$(A,\mathfrak{m})$ an Artin local ring, $E(A/\mathfrak{m})$ the injective hull of $A/\mathfrak{m}$. How do I see that $E(A/\mathfrak{m})$ is a finite $A$module?

2$\begingroup$ You can prove, for example, that $\ell(M)=\ell(\hom(M,E(A/\mathfrak{m}))$ for all $A$modules, and then specialize to $M=A$ to see that $\ell(E(A/\mathfrak{m})=\ell(A)$ (here $\ell$ is the length) $\endgroup$– Mariano SuárezÁlvarezCommented Aug 6, 2010 at 16:42

6$\begingroup$ You have been asking a stream of questions that are answered in most textbooks: you should really research a bit (even Google books should be of help!) before asking, really... $\endgroup$– Mariano SuárezÁlvarezCommented Aug 6, 2010 at 16:43

1$\begingroup$ Mariano, please feel free not to answer my questions if they seem unworthy of your time. $\endgroup$– ashpoolCommented Aug 7, 2010 at 3:45

$\begingroup$ Dear Kwan, you might want to think about the case when $A$ is a finitedimensional local $k$algebra. In this case, taking $k$vector space duals of the surjection $A \to k$ will give an injection $k \to A^*$ which is the injective hull of $k$. This is illustrative of the more general situation you consider. It is good to play with this, taking various choices of $A$ and then exploring the structure of $A^*$ as an $A$module; it might help in lessening the mystery of these injective envelopes (they are less of a black box than one might imagine from the general theory). $\endgroup$– EmertonCommented Aug 16, 2010 at 19:54
1 Answer
This is a proof of $\ell(M)=\ell(\mbox{Hom}(M,\mbox{E}(A/\mathfrak{m}))$ suggested by Mariano:
Induction on $\ell(M)\ $:
If $\ell(M)=0$, $M=0$ so obviously true. Suppose $\ell(M)=n\geq 1$. From a composition series of $M$ choose the submodule N right beneath M so that $\ell(N)=n1$ and $M/N\simeq A/\mathfrak{m}$. $0\rightarrow N\rightarrow M \rightarrow A/\mathfrak{m}\rightarrow 0$ induces $0\leftarrow \mbox{Hom}(N,E(A/\mathfrak{m}))\leftarrow \mbox{Hom}(M,E(A/\mathfrak{m}))\leftarrow \mbox{Hom}(A/\mathfrak{m},E(A/\mathfrak{m}))\leftarrow 0$.
Now $\mbox{Hom}(A/\mathfrak{m},E(A/\mathfrak{m}))\simeq A/\mathfrak{m}$ since $E(A/\mathfrak{m})$ is an essential extension of $A/\mathfrak{m}$, and $\ell(\mbox{Hom}(N,E(A/\mathfrak{m})))=\ell(N)=n1$ by the induction hypothesis. $\ell(A/\mathfrak{m})=1$ so $\ell(\mbox{Hom}(M,E(A/\mathfrak{m}))=(n1)+1=n$