Continuing from last time

We want to show that these sets are equal

{ The irreducible factors of x q−x over F=F p }={ The irreducible polynomials in F [ x ] whose degree divi
 ⊆ : Let g ( x ) ∈ F p [ x ] be an irreducible factor of x q−x . Let β be a root of g. Then since g
is irreducible, deg F ( β )=deg ( g ) .
p

Next, β ∈ K since g ∣ x q−x and so β is also a root of x q−x . So F p ( β ) ⊆K . Thus,

[ F p ( β ) :F p ] ∣ [ K : F p ]⇒ deg F ( β ) ∣ r⇒ deg ( g ) ∣ r
p

 ⊇ : Let g ∈ F p [ x ] be irreducible of degree k and suppose that k ∣r . Let β be a root of g in

some extension field of F p. Then [ F p ( β ) :F p ] =deg F ( β )=k since g=irr F ( β ) . Then
p p

F p ( β ) ≅ F p as they have the same size. Next, since k ∣r , it follows that F p ⊆ F p . Consider
k k r

the field homomorphism

φ : F p( β ) ≅ Fp k
→

Since β is a root of g, so is φ ( β ). But F p ⊆ F p so φ ( β ) ∈ F p =F q. Therefore, φ ( β ) is a

k r r

root of x −x . Since g is irreducible and has root φ ( β ), it follows that g ∣ x q−x .

q

Corollary: For all positive integer r , there exists an irreducible polynomial of degree r in F p [ x ]
for some p prime.

Proof:

Let q= pr . From last lecture, we know that there exists a field K of size q and contains F p as a
subfield and [ K : F p ] =r . We also know that K × is a cyclic group of order q−1 so suppose that it
is generated by α ∈ K ∖ {0 ,1 }. Since K= { 0 , 1, α ,α 2 ,… ,α q−1 }, it follows that K=F p ( α ) and so
[ F p ( α ) : F p ]=r as well. So deg F ( α )=r and so irr F ( α ) is the irreducible of degree r we’re looking
p p

for in F p [ x ].

Example:

1) Suppose we want to find an irreducible polynomial of degree 3 over F 2. Then we can

look at F 8 and x 8−x . This polynomial has factorisation
x ( x−1 ) ( x 3 + x 2+1 ) ( x 3 + x+ 1 )
Thus, we get our degree 3 polynomials x 3+ x2 +1 and x 3+ x+1. In fact, a theorem from
last lecture tells us that these are all the irreducible polynomials of degree 3 in F 2 [ x ].

2) Another way to restate what we’ve said in the above corollary is, if we want to find an
r
irreducible polynomial of degree r in F p [ x ], we just need to factorise x p −x into
irreducible factors.
Primitive element
Definition: Let K / F be a field extension. A primitive element for the extension is an element
α ∈ K that generates K over F . That is, K=F ( α ).

Lemma: Let F be a field of characteristic 0 . Suppose that K / F is a field extension generated by

two elements α , β ( K=F ( α , β )). Then for all except finitely many c ∈ F , γ=β +cα ∈ K is a
primitive element for K over F . That is, K=F ( γ ).

Proof: Let f =irr F ( α ) ∈ F [ x ] and g=irr F ( β ) ∈ F [ x ] . Let L / K be a field extension such that f , g
split completely in L. Suppose that for some m , n>0, the roots of f are α 1 , … , α m ∈ L and roots
of g are β 1 , … , β n ∈ L. Now, α , β are one of these roots so we’ll assume w.l.o.g that α =α 1 , β=β 1
.

Next, we saw that since char ( F )=0 and f , g are irreducible in F [ x ], it follows that f , g have no
multiple roots in L. Thus α i ≠ α j and β i ≠ β j for any i≠ j.

β j−β 1 β j −β 1
Fix c ∈ F ∖ { |
α 1−α i }
i≤ m , j≤ n . Note that F ∖
α 1−α i { | }
i≤ m , j≤ n is nonempty because F ⊆Q

which is an infinite set.

Claim: γ =β 1 +c α 1 is a primitive element of K=F ( α 1 , β 1 ) over F .

Let L1=F ( γ ) . We want to show that L1=K . Since γ ∈ K and F ∈ K , it follows that L1 ⊆ K . For
the converse, note that it suffices to show that α 1 ∈ L1 (since if so, β 1=γ −c α 1 which is also in L1
and thus K=F ( α 1 , β 1 ) ⊆ L1).

Let h ( x )=g ( γ −cx ) ∈ L1 [ x ] . Note that L is the extension in which g splits completely, so L1 ⊆ L.
Therefore, from tutorial 9, we know that gcd
L [x]
( h , f )=gcd ( h , f ). It suffices now to show that
L[ x]
1

gcd ( h , f )=x−α 1 (then gcd ( h , f )=x−α 1 ⇒ x−α 1 ∈ L1 [ x ] ⇒ α 1 ∈ L1).

L[ x] L 1[ x ]

Since g splits completely in L ⊇ L1, we have

g= ( x−β 1 )( x −β2 ) … ( x−β n ) ∈ L [ x ]

So we have the following split for hh=g ( γ −cx )¿ ( γ−cx−β 1 )( γ −cx−β 2 ) … ( γ−cx−β n ) ∈ L [ x ]
γ− β1 γ −β 2 γ −β n

( )( ) ( )
n
¿ (−c ) x− x− … x−
⏟c c c
α1

n
( γ −β −βc + β ) …( x− γ −β −βc + β )
¿ (−c ) ( x−α 1 ) x− 2 1 1 n 1 1

γ −β β −β γ −β β −β
¿ (−c ) ( x−α ) ( x− ) … ( x−
c )
n 1 2 1 1 n 1
1 + +
c c c
 β 2−β 1 β −β
n
(
¿ (−c ) ( x−α 1 ) x−α 1+
c ) (
… x−α 1 + n 1
c )
β 2−β 1 β −β

[(
¿ (−c )n ( x−α 1 ) x − α 1−
⏟ c
~
β2
⏟ )] [ (
… x− α n− n 1
c
n

)]
~ ~
¿ (−c ) ( x−α 1 ) ( x− β 2) … ( x− β n ) ∈ L [ x ]
~
βn

~
Now, in L [ x ] , f ( x ) =( x−α 1 )( x −α 2 ) … ( x−α n ) and ∀ j ≥2 , β j ≠ α j so gcd ( h , f )=x−α 1. To see
L[ x]

~ βi −β 1
why   β j ≠ α j , ∀ j≥ 2, note that c ≠ , ∀ i≤ m, j≤ n, it follows that
α 1−α j

~ β −β α −α j
β j=α 1− j 1 ≠ α 1− ( β j−β 1 ) 1 =α 1−α 1 +α j=α j
c β i−β 1

Theorem: (Primitive element theorem)

Every finite extension K of a field F of characteristic 0 contains a primitive element. That is,
∃ γ ∈ K : K=F ( γ ).
Proof:

Let K / F be a finite extension. Then K=F ( β 1 , … , β k ) for some β 1 , … β k ∈ K . We proceed by

using induction on k .

 k =1: Then K=F ( β k ) and thus γ =β k is the primitive element for the extension.
 Suppose that for any finite extension of F of degree k −1 contains a primitive element.
Now, K=F ( β 1 , … , β k−1 )( βk ) =F ( δ ) ( β k ) , δ ∈ F ( β 1 ,… , β k−1 ) and thus K=F ( δ , β k ). Using
our previous lemma, ∃ γ ∈ K : F ( δ , βk ) =F ( γ ). Therefore, γ ∈ K exists.

Story: These finite fields are sometimes called Galois fields because he prove the existence of
such fields as well as basic properties. Finite fields of order q= pr are elements of the set GF ( q )
of the Galois fields.

Galois Theory
Motivation: The idea was that if we have these irreducible polynomials, they have various roots.
Galois wanted to study the relation between these roots using symmetries which are groups.
Moreover, he wanted to study ways of solving equations involving 5 th degree polynomials. Now,
Abel proved that a general 5th degree polynomials cannot be solved by radicals (roots). Galois
explained why this is the case using the Galois groups, but he explained this way before groups
were invented.

Splitting fields
Definition: Let f ∈ F [ x ] where F is a field. ( f not necessarily irreducible) A splitting field f over
F is an extension field K / F such that

1) f splits completely in K ( f =( x−α 1 ) … ( x−α n ) , ∀ i ≤n , α i ∈ K )

2) K=F ( α 1 , … , α n ) ( K is generated by the roots of f )