5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Perhaps springerlink.com/content/q2184g8975054112 $\endgroup$ Apr 18, 2010 at 23:10
  • $\begingroup$ you may post this as an answer ;) $\endgroup$ Apr 19, 2010 at 7:42
  • $\begingroup$ @MartinBrandenburg Have you advanced so far? I know that the question is some 9 years old, but it seems that no final solution has been posted here and I wonder if by now you can describe the cokernel (and not only the "naive" cokernel). $\endgroup$ Oct 4, 2019 at 6:56
  • $\begingroup$ @FilippoAlbertoEdoardo Thanks for your interest in the problem! I did not really make any advance on that, but also didn't think that much about the problem either. $\endgroup$ Dec 22, 2019 at 14:25

3 Answers 3

4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Apr 19, 2010 at 11:07
  • $\begingroup$ 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? $\endgroup$ Apr 19, 2010 at 11:12
0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I don't assume any comaximality-condition. Please read the question carefully. $\endgroup$ Apr 19, 2010 at 6:31
  • $\begingroup$ To be honest, I wasn't sure exactly what you were assuming. Do you mean R to be commutative and unital? $\endgroup$ Apr 19, 2010 at 6:48
  • $\begingroup$ this is not important. $\endgroup$ Apr 19, 2010 at 7:33
0
$\begingroup$

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...

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Apr 19, 2010 at 13:41
  • $\begingroup$ 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. $\endgroup$
    – Bugs Bunny
    Apr 28, 2010 at 14:31
  • $\begingroup$ No the equality is also wrong in PIDs. $\endgroup$ Oct 6, 2010 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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