socle(M) being simple gives an upper bound for the dimension of End(M)?

Question

Suppose $k$ is a algebraically closed field of arbitrary characteristic. Let $A$ be a finite dimensional $k$-algebra and $M$ an $A$-module with finite dimension with respect to $k$.

I have seen it claimed that:

If the socle $\text{soc}(M)$ of $M$ is simple, then $\dim_k(\text{End}_A(M))$ is at most the multiplicity of $\text{soc}(M)$ as a composition factor for $M$.

So far I have not had much success proving this statement and any help would be greatly appreciated.

Edit: This claim comes from Chuang and Rouquier's paper Derived equivalences for the symmetric groups and $\mathfrak{sl}_2$-categorification. In particular, inside the proof of proposition 5.20. Above is a simplified version. There is also an extra bit of data that head($M$) is simple which may be needed for this to be true.

Answer 1

More generally, if $N$ is a finite dim'l $A$-module whose socle $S: = $ soc($N$) is simple, and $M$ is any finite dim'l $A$-module, we have $\dim \mathrm{Hom}_A(M,N) \leq m$, where $m$ denotes the mult. of $S$ as a comp. factor of $M$.

Proof: Let $M'$ be the maximal submodule of $M$ that does not contain $S$ as a comp. factor. Then since $S$ is the socle of $N$, we see that any homomorphism $S \to N$ must vanish. Thus we may replace $M$ by $M/M'$, and so assume that there is a copy of $S$ in soc$(M)$; call it $S_1$.

Now consider the exact sequence $$0 \to \mathrm{Hom}_A(M/S_1,N) \to \mathrm{Hom}_A(M,N) \to \mathrm{Hom}_A(S_1,N) = \mathrm{Hom}_A(S_1,S).$$ Since $S_1 \cong S$ are isomorphic simple modules, the right-hand space is one dimensional, and so we find that $$\dim \mathrm{Hom}_A(M,N) \leq \dim \mathrm{Hom}_A(M/S_1,N) + 1.$$ The mult. of $S$ as a comp. factor of $M/S_1$ is $m-1$, and so by induction we get the desired result.

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Taking $M = N$ gives the result you asked about.

If instead we choose $N$ to be an injective envelope of $S$, then the above exact sequence is exact on the right too, so we get an equality rather than just an inequality, and so obtain the useful formula that in this case $\dim \mathrm{Hom}_A(M,N) = m.$

Answer 2

If $N$ is a simple module, then $$\dim_k\operatorname{Hom}_A(N,M)= \begin{cases} 1 &\mbox{if $N\cong\operatorname{soc}(M)$}\\ 0 &\mbox{otherwise} \end{cases}$$ By induction on the composition length of $N$, it follows that $\dim_k\operatorname{Hom}_A(N,M)$ is at most the multiplicity of $\operatorname{soc}(M)$ as a composition factor of $N$, for any finite length module $N$. The inductive step uses that if $0\to N'\to N\to N''\to0$ is a short exact sequence of modules then there is an exact sequence $$\operatorname{Hom}_A(N'',M)\to\operatorname{Hom}_A(N,M)\to\operatorname{Hom}_A(N',M).$$