Show that $(A / \mathfrak{a}) \otimes_A M \cong M / \mathfrak{a} M$ for a ring $A$, ideal $\mathfrak{a}$, $A$-module $M$. 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.
 A: 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 $A / \mathfrak a \otimes_A M \cong M / \mathfrak a M$.
A: I would like to post an alternative proof:
By Proposition 2.12 of the ops answer (in other words by the universal property of tensor product)  the following bilinear map
\begin{eqnarray*}
\psi:A/\mathfrak{a}\times M & \to & M/\mathfrak{a}M \\
            (x+\mathfrak{a},M) & \mapsto & xm+\mathfrak{a}M
\end{eqnarray*}
induces the unique linear map
\begin{eqnarray*}
\Psi:A/\mathfrak{a}\otimes_A M & \to & M/\mathfrak{a}M \\
            x+\mathfrak{a}\otimes M & \mapsto & xm+\mathfrak{a}M
\end{eqnarray*} 
(that is $\Psi\otimes=\psi$, where $\otimes:A/\mathfrak{a}\times M\to A/\mathfrak{a}\otimes M$ is the tensor map with the universal property)
On the other hand, there is a linear map
\begin{eqnarray*}
\phi: &  M \to & A/\mathfrak{a}\otimes_A M \\
            & m  \mapsto & (1+\mathfrak{a})\otimes m
\end{eqnarray*}
with $\mathfrak{a}M\subseteq$ ker$\phi$, since for $x\in\mathfrak{a}$ and $m\in M$ we have  $\phi(xm)=(1+\mathfrak{a})\otimes xm=(x+\mathfrak{a})\otimes m=0$
Therefore, $\phi$ induces the following linear map 
$$\Phi:M/\mathfrak{a}M \to  A/\mathfrak{a}\otimes_A M$$
with the property $\Phi \pi=\phi$, where $\pi: M\to M/\mathfrak{a}M$ is the natural projection map. 
Next we show that $\Psi$ and $\Phi$ are inverses of each other.
$\Psi\Phi(m+\mathfrak{a}M)=\Psi((1+\mathfrak{a})\otimes m)= m+\mathfrak{a}M$
and
$\Phi\Psi((\sum_i\alpha_ix_i+\mathfrak{a})\otimes\sum_j\beta_jm_j)=\Phi\Psi((1+\mathfrak{a})\otimes\sum_{i,j}\alpha_i\beta_jx_im_j)=\Phi(\sum_{i,j}\alpha_i\beta_jx_im_j+\mathfrak{a}M)=$$(1+\mathfrak{a})\otimes \sum_{i,j}\alpha_i\beta_jx_im_j=(\sum_i\alpha_ix_i+\mathfrak{a})\otimes\sum_j\beta_jm_j$, where $\alpha_i,\beta_j\in \mathbb{N}$ and they are zero for all but finitely many $i,j$.
A: This is mostly similar to your argument, but much easier.

From the exact sequence $0\to\mathfrak{a}\to A\to A/\mathfrak{a}\to 0$ we get the commutative diagram with exact rows
$$\require{AMScd}\def\ma{\mathfrak{a}}
\begin{CD}
{} @. \ma\otimes_AM @>>> A\otimes_AM @>>> A/\ma\otimes_AM @>>> 0 \\
@. @VVV @VVV @VVV \\
0 @>>> \ma M @>>> M @>>> M/\ma M @>>> 0
\end{CD}
$$
where the leftmost vertical arrow is surjective and the middle vertical arrow is an isomorphism. A standard diagram chasing shows that the arrow
$$
A/\ma\otimes_AM\to M/\ma M,
$$
defined by $(a+\ma)\otimes x=ax+\ma M$ so as to make the diagram commutative, is an isomorphism too.
A: Proposition 2.12: Let $M, N$ be $A$-modules. Then there exists a pair $(T, g)$ consisting of an $A$-modules $T$ and an $A$-bilinear mapping $g : M \times N \to T$, with the following property:
Given any $A$-module $P$ and any $A$-bilinear mapping $f : M \times N \to P$, there exists a unique $A$-linear mapping $f' : T \to P$ such that $f = f' \circ g$ (in other words, every bilinear function on $M \times N$ factors through $T$).
Moreover, if $(T, g)$ and $(T', g')$ are two pairs with this property, then there exists a unique isomorphism $j : T \to T'$ such that $j \circ g = g'$.

Proposition 2.18: Let $$M' \xrightarrow f M \xrightarrow g M'' \to 0$$ be an exact sequence of $A$-modules and homomorphisms, and let $N$ be any $A$-module. Then the sequence $$M' \otimes N \xrightarrow{f \otimes 1} M \otimes N \xrightarrow {g \otimes 1} M'' \otimes N \to 0$$ (where $1$ denotes the identity mapping on $N$) is exact.

Proof of Exercise 2.2:
$\newcommand{im}{{\operatorname{im}}}$
Consider $$ 0 \to \mathfrak a \xrightarrow i A \xrightarrow \pi A / \mathfrak a \to 0$$ where $i : \mathfrak a \to A$ is the inclusion map: $x \mapsto x$, and $\pi : A \to A / \mathfrak a$ is the quotient map: $x \mapsto x + \mathfrak a$. This is a short exact sequence. By Proposition 2.18, $$\mathfrak a \otimes M \xrightarrow{i \otimes 1} A \otimes M \xrightarrow{\pi \otimes 1} (A/\mathfrak a) \otimes M \to 0$$ is exact. This means that $\ker \pi \oplus 1 = \im i \otimes 1$ and $\pi \otimes 1$ is surjective. By the fundamental theorem of module homomorphisms , $(A \otimes M) / \ker (\pi \otimes 1) \cong (A/\mathfrak a) \otimes M$. We show that $(A \otimes M) / \ker (\pi \otimes 1) \cong M/\mathfrak a M$.
Define $f : A \times M \to M/\mathfrak aM$ by $(a, m) \mapsto am + \mathfrak aM$ and $g : A \times M \to A \otimes M$ by $(a, m) \mapsto a \otimes m$. By Proposition 2.12, there exists a unique $f'$ such that $f = f' \circ g$: $$%\xymatrix{A \times M \ar[r]^g \ar[dr]_{f} & A \otimes M \ar[d]^{f'}\\ & M / \mathfrak aM }$$ Observe that $f'$ is surjective, since given any $\overline m = m + \mathfrak aM \in M/ \mathfrak aM$, we have $f'(1 \otimes m) = m + \mathfrak aM$. By the fundamental theorem of module homomorphisms, it follows that $A \otimes M / \ker f' \cong M/ \mathfrak aM$. 
We show that $\ker f' = \ker (\pi \otimes 1)$: Let $\sum_{k = 1}^n a_k \otimes m_k \in \ker (\pi \otimes 1) = \im (i \otimes 1)$ and observe that $$f' \bigg( \sum_{k = 1}^n a_k \otimes m_k \bigg) = \sum_{k = 1}^n f'(a_k \otimes m_k) = \sum_{k = 1}^n a_k m_k + \mathfrak aM = 0$$ and therefore $\sum_{k = 1}^n a_k \otimes m_k \in \ker f'$. Hence we have $\ker(\pi \otimes 1) \subseteq \ker f'$. Let $x \in \ker f'$ and suppose, without loss of generality, that the element we chose is $1 \otimes m$ (we can do this since $\sum a_k \otimes m_k = 1 \otimes \sum a_k m_k$). Hence we have $f'(1 \otimes m) = 1\cdot m + \mathfrak aM = m + \mathfrak aM = 0$. Thus $m \in \mathfrak aM$, so $m = \sum_{k = 1}^n a_i m_i$ where $a_i \in \mathfrak a$ and $m_i \in M$. It follows immediately that $1 \otimes m = 1 \otimes \sum_{k = 1}^n a_k m_k = \sum_{k = 1}^n a_k \otimes m_k \in \im (i \otimes 1) = \ker (\pi \otimes 1)$. Hence we have $\ker f' \subseteq \ker (\pi \otimes 1)$ and thus $\ker f' = \ker (\pi \otimes 1)$.
