How do I find all prime ideals of the localization $S^{-1}\Bbb{C}[X,Y]$ where $S=\{X^iY^j:i,j\in \Bbb{N}\}$?

Question

Let $S=\{X^iY^j:i,j\in \Bbb{N}\}$ and $R=\Bbb{C}[X,Y]$. I need to find all prime ideals of $S^{-1}R$. My idea was the following:

We know that the prime ideals of $S^{-1}R$ correspond uniqly to prime ideals of $R$ disjoint from $S$. We also know that the only prime ideals of $R$ are

1. $(0)$
2. $(P)$ where $P$ is irreducible in $R$
3. $(X+a,Y+b)$ for $a,b\in \Bbb{C}$

Now we only need to check which of them does not intersect $S$.

1. $(0)\cap S=\emptyset$ since $0\notin S$. Hence $(0)S^{-1}R=(0)$ is a prime ideal os $S^{-1}R$.
2. Let $P$ be irreducible in $R$. Here I have some problems since I don't understand how $P$ looks like.
3. $(X+a,Y+b)\cap S\neq \emptyset$ iff $a=0$ or $b=0$. So $(X+a,Y+b)S^{-1}R$ is a prime ideal if $a,b\neq 0$.

Now my first question is, is this correct so? And my second one is, how do I work out point $2$.

Thanks for your help

In my opinion with $(P)$ never intersects $S$ trivially but I'm not sure.

Answer 1

What you've written is all correct! For point 2, here's the best way to approach it. Suppose $(P)\cap S\neq\varnothing$; say $X^iY^j\in (P)$. This means in other words that $P$ divides $X^iY^j$. But $\mathbb{C}[X,Y]$ is a UFD, and so – since $P$ is irreducible – this means that $P$ is associate to one of the irreducible factors of $X^iY^j$. Up to units, the only such factors are $X$ and $Y$, so we conclude that either $(P)=(X)$ or $(P)=(Y)$. All other principal ideals of $\mathbb{C}[X,Y]$ are thus disjoint from $S$.