# Localization at maximal ideal of tensor product of algebras

Let $$k$$ be an algebraically closed field. Set $$R=k[x_1,...,x_n]/I$$ and $$S=k[y_1,...,y_m]/J$$ finite generated $$k$$-algebras.

By Hilbert Nullstellensatz, the maximal ideals of $$R\otimes_k S$$ are of the form $$m\otimes_k S+R\otimes_k n$$ with $$I\subseteq m=(x_1-a_1,...,x_n-a_n)$$ and $$J\subseteq n=(y_1-b_1,...,y_m-b_m)$$, where $$a_i,b_i\in k$$.

Is the natural map $$R_m\otimes_k S_n\rightarrow (R\otimes_kS)_{m\otimes_k S+R\otimes_k n}$$ an isomorphism?

Also, is $$R_m\otimes_k S_n$$ local ring?

Similarly, we can consider prime ideals $$I\subseteq p$$ and $$J\subseteq q$$. Then $$p\otimes_k S+R\otimes_k q$$ is prime ideal of $$R\otimes_k S$$ since $$k$$ is algebraically closed, see Tensor product of domains is a domain. What is the localization of $$R\otimes_k S$$ at this prime ideal?

This is not true. We just need to show in general, $$R_m\otimes_k S_n$$ is not local.
Assume $$k$$ is algebraically closed. Let $$R=k[x]$$ and $$S=k[y]$$. Set $$m=(x)$$ and $$n=(y)$$. Then we have
$$R_m\otimes_k S_n\cong T^{-1}k[x,y]$$, where $$T=\{fg\mid f\in k[x],g\in k[y], f(0)\neq 0,g(0)\neq 0\}$$
Claim: $$T^{-1}(x+y+1)$$ is maximal ideal of $$T^{-1}k[x,y]$$. The spectrum of $$T^{-1}k[x,y]$$ corresponds to the prime ideal of $$k[x,y]$$ which has empty intersection with $$T$$. If $$T^{-1}(x+y+1)$$ is not maximal, there exists prime ideal $$p$$ containing $$(x+y+1)$$ such that $$p\cap T=\emptyset$$. Note that $$p=(x-a,y-b)$$ since $$\dim(k[x,y])=2$$. $$p\cap T=\emptyset$$ implies that $$a=b=0$$. This contradicts with $$(x+y+1)\subseteq p$$. The claim follows.
Since $$T^{-1}(x,y)$$ is also maximal ideal of $$T^{-1}k[x,y]$$. We get $$T^{-1}k[x,y]$$ is not local.
The above example that $$k[x]_{(x)}\otimes_k k[y]_{(y)}$$ is not local works for any field. Assume that $$T^{-1}k[x,y]$$ is local. We have the maximal ideal of $$T^{-1}k[x,y]$$ is $$T^{-1}(x,y)$$. It is clear that $$x+y+1\notin T^{-1}(x,y)$$. Hence it is invertible in $$T^{-1}k[x,y]$$. Thus we get $$(x+y+1)\alpha/fg=1/1, \text{ for some }\alpha\in k[x,y] \text{ and } fg\in T.$$ Then we get $$(x+y+1)\alpha=fg$$, this contradicts with $$k[x,y]$$ is UFD.