# Degree of $\mathbb{Q}(\sqrt[3]{2}, i, \sqrt{2})$ over $\mathbb{Q}$

As the question indicates, I am trying to find the degree of $$\mathbb{Q}(\sqrt[3]{2}, i, \sqrt{2})$$ over $$\mathbb{Q}$$. I am told that it is 12, but I don't understand why.

I have $$[\mathbb{Q}(\sqrt[3]{2}, i, \sqrt{2}):\mathbb{Q}(i,\sqrt{2})][\mathbb{Q}(i,\sqrt{2}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$$, where the last two factors are $$4$$ and $$2$$ respectively, meaning the total degree could not be $$12$$.

What am I getting wrong?

• How do you get the degree $4$? – G. Chiusole Aug 8 at 16:13
• It should be $2$, as the minimal polynomial for $i$ over $\mathbb{Q}$ is $x^2+1$. – ponchan Aug 8 at 16:16
• Exactly. Is it clear now? – G. Chiusole Aug 8 at 16:17
• @ponchan Actually, it should be the minimal polynomial of $i$ over $\Bbb Q(\sqrt 3)$. But the polynomial is the same. – Arthur Aug 8 at 16:17
• $\sqrt3\notin\Bbb Q(\sqrt[3]2,i\sqrt2)$. – Lord Shark the Unknown Aug 8 at 16:22

It's easier to start with $$[\mathbb{Q}(i,\sqrt[3]{2},\sqrt{2}):\mathbb{Q}(\sqrt[3]{2},\sqrt{2})]$$ which is $$2$$ because the smaller field is contained in the real numbers, so $$x^2+1$$ is the minimal polynomial of $$i$$ over it.
Now note that $$x^3-2\in\mathbb{Q}(\sqrt{2})$$ has $$\sqrt[3]{2}$$ as root and the other two roots are nonreal. Thus it's reducible over $$\mathbb{Q}(\sqrt{2})$$ if and only if $$\sqrt[3]{2}\in\mathbb{Q}(\sqrt{2})$$.
I think it’s easiest of all not to treat the extension as a tower but as a parallelogram of fields. The generators are $$\sqrt2$$, $$i$$, and $$\sqrt[3]2$$, of degrees $$2$$, $$2$$, and $$3$$ over $$\Bbb Q$$ respectively.
It’s easily seen that the two quadratic irrationalities $$i$$ and $$\sqrt2$$ generate a quartic field, and you put this field and $$\Bbb Q(\sqrt[3]2\,)$$ at the two corners of the parallelogram adjacent to the $$\Bbb Q$$-corner, so that the full field has subfields of degrees $$4$$ and $$3$$. The smallest integer divisible by these is $$12$$, and since the generators have the degrees they do, the degree is no bigger than $$12$$, either.