Math 503, Spring 2015: Homework Assignment #2
1. Let p be prime. Find the splitting field F for xp − 2 over Q. What is [F : Q]?
√
The roots of xp − 2 are ζpk p 2, where ζp := e2πi/p and 0 ≤ k < p. These are all contained
√
all √
these roots, so
in the field √
Q( p 2, ζp ). Conversely, the splitting field of xp − 2 contains
√
it contains p 2. Furthermore, since the splitting field contains ζp p 2 and ζp2 p 2, it contains
√
√
their quotient, ζp . Therefore the splitting field contains Q( p 2, ζp ). So Q( p 2, ζp ) is the
splitting field of xp − 2. Now
√
√
√
√
p
p
p
p
[Q( 2, ζp ) : Q] = [Q( 2, ζp ) : Q( 2)][Q( 2 : Q].
√
Then xp − 2 is irreducible over Q by Eisenstein’s criterion, so [Q(√p 2 : Q] = p. Now ζp
satisfies the cyclotomic polynomial Φp (x)
of degree p − 1 over Q( p 2). But since all its
√
roots
are complex
roots of √
unity, and Q( p 2) is totally real,
√
√
√ then Φp (x) is irreducible over
Q( p 2), so [Q( p 2, ζp ) : Q( p 2)] = p − 1. Therefore, [Q( p 2, ζp ) : Q] = p(p − 1).
2. Let K1 and K2 be finite extensions of a field F contained in a field K, and suppose both
are splitting fields over F . Prove that their composite, K1 K2 , and also their intersection,
K1 ∩ K2 are each splitting fields over F . (The composite of K1 and K2 is the smallest
subfield of K containing both K1 and K2 .)
Let Ki be the splitting field of fi (x) ∈ F [x]. We have K1 = F (α1 , ..., αn ), where the αi
are the roots of f1 (x), and K2 = F (β1 , ..., βm ) where the βi are the roots of f2 (x). Then
K1 K2 = F (α1 , ..., αn , β1 , ..., βm ). Note that f1 (x)f2 (x) ∈ F [x] and it splits over K1 K2 . If
K1 K2 is not the splitting field, then there is some field E with F ⊂ K1 ⊂ E ⊂ K1 K2 and
F ⊂ K2 ⊂ E ⊂ K1 K2 , which violates the minimality of K1 K2 .
For the intersection, we first prove a lemma.
Lemma: Let K be a splitting field over F . Then every irreducible polynomial in F [x]
with a root in K splits completely in K[x].
Proof. Suppose K is a splitting field for f (x) ∈ F [x]. Let p(x) be irreducible over F [x]
and let α ∈ K be a root of p(x). If p(x) has only one root, then we are done. Otherwise,
let β be a different root of p(x). Then by the General Conjugate Theorem, there is an
isomorphism σ : F (α) → F (β) which fixes F and with σ(α) = β. Now f (x) ∈ F (α)[x]
and σ(f (x)) = f (x) is in F (β)[x]. We still have K a splitting field for f (x) over F (α)
and then K(β) is a splitting field for f (x) over F (β). Thus σ extends to an isomorphism
σ : K → K(β). Since σ is an isomorphism and fixes F , then both K and K(β) must be
vector spaces of F of the same dimension. Therefore K(β) = K, so β ∈ K. Since β was
arbitrary, the result follows.
Now since K1 ∩ K2 is finite over F we can write K1 ∩ K2 = F (γ1 , ..., γs ), where, say, γs
is not contained in any proper subfield of K1 ∩ K2 . Let p(x) be the minimal polynomial
for γs over F . Then p(x) is irreducible, and γs ∈ K1 , so by the lemma, all the roots of
p(x) lie in K1 . Similarly, they all lie in K2 , so they are all in K1 ∩ K2 . Thus p(x) splits
completely over K1 ∩ K2 . Since γs is not contained in any proper subfield of K1 ∩ K2 ,
then K1 ∩ K2 is a splitting field for p(x) over F .
3. Find the splitting field F for x4 + x2 + 1 over Q.
We can factor x4 + x2 + 1 = (x2 + x + 1)(x2 − x + 1) = Φ3 (x)Φ6 (x) over Q. The roots
of these two polynomials are all 6th roots of unity, generated by ζ6 . Hence the splitting
field for x4 + x2 + 1 over Q is Q(ζ6 ), which has degree 2.
√
4. Let K = Q( 2, i), and let G = Gal(K/Q).
(a) Compute G.
√
Any automorphism σ of K is determined by
σ(
2) √
and σ(i), which must be roots
√
of x2 − 2 and x2 + 1, respectively. Thus σ( 2) = ± 2 and σ(i) = ±i. This yields
four possible automorphisms. Each can be shown to be a valid automorphism via the
General Conjugate
Extension
Theorem.
Therefore we
√
√Theorem and√the Isomorphism
√
√
√
have σ : 2 7→ − 2, i 7→ i, τ : 2 7→ 2, i 7→ −i, στ : 2 7→ − 2, i 7→ −i, and the
identity map. Since these each have order 2, we have G ∼
= C2 × C2 .
(b) Find all subgroups of G, and write each one in the form Gal(K/E) where Q ⊂ E ⊂ K.
We have the following subgroups of G:
{1} = Gal(K/Q)
hσi = Gal(K/Q(i))
√
hτ i = Gal(K/Q( 2))
√
hστ i = Gal(K/Q(i 2)).
5. Let K be a splitting field over a field F for a polynomial f (x) ∈ F [x]. Prove that
|Gal(K/F )| ≤ [K : F ]. When are the two equal? (Hint: recall that we proved the
uniqueness of splitting fields by taking an isomorphism σ : F → F 0 and lifting it to an
isomorphism σ : E → E 0 , where E/F is a splitting field for f (x) ∈ F [x] and E 0 /F 0 is a
splitting field for σ(f (x)) ∈ F 0 [x]. Use a similar proof by induction to count the number
of such lifts of σ, then specialize to the case F = F 0 and E = E 0 .)
We will prove the following generalization.
Theorem: Let σ : F → F 0 be an isomorphism, let E/F be the splitting field for f (x) ∈
F [x], and let E 0 be the splitting field for σ(f (x)) over F 0 . Then the number of lifts of σ
to an isomorphism φ : E → E 0 is at most [E : F ] with equality if f (x) is separable over
F.
We induct on [E : F ]. If [E : F ] = 1, then E = F , E 0 = F 0 , and φ = σ, hence there
is one lift. Now let n > 1 and suppose that the result holds when [E : F ] < n. Now let
[E : F ] = n. Then f (x) must have an irreducible factor p(x) of degree greater than 1. Let
α be a root of p(x). Then by the Isomorophism Extension Theorem, every isomorphism of
F (α) which restricts to σ is of the form τ : F (α) → F (β) where τ (α) = β and β is a root
of p0 (x) = σ(p(x)) ∈ F 0 [x]. So the number of such isomorphisms is equal to the number
of distinct roots of p0 (x), which is bounded above by deg p0 (x) = deg p(x) = [F (α) : F ],
with equality when p(x) is separable.
If f (x) splits in F (α) we are done, otherwise [E : F (α)] < [E : F ] = n. Then E is a
splitting field for f (x) over F (α) and E 0 is a splitting field for τ (f (x)) over F 0 (β), where
τ : F (α) → F 0 (β) is an isomorphism. We know by our previous theorem that τ extends
to an isomorphism φ : E → E 0 . We can apply our induction hypothesis to conclude that
the number of such extensions is at most [E : F (α)] with equality when f (x) is separable.
Therefore the total number of extensions of σ : F → F 0 to some φ : E → E 0 is bounded
above by [F (α) : F ][E : F (α)] = [E : F ] with equality when both p(x) and f (x) are
separable. Since p(x) | f (x), then we only need f (x) separable.
Finally, to answer the original question, let F 0 = F , let σ be the identity on F , and
let K = E 0 = E be a splitting field for f (x) ∈ F [x]. Then the number of lifts of σ to an
automorphism on K is bounded above by [K : F ] with equality when f (x) is separable.
Therfore |Gal(K/F )| ≤ [K : F ].
n
6. Let Fpn be the field of pn elements. Recall that Fpn is the splitting field over Fp of xp − x.
Prove that Gal(Fpn /Fp ) is cyclic of order n, generated by the Frobenius map σp : α 7→ αp .
n
First, we note that since Fpn is the splitting field for xp − x, which is separable of
degree pn over Fp , then the elements of Fpn are exactly the roots of this polynomial. Thus
n
every a ∈ Fpn satisfies ap = a. Now |Fpn | = pn and |Fp | = p. Since Fpn ⊃ Fp , it is
a vector space over Fp of degree n. Thus [Fpn : Fp ] = n, so by the previous problem,
Gal(Fpn /Fp ) ≤ n. Now we will show that σp ∈ Gal(Fpn /Fp ) and has order n. This will
show that Gal(Fpn /Fp ) must be cyclic of order n.
To see that σp is an automorphism of Fpn , we first show that σp maps into Fpn . Let
n
n
n
a ∈ Fpn . Then (ap )p = (ap )p = ap , so σ(a) = ap is a root of xp − x and thus is in
Fpn . Next, for a, b ∈ Fpn , we have σp (a + b) = (a + b)p = ap + bp = σp (a) + σp (b) by the
Freshman’s Dream. Also, σp (ab) = (ab)p = ap bp = σp (a)σp (b), so σp preserves operations.
Next, suppose that σp (a) = σp (b). Then ap = bp , so 0 = ap −bp = (a−b)p , hence a−b = 0,
so a = b. Thus σp is injective from Fpn to itself. Since Fpn is finite, then σp is a bijection.
Then if a ∈ Fp , then σp (a) = ap = a by Fermat’s Little Theorem, so σp fixes Fp , and thus
σp ∈ Gal(Fpn /Fp ).
n
Finally, for a ∈ Fpn , we have σpn (a) = ap = a, so σpn is the identity. Suppose the order
k
of σp in the Galois group is k < n. Then ap = a for all a ∈ Fpn . But this implies that
k
there are more than pk roots of xp − x. Therefore σp has order n.