Let $U,V$ be submodules of a $R$-module $M$. Then the diagonal induces an isomorphism
$M/(U \cap V) \to M/U \times_{M/(U+V)} M/V.$
This is a (useful!) generalization of the Chinese Remainder Theorem and the proof is very easy. But I'm interested what happens when we take finitely many submodules $U_1,…,U_n$. How can we relate $M/(U_1 \cap … \cap U_n)$ with the $M/U_i$? I think the case $n=2$ can not be used for an induction, there are more compatiblities to check for an element in $\prod_i M/U_i$ to come from $M$. I wonder if there is a nice description.
For $M=R$, this question asks for a sort of sheaf condition for sections on closed subschemes.
Best Answer
So this is what's in Kleinert's paper "Some remarks on the Chinese Remainder Theorem" that I mentioned in the comments.
If $\mathcal F=\{U_1,U_2,\dots,U_n\}$ is a family of submodules of the $R$-module M, then there is an embedding $\phi(\mathcal F)$ of $M/U_1\cap \cdots \cap U_n$ into $$M(\mathcal F):= \{(u_i)\in \prod M/U_i \quad \rvert u_i\equiv u_j \mod (U_i+U_j),\forall i,j\}.$$ Let the cokernel of $\phi$ be $$O(\mathcal F)=M(\mathcal F)/\phi(M/U_1\cap \cdots \cap U_n).$$ $O(F)$ is thought of as the obstruction against the ability to solve simultaneous congruences, and so we say that the generalized Chinese Remainder Theorem holds if $O(\mathcal F)=0$. He proceeds to the following sheaf-theoretical interpretation of the problem:
Let $X$ be the discrete topological space $\{1,2,\dots,n\}$, and define the presheaf $\mathcal P(\mathcal F)$ on $X$ by $\mathcal P(V)=M/\sum_{i\notin V}U_i$, for $V\subset X$. If $V\subset W$ the restriction map is given by the residue map $$\mathcal P(W)=M/\sum_{i\notin W}U_i\to M/\sum_{i\notin V}U_i=\mathcal P(V).$$ Now let $\mathcal U$ be the covering $\{X/\{i\}\}$. It follows that $M(\mathcal F)$ is the set of cocycles $C^0(\mathcal U,\mathcal P)$ and that $O(\mathcal F)=0$ iff $\mathcal P$ satisfies the second sheaf axiom with respect to $\mathcal U$. He also makes the remark that when $n=2$ , which you described in the question, this is always the case and so the generalized Chinese Remainder Theorem always holds, even though it doesn't always in the general case.