For ideals of $R$, $I_1, I_2$, Is it true that $(I_1 \cap I_2)M \cong I_1 M \cap I_2 M$ for any $R$-module $M$?

Question

It seems true if $M$ is finitely generated module over PID $R$ because we can take a linearly independent finite generator and so we can use $$ \sum_{i=1}^n c_i m_i = 0 \Leftrightarrow \forall i \in \{1,2, \dots, n\}: c_i = 0 $$

But, in general case, Is it true that $(I_1 \cap I_2)M \cong I_1 M \cap I_2 M$? Then, how to prove this?

Of course, it's enough to prove that $$ I_1 M \cap I_2 M \le (I_1 \cap I_2) M $$ since the opposite direction is obvious.

Answer 1

Let $R=\Bbb Z[X,Y]$, $I_1=(X)$ and $I_2=(Y)$. Then $I_1\cap I_2=(XY)$.

Now let $M=\Bbb Z[T]/(T^2)$. Make it into an $R$-module by insisting that $Xf(T)=Yf(T) =Tf(T)$. Then $I_1M=I_2M=(T)$ which is a copy of $\Bbb Z$. But $(I_1\cap I_2)M =XYM=\{0\}$.