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Let $A$ be a local ring and $\mathcal m$ the maximal ideal, considered as an $A$-module.

Is then every $A$-module-homomorphism $\mathcal m \rightarrow A/\mathcal m$ equal to zero?

Remark: I pose this question because I read that

$Hom_A(A/\mathcal m, A/\mathcal m)$ is $A/\mathcal m$.

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No. E.g. if $A = \mathbb C[[T]],$ then $m = T \mathbb C[[T]]$ and so is isomorphic to $A$ as an $A$-module. Hence $Hom_A(m,A/m) = A/m = \mathbb C$.

In general, $Hom(m,A/m) = Hom(m/m^2,A/m)$, and so this $A$-module is in fact an $A/m$-vector space, of dimension equal to the dimension of $m/m^2$ (assuming that the latter is finite).

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But is the above statement in the remark about $Hom(A/\mathcal m, A/\mathcal m)$ right? For example, Mariano used it in an answer to a question of mine here math.stackexchange.com/questions/75673/… –  Veen Oct 29 '11 at 21:19
1  
Dear Veen, yes, it is right. More generally if $M,N$ are $A/m$-modules you can look at them also as $A$-modules and you have a canonical identification $Hom_A(M,N)=Hom_{A/m}(M,N)$. If $M=N=A/m$, you see that $Hom_A(A/m,A/m)=Hom_{A/m}(A/m,A/m)=A/m$, the last equality being a triviality about vector spaces of dimension one. –  Georges Elencwajg Oct 29 '11 at 21:40
    
I got the point. Thanks a lot, Georges! –  Veen Oct 29 '11 at 22:16

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