Example of prime, not maximal ideal

Question

This is a homework from videolecture:

Show that $(x^2-y)$ is prime but not maximal in $\mathbb C[x,y]$".

Linked SE pages offer to approach this by proving that $\mathbb C[x,y]/(x^2-y)$ is an integral domain but not field. However, I feel that exhibiting an ideal strictly containing $(x^2-y)$ is easier, and $(x^2+y^2)$ seems to fit the bill. It also seems that proving that $(x^2-y)$ is prime directly is easier, because the polynomial $x^2-y$ can't be factored.

Answer 1

It may be worth saying that since $\Bbb C$ is algebraically closed, the maximal ideals in $\Bbb C[X,Y]$ are precisely the ideals of the form $(X-\zeta_1,Y-\zeta_2)$ where $\zeta_1$ and $\zeta_2$ are arbitrary complex numbers.

Moreover, the maximal ideal $(X-\zeta_1,Y-\zeta_2)$ contains the ideal $(P(X,Y))$ for a (not necessarily irreducible) polynomial $P(X,Y)$ if and only if $P(\zeta_1,\zeta_2)=0$.

Answer 2

An ideal $(x^2+y) \subset \mathbb{C}[x,y]$ is prime because a polynomial $x^2+y$ is irreducible. Indeed, if we assume that $x^2+y = a \cdot b,\ a,b \in \mathbb{C}[x,y]$, then if $y$-degree of $a$ is $1$, y-degree of $b$ must be $0$. If $x$-degree of $a$ is $2$, then $x$-degree of $b$ is $0$, and $b$ is just a complex number. Otherwise, if $x$-degree of $a$ is not $2$, then $a \cdot b$ contains $x^2y$ or $xy$ terms, that are not contained in $x^2+y$.

The ideal is not maximal, because it's contained in an ideal $(x,y) \subset \mathbb{C}[x,y]$.