TITLE: Isomorphism of modules + tensor product
QUESTION [8 upvotes]: Is it true that:
$$M{\otimes}_{A}(A/I) \cong M/IM$$
and
$$IM \cong I {\otimes}_AM$$
where $A$ is a commutative ring, $M$ an $A$-module, and $I \subset A$ an ideal.

REPLY [3 votes]: There is a proof of your first equation given in page 17 of Osborne's Basic Homological Algebra book (presuming you have defined tensor product in terms of the usual universal property). I'll give an outline here
Define $\phi':M \times A/I \to M/IM$ by $\phi'(m,a+I) = am + IM.$ Check that this is well defined. Suppose there is a bilinear $\psi:  M \times A/I \to G$. Using the fact that $A \otimes M \simeq M$ we have the following diagram
[image lost]
$\theta$ and $\phi$ come from the above tensor product ($A \otimes M \simeq M$). Convince yourself that $\theta'$ is induced from $\theta$, that the diagram commutes and that $\theta'$ is unique.