For $x+y \in \mathbb{Q}[x,y]$, show that $\langle x+y\rangle$ is maximal among all principal ideals and show that it is not maximal among all ideals

Question

This is my approach:

So, I know that $\langle x+y\rangle$ consists of all polynomials with $x+y$ as a factor in $\mathbb{Q}[x,y]$. I believe it must be maximal among all principal ideals because $x+y$ is prime in $\mathbb{Q}[x,y]$, thus rendering $\langle x+y\rangle$ a prime ideal meaning it must be maximal. I know my logic is flawed, but I am not sure where to go from here. Furthermore, I am not too sure how to show that it is not maximal among all other ideals.

Any help would be appreciated.

Answer 1

$I=\left<x+y\right>$ is maximal amongst proper principal ideals, as $x+y$ is irreducible. A principal ideal strictly containing $I$ would be generated by a proper factor of $x+y$, and as $x+y$ is irreducible, this ideal could only be the whole ring.

Geometrically, the ideal $\left<x+y\right>$ corresponds to the line $x+y=0$ in the affine plane. Maximal ideals correspond to points in the affine plane (well, at least if the ground field is algebraically closed). So pick a point on this line, say $(0,0)$ and let $J$ be the ideal of polynomials vanishing there, so $J=\left<x,y\right>$. Then $J$ is a proper ideal strictly containing $I$.

Answer 2

Let $\,D=\Bbb Q[y].\,$ $\,(x\!+\!y) \subseteq D[x]\,$ is prime (thus $\rm\color{#c00}{irred}$) since $\,Q := D[x]/(x\!+\!y) \cong D\,$ is a domain.

So $\,(z) \supsetneq (x\!+\!y) \iff z\mid x\!+\!y\,$ properly $\color{#c00}\iff z\mid 1\iff (z) = (1),\,$ by contains = divides

And $(x+y)$ is not maximal since its quotient $\,Q \cong \Bbb Q[y]\,$ is not a field.