Is $(X^2+Y,Y^2-2)$ maximal in $\mathbb{Q}[X,Y]$? I'm trying to determine whether the generated ideal $(X^2+Y,Y^2-2)$ is maximal in $\mathbb{Q}[X,Y]$.
I take nonzero $f\in \mathbb{Q}[X,Y]/(X^2+Y,Y^2-2)$. In this quotient, I think $Y=-X^2$, and $Y^2=2$, and $X^4=(-Y)^2=2$. So I think $f$ can be viewed as having form $a_0+a_1X+a_2X^2+a_3X^3$. Then $f$ is nonzero in the quotient $\mathbb{Q}[X]/(X^4-2)$, which is a field since $X^4-2$ is irreducible. So there exists $g\in\mathbb{Q}[X]/(X^4-2)$ such that $fg=1$ in $\mathbb{Q}[X]/(X^4-2)$, and hence also in $\mathbb{Q}[X,Y]/(X^2+Y,Y^2-2)$?
I'm not sure exactly what I'm doing since I haven't dealt with ideals of polynomial rings in multiple variables before. Is there a less handwavey approach to this?
 A: I think your work is good, but I might take a few points off if I were grading this on a homework because I don't think you justified entirely why
$$\mathbb{Q}[X,Y]/(X^2+Y,Y^2-2)\cong\mathbb{Q}[X]/(X^4-2).$$
I suppose I'm not convinced (by reading your argument) that there may not have been other relations between $X$ and $Y$ that were forgotten after you passed to $\mathbb{Q}[X]/(X^4-2)$.
Here is how I would approach the problem.
Note that $\mathbb{Q}[Y]/(Y^2-2)\cong\mathbb{Q}(\sqrt{2})$, with the equivalence class $Y+(Y^2-2)$ corresponding to the number $\sqrt{2}$ (or to $-\sqrt{2}$, it's ultimately the same). Therefore
$$\mathbb{Q}[X,Y]/(X^2+Y,Y^2-2)\cong\mathbb{Q}(\sqrt{2})[X]/(X^2+\sqrt{2}).$$
Therefore
$$\mathbb{Q}[X,Y]/(X^2+Y,Y^2-2)\,\text{ is a field}\iff \mathbb{Q}(\sqrt{2})[X]/(X^2+\sqrt{2})\,\text{ is a field}$$
and therefore
$$(X^2+Y,Y^2-2)\,\text{ is maximal in }\mathbb{Q}[X,Y]\iff (X^2+\sqrt{2})\,\text{ is maximal in }\mathbb{Q}(\sqrt{2})[X]$$
Observe that $\mathbb{Q}(\sqrt{2})$ is a field, hence $\mathbb{Q}(\sqrt{2})[X]$ is a principal ideal domain (PID). Therefore a non-zero ideal $(f)\subset\mathbb{Q}(\sqrt{2})[X]$ is a maximal ideal if and only if $f$ is irreducible in $\mathbb{Q}(\sqrt{2})[X]$.
Suppose that $X^2+\sqrt{2}$ factored in $\mathbb{Q}(\sqrt{2})[X]$. Its factors would have to be of degree $1$, since it itself is only of degree $2$. But a degree $1$ polynomial $X-\alpha\in\mathbb{Q}(\sqrt{2})[X]$ is a factor of $X^2+\sqrt{2}$ if and only if $\alpha^2+\sqrt{2}=0$, and there are no such $\alpha\in\mathbb{Q}(\sqrt{2})$ (observe that $\alpha$ would have to be imaginary, while $\mathbb{Q}(\sqrt{2})\subset\mathbb{R}$).
Note that if we had chosen $Y+(Y^2-2)$ to map to $-\sqrt{2}$, we'd have to argue why $X^2-\sqrt{2}$ was irreducible in $\mathbb{Q}(\sqrt{2})[X]$. Of course this is still true, but the argument would be more of a pain unless you're familiar with Eisenstein's criterion.
A: Your intuition is  correct: $$\mathbb Q[X,Y]/(Y+X^2)\simeq \mathbb Q[X]$$ by $Y\mapsto-X^2$, and then factorize the last ring through $X^4-2$ (which corresponds to $Y^2-2$ in the above isomorphism). Thus you get $$\mathbb{Q}[X,Y]/(Y+X^2,Y^2-2)\simeq \mathbb Q[X]/(X^4-2).$$ Since $X^4-2$ is irreducible over $\mathbb Q$ (by elementary arguments or by Eisenstein) the quotient ring $\mathbb Q[X]/(X^4-2)$ is a field.
A: Note $\,\ I = (y^2\!-\!2,y\!+\!x^2) = (x^4\!-2,y\!+\!x^2)\,\ $ by $\ {\rm mod}\,\ y\!+\!x^2\!:\,\  y\equiv -x^2\,\Rightarrow\, y^2\!-2\equiv x^4\!-2.$
Hence $\ \Bbb Q[x,y]/I \,\cong\, \Bbb Q[x,y]/(x^4\!-2,y\!+\!x^2)\,\cong\, \Bbb Q[x]/(x^4\!-2)\ $ by $\,R[y]/(y\!-\!r)\cong R.\ \ $ QED
