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I'm a little unsure of why the following module isomorphism holds. Say $R$ is a commutative, unital ring and $\mathcal{M}$ an $R$-module. If $I_1,\dots,I_n$ are pairwise comaximal ideals of $R$, then $$ M/(I_1\cdots I_n)M\cong\bigoplus_{i=1}^n M/I_iM. $$

This seems like it would fall out from the isomorphism theorems, for some suitable surjective homomorphism $\psi\colon M\to \bigoplus_{i=1}^n M/I_iM$ with $\ker\psi=(I_1\cdots I_n)M$.

Such a map is not obvious to me. For instance, where does the pairwise comaximality of the ideals come into play? If this is indeed the right direction, what could $\psi$ be?

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This is the Chinese Remainder Theorem. – Mariano Suárez-Alvarez Oct 14 '11 at 4:45
up vote 7 down vote accepted

This is just the Chinese Remainder Theorem applied to modules. The Isomorphism Theorems help, but you do indeed need to go beyond the Isomorphism Theorems and use the fact that the $I_j$ are pairwise comaximal in some way.


Essentially, the isomorphism theorems and the universal property of the direct sum let you define a homomorphism $$M/(I_1\cap\cdots\cap I_n)M\to\bigoplus_{j=1}^n M/I_jM$$ by considering the family of canonical projections $M\to M/I_jM$, which give a map into the direct sum, and then noting that the intersection equals the kernel of the induced map.

But this is as far as the isomorphism theorems take you. Here, you have to make no assumptions about the $I_j$ other than that they are ideals.

Then you need to actually get into the brass tacks. The fact that the $I_j$ are comaximal is used for two things:

  1. To show that $I_1\cap\cdots\cap I_n = I_1\cdots I_n$; and
  2. To show that the induced map is onto the direct sum (the isomorphism theorems only guarantee that the image of $M$ is a subdirect product, that is, that the composition of the map with the projections are onto).

The arguments here are the same as for the usual Chinese Remainder Theorem.

First: $I_1I_2\cdots I_n\subseteq I_1\cap\cdots\cap I_n$. For the converse inclusion, first note that if $I$ and $J$ are comaximal, then $IJ=I\cap J$, since given $a\in I\cap J$, write $1=a_i+a_j$, with $a_i\in I$, $a_j\in J$, and we have $a=a(a_i+a_j) = aa_i+aa_j\in IJ=JI$. Then note that $I_1\cap\cdots I_{n-1}$ is comaximal with $I_n$, since $$R=(I_1+I_n)(I_2+I_n)\cdots(I_{n-1}+I_n) \subseteq I_1I_2\cdots I_{n-1}+I_n\subseteq R.$$ Now use an inductive argument to show that $I_1\cap\cdots\cap I_n\subseteq I_1\cdots I_n$.

To show the induced map is onto, you show that for each $j$ you can find $r\in R$ such that $r\equiv 1 \pmod{I_j}$ and $r\equiv 0\pmod{I_k}$ for $k\neq j$ (use the fact that $I_j$ is comaximal with $\cap_{k\neq j}I_k$). Then given $m_1,\ldots,m_n\in M$, show that $r_1m_1+\cdots +r_nm_n\equiv m_k\pmod{I_kM}$ for each $k$ to get that the map $M\to\oplus M/I_jM$ is onto.

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Thanks for your post! – Danielle Intal Oct 14 '11 at 23:12

If you have the Chinese remainder theorem in hand, then you can tensor the isomorphism $$ A/(I_1 \cdots I_n) \to A/I_1 \times \cdots \times A/I_n $$ and its inverse with $M$, using that $A/I \otimes M \approx M/IM$ and that the tensor product distributes across direct sums.

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Thanks! I don't know how to tensor, but maybe someday. – Danielle Intal Oct 14 '11 at 23:12

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