Math 901
Solutions to Homework # 3
1. Let E be the splitting field of f (x) = x6 + 3 ∈ Q[x] and let α ∈ E be a root of f (x).
(a) Prove that E = Q(α). (Hint: Note that
√
1+ 3i
2
is a primitive 6th root of unity.)
(b) Let G be the Galois group of E/Q. Determine, with justification, whether G is
abelian.
Solution: For part (a), the roots of f (x) are ω i α for i = 0, . . . , 5, where
ω is a primitive
√
3 2
3
6th root of unity. Then E = Q(ω, α). But
(α
)
=
−3,
so
α
=
±
3i.
Without
loss of
√
√
1+ 3i
3
generality, assume α = 3i. Since 2 is a primitive 6th root of unity, we see that
ω ∈ Q(α). Hence, E = Q(α).
For part (b), note that as E = Q(α) and x6 + 3 is irreducible over Q (by Eisenstein),
we have [E : Q] = 6. Thus G = Gal(E/Q) has order 6. For each i = 0, . . . , 5, let
σi : E → E be given by σi (α) = ω i α. These are well-defined since α and ω i α are both
roots of x6 + 3. Thus, G = {σi | i = 0, . . . , 5}. Note that for each j ≥ 0:
√
1 + 3i
)
σj (ω) = σj (
2
1 + α3
= σj (
)
2
1 + σj (α)3
=
2
1 + ω 3j α3
=
2
√
1 + (−1)j 3i
=
.
2
Hence, σj (ω) = ω 5 if j is odd and σj (ω) = ω if j is even. Now,
σ2 σ1 (α) = σ2 (ωα) = σ2 (ω)σ2 (α) = ω · ω 2 α = ω 3 α.
Also,
σ1 σ2 (α) = σ1 (ω 2 α) = σ1 (ω)2 σ1 (α) = (ω 5 )2 · ωα = ω 5 α.
Hence, σ1 σ2 6= σ2 σ1 and G is not abelian.
2. Let E/F be a finite Galois field extension with Galois group G. Let α ∈ E and H the
Galois group of E/F (α). Let σ1 , . . . , σn be a complete set of coset representatives for
H in G. (I.e., n = [G : H] and σi H 6= σj H for all i 6= j.) Prove that the minimal
polynomial of α over F is
n
Y
(x − σi (α)).
i=1
Q
Solution: Let f (x) = ni=1 (x − σi (x)). Clearly, f (x) is monic and deg f = [G : H] =
[F (α) : F ]. Let σ1 be the coset representative for H; i.e., σ1 ∈ H. Then σ1 (α) = α,
so α is a root of f (x). It suffices to show that f (x) ∈ F [x]. It suffices to show
that f τ (x) = f (x) for every τ ∈ G. Note that σi (α) 6= σj (α) for i 6= j. Otherwise,
σj−1 σi (α) = α and hence σj−1 σi ∈ H, contradicting our assumption that σi H 6= σj H.
Thus, σ1 , . . . , σn restricted to F (α) are the distinct embeddings of F (α) into F , as
[F (α) : F ]s = [F (α) : F ] = n. As τ is injective, τ σ1 , . . . τ σn (restricted to F (α)) is a
permutation of σ1 , . . . , σn (restricted to F (α)). Hence, f τ (x) = f (x) for every τ ∈ G.
Thus, f (x) ∈ F [x].
3. Let E be the splitting field of x4 − 2 over Q. Find a presentation (with justification)
for Gal(E/Q).
√
√
√
Solution: The roots of x4 − 2 are ± 4 2 and ±i 4 2. Thus, E = (i, α) where α = 4 2.
Note that [Q(α) : Q] = 4 as x4 − 2 is irreducible by Eisenstein. Since i 6∈ Q(α) then
[E : Q] = 8. Let G be the Galois group of E/F and let σ : E → E be complex
conjugation (so σ(i) = −i and σ(α) = α). Clearly, σ has order 2. Since [E : Q(i)] = 4,
we have that x4 − 2 is irreducible over Q(i). Thus, there exists τ : E → E such that
τ (α) = iα and τ (i) = i. It is easily seen that τ has order 4. Now, στ (i) = −i and
στ (α) = −iα. Also, τ 3 σ(i) = −i and τ 3 σ(α) = −iα. Hence, στ = τ 3 α. Thus, a
presentation for G is hσ, τ | σ 2 = 1, τ 4 = 1, στ = τ 3 αi.
4. Let E/F be a finite Galois extension and K an intermediate field. Let G = Gal(E/F )
and H = Gal(E/K). Prove that NG (H) = {g ∈ G | g(K) = K} and NG (H)/H ∼
=
Aut(K/F ).
Solution: Suppose g ∈ NG (H) and h ∈ H. Since g −1 hg ∈ H then g −1 hg restricted to
K is the identity map. For k ∈ K, k = (g −1 hg)(k), so g(k) = h(g(k)). Since this holds
for all h ∈ H, we see that g(k) ∈ EH = K for all k ∈ K. Thus, g(K) ⊆ K. As K/F
is algebraic, we have g(K) = K. Conversely, suppose g(K) = K. Let h ∈ H. To show
g −1 hg ∈ H it suffices to show that (g −1 hg)(k) = k for all k ∈ K. But as g(k) ∈ K,
h(g(k)) = g(k). Hence, g −1 (hg)(k) = (g −1 g)(k) = k. Thus, g ∈ NG (H).
For the second statement, define a map φ : NG (H) → Aut(K/F ) by restriction to
K. This map is well-defined by the first part of the problem. Clearly, φ is a group
homomorphism. Any σ ∈ Aut(K/F ) can be extended to an element τ ∈ G. Then
φ(τ ) = σ and we have that φ is surjective. Finally, σ ∈ ker φ if and only if σ fixes K,
which is if and only if σ ∈ H. The desired isomorphism now follows.
5. Let K ⊆ E, F ⊆ L are fields and suppose E/K is finite and Galois. Prove that EF/F
is Galois and Gal(EF/F ) is isomorphic to a subgroup of Gal(E/K).
Solution: As E/K is finite and separable, E = K(α). Let f (x) = Min(α, K). Then
E is the splitting field of f (x) over K. Now EF = F (α). As α is separable over K, α
is separable over F . Thus, EF/F is separable. Also, EF is the splitting field for f (x)
over F , so EF/F is normal and hence Galois. Now define φ : Gal(EF/F ) → Gal(E/K)
by restriction to E. Then φ is a well-defined group homomorphism. If σ ∈ ker φ then
σ restricted to E is the identity map. Since σ restricted to F is the identity map,
we obtain that σ is the identity map on EF . (See, for example, Problem # 3 from
Homework # 1.) Thus, ker φ = {1} and φ is injective.
6. Let F be a field and f (x) ∈ F [x] a separable irreducible polynomial of prime degree.
Let α be a root of f (x) and suppose f (x) has at least two roots in E = F (α). Prove
that E is the splitting field for f (x) and that E/F is cyclic. (Hint: Let L be the normal
closure of E/F , G = Gal(L/F ), and H = Gal(L/E). Prove that H 6= NG (H).)
Solution: Let β 6= α be a root of f (x) in E = F (α). As f (x) is irreducible, E = F (β)
as well. Also, there exists an automorphism σ of E which fixes F and sends α to
β. Extend σ to an element τ ∈ G. As τ (E) = E, τ ∈ NG (H) by Problem #4.
Moreover, as τ does not fix E, τ 6∈ H. Hence, H is a proper subgroup of NG (H). Now,
[G : H] = [E : F ] = p (the degree of f (x)), so [G : NG (H)] = 1. Hence, H is normal
in G, which implies E is normal over F . Hence, E = L and G has prime order.
E
7. Let E be a finite extension of a finite field F . Prove that TrE
F and NF are surjective
(as maps from E to F ). (Recall from Math 818 that E/F is a cylic extension.)
E
Solution: As finite fields are perfect, E/F is separable. Thus, TrE
F is nonzero. As TrF
is a linear transformation of F -vector spaces from E to F , and F is one-dimensional,
E
we see that TrE
F is surjective. For the norm, clearly NF (α) = 0 if and only if α = 0.
∗
∗
∗
Let φ be NE
F restricted to E = E \ {0}. It suffices to prove that φ : E → F is
∗
surjective. As φ is a group homomorphism, it is enough to prove that |E |/|K| = |F ∗ |,
α
| α ∈ E ∗ }. Now, K
where K = ker φ. Let G = hσi. By Hilbert’s Satz 90, K = { σ(α)
α
is the image of the group homomorphism f : E ∗ → E ∗ given by f (α) = σ(α)
. Note
∗
that α ∈ ker f if and only if σ(α) = α, which holds if and only if α ∈ F (as F = Eσ ).
Hence, |K| = |E ∗ |/|F ∗ | and so |E ∗ |/|K| = |F ∗ |.