Example to show that multiplication by ideals and intersection of submodules do not commute

Question

The key point of question about typical proof of Krull Intersection Theorem is that multiplication by ideals and intersection of submodules do not commute. Can anyone give me an example of this?

That is,

I'm looking for $A$-modules $M, N, P$ and an ideal $\mathfrak a \subset A$ such that $M \subset P$, $N \subset P$ and $\mathfrak a (M \cap N) \neq \mathfrak a M \cap \mathfrak a N$.

Answer 1

A bit contrived, but take $A = \mathbb{C}[x,y,z]$ and let $I = (x^2, xy-xz)$. Set $P = A / I$. Then take $M = (x+ I,y+ I)$ and $N = (x+ I,z+I)$, $\mathfrak{a} = (x)$. We have $\mathfrak{a}M = (xy+I)$ and $\mathfrak{a}N = (xz+I)$, so
$$\mathfrak a M \cap \mathfrak a N = (xy+I) = (xz+I)$$ because we have identified $xy - xz \in I$. On the other hand, $M \cap N = (x+ I)$, so $$ \mathfrak a (M \cap N) = 0.$$

Answer 2

Hint $\ $ Consider $\,(x,y)((x)\cap(y)) = (x,y)(xy)$ in $\,\Bbb Z[x,y]$

But distributed it is $\ (x^2,xy)\cap (xy,y^2) = (xy) $ since

$\quad ax^2\!+bxy = cxy + dy^2\ \overset{\large y=0,}{\underset{\large x=1}\Rightarrow}\,a=0;\ \overset{\large x=0}{\underset{\large y=1}\Rightarrow}\,d=0 $