0
$\begingroup$

Let $R$ be a commutative ring and let $M$ be an $R$-module. Let $M_1,\dots, M_n$ be submodules of $M$. Is there a short exact sequence of the form

$$0\to \bigcap_{i=1}^nM_i\xrightarrow{u}\bigoplus_{i=1}^nM_i\xrightarrow{v}\sum_{i=1}^nM_i\to0~~~~~~~~~~~(1)$$

where if $f_{j}$ is the obvious composition of inclusions $\bigcap_{i=1}^nM_i\to M_j\to \bigoplus_{i=1}^nM_i$ for every $j=1, \dots, n$, then $u=\sum_{j=1}^nf_j$.

I am not sure how to construct the map $v$ so that short exact sequence $(1)$ exists? In the case $n=2$, then the map is defined in Bourbaki's book Algebra $I$ as $v=i-j$ where $i: M_1\to M_1+M_2$ and $j:M_2\to M_1+M_2$. How do we generalise this to an arbitrary $n$ if possible at all?

$\endgroup$
1
  • 1
    $\begingroup$ The "natural" exact sequence in this scenario is not a short exact sequence if $n>2$, see e.g. this answer. $\endgroup$
    – Thorgott
    Mar 24, 2023 at 0:51

1 Answer 1

1
$\begingroup$

What one would expect to have is a long exact sequence with $j$th term given by $C_j = \bigoplus_{A\subseteq [n], |A| = j} M_A$ where $M_A=\bigcap_{a\in A} M_a$ and the differential is given by sending an element in $M_A$ to a signed sum in $M_{A\smallsetminus a}$ for $a\in A$.

But this is not exact in general. The key word you want is distributivity, see for example here. See also this post. In particular, try to check by hand that exactness is non-trivial already when $n=3$, i.e. there is a complex

$$0 \longrightarrow V_1\cap V_2 \cap V_3 \longrightarrow \bigoplus_{1\leqslant i<j\leqslant 3} V_i \cap V_j \longrightarrow \bigoplus_{1\leqslant i\leqslant 3} V_i \longrightarrow \sum_{1\leqslant i\leqslant 3} V_i \longrightarrow 0$$

which is not exact in general.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .