This is an old homework question in which I was unable to complete.

I feel like the following theorem may be of some help to complete the problem.

Find all prime ideals of the ring $\mathbb{Q}[X,Y] / (X^2 Y^2)$.

$\textbf{Theorem :}$ Let $R$ be a principal ideal domain and let $p \in R$ be nonzero. The the following are equivalent:

1) $p$ is irreducible,

2) $p$ is prime,

3) The ideal $p$ is maximal.

I know that $\mathbb{Q}[X,Y]$ is the polynomial ring of two indeterminates $X$ and $Y$. Also, that we are modding out by the idea $(X^2 Y^2)$. I am not sure what to do next?

Just some random thoughts:

Since $(X^2) \subseteq (X^2Y^2)$ then $(X) \subseteq (X^2Y^2)$. I think that when we mod out by some ideal we are "essentially" setting something equal to $0$, but what is equal to $0$? Should the First Isomorphism Theorem for Rings be used?

Will someone please be willing to work this problem out with a detailed explanation? I would like to understand this problem, and any problems in the future that involve anything similar.

Thank you for your help!!!


A prime ideal of $\mathbb Q[X,Y]/(X^2Y^2)$ is of the form $P/(X^2Y^2)$ with $P\subset\mathbb Q[X,Y]$ prime containing $X^2Y^2$. Then $X\in P$, or $Y\in P$. If $X\in P$ then $P/(X)$ is a prime ideal of $\mathbb Q[X,Y]/(X)\simeq\mathbb Q[Y]$, so $P/(X)=(f(Y))$ with $f\in\mathbb Q[Y]$ irreducible or $P/(X)=(0)$. Thus $P=(X,f(Y))$ or $P=(X)$. Similarly $P=(Y,g(X))$ with $g\in\mathbb Q[X]$ irreducible or $P=(Y)$.

  • $\begingroup$ So a prime ideal is either the ideal $(X,f(Y))$ OR $(Y,g(X))$ but not both? Are these two prime ideals equivalent? $\endgroup$ – mathamphetamines Feb 21 '16 at 8:33
  • $\begingroup$ @mathamphetamines What means "equivalent"? $\endgroup$ – user26857 Feb 21 '16 at 8:55
  • $\begingroup$ Are they equal? Or are there two different prime ideals? $\endgroup$ – mathamphetamines Feb 21 '16 at 8:55
  • 1
    $\begingroup$ In general, they are not equal, although $(X,Y)$ can be obtained by both procedures. $\endgroup$ – user26857 Feb 21 '16 at 8:56
  • $\begingroup$ Got it. Thank you very much for your detailed explanation. It is greatly appreciated. $\endgroup$ – mathamphetamines Feb 21 '16 at 8:57

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