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 $P=(Y)$.