# Motivation for looking at the product module

Let $$M$$ be an $$A-$$module, $$A$$ a commutative ring with identity. Then we have the following result that , $$\frac{A}{I} \otimes M \simeq \frac{M}{IM}$$ for any ideal $$I$$ in $$A$$.

This simply follows by the observation that, $$IM \le \text{Ker}(\psi)\le M$$, where $$\psi:M \to \frac{A}{I} \otimes M$$ can be extended to a map $$\phi:\frac{M}{IM}\to \frac{A}{I} \otimes M$$ and showing $$\phi$$ is indeed an isomorphism.

My question is, why is it natural to look at the submodule $$IM$$? What intuition lies behind this choice?

It comes from the short exact sequence $$0 \rightarrow I \rightarrow A \rightarrow A/I \rightarrow 0$$ of $$A$$-modules. As you might know, tensoring with $$M$$ is a right exact functor, from which we get the following exact sequence $$I\otimes M \rightarrow A\otimes M \rightarrow (A/I) \otimes M\rightarrow 0.$$ The middle term $$A \otimes M$$ is canonically isomorphic to $$M$$, thus $$(A/I) \otimes M$$ is isomorphic to the quotient of $$M$$ by the image of $$I \otimes M$$ in $$M$$.
Now if you follow the definitions closely, then with a moment of reflection you will realize that the image of $$I \otimes M$$ in $$M$$ is exactly what we call $$IM$$.