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?


1 Answer 1


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 $IM$.


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