How to "Hom" the canonical sequence in a ring I'm stuck with the dual of an exercise in Atiyah-MacDonald.
It's easily seen that tensoring over $R$ with $-\otimes M$ the sequence
$$
0\to \mathfrak a\to R\to R/\mathfrak a\to 0
$$
one gets the isomorphism $R/\mathfrak a\otimes M\cong M/\mathfrak aM$.
What can I deduce from "Hom(-,M)ing" the same sequence? I have, let's say
$$
0\to \mathfrak a\xrightarrow{\alpha} R\xrightarrow{\beta} R/\mathfrak a\to 0
$$
and 
$$
0\to \hom(R/\mathfrak a,M) \xrightarrow{\beta^*} \hom(R,M)\cong M\xrightarrow{\alpha^*} \hom(\mathfrak a,M)
$$
but now what can I do? What is $\hom(R/\mathfrak a,M)$ like?
I know, it's easy, but...

Hint: Any $R$-module homomorphism $f$ from $R/\mathfrak{a}$ to $M$ is determined once we know $f(1)$. But every such homomorphism is killed by $\mathfrak{a}$ since $\mathfrak{a}f(1)=f(\mathfrak{a})=f(0)=0$. So one can show that, $Hom_R(R/\mathfrak{a},M)\cong Ann_{\mathfrak{a}}(M)$ 
