# Intermediate field and Galois extension.

I have the following problem:

Let $$p(x)=(x^{12}-16)(x^2-3)$$. Show that $$K=\mathbb{Q}(\sqrt{2},\sqrt{3},i)$$ is the splitting field of $$p$$ over $$\mathbb{Q}$$, $$[K:\mathbb{Q}]=12$$ and show that exists a Galois extension $$E$$ of $$\mathbb{Q}$$ such that $$\mathbb{Q} and $$[E:\mathbb{Q}]=6.$$

I've done the first two parts already. For the third, I've tried to get explicitly the field $$E$$ but i haven't succeded. If you draw the diagram you have two options for $$E$$, $$\mathbb{Q}(\sqrt{2},i)$$ or $$\mathbb{Q}(\sqrt{2},\sqrt{3})$$. The problem is that I can't conclude with these two options.

What I'm missing?

Let $$E$$ be the Galois field $$\Bbb Q(\sqrt2,\zeta_3)\ ,$$ where $$\zeta_3$$ is a primitive root of unity, for instance $$\zeta_3=(-1+\sqrt{-3})/2$$. Note that $$\zeta_3$$ is an element of the given field $$K$$. This solves the remaining part.
($$\Bbb Q(\zeta_3)$$ is Galois over $$\Bbb Q$$, degree $$2$$, and further adjoining $$a=\sqrt2$$ to get $$E$$ also leads to a Galois extension, because the conjugates $$a\zeta_3^k$$, $$k=0,1,2$$, are in this extension $$E$$.)