4

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.

Hailie Mathieson
  • 1,950

1 Answers1

6

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.

Bruno Joyal
  • 54,711
  • Hey thanks Bruno. So for any prime $P$ in $A$, $0=XY\in P$, so either $X\in P$ or $Y\in P$. So there are at most two minimal primes $(X)$ and $(Y)$ in $A$. Now $(X)$ and $(Y)$ are primes in $k[X,Y]$, since $X$ and $Y$ are irreducible. Since $(X)\supset (XY)$, $(X)$ is a prime in $A$, and likewise for $(Y)$. So these are the two minimal primes? – Hailie Mathieson Mar 06 '12 at 02:12
  • @hmIII You're welcome! I added more meat to my answer. – Bruno Joyal Mar 06 '12 at 02:17
  • 1
    Where is the third isomorphism theorem used? If $k[X,Y]/(X)\cong k[Y]$, then $k[X,Y]/(X)$ is an integral domain since $k[Y]$ is, so $(X)$ is prime? – Hailie Mathieson Mar 06 '12 at 02:23
  • Exactly - $(k[X,Y]/(XY))/((X)/(XY)) \cong k[X,Y]/(X)$ – Bruno Joyal Mar 06 '12 at 02:26
  • Ah, thanks. By the way, is there good explanantion of why $k[X,Y]/(X)\cong k[Y]$? Intuitively it's clear since $X=0$ in the quotient, so you just have polynomials in $Y$. Is there a better justification using the isomorphism theorems or something? – Hailie Mathieson Mar 06 '12 at 02:31
  • Consider the map $k[X,Y] \to k[Y]$ which takes $f(x,y)$ to $f(0,y)$. It's a surjective ring homomorphism. What is its kernel? Now use the first isomorphism theorem. – Bruno Joyal Mar 06 '12 at 02:34
  • 1
    Oh right, the kernel is just $(X)$, and map is surjective onto $k[Y]$. – Hailie Mathieson Mar 06 '12 at 02:37