# Generalized Chinese Remainder Theorem

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.

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– Gjergji Zaimi Apr 18 '10 at 23:10
you may post this as an answer ;) – Martin Brandenburg Apr 19 '10 at 7:42

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.

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Thanks. The paper studies the obstruction from the image to the "naive" image (pairwise compatibility). I'm interested in the image of $M$ in $\prod_i M/U_i$ itsself. – Martin Brandenburg Apr 19 '10 at 11:07
In the remark about the surjectivity of q, I think that (2) is wrong and should be replaced by $b_i \equiv a_i$ mod $A_i + A_{n+1}$. Other opinions? – Martin Brandenburg Apr 19 '10 at 11:12

If we consider submodules generated by applying ideals to $M$, ie. $I \ M$, where $I \subseteq R$ is an ideal, we can generalize the two submodule case to any finite number of modules by induction from n = 2. All we need to check is that if $A_{i} \ M = U_{i}$ and the $A_{i}$'s are pairwise comaximal, then $A_{1} ... A_{n-1}$ and $A_{n}$ are comaximal.

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I don't assume any comaximality-condition. Please read the question carefully. – Martin Brandenburg Apr 19 '10 at 6:31
To be honest, I wasn't sure exactly what you were assuming. Do you mean R to be commutative and unital? – Ryan Thorngren Apr 19 '10 at 6:48
this is not important. – Martin Brandenburg Apr 19 '10 at 7:33

Sorry, no clue but I will take a wild guess. Your Chinese theorem can be stated as exactness of the sequence $$0 \rightarrow M/A\cap B \rightarrow M/A \times M/B \rightarrow M/A+B \rightarrow 0$$ The third arrow is $(x+A,y+B)\mapsto (x-y)+(A+B)$. I can envision that you may be able to produce a long exact sequence starting with $$0 \rightarrow M/\cap_i A_i \rightarrow$$ as soon as the lattice generated by $A_i$-s is distributive. Notice that distributivity will ensure that you have $2^n$ submodules to work with...

Sorry if I misunderstood anything or said something completely ridiculous...

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Assuming $A_1 \cap (A_2 + A_3) = A_1 \cap A_2 + A_1 \cap A_3$ etc. makes an induction easy. But I think this is almost never the case? – Martin Brandenburg Apr 19 '10 at 13:41
What do you mean? Off course, it is rare but it holds for PID-s (where you have usual Chinese RT holds) and for your case as 2 subs generate a distributive lattice. – Bugs Bunny Apr 28 '10 at 14:31
No the equality is also wrong in PIDs. – Martin Brandenburg Oct 6 '10 at 17:10