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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.

Martin Sleziak
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Tegiri Nenashi
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    $(x,y)$ 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. – Mariano Suárez-Álvarez Oct 18 '11 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. – Dylan Moreland Oct 18 '11 at 18:11
  • @Mariano: Thank you: so it is not just sufficient to check if one variety -- ${(x=0,y=0)}$ -- is subset of the other! – Tegiri Nenashi Oct 18 '11 at 18:17
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    You should edit your question and specify which ring you are working in. $\mathbb{Z}[x,y]$? $\mathbb{Q}[x,y]$? ... – Bill Cook Oct 18 '11 at 18:17
  • @TegiriNenashi If you have the setting and theorems of Hartshorne Ch 1, then things become much easier. From these it follows that for $x^2 - y$ to be maximal it would have to have a zero set consisting of one point. – Dylan Moreland Oct 18 '11 at 18:19
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    @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. – Dylan Moreland Oct 18 '11 at 18:32
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    Perhaps the time has come to gather these comments into an answer? – Gerry Myerson Oct 19 '11 at 02:52
  • Let $R=\mathbb{C}[x,y]$, let $(x)$ be an ideal. It is prime since $R/(x)$ is a domain, and it is not maximal since $R/(x)=\mathbb{C}[y]$ is not a field. – José Alejandro Aburto Araneda Jan 02 '19 at 00:45

2 Answers2

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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 $P(\zeta_1,\zeta_2)=0$.

Andrea Mori
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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 $(x,y) \subset \mathbb{C}[x,y]$.

Sergey Filkin
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    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 $b \in (x^2+y)$. – Pranav Bisht Jan 01 '17 at 12:24
  • In any UFD a principal ideal is prime iff it is generated by an irreducible element. –  May 05 '21 at 16:45