TITLE: Show that $(A / \mathfrak{a}) \otimes_A M \cong M / \mathfrak{a} M$ for a ring $A$, ideal $\mathfrak{a}$, $A$-module $M$.
QUESTION [6 upvotes]: This is Atiyah-Macdonald Exercise 2.2

Exercise: Let $A$ be a ring, $\mathfrak a$ an ideal, $M$ an $A$-module. Show that $(A/\mathfrak a) \otimes_A M$ is isomorphic to $M/\mathfrak aM$. [Tensor the exact sequence $0 \to \mathfrak a \to A \to A/\mathfrak a$ with $M$.]

I would like to verify my proof. It will be posted as a community wiki. Leave comments or edit, but if you edit, make the text red or at least make the edit obvious to distinguish my original approach, with what is incorrect or can be better written, etc.
Also, feel free to post alternate proofs.

REPLY [5 votes]: Reposting from my comments since other proofs have already been posted.
By tensoring the exact sequence $0 \to \mathfrak a \to A \to A / \mathfrak a \to 0$ with $M$, we get the exact sequence
$$
\mathfrak a \otimes_A M \to A \otimes_A M \to A / \mathfrak a \otimes_A M \to 0.
$$
Consider the canonical isomorphism $A \otimes_A M \cong M$. By composing the arrows going in and out of $A \otimes_A M$ in the exact sequence above with the arrows of this isomorphism, we get the exact sequence
$$
\mathfrak a \otimes_A M \xrightarrow{\varphi} M \xrightarrow{\psi} A / \mathfrak a \otimes_A M \to 0.
$$
It is straightforward to verify that this sequence is indeed exact. We have
$$
\ker \psi = \operatorname{im} \varphi = \mathfrak a M.
$$
Thus, the first isomorphism theorem gives $A / \mathfrak a \otimes_A M \cong M / \mathfrak a M$.