Is $\bigcap_{i\in I}(M_i/N)=\bigl(\bigcap_{i\in I}M_i\bigr)/N$?

Question

Let $M$ be an $R$-module, $N$ a submodule of $M$ and $\{M_i\}_{i\in I}$ a family of submodules of $M$ such that $N\subseteq M_i$ for all $i\in I$. Is this true that $\bigcap_{i\in I}(M_i/N)=\bigl(\bigcap_{i\in I}M_i\bigr)/N$? Thanks a lot.

Answer 1

Consider the canonical map $\pi\colon M\to M/N$; then $$ \pi^{-1}\bigl(\bigcap_i(M_i/N)\bigr)=\bigcap_i M_i $$ Since also $$ \pi^{-1}\bigl(\bigl(\bigcap_i M_i\bigr)/N)\bigr)=\bigcap_i M_i $$ you can easily end.

Answer 2

Yes, you can show inclusion both ways as follows:

Let $m\in M$ and $mN\in M/N$ be a coset.

If $mN\in \bigcap_{i\in I}(M_i/N)$ then $\forall n\in N:m+n\in M_i\implies m=m+0\in M_i$, so $$m\in \bigcap_{i\in I}M_i\implies mN\in\bigl(\bigcap_{i\in I}M_i\bigr)/N$$

Conversely, if $mN\in \bigl(\bigcap_{i\in I}M_i\bigr)/N$ then

$$m\in\bigcap_{i\in I}M_i\implies mN\in\bigcap_{i\in I}(M_i/N)$$