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

Question

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?

Answer 1

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 $\sqrt[3]{2}\in\mathbb{Q}(\sqrt{2})$.

Answer 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.