Prime ideals in $\mathbb{C}[x,y]/\langle x^2-y^2\rangle$.

Question

I want to find all the prime ideals in the quotient ring $R:=\mathbb{C}[x,y]/\langle x^2-y^2\rangle$.

Should I start trying to show that $R \simeq \mathbb{C}[x]$ or something, and then trying to find prime ideals here?

I've read some similar examples, but couldn't find out any sort of "trick" to solve this kind of problem.

Is there a general way to find prime ideals in polynomial quotient rings?

On the second part of the question, I need to determine which of them are minimal ideals. I've shown the existence of minimal ideals using Zorn's Lemma, but it doesn't give me any clue of how to find them. So the question is:

How can I describe minimal ideals?

Thanks in advance!

Answer 1

The prime ideals in $\mathbf C[x,y]/(x^2-y^2)$ bijectively corresponds to the prime ideals of $\mathbf C[x,y]$ which contain $x^2-y^2=(x-y)(x+y)$, i.e. which contain either $x-y$ or $x+y$.

$(x-y)/(x^2-y^2)$ and $(x+y)/(x^2-y^2)$ are prime ideals of $\mathbf C[x,y]/(x^2-y^2)$, since by the 3rd isomorphism theorem, the quotients are isomorphic to $\mathbf C[x,y]/(x-y)$ and $\mathbf C[x,y]/(x+y)$ respectively, and both are isomorphic to the integral domain $\mathbf C[t]$.

Thus, (the images of) $(x-y)$ and $(x+y)$ are the minimal prime ideals of $\mathbf C[x,y]/(x^2-y^2)$.

The non-minimal prime ideals are necessarily maximal, since every non-zero prime ideal in the P.I.D. $\mathbf C[t]$ is maximal, generated by an irreducible element. As $\mathbf C$ is algebraically closed these prime ideals have the form $(t-\alpha), \quad\alpha\in \mathbf C$. It is easy to deduce these maximal ideals are $(x-\alpha, y-\alpha)\;$ and $\;(x-\alpha, y+\alpha),$ $\;\alpha\in \mathbf C$.

Note this is in accordance with Hilbert's Nullstellensatz.

Answer 2

Hint: the general problem is a difficult one, so let's look at your specific problem: in your ring $R = \Bbb{C}[x, y]\langle x^2 - y^2\rangle$, let me write $x$ for its image in $R$ and likewise for $y$. So that in $R$, $x^2 - y^2 = (x - y)(x + y) = 0$, and both $x - y$ and $x + y$ are zero-divisors.. If $P$ is a prime ideal of $R$, $R/P$ is an integral domain, i.e., it has no non-trivial zero divisors. So if $P$ is a prime ideal of $.R$, either $x - y$ or $x + y$ must belong to $P$. This means that the natural homomorphism from $R$ to $R/P$ must factor through a homomorphism of $R$ onto $\Bbb{C}[z]$ that either maps both $x$ and $y$ to $z$ or maps $x$ to $z$ and $y$ to $-z$. Hence the prime ideals of $R$ are the inverse images of prime ideals of $\Bbb{C}[z]$ under one of these two homomorphisms.