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Let $R$ be a ring and $k$ be a field such that $k\hookrightarrow R$. Thus, given two $R$-modules $M$ and $N$, we can regard $\operatorname{Hom}(M,N)=\operatorname{Hom}_R(M,N)$ as a vector space over $k$. The following facts are given:

  1. If $E$ and $E'$ are two simple $R$-modules, we know that the dimension of $\operatorname{Hom}(E,E')$ is at most $1$.

  2. Both $M$ and $N$ have finite length, i.e. they have a finite sequence $0=M_0\subsetneq M_1\subsetneq\dots\subsetneq M_n=M$ such that $M_i/M_{i-1}$ is simple, and similarly for $N$.

Now, I should prove by induction over the length of $M$ that the dimension of $\operatorname{Hom}(M,E'')$ is finite for every simple $R$-module $E''$. The case length $=1$ is obvious, but how to do the induction step?

Ultimately, I want to prove that the dimension of $\operatorname{Hom}(M,N)$ is finite, but I think that I will be able to do it myself if I understood the $\operatorname{Hom}(M,E'')$-case.

Thank you!

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1 Answer 1

up vote 2 down vote accepted

Suppose you know $\text{Hom}(M_i, E')$ is finite dimensional for $i<k$.

Then $0 \rightarrow M_{k-1} \rightarrow M_k \rightarrow M_{k}/M_{k-1} \rightarrow 0$ is exact, whence

$0 \rightarrow \text{Hom}(M_{k}/M_{k-1}, E') \rightarrow \text{Hom}(M_k, E') \rightarrow \text{Hom}(M_{k-1}, E')$

is exact. But then $Hom(M_k, E')/\text{Hom}(M_k/M_{k-1}, E')$ is a subspace of a finite-dimensional vector space, so it's finite dimensional, and we also know the bottom guy is finite dimensional by assumption, so we're done.

Now you know it for $Hom(M, E')$, and the case for $Hom(M, N)$ from here is even easier I think.

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Ah - so basically, I have to use the right-exactness of the functor $\operatorname{Hom}(\cdot, E')$, which I had forgotten. Thank you for pointing that out! :) The other part of my question is proven similarly by the left-exactness of $\operatorname{Hom}(N,\cdot)$... –  Sh4pe Jan 29 '13 at 15:58
    
Yup! No problem –  Dylan Wilson Jan 29 '13 at 17:58

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