Galois Theory
Solution Sheet 6
1. Let K = Z/2Z.
(a) Find an irreducible polynomial of degree 4 in K[X]. (Hint: See
Sheet 2, Exercise 3d.)
Solution: The solution to Sheet 2, Exercise 3d) shows that X 4 +
X + 1 is irreducible in K[X]. (As usual, I’ll omit the [·] denoting
congruence classes since this is clear from the context and would
clash with other notation later.)
(b) Use the solution to (a) to construct a field L with 16 elements.
Solution: Let f (X) = X 4 +X +1 ∈ K[X]. Since f is irreducible
K[X]/(f ) is a field. Since deg f = 4, we have [L : K] = 4. Since
L is a 4-dimensional vector space over K = Z/2Z, it has 24 = 16
elements (see Sheet 2, Exercise 6b).
(c) For the field L in (b), find a generator for the group L× .
Solution: Letting α = [X] ∈ L, the 16 elements of L are:
a0 + a1 α + a2 α 2 + a3 α 3
where ai ∈ {0, 1} = K for i = 0, 1, 2, 3. Since the minimal polynomial of α over K is X 4 +X +1, the above elements are distinct.
Note also that α4 = α + 1 since α4 + α + 1 = 0 (recall that −1 = 1
in K).
Since L× has order 15, it suffices to find an element whose order
doesn’t divide 3 or 5. I claim that α is such an element; it suffices
to check that α3 6= 1 and α5 6= 1. We clearly have α3 6= 1 from
the above discussion, and furthermore
α5 = αα4 = α(α + 1) = α2 + α 6= 1.
Therefore α is a generator of L× .
√ √
2. Let p and q be distinct prime numbers, let K = Q( p, q) and let
√
√
α = p + q. (See Sheet 4, Exercise 3; in particular [K : Q] = 4.)
(a) Prove that K = Q(α).
Solution: From the proof of the Primitive Element Theorem,
it suffices to prove the τi (α) are distinct, where τ1 , τ2 , τ3 , τ4 are
the 4 embeddings K → C. These embeddings are given by (the
solution of) Sheet 4, Exercise 3d), and their values on α are:
√
√
√
√
τ1 (α) = p + q,
τ2 (α) = p − q,
√
√
√
√
τ3 (α) = − p + q, τ4 (α) = − p − q.
√
If i 6= j, then the difference between τi (α) and τj (α) is ±2 p,
√
±2 q, or ±2α, none of which is 0.
(b) Find the minimal polynomial of α over Q.
Solution: The values of τi (α) are the 4 roots of mα,Q , so
mα,Q =√
√
√
√
√
√
√
√
(X − p − q)(X − p + q)(X + p − q)(X + p + q)
√
√
= (X 2 − 2 pX + p − q)(X 2 + 2 pX + p − q)
= X 4 − 2(p + q)X 2 + (p − q)2 .
(c) Show that K is a normal extension of Q.
Solution: Since K is the splitting field over Q of (X 2 −p)(X 2 −q),
it is a normal extension of Q.
3. Let m and n be integers such that m ≥ 1 and n ≥ 2, and let
√
K = Q( n m, e2πi/n ).
(a) Prove that K is a splitting field over Q for the polynomial X n −
m ∈ Q[X], and that [K : Q] ≤ n(n − 1).
√
Solution: Let β = n m, ζ = e2πi/n and βk = βζ k for k =
0, 1, . . . , n − 1. Then for each k = 0, 1, . . . , n − 1, we have
βkn = (βζ k )n = β n ζ kn = m.
Since the βk for k = 0, 1, . . . , n − 1 are n distinct roots (in C) of
X n − m, it follows that
X n − m = (X − β0 )(X − β1 ) · · · (X − βn−1 )
in C[X].
Therefore L = Q(β0 , β1 , . . . , βn−1 ) is a splitting field for X n − m
over Q. Note that βk = βζ k ∈ K for each k, so L ⊂ K. On the
other hand β = β0 and ζ = β1 /β0 , so K ⊂ L, and we conclude
that K = L is a splitting field for X n − m over Q.
To bound the degree, note that since β is a root of X n − m, the
minimal polynomial of β over Q has degree at most n. Therefore
[Q(β) : Q] ≤ n. Since ζ is a root of
(X n − 1)/(X − 1) = X n−1 + X n−2 + · · · + 1,
its minimal polynomial over Q(β) has degree at most n − 1. Since
L is generated over Q(β) by ζ, it follows that [L : Q(β)] ≤ n − 1.
Therefore by the Tower Law
[L : Q] = [L : Q(β)][Q(β) : Q] ≤ (n − 1)n.
(b) Prove that if n is prime, then [K : Q] is divisible by n − 1. (Hint:
See Sheet 4, Exercise 1)
Solution: We have Q(ζn ) ⊂ K, and by Sheet 4, Exercise 1, the
minimal polynomial of ζ (which is there denoted α) has degree
n − 1. Therefore [Q(ζn ) : Q] = n − 1. So by the Tower Law,
[K : Q] = [K : Q(ζn )][Q(ζn ); Q] is divisible by n − 1.
(c) Prove that if m is prime, then [K : Q] is divisible by n.
Solution: If m is prime, then X n − m satisfies Eisenstein’s Criterion, so it is the minimal polynomial of β, and [Q(β) : Q] = n.
Since Q(β) ⊂ K, we use the Tower Law again to conclude that
[K : Q] = [K : Q(β)][Q(β) : Q] is divisible by n.
(d) Deduce from (a), (b) and (c) that if m and n are both prime,
then [K : Q] = n(n − 1). In this case, describe all the embeddings
K → C, and find an element α ∈ K such that K = Q(α).
Solution: By (a), we have [K : Q] ≤ n(n − 1), and by (b) and
(c) we have that [K : Q] is divisible by n and by (n − 1). Since n
is prime, the least common multiple of n and n − 1 is n(n − 1),
so [K : Q] is at least n(n − 1). Therefore [K : Q] = n(n − 1).
Since X n −m is the minimal polynomial of β over Q, the n embeddings Q(β) → C, say σk for k = 0, 1, . . . , n − 1, are determined
by σk (β) = βk . Since [K : Q(β)] = n − 1 and ζ is a root of
f (X) = X n−1 + X n−2 + · · · + 1, it follows that this is in fact the
minimal polynomial of ζ over Q(β). So for each k, the extensions of σk to embeddings τ : K → C correspond to the roots of
σ̃k (f ) = f in C, namely ζ j for j = 1, 2, . . . , n − 1 (see Sheet 4, Exercise 1c). The n(n−1) embeddings, say τkj for k = 0, 1, . . . , n−1
and j = 1, 2, . . . , n − 1, are therefore determined by the formulas:
τkj (β) = βk = βζ k ,
τkj (ζ) = ζ j .
We claim that letting α = β + ζ gives K = Q(α). From the proof
of the Primitive Element Theorem, it suffices to show that the
elements τkj (α) for k = 0, 1, . . . , n − 1, j = 1, 2, . . . , n − 1 are
distinct; i.e., if τkj (α) = τk0 j 0 (α) for two such pairs k, j and k 0 , j 0 ,
then k = k 0 and j = j 0 . So suppose that τkj (α) = τk0 j 0 (α); i.e.,
that
0
βζ k + ζ j = τkj (β + ζ) = τk0 j 0 (β + ζ) = βζ k + ζ j
0
(1)
for some k, k 0 ∈ {0, 1, . . . , n−1}, j, j 0 ∈ {1, 2, . . . , n−1}. If k 6= k 0 ,
then equation (1) implies that
0
ζj − ζj
β = k0
∈ Q(ζ),
ζ − ζk
contradicting that L = Q(ζ, β) has degree n over Q(ζ). Therefore
0
k = k 0 , in which case (1) implies that ζ j = ζ j , and hence also
j = j0.
4. Let K be a field of characteristic p, let L = K(Y1 , Y2 ) be the field of
rational functions over K in the variables Y1 , Y2 , and similarly define
M = K(Z1 , Z2 ).
(a) Prove that there is a unique K-embedding ι : L → M such that
ι(Y1 ) = Z1p and ι(Y2 ) = Z2p .
Solution: First define ι : K[Y1 , Y2 ] → K[Z1 , Z2 ] by setting
(ι(f ))(Z1 , Z2 ) = f (Z1p , Z2p ). Then ι is a homomorphism of rings
since ι(1) = 1,
(ι(f + g))(Z1 , Z2 ) = (f + g)(Z1p , Z2p ) = f (Z1p , Z2p ) + g(Z1p , Z2p )
= (ι(f ))(Z1 , Z2 ) + (ι(g))(Z1 , Z2 ) = (ι(f ) + ι(g))(Z1 , Z2 )
shows ι(f + g) = ι(f ) + ι(g), and similarly ι(f g) = ι(f )ι(g).
Then ι extends to a homomorphism of fraction fields K(Y1 , Y2 ) →
K(Z1 , Z2 ) by defining ι(f /g) = ι(f )/ι(g). By construction, we
have ι(r) = r for all r ∈ K, ι(Y1 ) = Z1p and ι(Y2 ) = Z2p , so
it has the required properties. Moreover it is the unique such
homomorphism since K(Y1 , Y2 ) is generated over K by Y1 and
Y2 .
(b) Identifying L with a subfield of M via ι, and renaming Z1 as α1
and Z2 as α2 , prove that [M : L(α1 )] = [L(α1 ) : L] = p. (Hint:
See Sheet 5, Exercise 5d.)
Solution: We may view L as the field of rational functions over
K(Y2 ) in the variable Y1 , and apply Sheet 5, Exercise 5d) to
conclude that X p − Y1 is irreducible over L. Since α1p = Y1 , we
see that α1 is a root of this polynomial, so it must be the minimal
polynomial of α1 over L. Therefore [L(α1 ) : L] = p.
Similarly L(α1 ) = K(Z1 , Y2 ) is the field of rational functions over
K(Z2 ) in the variable Y2 , so X p −Y2 is irreducible over L(α1 ), and
is therefore the minimal polynomial of α2 over L(α1 ). Therefore
the degree of M = L(α1 , α2 ) over L(α1 ) is p.
(c) Prove that if β ∈ M , then β p ∈ L.
Solution: Since α1 and α2 are algebraic over L, every element
β ∈ M can be written as a finite sum
X
β=
rmn α1m α2n
m,n
for some elements rmn ∈ L. Since M is a field of characteristic p,
we have, by Sheet 5, Exercise 5a), that (s + t)p = sp + tp for all
s, t ∈ M , and it follows that
!p
X
X
X
p
p
βp =
rmn α1m α2n
=
rmn
Y1m Y2n ∈ L.
rmn
α1mp α2np =
m,n
m,n
m,n
(d) Deduce from (b) and (c) that M is a finite extension of L which
is not a simple extension.
Solution: By (b) and the Tower Law, we have [M : L] = p2 ; in
particular M is a finite extension of L.
By (c), if β ∈ L, then β p ∈ L, so β is a root of the polynomial
X p − β p ∈ L[X]. Therefore the minimal polynomial of β over L
has degree at most p, so [L(β) : L] ≤ p. Since [M : L] = p2 , we
cannot have M = L(β), so M is not a simple extension of L.