MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Perhaps – 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.

share|cite|improve this answer
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.

share|cite|improve this answer
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...

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.