Showing that $(\mathbb{C}[x,y]/(xy))_x\cong \mathbb{C}[x]_x$ I'd like to understand rigorously why $(\mathbb{C}[x,y]/(xy))_x\cong \mathbb{C}[x]_x$.
Intuitively, it seems reasonable. However I'd like to know whether the kernel of the map I've constructed for the isomorphism is correct.
We can start with a map $f:\mathbb{C}[x,y]\rightarrow \mathbb{C}[x]_x$ mapping $f(x,y) \mapsto f(x,0)\mapsto f(x,0)/1$. If the kernel of this map is $(xy)$,we get a map $f:\mathbb{C}[x,y]/(xy)\rightarrow \mathbb{C}[x]_x$.
This gives us an induced map from $(\mathbb{C}[x,y]/(xy))_x\rightarrow \mathbb{C}[x]_x$ by the universal property of localization, and it is not hard to show that this is a bijection.
However, is the kernel in fact $(xy)$?
We have $\rm ker f= \{f(x,y)\in \mathbb{C}[x,y]\mid f(x,0)/1=0/x^n \text{ for some $n$}\}$, and so the kernel consists of elements of the form $x^nf(x,y)$ where $f(x,0)=0$ or $x=0$. Does this imply that the kernel is $(xy)$?
 A: We have a short exact sequence of $\mathbb{C}[x,y]$-modules
$$0\rightarrow (xy)\rightarrow \mathbb{C}[x,y]\rightarrow \mathbb{C}[x,y]/(xy)\rightarrow 0.$$
Then note that $\mathbb{C}[x,y]_x$ is a flat $\mathbb{C}[x,y]$-module. In particular, the functor $(-)\otimes_{\mathbb{C}[x,y]}\mathbb{C}[x,y]_x$ is exact, not just right exact. Hence if we tensor the above short exact sequence, we get
$$0\rightarrow (xy)_x\rightarrow \mathbb{C}[x,y]_x\rightarrow \left(\mathbb{C}[x,y]/(xy)\right)_x\rightarrow 0.$$
Note that the ideal$(xy)_x$ is just the principal ideal $(y)\subset \mathbb{C}[x,y]_x$. Thus rewriting the above exact sequence using this isomorphism, we get
$$0\rightarrow (y)\rightarrow \mathbb{C}[x,y]_x\rightarrow \left(\mathbb{C}[x,y]/(xy)\right)_x\rightarrow 0.$$
However we also have a short exact sequence
$$0\rightarrow (y)\rightarrow \mathbb{C}[x,y]\rightarrow \mathbb{C}[x]\rightarrow 0.$$
And thus after tensoring with $\mathbb{C}[x,y]_x$, we get a short exact sequence
$$0\rightarrow (y)\rightarrow \mathbb{C}[x,y]_x\rightarrow \mathbb{C}[x]_x\rightarrow 0.$$
Since the cokernel is unique up to isomorphism, we see that
$$\left(\mathbb{C}[x,y]/(xy)\right)_x\cong \mathbb{C}[x]_x.$$
