Factoring $x^4-2$ over intermediate subfields

Question

Let $\alpha = 2^{1/4}$. Factor the polynomial $x^4-2$ into irreducible factors over each of the fields, $\mathbb{Q},\mathbb{Q}[\sqrt{2}],\mathbb{Q}[\sqrt{2},i],\mathbb{Q}[\alpha],\mathbb{Q}[\alpha,i]$.

I'm having some issues executing this the way my professor wants us to, I believe(emphasis on believe) that he wants us to consider intermediate fields between $\mathbb{Q}$ and $\mathbb{Q}[\alpha,i]$ and then write out the subgroup of the galois group that fixes elements in the intermediate field and use that subgroup to find the irreducible polynomial of a certain element. But it just isn't clicking. I know at least what a few of them should end up being. in $\mathbb{Q}[\alpha,i]$ the polynomial splits completely into $(x-\alpha)(x+\alpha)(x-i\alpha)(x+i\alpha)$, in $\mathbb{Q}[\sqrt{2}]$ the polynomial splits into $(x^2-\sqrt{2})(x^2+\sqrt{2})$,in $\mathbb{Q}[\alpha]$ the polynomial splits into $(x-\alpha)(x+\alpha)(x^2+\sqrt{2})$ but I got those answers based on pure intuition as opposed to some systematic method using galois theory. any help would be appreciated, thanks!

Answer 1

You know the full splitting of the polynomial is $$x^4-2=\left(x-2^{1/4}\right)\left(x+2^{1/4}\right)\left(x-i2^{1/4}\right)\left(x+i2^{1/4}\right)$$ Of course, that is the factorization in $\mathbb{Q}\left(2^{1/4},i\right)$, and the polynomial doesn't factor at all over $\mathbb{Q}$.

If we multiply the first two factors and the second two factors, we get

$$x^4-2=\left(x^2-\sqrt{2}\right)\left(x^2+\sqrt{2}\right)$$ Both of these are polynomials in $\mathbb{Q}(\sqrt{2})$, and you can see that they are irreducible.

In $\mathbb{Q}(2^{1/4})$ you can go one step further,

$$x^4-2=\left(x-2^{1/4}\right)\left(x+2^{1/4}\right)\left(x^2+\sqrt{2}\right)$$ but you can't factor the second part because $i2^{1/4}$ is not in $\mathbb{Q}(2^{1/4})$. (Notice, by the way, that $\sqrt{2}\in \mathbb{Q}(2^{1/4})$, so its still okay for the first quadratic factor to be in there.)

In $\mathbb{Q}(i,\sqrt{2})$, it factors to

$$x^4-2=\left(x^2-\sqrt{2}\right)\left(x^2+\sqrt{2}\right)$$ because $2^{1/4}$ is not in $\mathbb{Q}(i,\sqrt{2})$ and any product among the four factors aside from this will leave you with a stray $2^{1/4}$. It's okay that this is the same as the first example; though the factorization is the same, the Galois group structure will be affected by the difference in subfields once you get to computing that.

As far as a purely systematic method goes, I think you will find that there isn't one for the general case. Usually with problems like this it's best to use known factorizations to break the polynomial down to bare bones and then reconstruct it gradually.