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?

Thank you in advance.


1 Answer 1


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.


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