- #1

PsychonautQQ

- 784

- 10

## Homework Statement

Let f(x) = x^4 - 6x^2 - 2. Let K be a splitting field for this polynomial over Q, show that Gal(K:Q) is non-abelian of order 8.

## Homework Equations

## The Attempt at a Solution

So I calculated the roots of this polynomial, one root was r = (3+(11^1/2))^1/2, and the others were -r, ir and -ir where i=(-1)^1/2. Thus a Q(i,r) is a splitting field of f(x) over Q.

Now time to calculate the Galois group. The minimal polynomial of r over Q is f(x). I know this because it is a root of f(x) and f(x) is irreducible over Q by Eisenstein criteria with p=2. Thus there is an element of Gal(K:Q) that takes r to -r, and another that takes r to ir, and another that takes r to -ir, because it must take r to the other roots of it's minimal polynomial in Q.

The minimal polynomial of i over Q(r) is x^2+1, and so there is are elements of the Galois group that take i to i and i to -i.

We also know that [Q(r,i):Q] = [Q(r,i):Q(r)] * [Q(r):Q]. Since the minimal polynomial of i over Q(r) is x^2 + 1 we know that [Q(r,i):Q(r)] = 2 and thus [Q(r,i):Q] = 8.

Since Char(Q)=0 and Q(r,i) is a finite splitting field of this polynomial over Q, [Q(r,i):Q] is a galois extension and so Gal(Q(r,i):Q) = [Q(r,i):Q] = 8.

So any element in the Galois group can take r to one of four different places and i to 1 of 2 different places, and there is an element for each of these possibilities, and so these are the 8 elements in the Galois group. (is this logic shaky?)

Now, to show it's non abelian,. Let b the map that permutes the roots of the minimal polynomial of i over Q(r) and c be the map that takes r to ir, (which are both roots of its minimal polynomial over Q). Then bc(r) = b(ir) = -ir and cb(r) = c(r) = ir. Thus cb(r) does not equal bc(r) so this group is not abelian.