# Example of prime, not maximal ideal

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.

• $(x,y)$ is a example of an ideal which contains yours. $(x^2+y^2)$, on the other hand, is not. Indeed, as you observe, $x^2-y$ is irreducible, so the ideal $(x^2-y)$ is not contained non-trivially in any principal ideal. Oct 18, 2011 at 18:10
• When you say primary, do you mean prime? Showing that $x^2 - y$ is irreducible (I assume we are working over a field) would be enough to show that the ideal is prime, but you'd have to prove that. And you would still have to show that it isn't maximal, somehow. Oct 18, 2011 at 18:11
• You should edit your question and specify which ring you are working in. $\mathbb{Z}[x,y]$? $\mathbb{Q}[x,y]$? ... Oct 18, 2011 at 18:17
• @TegiriNenashi Over $\mathbf{C}$ there are far more solutions to $x^2 + y^2 = 0$ than the one you've written down; $(1, i)$ is one example. Oct 18, 2011 at 18:32
• Perhaps the time has come to gather these comments into an answer? Oct 19, 2011 at 2:52

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 $P(\zeta_1,\zeta_2)=0$.
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 $(x,y) \subset \mathbb{C}[x,y]$.
• But this is not quite the definition of prime ideal. I mean ideal generated by an irreducible element may not be a prime ideal, for example in $Z[\sqrt{-5}] \,\, (3)$ is not a prime ideal since, $(1+\sqrt{-5}).(1-\sqrt{-5}) = 6 \in (3)$ but neither $1+\sqrt{-5}$ nor $1-\sqrt{-5}$ are in $(3)$. Can you prove to me in this fashion: Show that if $a.b \in (x^2+y)$ then either $a \in (x^2+y)$ or $b \in (x^2+y)$. Jan 1, 2017 at 12:24