The Chinese Remainder Theorem is a geometric fact (more general version)

Question

The idea that the Chinese Remainder Theorem is a geometric fact is brought up in 4.4.11 of Vakil's notes on algebraic geometry. The special case discussed there is described here. Following this idea, I'm trying to generalize it to the general Chinese remainder theorem in ring theory. For simplicity, let me restrict to two ideals. Then the theorem states that if $R$ is a (commutative) ring and $I,J$ are ideals such that $I+J=R$, then there is a ring isomorphism $R/(I\cap J)\to R/I\times R/J$. Taking the quotient by $I\cap J$, we may assume $I\cap J=0$. To interpret this in terms of affine schemes, consider the closed subsets $V(I),V(J)\subset\operatorname{Spec}R$. The condition $I+J=R$ is equivalent to $V(I)\cap V(J)=\varnothing$, in which case they are both open sets. Then $I\cap J=0$ says $V(I)\cup V(J)=\operatorname{Spec}R$. Thus $\operatorname{Spec}R$ is the disjoint union of $V(I)$ and $V(J)$ as a topological space. I need to show that $\mathcal{O}_{\operatorname{Spec}R}(V(I))\cong R/I,\mathcal{O}_{\operatorname{Spec}R}(V(J))\cong R/J$, so that $$R\cong\mathcal{O}_{\operatorname{Spec}R}(\operatorname{Spec}R)=\mathcal{O}_{\operatorname{Spec}R}(V(I))\times\mathcal{O}_{\operatorname{Spec}R}(V(J))\cong R/I\times R/J.$$ This is where I got stuck. I tried to prove it by showing $\mathcal{O}_{\operatorname{Spec}R}|_{V(I)}\cong\mathcal{O}_{\operatorname{Spec}R/I}$, but their stalks $R_\mathfrak{p}$ and $(R/I)_{\mathfrak{p}/I}\cong R_\mathfrak{p}/I_\mathfrak{p}$ ($I\subset\mathfrak{p}$) do not seem to be isomorphic. What is wrong here?

Answer 1

We have $R=S \times T$ for rings $S,T$. Also $I=T$ an ideal in $R$, and $\mathfrak{p}\supseteq T$ a prime ideal. We will show: $$R_{\mathfrak{p}}\cong(R/I)_{\mathfrak{p}/I} \cong R_{\mathfrak{p}}/I_{\mathfrak{p}}$$

We have $\mathfrak{p}=p\times T$ for some prime ideal $p\subset S$.

We have:\begin{eqnarray*} R/I&=&S,\\ \mathfrak{p}/I&=&p,\\ R_{\mathfrak{p}}&=& S_p,\\ I_\mathfrak{p}&=&0. \end{eqnarray*} To see why the last two statements hold, note that $(0,1)(1,0)=(0,0)$ and $(1,0)\notin \mathfrak{p}$.

The result follows immediately:$$R_{\mathfrak{p}}\cong S_p\cong(R/I)_{\mathfrak{p}/I} \cong R_{\mathfrak{p}}/I_{\mathfrak{p}}$$