Math 901
Solutions to Exam # 1
Section III
5. Let E be the splitting field for x8 − 1 over Q. Find (explicitly) the elements of the
Galois group of E/Q and find (with justification) primitive elements for each of the
intermediate fields of E/Q.
Solution: The roots of x8 − 1 are just the 8th roots of unity. Let ω be a primitive
8th root of unity. Then E = Q(ω). We also know from a result in class that G =
Gal(E/Q) ∼
= Z∗8 . Let φ be an element of G. Then φ is determined by φ(ω). Further,
φ(ω) must be a primitive 8th root of unity, which are ω, ω 3 , ω 5 , and ω 7 . Since |G| = 4,
all four primitive 8th roots of unity determine legitimate automorphisms of E/Q. Let σ
be the element of G defined by σ(ω) = ω 3 and τ the element of G given by τ (ω) = ω 5 .
Then (στ )(ω) = (τ σ)(ω) = ω 7 . Further, σ 2 = τ 2 = (στ )2 = 1, the identity map
on E. Thus, G = {1, σ, τ, στ }, the Klein-4 group. The intermediate fields of E/Q
correspond to the subgroups of G. The nontrivial subgroups of G are hσi, hτ i, and
hστ i. Note that as |σ| = 2, ω + σ(ω) = ω + ω 3 ∈ Eσ . Thus, Q(ω + ω 3 ) ⊆ Eσ .
Further, ω is a root of (x − ω)(x − ω 3 ) = x2 − (ω + ω 3 ) − 1 ∈ Q(ω + ω 3 )[x]. Hence,
[E : Q(ω + ω 3 )] ≤ 2. Since, [E : Eσ ] = |σ| = 2, we see that Eσ = Q(ω + ω 3 ). In
almost identical fashion, one can show that Eστ = Q(ω + ω 7 ). For hτ i, notice that
τ (ω 2 ) = (τ (ω))2 = ω 10 = ω 2 . Thus, ω 2 ∈ Eτ . If one notices that ω 2 = ±i, then one
sees that [Q(ω 2 ) : Q] = 2 and hence Eτ = Q(ω 2 ). Alternatively, one can observe that
ω is a root of (x − ω)(x + ω) = x2 − ω 2 ∈ Q(ω 2 )[x]. Thus, as above, we have that
[E : Q(ω 2 )] ≤ 2 and Eτ = Q(ω 2 ).
6. Let f (x) ∈ Q[x] be an irreducible polynomial of degree 7 and E its splitting field. Let
G be the Galois group of E/Q and suppose [E : Q] = 21. Prove that Q(α) 6= Q(β) for
every pair of distinct roots α, β of f (x). (Hint: First argue that G cannot be abelian.)
Solution: Let α ∈ E be a root of f (x). Then [Q(α) : Q] = 7. If G is abelian, then
every intermediate field of E/Q is normal over Q, and so Q(α) would be the splitting
field of f (x). This contradicts that E is the splitting field and [E : Q] = 21. So G
is nonabelian. Let P be a Sylow 3-subgroup of G and Q a Sylow 7-subgroup of G.
By Sylow’s Theorems, we have that Q is normal in G. If P is also normal, then G
would be cyclic, a contradiction. Hence, P is not normal. Since [G : P ] = 7 and
P ⊆ NG (P ) 6= G, we must have NG (P ) = P . Now, suppose L = Q(α) = Q(β)
for two distinct roots α and β of f (x). Then Aut(L/Q) 6= {1}, as there exists an
automorphism of L sending α to β. Let H = Gal(E/L). Then |H| = 3, a Sylow
3-subgroup. By a homework problem (in fact, Problem # 3 on this exam), we have
NG (H)/H ∼
= Aut(L/Q). But this contradicts that NG (H) = H for every Sylow 3subgroup of G. Thus, Q(α) 6= Q(β) for every pair of distinct roots α and β of f (x).
7. Let E/Q be an algebraic field extension and suppose every irreducible polynomial in
Q[x] has a root in E. Prove that E is algebraically closed. (Hint: Use the primitive
element theorem on an appropriate splitting field.)
Solution: Fix some algebraic closure E of E. It suffices to show that E = E. Let
α ∈ E. Then α is algebraic over Q. Let f (x) ∈ Q[x] be the minimal polynomial of α
over Q and L the splitting field for f (x) over Q. Note that α ∈ L. As L/Q is finite and
separable (characteristic zero), there exists β ∈ L such that L = Q(β) by the Primitive
Element Theorem. Finally, let h(x) be the minimal polynomial of β over Q. Note that
since L is normal over Q, h(x) splits in L and in fact L = Q(γ) for every root γ of
h(x). (Note that as h(x) is irreducible, [L : Q] = [Q(β) : Q] = [Q(γ) : Q] for every
root γ of h(x).) But by hypothesis, h(x) has a root in E. Hence, α ∈ L ⊆ E.
8. Let E/Q be a finite Galois extension whose Galois group is simple and nonabelian.
Suppose there exists a prime p and an element α ∈ E such that αp ∈ Q. Prove that
α ∈ Q. (Hint: Let β = αp . Then the minimal polynomial of α over Q divides xp − β.
Find the roots of this polynomial.)
Solution: Let β = αp as in the hint. The roots of f (x) = xp − β are ω i α, for
i = 0, . . . , p − 1, where ω is a primitive pth root of unity. Let g(x) be the minimal
polynomial of α over Q. If deg g(x) = 1 then α ∈ Q and we are done. So suppose
deg g ≥ 2. As g(x) divides f (x), we must have ω i α is a root of g(x) for some i =
1, . . . , p − 1. Since E/Q is normal and g(x) has a root (namely, α) in E, g(x) splits in
E. Thus, ω i α ∈ E. Since α ∈ E, we obtain that ω i ∈ E. Now, ω i is also a primitive
pth root of unity (as gcd(i, p) = 1), so Q(ω i ) = Q(ω) is a subfield of E. Since Q(ω)
is the splitting field for xp − 1 over Q, we have that Q(ω)/Q is a normal extension.
Then H = Gal(E/Q(ω)) is a normal subgroup of G = Gal(E/Q). Since G is simple,
we must have H = {1} or H = G.
Suppose first that H = {1}. Then E = Q(ω). But then G = Gal(Q(ω))/Q) ∼
= Z∗p ,
which is abelian, a contradiction.
Now suppose H = G. Then Q(ω) = Q, which implies p = 2 (as [Q(ω) : Q] = φ(p) =
p − 1). Then α is a root of x2 − β ∈ Q[x]. As we are assuming α 6∈ Q, we have
[Q(α) : Q] = 2. Thus, Q(α)/Q is a normal extension. Since G is simple, there are
no intermediate fields which are normal over Q except E and Q. Thus, we must have
E = Q(α). But then |G| = 2 and G is abelian, a contradiction.
Hence, deg g = 1 and α ∈ Q.