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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...

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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, $Hom_R(R/\mathfrak{a},M)\cong Ann_{\mathfrak{a}}(M)$

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    @Mariano: Thanks, modified. @Matt: I have not really seen that notation before. I have only seen the one I used in commutative algebra books and papers, though sometimes the positions of $a$ and $M$ are switched.2010-11-15
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Asking «what is $M/\operatorname{im}\alpha^*$?» is not the dual of the A-M exercise.

If you really want a dual statement, you should be looking for a description of $\hom(R/\mathfrak a,M)$.

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    Sure. The two comments are meaningless, I remove them.2010-11-15