Math 662
Homework #3
October 5, 2005
Due Wednesday, October 12.
1. Let E be the splitting field of f = x4 − 2 over Q. Determine the Galois group Gal(E/Q)
and show explicitly how elements of the Galois group act on elements of E. Determine the
bijection of subgroups of Gal(E/Q) and intermediate fields Q ⊆ K ⊆ E as guaranteed by the
Fundamental Theorem of Galois Theory.
2. Cubic extensions. We know that every extension of Q of degree 2 is Galois. Here and in the
next problem we will look at extensions of degree 3 and 4. It is worth noting that there is
nothing particularly special about using Q as the base field—any field of characteristic not 2
or 3 is fine here.
Let f ∈ Q[x] be a monic irreducible polynomial of degree 3 with roots α, β, γ ∈ C, and let
L = Q(α, β, γ) be the splitting field of f .
(a) Let
∆ = (α − β)2 (β − γ)2 (α − γ)2
be the discriminant of f . Show that Gal(L/Q) ∼
= Z/3Z if and only if ∆ is a square in
Q. (Hint: We know that Gal(L/Q) is isomorphic to a subgroup of S3 . Which subgroups
are possible given that f is irreducible? Consider the possible actions of elements of
Gal(L/Q) on the element δ = (α − β)(β − γ)(α − γ) ∈ L.)
(b) If f = x3 + Ax + B, then ∆ can be expressed as a polynomial in A and B. Find this
expression. (Hint: ∆, A, and B are all symmetric in α, β, γ, and each is expressible in
terms of elementary symmetric polynomials.)
(c) Use parts (a) and (b) to determine the Galois groups (of the splitting fields) of the
polynomials f = x3 + x + 1 and g = x3 − 9x + 9.
√
Q2 := {r2 | r ∈ Q}.) Let K = Q( d). Let
3. Quartic extensions.
Let d ∈ Q, d ∈
/ Q2 . (Here
√
√
/ K 2 , and let L = K( θ). Thus [L : Q] = 4.
θ = a + b d with θ ∈
√
(a) Let θ0 =√a − b d, and consider the following three conditions: (1) L/Q is Galois; (2)
L = K( θ0 ); (3) Either θθ0 ∈ Q2 or dθθ0 ∈ Q2 . We will show that they are equivalent,
but first show that (1) ⇔ (2) and that (3) ⇒ (2).
∼
(b) Assuming that b = 0, show that (1), (2), and (3) all hold. Show that Gal(L/Q) =
Z/2Z ⊕ Z/2Z in this case.
2
2
0
(c) In parts (c)–(d) assume
√ that b 6= 0 and that (1)√& (2) hold. Let f (x) = (x − θ)(x − θ ).
Show that f = Irr( θ, Q, x), and thus L = Q( θ)
(d) We know that # Gal(L/K) = 2. Let τ ∈ Gal(L/K) be the nontrivial automorphism.
Now we can pick σ ∈ Gal(L/Q) so that
√
√
σ( θ) = θ0 .
Squaring both sides implies that σ(θ) = θ0 6= θ, and so
σ|K 6= idK .
But σ 2 |K = idK , and so σ 2 ∈ Gal(L/K). There are two possibilities:
Case 1: σ 2 = 1. Prove in this case that θθ0 ∈ Q2 .
Case 2: σ 2 = τ . Prove in this case that dθθ0 ∈ Q2 .
(e) Show that under the conditions of part (a) that Gal(L/Q) ∼
= Z/2Z ⊕ Z/2Z if and only
if θθ0 ∈ Q2 and that Gal(L/Q) ∼
= Z/4Z if and only if dθθ0 ∈ Q2 .
p
√
√ √
(f) For L = Q( i) (where i = −1) and L = Q −2 + 2 show that L/Q is Galois and
determine Gal(L/Q).
√
(g) Suppose that K = Q( d) is an imaginary quadratic field (d < 0). Show that there is no
field L ⊆ C with K ⊆ L and [L : K] = 2 so that L/Q is Galois and Gal(L/Q) ∼
= Z/4Z.