Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A/\mathcal m$.

share|cite|improve this question
up vote 1 down vote accepted

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).

share|cite|improve this answer
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… – Veen Oct 29 '11 at 21:19
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.