$(A,\mathfrak{m})$ 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?

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 $\ell(\mbox{Hom}(M,E(A/\mathfrak{m}))=(n1)+1=n$ 

