Short proof of $\mathbb{Q}[x,y]/\langle x^2+1, y^4-2\rangle \equiv\mathbb{Q}[\sqrt[4]{2}, i]$

Question

I am looking for an indirect proof of $$E = \mathbb{Q}[x,y]/_{\langle x^2+1, y^4-2\rangle}\cong\mathbb{Q}[\sqrt[4]{2}, i],$$ much preferably using module homomorphism theorems.

To be more specific, above problem was part of a multi-part, algebra qual problem determining Galois group of $x^4-2$ over $\mathbb{Q}.$ I can establish the above equivalence by identifying an explicit isomorphism $\phi$ defined by: $\phi(X)=i$ and $\phi(Y) = \sqrt[4]{2}$ by manually checking that this is indeed an isomorphism, where $X,Y$ are the equivalence classes of $x,y$ in $E.$

But I feel like there must be some slick ring/module theory proof using homomorphism theorems and maybe other theorems about prime/maximal ideals.

Answer 1

Looks like you got it all right. I might organize the details as follows:

1. Given a field $K$, an extension field $L$, and a point $(a,b)\in L^2$, evaluating a polynomial gives a homomorphism of rings $\phi:K[x,y]\to L, f(x,y)\mapsto f(a,b)$.
2. By the first isomorphism theorem we get an isomorphism of rings $$K[x,y]/\operatorname{Ker}(\phi)\simeq \operatorname{Im}(\phi).$$
3. We apply this to $K=\Bbb{Q}$, $L=K(i,\root4\of2)$, $a=i,b=\root4\of2$.
4. As an integral domain, f.d. over $K$, we see that $\operatorname{Im}(\phi)$ is a field. Because it contains $i$ and $\root4\of2$ it must be all of $L$. In this step we can also appeal to $L/K$ being algebraic, also allowing us to conclude that $\phi$ is surjective.
5. By Eisenstein $[K(\root4\of2):K]=4$. Because $K(\root4\of2)$ is a real field $[L:K(\root4\of2)]=2$. Therefore $[L:K]=8$.
6. Clearly $x^2+1$ and $y^4-2$ are both in $\operatorname{Ker}(\phi)$. Denote the ideal of $K[x,y]$ they generate by $I$. It is easy to see that $K[x,y]/I$ is spanned, as a vector space over $K$, by the cosets of $x^my^n, 0\le m<2, 0\le n<4$. Replace higher powers of $x$ or $y$ in the obvious way.
7. We know that $K[x,y]/\operatorname{Ker}(\phi)$ is 8-dimensional, and that $I\subseteq \operatorname{Ker}(\phi)$, so there is no wiggle room. We must have $I=\operatorname{Ker}(\phi)$.

Nothing much to it. Variations are possible, my habit is to reduce it to first principles.

Answer 2

In short, we will show that the left side is a field and can then check directly that both fields are a splitting field of the same polynomial.

Consider the proof of the existence of a splitting filed. Start with $F_0=F$ and $p(x) \in F[x]$. Then, take one of the irreducible polynomials that divides it (say $f_1(x)$) and then make the field extension $F_1 = F[x]/(f_1(x))/F$. The polynomial will have at least one linear factor, and you can work with a lower degree polynomial. Then you proceed by induction. If we do this explicitly for the polynomial $(t^2+1)(t^4-2) \in \Bbb{Q}[t]$ one way of going about this is to get $F_1 = \Bbb{Q}[x]/(x^2+1)$ and $F_2 = F_1[y]/(y^4-2)$ (Note: In the first one, the ideal is generated in $\Bbb{Q}$ and in the second one, the ideal is generated in $F_1$). This ring will be isomorphic to $\Bbb{Q}[x,y]/(x^2+1,y^4-2)$ meaning that it is a field. You can then check directly that both $\Bbb{Q}[x,y]/(x^2+1,y^4-2)$ and $\mathbb{Q}(\sqrt[4]{2}, i)$ are splitting fields of $(t^2+1)(t^4-2)$ meaning that they are isomorphic (For the left side, you will get that the roots are $\pm X, \pm Y$, and $\pm XY$ and that when you adjoin them, you get the whole field).