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I working on a problem for practice. For $k$ a field, I was able to show that any element of $A=k[X,Y]/(XY)$ has a unique representation in form $a+f(X)X+g(Y)Y$ for $a\in k$, $f(X)\in k[X]$ and $g(Y)\in k[Y]$.

Why does $k[X,Y]/(XY)$ have exactly two minimal prime ideals? I know the primes of $k[X,Y]$ are precisely $(0)$, $(f)$ for $f$ irreducible, and the maximal ideals. Also, the primes of $A$ are those primes of $k[X,Y]$ containing $(XY)$. But how can one tell there are precisely two such minimal primes?

My hunch is something like $(X)$ and $(Y)$ are two minimal primes containing $(XY)$, and then the correspondence theorem comes in somehow, but I haven't gotten very far on this idea yet.

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Hint: any prime ideal contains either $(X)$ or $(Y)$, since $XY=0$.

Edit: Since you've made an effort, I will finish.

First, to see that $(X)$ and $(Y)$ are prime, use the third isomorphism theorem for rings, and the fact that $k[X,Y]/(X) \cong k[Y]$ (make sure this is obvious to you). If you're going to use the fact that $X$ and $Y$ are irreducible, you have to prove this somehow. Sometimes in quotient rings, elements can look irreducible without being so. Using the isomorphism theorems is really the way to go.

Now $(X)$ and $(Y)$ are the only minimal primes since any prime must contain one of them.

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    Oh right, the kernel is just $(X)$, and map is surjective onto $k[Y]$.2012-03-06