The Galois Group of f = x3 − 2 over Q
As noted in class, f is irreducible over Q by
Eisenstein and its splitting field is F = Q(ζ,
√
ζ = ei2π/3 = cos(2π/3) + i sin(2π/3) = − 21 + i 23 .
√
3
2), where
Since the irreducible polynomial of ζ over Q√is x2 + x + 1, Q(ζ) is a two-dimensional vector space over Q with
basis {1, ζ}. The irreducible polynomial of 3 2 over Q is f = x3 − 2, which is still irreducible over Q(ζ) since
√ √ 2
it has no root in that field. Thus F is a three-dimensional vector space over Q(ζ) with basis {1, 3 2, 3 2 }.
√ √
√ 2 √ 2
It follows then that F is a six-dimensional vector space over Q with basis {1, ζ, 3 2, 3 2ζ, 3 2 , 3 2 ζ}.
The elements of the Galois group G = AutQ F are vector space isomorphisms in addition to being field
isomorphisms. Thus they can be represented relative to a fixed basis
invertible matrices over Q. I
√ by √
discussed one element σ ∈ G with the property that σ(ζ) = ζ 2 and σ( 3 2) = 3 2. (I actually called this σ̂ in
class.) Using the fact that ζ 2 = −1 − ζ, we get σ in matrix form as


1 −1 0 0 0 0
0 −1 0 0 0 0 


0 0 1 −1 0 0 
.

(1)
Mσ = 

0 0 0 −1 0 0 
0 0 0 0 1 −1
0 0 0 0 0 −1
Verify from (1) that Mσ2 = I , hence σ 2 = id.
√
√
I began discussing τ ∈ G defined by taking τ ( 3 2) = 3 2ζ. I said we had two possible cases, one where
τ (ζ) = ζ and one where τ (ζ) = ζ 2 . Both cases actually lead to elements of G. If we use the case where
τ (ζ) = ζ, we get the matrix for τ as


1 0 0 0
0 0
0 1 0 0
0 0


0 0 0 −1 0 0

(2)
Mτ = 
0 0 1 −1 0 0 .


0 0 0 0 −1 1
0 0 0 0 −1 0
Verify from (2) that Mτ3 = I , hence τ 3 = id.
√
√
√ 2
If we label the roots of f as u1 =√3 2, u2 √
= 3 2ζ,√u3 = 3 2ζ
to the permutation
√ , we see that
√ τ corresponds
√
(123) ∈ S3 . What about σ? σ( 3 2) = 3 2, σ( 3 2ζ) = 3 2ζ 2 , and σ( 3 2ζ 2 ) = 3 2ζ—since ζ 3 = 1—so σ
corresponds to (23) ∈ S3 . With this information, we can write S3 = {id, σ, τ, τ 2 , στ, στ 2 }. But now we can
calculate the fixed fields of the various subgroups of S3 by using the matrix representations of σ and τ from
(1) and (2) and their products and then letting MATLAB, for example, do the hard work. For example, to
find hσi, we would solve (Mσ − I )x = 0.
Remark. We use the fact that all the subgroups of S3 are cyclic and that the fixed field of a cyclic group hµi
is simply the set of those elements in F that are fixed by µ. Also, don’t forget that ζ 2 = −1 − ζ.
√
√ 2
√
Case (hσi). hσi0 = h(23)i0 = {a0 + a2 3 2 + a4 3 2 | ai ∈ Q} = Q( 3 2).
Case (hτ i). hτ i0 = h(123)i0 = {a0 + a1 ζ | ai ∈ Q} = Q(ζ).
√ 2 √ 2
√
√
Case (hστ i). hστ i0 = h(13)i0 = {a0 + a3 3 2ζ + a4 ( 3 2 + 3 2 ζ) | ai ∈ Q} = Q( 3 2ζ).
√
√
√ 2
√ 2
Case (hστ 2 i). hστ 2 i0 = h(12)i0 = {a0 + a2 ( 3 2 + 3 2ζ) + a4 3 2 ζ | ai ∈ Q} = Q( 3 2 ζ).
√
√
√ 2
Remarks.
i. Since only hτ i is normal in S3 , none of the fields Q( 3 2), Q( 3 2ζ), and Q( 3 2 ζ) is Galois
over Q.
ii. There are lots of different ways of representing the intermediate fields given above. For example
√ 2
√
√ 2
√
√
√ 2
Q( 3 2 ζ) = Q( 3 2ζ 2 ) since 3 2 ζ = ( 3 2ζ 2 )2 and 3 2ζ 2 = ( 3 2 ζ)2 /2.
c R. Kubelka 2014