I have the following problem:

Let $p(x)=(x^{12}-16)(x^2-3)$. Show that $K=\mathbb{Q}(\sqrt[3]{2},\sqrt{3},i)$ 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 $\mathbb{Q}<E<K$ 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 $E$, $\mathbb{Q}(\sqrt[3]{2},i)$ or $\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$. The problem is that I can't conclude with these two options.

What I'm missing?

Thanks in advance and sorry for my bad english.


1 Answer 1


Let $E$ be the Galois field $$ \Bbb Q(\sqrt[3]2,\zeta_3)\ , $$ 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 $a=\sqrt[3]2$ to get $E$ also leads to a Galois extension, because the conjugates $a\zeta_3^k$, $k=0,1,2$, are in this extension $E$.)


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