2

Find Galois group of $Q=\mathbb{Q}\sqrt{(2+\sqrt2)(3+\sqrt3)}$.

I know the minimal polynomial is $f(x)=x^8 -24x^6+144x^4-288x^2+144$. It's irreducible over $\mathbb{Q}$, hence, $[Q:\mathbb Q]=8$.

Let $g(x^2)=f(x)$. It's splitting field is $K=\mathbb Q(2+\sqrt2)(3+\sqrt3)$. One can show that $K\simeq\mathbb{Q}(\sqrt2,\sqrt3)$. Obviously, $[K:\mathbb{Q}]=4$ and, since $K$ contains $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$, $Gal(K,\mathbb Q)\simeq \mathbb Z_2^2$.

Since $[Q:K]=2$, $Gal(Q,K)\simeq \mathbb Z_2$.

I suppose that $Gal(Q,\mathbb Q)\simeq \mathbb Z_2^3$ but I don't really know how to prove that (and not 100% sure if it's true).

Could you please help me?

Maxim Nikitin
  • 1,323
  • 1
    It's not immediately clear to me that $Q$ is even a Galois extension. Factoring modulo $5$ it looks the group has an element of order $4$. I would guess that it is not abelian. –  Jun 03 '20 at 22:37
  • There is an outline of the proof in Dummit and Foote section 14.2 exercise 27. – TAPLON Jun 03 '20 at 22:38
  • It is the quaternion group of order $8$, see https://math.stackexchange.com/questions/2298679/prove-that-mathbbq-sqrt2-sqrt23-sqrt3-is-galois-over-mathbb?rq=1 – GreginGre Jun 03 '20 at 23:33

0 Answers0