I've met a problem in my algebra exercises.

Let $M_1,M_2,\ldots,M_n$ be submodules of $M$ such that each $M/M_i$ has finite length. Prove that $M/(M_1\cap M_2 \cap \cdots\cap M_n)$ has finite length. Moreover, determine a formula for computing this length.

I've tried some examples for this. Let $$ M=\mathbb{Z},M_1=4\mathbb{Z},M_2=9\mathbb{Z},M_3=6\mathbb{Z} $$ Denote $c(M)$ be the length of module $M$.
We have, $$ c(M/M_1)=c(M/M_2)=c(M/M_3)=2 $$ $$ c(M/(M_1\cap M_2))=4,c(M/(M_1\cap M_3))=3 $$

It seems that it should be the $\sum_{i=1}^nc(M/M_i)$ minus something showing the relations between $M_i$. But I cant figure it out.

Can you please help? Thank you!

A summary of the answer:
Thanks to Georges Elencwajg's help, the formula should be: $$ \begin{align*} \phantom{=}&c(M/(M_1\cap M_2 \cap \cdots\cap M_n))\\ &=c(M/M_1)+\cdots+c(M/M_n)\\ &\quad{}-(c(M/(M_1+M_2))+c(M/(M_1+M_3))+\cdots+c(M/(M_{n-1}+M_n)))\\ &\quad{}+(c(M/(M_1+M_2+M_3))+\cdots+c(M/(M_{n-2}+M_{n-1}+M_n)))\\ &\qquad{}\vdots\\ &\quad{}+(-1)^{n-1}c(M/(M_1+\cdots+M_n)) \end{align*} $$

  • 1
    $\begingroup$ You are on the right track. The formula for the length is along the lines of inclusion-exclusion, using the lengths of those quotients by intersections as the ingredients. Such a formula will only prove finite length with a strong induction, since you'll need to know all the smaller intersections are also finite length. $\endgroup$ – Jack Schmidt Nov 17 '11 at 20:03

First, a reminder

Key result on finite-length modules
Given a short exact sequence of modules $0\to N'\to N\to N''\to 0$, one has the equivalence $$N \text{ has finite length } \iff N'\text{ and } N'' \text{ have finite length } $$ and if this is the case $$\text{ length } (N)= \text{ length } (N')+\text{ length } (N'') $$

And now, back to your question. Consider the exact sequence
$ 0\to M_1/M_1\cap M_2 \to M/M_2 $
The implication $\Rightarrow$ tells you that $M_1/M_1\cap M_2$ has finite length, since by assumption$M/M_2$ has finite length.

Now contemplate the exact sequence
$0\to M_1/M_1\cap M_2 \to M/M_1\cap M_2 \to M/M_1 \to 0$.
Since we just proved that the leftmost module has finite length and since the rightmost module has finite length by assumption , the implication $\Leftarrow$ of the key result tells us that the middle module $M/M_1 \cap M_2$ has finite length: we have just proved the case $n=2$ of the question.
A trivial induction yields the general case.

  • $\begingroup$ Thank you for your answer! I've tried to deduce the length formula from your result, getting that $c(M/(M_1\cap M_2))=c(M/M_1)+c(M/M_2)-c(M/(M_1+M_2)). However, I got stuck when I tried to generalize it for $n>2$. (I want to find a formula symmetric in $M_i$) $\endgroup$ – Roun Nov 18 '11 at 3:16
  • $\begingroup$ Ah…Now I think I've worked the formula out and it just looks like inclusion-exclusion principle, as Jack Schmidt commented. I'll post it here later. Thanks again! $\endgroup$ – Roun Nov 18 '11 at 3:41
  • $\begingroup$ @GeorgesElencwajg: You've shown that $lngt\frac{M}{A\cap B}=lngt\frac{M}{A}+lngt\frac{M}{B}-lngt\frac{M}{A\cap B}$. How is the induction trivial? Taking $A=M_1$ and $B=M_2\cap\ldots\cap M_n$, the obtained formula includes $lngt\frac{M}{A_1+A_2\cap\ldots\cap A_n}$, which I don't know how to handle. $\endgroup$ – Leo Jan 8 '14 at 19:50

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