Math 306, Spring 2012

Homework 5 Solutions
(1) (a) (3 pts) Find a linearly dependent set of three vectors in R3 , but such that any set of two of them is linearly
independent.
(b) (5 pts) Let V be a vector space over C. Suppose that B = {v1 , v2 , v3 } is a linearly independent subset of
V . Prove that the set B 0 = {v1 + v2 , v2 + v3 , v1 + v3 } is a linearly independent subset of V .
Solution:
(a) Take, for example, {(1, 0, 0), (0, 1, 0), (1, 1, 0)}
(b) Suppose that there are c1 , c2 , c2 ∈ C such that c1 (v1 + v2 ) + c2 (v2 + v3 ) + c3 (v1 + v3 ) = 0. Then
(c1 + c3 )v1 + (c1 + c2 )v2 + (c2 + c3 )v3 = 0. Since B = {v1 , v2 , v3 } is linearly independent, we conclude
that c1 + c3 = 0, c1 + c2 = 0 and c2 + c3 = 0. Then we can write
    
1 0 1 c1 0
 1 1 0   c2  =  0  .
0 1 1 c3 0
Since the determinant of this 3 × 3 matrix is 2, it is invertible, so
   −1    
c1 1 0 1 0 0
 c2  =  1 1 0   0  =  0 ,
c3 0 1 1 0 0
as required.
(2) (5 pts/part) Let K be a field and let K n be the vector space of n-tuples over K. For all j ∈ {1, . . . , n}, let
ej = (0, . . . , 0, 1, 0, . . . , 0, . . . , 0),
where the 1 occurs in the j-th position. Let fj = e1 + · · · + ej for all j ∈ {1, . . . , n}.
(a) Prove that B1 = {e1 , e2 , . . . , en } is a basis for K n .
(b) Prove that B2 = {f1 , f2 , . . . , fn } is a basis for K n .
(Hint: you may assume that the familiar result which says that rows or columns of a square matrix are linearly
independent iff matrix is invertible extends to matrices over general fields K.)
Solution:
(a) Suppose that there are a1 , . . . , an ∈ K such that a1 e1 +· · ·+an en = 0. Then we have (a1 , a2 , . . . , an ) = 0,
so a1 = a2 = · · · = an = 0. Hence B1 is linearly independent. To show that B1 spans K n , consider
v ∈ K n . Then v = (v1 , . . . , vn ) for some vi ∈ K. Therefore v = v1 e1 + · · · + vn en .
(b) Suppose that there are a1 , . . . , an ∈ K such that a1 f1 + · · · + an fn = 0. Then we have
(a1 , a1 + a2 , . . . , a1 + a2 + · · · + an ) = 0.
Equivalently we can write
    
1 1 ··· 1 a1 0
 0 1 ··· 1  a2   0 
= .
    
 .. .. . . ..  .. ..
 . . . .  .   . 
0 0 ··· 1 an 0
Since the determinant of this n × n matrix A is 1, it is invertible, so it follows that a1 = a2 = · · · = an = 0.
Hence B2 is linearly independent. One way to show that B2 spans Kn is to let B 0 = B2 ∪ {v} for some
v = (v1 , . . . , vn ) ∈ K n and find b1 , . . . , bn ∈ K such that v = b1 f1 + · · · + bn fn (this would show that B2 is
a maximal linearly independent set and hence spans the entire space). Equivalently, we want b1 , . . . , bn ∈ K
 such that
    
1 1 ··· 1 b1 v1
 0 1 ··· 1  b2   v2 
= .
    
 .. .. . . ..  .. ..
 . . . .  .   . 
0 0 ··· 1 bn vn
We can solve for such bi by multiplying each side by A−1 . Hence B 0 is linearly dependent, so B2 is maximal.
(3) (3 pts/part) Find an infinite linearly independent subset of the following vector spaces. No proof is required.
(a) R over Q.
(b) K(t) over K, where K is a field and t is an indeterminate.
(c) K S over K, where K is a field and S is an infinite set.
Solution:
√
(a) B = { p : p is prime}
(b) B = {tn : n ∈ Z≥0 }
(c) B = {ft : t ∈ S}, where ft : S → K is defined by

0 if s 6= t,
ft (s) =
1 if s = t,
for all s ∈ S.
(4) (3 pts/part) Compute the degree [L : K] for each of the field extensions below, and exhibit a basis for L as a
vector space
√ over K if the degree is finite.
3
(a) Q(√ 5) : Q
(b) R( 5 2) : R
(c) Q(e√2πi/5
√) : Q √
(d) Q( 3, 5) : Q( 3)
(e) C : Q
Solution: √ √
(a) The extension has degree 3 with basis B = {1, 3 5, 3 25}.
(b) The extension has degree 1 with basis B = {1}.
(c) The extension has degree 4 with basis B e2πi/5 , e4πi/5 , e6πi/5 }.
= {1, √
(d) The extension has degree 2 with basis B = {1, 5}.
(e) The extension has infinite degree.
(5) Use the Tower Law for the following.
(a) (5 pts) Prove that, if L : K is a field extension with [L : K] = 1, then L = K. (Hint: Show that K ⊆ L
and L ⊆ K.)
Remark: We have already used this result in case of simple extensions K(α) : K in class. The proof in
that case is that if the degree of this extension is 1, then the degree of the monic minimal polynomial over
K for α is 1, i.e. the polynomial must be x − α, which means that α is in K.
(b) (3 pts) If [L : K] is a prime integer, prove that there are no intermediate fields M strictly between L and
K.
Solution:
(a) Suppose that [L : K] = 1. Clearly K ⊆ L. Let B = {e} be a basis of L : K. We claim that e ∈ K.
Indeed, if α ∈ K ∗ , then α ∈ L, so there is β ∈ K such that α = βe. Clearly β 6= 0, so e = α/β. Therefore
e ∈ K. Now for all γ ∈ L, there is δ ∈ K such that γ = δe, so γ ∈ K. Therefore L ⊆ K, as required.
(b) Let [L : K] be prime. Suppose that M is a field with K ⊆ M ⊆ L. Then by the Tower Law, we have
[L : M ] = 1 or [M : K] = 1. So L = M or K = M , so M is not strictly contained between L and K.
(6) (5 pts/part) √ √
√ { √6, √
(a) Prove that B = 10} is a linearly independent subset of R as a vector space over Q.
(b) Prove that Q( 6, 10, 15) : Q has degree 4 and not 8. Exhibit a basis for this extension.
Solution: √ √
(a) Suppose that there are a, b ∈ Q such that a 6 + b 10 = 0. Suppose that b 6= 0. Then a 6= 0. Without
loss of generality, assume that a and b are coprime integers. Then 6a2 = 10b2 , i.e. 3a2 = 5b2 . Hence
3|5b2 , √
so 3|b. 2 2
√ Hence 9|3a , so 3|a . Therefore 3|a, a contradiction. Therefore b = 0, and so a = 0. So
B = { 6, 10} is a linearly independent subset.
√ √
60
√ √ √ √ √ √ √
(b) Since 15 = 2 ∈ Q( 6, 10), it follows that Q( 6, 10, 15) = Q( 6, 10). Clearly

√ √
[Q( 6, 10) : Q] = 4,

so we are done.
(7) (3 pts/part) The following statements are all false. Provide a counterexample or a counterproof. Recall that an
extension L : K is finite if the degree [L : K] is finite.
(a) Every field extension of R is a finite extension.
(b) Every field extension of a finite field is a finite extension.
(c) There is some element of C that is transcendental over R.
(d) If K is a field, then every algebraic extension of K is finite. √
(e) For all n ∈ Z≥2 , there are no intermediate fields properly between Q( n 2) and Q.
Solution:
(a) If t is an indeterminate, then R(t) is an infinite extension of R.
(b) If t is an indeterminate, then Z2 (t) is an infinite extension of Z2 .
(c) Note that [C : R] = 2 which is prime. Hence for any α ∈ C, we have [R(α) : R] = 1 or 2. Therefore α is
algebraic over R. √ √ √
(d) The extension √ Q( 2, 3, 5, . . .) : Q is algebraic
√ but infinite.
(e) The field Q( 2) is lies properly between Q( 4 2) and Q.
(8) (5 pts/part)
(a) Suppose that [L : K] is a prime number. Prove that L : K is a simple extension, i.e. there is α ∈ L such
that L = K(α). (Hint: Look at an earlier problem.)
(b) Let L : K be a finite extension, and let p be an irreducible polynomial in K[x] with deg p ≥ 2. Prove by
contradiction that, if deg p and [L : K] are coprime, then p has no zeros in L. (Hint: If α ∈ L is a root of
p, then consider the field K(α).)
Solution:
(a) Suppose that [L : K] = p is prime and let α ∈ L\K. Then [K(α) : K] divides [L : K], so [K(α) : K] = 1
or p. Since K(α) 6= K, it follows that [K(α) : K] = p and so [L : K(α)] = 1. Therefore L = K(α) and
L is a simple extension of K.
(b) Suppose that α ∈ L is a root of p. Then [K(α) : K] divides [L : K]. But [K(α) : K] = deg p, so deg p
divides [L : K]. Since these numbers are assumed to be coprime, it follows that deg p = 1, a contradiction.
(9) (5 pts/part) We say that a rational number a is a square in Q if there is b ∈ Q such that b2 = a. Let m, n ∈ Q
be non-squares. Prove the following. √ √
(a) If mn is a square in Q, then [Q( m, n) √: Q]√= 2.
(b) If mn is a non-square in Q, we have [Q( m, n) : Q] = 4.
Solution:
 (a) Suppose that m and n are nonsquares in Q and suppose that mn = a2 for some a ∈ Q. Notice that
2 √ √ √
neither m nor n is zero, so we can write n = am , or n = ± √am . Therefore n ∈ Q( m), so
√ √ √
[Q( m, n) : Q] = [Q( m) : Q] = 2.
√ √
(b) Suppose
√ √that mn √ is a nonsquare,
√ and suppose on the contrary that [Q( m, n) : Q] = 2. Since
Q( m, n) = (Q( m))( n), it follows that
√ √ √ √ √ √
[Q( m, n) : Q] = [Q( m, n) : Q( m)][Q( m) : Q].
√ √ √ √ √ √ √ √
Since [Q( m) : Q] = 2, it follows that
√ [Q( √m, n) : Q( m)] = 1, √ so n ∈ Q( m), i.e. n = a + b m
for some a, b ∈ Q. Therefore a2 = ( n − b m)2 = n + bm2 − 2b mn. Therefore mn is a square in Q, a
contradiction.
(10) (5 pts) Suppose that M : L : K is a tower of field extensions and let α ∈ M be algebraic over L. Assume that
[K(α) : K] and [L : K] are relatively prime. Prove that the minimum polynomial mL α of α over L actually has
its coefficients in K.
Solution: Let m = deg mK 0 L
α = [K(α) : K] and n = [L : K] and m = deg mα = [L(α) : L]. The hypotheses
give (m, n) = 1. We certainly know that m ≤ m and mα |mα . Hence [L(α) : K] = m0 n ≤ mn. However it is
0 L K

clear that both m and n divide [L(α) : K], and since they are relatively prime we must have mn = [L(α) : K].
In particular m = m0 and therefore mL K L
α = mα , so mα is really a polynomial with coefficients in K.