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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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    I got the point. Thank$s$ a lot, Georges!2011-10-29