Math 121 Homework 5 Solutions
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Section 14.2, Problem 5. Prove that the Galoisgroup of
x − 2 over Q
a b
for p a prime is isomorphic to the group of matrices
for a, b ∈ Fp ,
0 1
a 6= 0.
Solution. The polynomial is irreducible by Eisenstein’s criterion. So its
splitting field is generated by the roots of xp − 2 which are αζ i (0 6 i < p)
where α = 21/p and ζ = e2πi/p . Thus the splitting field is Q(α, i). Note that
Q(ζ)/Q has degree φ(p) = p − 1, while Q(α)/Q has degree p. These degrees
are coprime, so [Q(α, i) : Q] = p(p − 1) by Corollary 22 on page 529.
Let m ∈ Z be prime to p. We will show that there exists τα ∈ Gal(K/Q)
such that
τm (α) = α
.
(1)
τm (ζ) = ζ m
Since K/Q is Galois, K/Q(α) is also Galois and has degree p − 1. Indeed,
K = Q(α, ζ) so the effect of θ ∈ Gal(K/Q(α)) is determined by θ(ζ) which
must be a primitive p-th root of unity. There are p − 1 elements of the Galois
group and exactly p − 1 such p-th roots of unity, so Gal(K/Q(α)) must be
transitive on these. Therefore we can find τm as required.
Now we will show that Gal(K/Q) contains an automorphism σ such that
σ(α) = αζ,
σ(ζ) = ζ.
(2)
Indeed, α and αζ are roots of the same irreducible polynomial xp − 2,
so we may find an automorphism σ1 such that σ1 (α) = αζ. Now σ1 (ζ) is a
primitive p-th root of unity, say σ1 (ζ) = ζ k , p - k. Then τm (ζ k ) = ζ for some
α. (So if k is the image of k in F×
p , m = 1/k.) Then τm σ1 has the desired
effect (2).
1
km
m
The elements τm σ k and σ km τm have the same effect, α 7→ αζ
, ζ 7→
ζ .
a b
−1
= σ km . now the “affine group” G of matrices
for
So τm σ k τm
0 1
a, b ∈ Fp , a 6= 0 also has generators
m
1 1
1 k
k
tm =
,
s=
,
so
s =
.
1
1
1
km
. Therefore there is an isomorsubject to the same relations tm sk t−1
m = s
phism Gal(K/Q) → G in which τm 7→ tm and σ 7→ s.
√ Section 14.2, Problem 6. Let K = Q 8 2, i and let F1 = Q(i),
√ √ F2 = Q 2 , F3 = Q −2 . Prove that Gal(K/F1 ) ∼
= Z8 , Gal(K/F2 ) ∼
= D8
∼
and Gal(K/F3 ) = Q8 .
Solution. First let us√observe that K is Galois and of degree 16 over
Q. To begin with, α = 8 2 is the root of the polynomial x2 − 2 which is
irreducible by Eisenstein’s criterion, so [Q(α) : Q] = 8. Now i ∈
/ Q(α) since
Q(α) ⊆ R. Therefore [Q(α, i) : Q] = 2 and so [Q(α, i) : Q] = [Q(α, i) :
Q(α)][Q(α) : Q] = 16. √
The field Q(α, i) contains the primitive 8-th roots of
unity since it contains 2 = α4 and hence ε8 = √12 (1 + i) = e2πi/8 . Hence it
contains all the roots of
8
x −2=
7
Y
(x − αεk8 ).
k=0
It is thus a splitting field for this polynomial, and Galois over Q.
Lemma 1. Let β = αεk8 be any root of x8 − 2. Then there exist elements
σ, σ 0 ∈ Gal(K/Q) such that
σ(α) = β,
σ(i) = i,
and
σ 0 (α) = β,
σ 0 (i) = −i.
Proof. There exists φ : Q(α) −→ Q(β) such that α 7→ β by Theorem 8 on
page 519. By Theorem 27 on page 541, this extends to an automorphism of
the splitting field K.
We show that i ∈
/ Q(β). Otherwise φ−1 (i) ∈ Q(α) ⊆ R and φ−1 (i) is
a root of x2 + 1 = 0, that is, φ−1 (i) = ±i which is not real, so this is a
contradiction.
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Since Q(β) has degree 8 over Q and i ∈
/ Q(β), Q(β, i) has degree 16 over
Q and hence Q(β, i) = K. Since K is Galois over Q, it is Galois over Q(β)
and | Gal(K/Q(β))| = 2. Let ψ be the generator, so that ψ(β) = β and
ψ(i) = −i. Then φ and ψφ both take α to β, and one takes i to i, the other
i to −i. This gives us σ and σ 0 .
√ We may now handle
the
three
cases.
Each
of
Gal(K/Q(i)),
Gal
K/Q
2
√ and Gal K/Q −2 has order 8, which is the degree of the extension.
First, the Lemma gives us an element θ ∈ Gal(K/Q) such
√ = ε8 α
√that
θ(α)
and θ(i) = i. This is then in Gal(K/Q(i)). We find that θ 2 = − 2 since
√
√
2 = α4 and − 2 = (ε8 α)4 . Therefore
1
1
θ(ε8 ) = θ √ (1 + i) = √ (1 + i) = −ε8 .
2
− 2
We may now calculate that θ2 (α) = −ε28 α = −iα and θ2 (ε8 ) = ε8 so θ4 (α) =
−α and θ4 (i) = i. We see that θ has order 8 and therefore Gal(K/Q(i)) is
cyclic of order 8.
√ Next, let us show that Gal K/Q 2 ∼
= D8 . The Lemma gives us two
elements ρ, τ ∈ Gal(K/Q) such that
ρ(α) = iα,
ρ(i) = i,
and
τ (α) = α,
τ (i) = −i.
√
√ √
Since 2 = α4 , both of these fix 2 and so they are in Gal K/Q 2 . It
is easy to see that ρ has order 4 and τ has order 2, and τ ρτ −1 = ρ−1 , so they
generate a dihedral group of order 8. √ Finally, let us show that Gal K/Q −2 ∼
= Q8 . The Lemma gives us
two elements µ and ν such that
µ(α) = ε8 α
,
µ(i) = −i
and
ν(α) = ε−1
8 α
.
ν(i) = −i
We will show that µ, ν satisfy
νµν −1 = µ−1 .
√
−2 = iα4 , µ sends
These
relations define √
the quaternion group Q8 . Since
√
√
−2 to (−i)(ε8 α)4 = −2 and is in Gal K/Q −2 ; similarly ν is also
in this Galois group. Using ε8 = √1−2 (−1 + i) we also calculate
µ2 = ν 2 ,
µ4 = 1,
µ(ε8 ) = ν(ε8 ) = ε38 .
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Using this we calculate
µ2 (α) = −α
µ2 (i) = i
and ν has the same effect. Thus µ2 = ν 2 and µ, ν have order 4. The relation
νµν −1 = µ−1 may now be checked.
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√ Section 14.2, Problem 13. Show that Q
2 + 2 is a cyclic quartic
field, that is, it is Galois over Q of degree 4 with cyclic Galois group.
Consider the polynomial
q
q
q
q
√
√
√
√
x+ 2+ 2
x− 2− 2
x+ 2− 2 .
f (x) = x − 2 + 2
In every square root, the argument is a positive real number and as usual we
are taking the positive square root. This polynomial equals
√ 2 √ 2
x − 2+ 2
x − 2 − 2 = (x2 − 2)2 − 2 = x4 − 4x2 + 2.
p
p
√
√
2
+
2
and
β
=
2
−
2.
It is Eisenstein,
hence
irreducible.
Let
α
=
√
Then 2 = α2 − 2 ∈ Q(α), and since
r
q
q
√ √
√
√ √
2+ 2 2− 2 = 2
αβ = 2 + 2 · 2 − 2 =
we also have
√
2
β=
∈ Q(α).
α
The other two roots are −α, −β and so Q(α) is the splitting field of f . Since
f is separable Q(α) is Galois over Q. Its degree is 4, so the Galois group has
degree 4.
Now to compute the Galois group, Theorem 8 on page 519 guarantees
that
is an isomorphism σ : Q(α)
√ there
√ −→ Q(β) such that σ(α) =√β. Then
σ 2 = σ(α2 − 2) = β 2 − 2 = − 2. Remembering that β = 2/α we
√ √
then have σ(β) = σ 2 /σ(α) = − 2/β = −α. This σ permutes the roots
cyclicly:
α → β → −α → −β → α.
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This proves that the Galois group is cyclic of order 4.
Section 14.3, Problem 5. Exhibit an explicit isomorphism between
the splitting fields of x3 − x + 1 and x3 − x − 1 over F3 .
Solution. If x3 − x + 1 = 0 then y 3 − y − 1 = 0 where y = −x.
Section 14.4, Problem 8. Determine the splitting field of the polynomial xp − x − a where a ∈ Fp , a 6= 0.
Solution. The polynomial is irreducible and separable by Problem 5 in
Section 13.5, which was assigned in Homework 3. It has degree p, so every
root must lie in the unique field with pp elements. Thus Fpp is the splitting
field.
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