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.