MA 3419: Galois theory
Selected answers/solutions to the assignment due November 3, 2015
1. We have α = β3 − β, and α3 − α − 1 = 0. Substituting the expression for α, we obtain
− β)3 − (β3 − β) − 1 = 0, or β9 − 3β7 + 3β5 − 2β3 + β − 1 = 0.
An alternative solution (which is better for more general problems): let α1 , α2 , α3 be all
the three roots of x3 − x − 1. We consider the polynomial
(β3
(x3 − x − α1 )(x3 − x − α2 )(x3 − x − α3 ),
expand it, and use the Vieta theorem according to which
α1 + α2 + α3 = 0,
α1 α2 + α1 α3 + α2 α3 = −1,
α1 α2 α3 = 1.
This gives the polynomial above.
2. By Tower Law, we have
m[k(α, β) : k(α)] = [k(α, β) : k(α)][k(α) : k] = [k(α, β) : k] =
= [k(α, β) : k(β)][k(β) : k] = [k(α, β) : k(β)]n,
√
√
and the statement follows. For α = 3 2 and β = ω 3 2 we have [k(α) : k] = [k(β) : k] = 3,
but [k(α, β) : k(α)] = 2. (We have Q(α, β) = Q(α, ω), the minimal polynomial of ω over Q
is x2 + x + 1, and this polynomial
has no √
real roots so it cannot split in Q(α)).
√
√
4
4
4
3. The roots
of x −2 are ± 2 and ±i 2, so the field generated by those roots if Q( 4 2, i).
√
Note that [Q( 4 2) : Q] = 4√since x4 −√
2 is irreducible by Eisenstein, and this extension
contains
√
4
4
4
only real numbers, so [Q( 2, i) : Q( √
2)] =
and hence
√ 2, √
√ by
√ Tower
√ Law [Q( 2, i) : Q] = 8. As
a basis we can take the elements 1, 4 2, 4 4, 4 8, i, i 4 2, i 4 4, i 4 8; these elements manifestly
form a spanning set, and the degree computation shows that they are linearly independent.
4. That splitting field is generated by any primitive 12-th root of 1. Note that
x12 − 1 = (x6 − 1)(x6 + 1) = (x6 − 1)(x2 + 1)(x4 − x2 + 1). From this formula, we see
that the primitive roots of 1 of degree 12 are roots of x4 − x2 + 1. If we denote e2πi/12 = ξ,
then the primitive roots are ξ, ξ5 , ξ7 , and ξ11 . Moreover, this polynomial can easily be
shown to be irreducible over Q, as it has no rational roots, and cannot be factorised as a
product of two quadratic polynomials with integer coefficients (to ensure that their constant
terms are integers, one must have roots ξ and ξ11 , and the other ξ5 and ξ7 , but then the
sum ξ + ξ11 = 2 cos(2π/11) is not an integer. Thus, [Q(ξ) : Q] = 4. For a basis, we can take
1, ξ, ξ2 , ξ5 . Indeed, ξ3 = ξ + ξ5 , ξ4 = ξ2 − 1, and ξ6+k = −ξk for 0 6 k 6 5.
Also, note that the splitting field of (x4 − 1)(x3 − 1) is Q(i, ω), and iω is a primitive 12-th
7
root of 1: 2πi(1/4 + 1/3) = 2πi 12
, so iω = ξ7 .
5. Let p be a prime number, let k be a field, and let a ∈ k.
(a) We may assume a 6= 0 since otherwise the claim is trivial. In the splitting field of this
√
polynomial the roots of xp − a are of the form p aξ, where ξ is a p-th root of 1 (these may be
distinct or not, depending on char(k). Thus, if xp − a is not irreducible, and factorizes over k
√ d
as f(x)g(x), the constant term of f(x) is p a ζ, where 0 < d < p is the degree of f(x), and ζ is
√ d
√ 1−py x √
a p-th root of 1. We have dx+py = 1 for some x, y ∈ Z, so ( p a ζ)x = p a
ζ = p aa−y ζx
√
is an element of k, and therefore p aζx is an element of k, that is xp − a has a root in k.
(b) If a is a p-th power in k, then it certainly is a p-th power in K. Suppose that a is a
p-th power in K, but it is not a p-th power in k. In that case, xp − a is irreducible over k,
√
so K contains the extension k( p a) of degree p, and by Tower Law [K : k] is divisible by p, a
contradiction.
6. (a) Since C = R(i), any R-automorphism of C is determined by the image of i which
2 + 1 = 0. Thus, the group is Z/2Z.
must be ±i, since i√
√
√
√
(b) Note that √3 ∈
/ Q( 2), for otherwise √
we have p
3 = a + b 2 with a, b ∈ Q, so
2
3 =√a2 +
√2b + 2ab
√ 2. This implies ab =√0, so
√ 3 ∈ Q or 3/2 ∈ Q, a contradiction. Thus,
[Q(
2,
3)
:
Q(
2)]
=
2,
and
hence
[Q(
2,
3) : Q] = 4 by Tower Law. This implies that
√ √ √
√ √
1, 2, 3, 6 form
√ a basis√ of Q( 2, 3) over Q, and every Q-automorphism is determined
by the images of 2 and 3, each of which can remain fixed or change the sign. Hence, the
group is (Z/2Z)2 .
7. Yes, since F8 is normal (it is the splitting field of x8 − x over F2 ) and separable (since
every element of a finite field of characteristic p is a p-th power, so a result from class applies).
The Galois group of this extension is cyclic of order 3, generated by the automorphism x 7→ x2 .
(The degree of the extension is 3, hence the group is cyclic of order 3, hence we just need to
find one nontrivial automorphism to generate it).
8. As we saw above, the degree of the splitting field is 4, so the Galois group is of order 4; it
has the elements σ1 (ξ) = ξ5 , σ2 (ξ) = ξ7 , and σ3 (ξ) = ξ11 . We have σ21 (ξ) = (ξ5 )5 = ξ25 = ξ,
so σ21 = e. Similarly, σ22 (ξ) = (ξ7 )7 = ξ49 = ξ, and σ23 (ξ) = ξ121 = ξ. Hence, the square of
each element is the identity element, so the group is (Z/2Z)2 .