Given ring $A$, ideal $I$, and $A$-module $M$, show that $A/I \otimes_A M$ is isomorphic to $M/IM$.

Question

The question is stated as in the title; the hint I am given is to "tensor the exact sequence" $0 \rightarrow I \rightarrow A \rightarrow A/I \rightarrow 0$, which I take to mean using that sequence and a theorem from the book to say that the sequence $I ⊗ M → A ⊗ M → A/I ⊗ M → 0$ is exact. Having done this, I have no idea how to proceed to the desired fact. The trouble is that I'm just not terribly familiar with either exact sequences or tensor products.

I'm guessing the most obvious homomorphisms are in play for the original sequence, i.e., $I \rightarrow A$ is the homomorphism mapping the elements of $I$ to themselves within $A$, and $A \rightarrow A/I$ is the natural projection $\pi(a) = a + I$. The theorem from the book indicates that given an exact sequence with homomorphisms, for example, f and g, the new homomorphisms for the "tensored" sequence are "$f \otimes 1$" and "$g \otimes 1$", but I don't see any explanation for what those homomorphisms actually mean in terms of the elements they're acting on. In particular, I think I'm meant to use the fact that $\pi \otimes 1:A \otimes M \rightarrow A/I \otimes M$ is onto, but I don't know what that really means.

(A is commutative with identity, in case that isn't clear from the context/tags)

Answer 1

As you say, start with the tensored exact sequence $$I\otimes_A M\to A\otimes_A M \to (A/I)\otimes_A M\to 0.$$ The image of $I\otimes_A M$ under the first map is clearly sums of elements of the form $im$ for $i\in I$, $m\in M$, which is just $IM$. Next, note that $A\otimes_A M \cong M$ via $a\otimes m \mapsto am$.

Can you take it from there?