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Let $(R,m,k)$ be a local ring, and $M$ be an $R$-module of finite length. Suppose the minimal number of generators $\mu(M)$ of $M$ is $n$. What is the relation of the minimal number of generators $\mu(Hom_R(M,M))$ of $Hom_R(M,M)$ with $n$? Since $k$ is not necessarily flat over $R$, it's not necessarily true that $Hom_R(M,M)\otimes k\cong Hom(M\otimes k, M\otimes k)\cong k^{n^2}$. How does $\mu(Hom_R(M,M))$ compared to $n^2$ in general?

Flying pencil
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1 Answers1

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They don’t really compare in general. Let $R=k[x,y]/(x,y)^n$. Then $(R,\mathfrak{m},k)$ is an Artinian local ring, so every finitely generated module over $R$ has finite length.

First let $M=E_R(k)$, the injective hull of $k$ over $R$. It is well known from the theory of Matlis duality, that $\operatorname{\operatorname{Hom}}_R(M,M) \cong R$, so that $\mu_R(\operatorname{Hom}_R(M,M))=1$, but it can be shown that $\mu_R(M)=n$.

On the other hand, we could keep the same ring but take $M=(x,y)$. Of course $\mu_R(M)=2$, but one can show $\mu_R(\operatorname{Hom}_R(M,M))=n+2$.

These examples show that $\mu_R(\operatorname{Hom}_R(M,M))$ may be as far away as you like, in either direction, from $\mu_R(M)^2$. Both examples require some work/theory to show the relevant claims, but I will leave this as an exercise.

What we do always have is a surjection $R^{\oplus \mu_R(M)} \twoheadrightarrow M$ induces which induces an injection $\operatorname{Hom}_R(M,M) \hookrightarrow M^{\oplus \mu_R(M)}$. This implies $$\mu_R(\operatorname{Hom}_R(M,M)) \le l_R(\operatorname{Hom}_R(M,M)) \le \mu_R(M)l_R(M).$$ This can be refined to some degree, but it’s hard to say anything more concrete without additional hypotheses.

metalspringpro
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  • I get a set of minimal generators for $Hom R(M,M)$ in the second example to be those f such that: 1. $f(x)=x,f(y)=y$, 2. $f(x)=0, f(y)\in R{n-1}$, 3. $f(y)=0, f(x)\in R_{n-1}$. There are $2n+1$ many. – Flying pencil Jun 06 '23 at 15:16
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    @Flyingpencil A general observation: there is an injection $\operatorname{Hom}_R(\mathfrak{m},\mathfrak{m}) \to \operatorname{Hom}_R(\mathfrak{m},R)$ which precomposes with the natural injection $\mathfrak{m} \to R$. This map is an isomorphism. Indeed, if $f:\mathfrak{m} \to R$ maps to an element outside of $\mathfrak{m}$ then this element is a unit, so $f$ would be surjective. But $l_R(\mathfrak{m})=l_R(R)-1$, so this cannot be. – metalspringpro Jun 06 '23 at 16:26
  • @Flyingpencil Now, let’s look for example to the $n=2$ case. Then $\mathfrak{m} \cong k^{\oplus 2}$. So, $\operatorname{Hom}_R(\mathfrak{m},R) \cong \operatorname{Hom}_R(k^{\oplus 2},R) \cong \operatorname{Hom}_R(k,R)^{\oplus 2}$. But $\operatorname{Hom}_R(k,R) \cong k^{\oplus 2}$; $\operatorname{Hom}_R(k,R)$ can be identified with the socle of $R$. So $\mu_R(\operatorname{Hom}_R(\mathfrak{m},\mathfrak{m}))=4 \ne 2n+1$. – metalspringpro Jun 06 '23 at 16:33
  • But for $n=3$, isn't it $m\cong k^{\oplus 5}$? $Hom_R(m,R)\cong Hom_R( k^{\oplus 5},R)$ and $Hom_R(k,R)\cong k^{\oplus 3}$ gives $\mu_R= 15$? – Flying pencil Jun 06 '23 at 16:49
  • @Flyingpencil The situation is a bit trickier when $n=3$. In this case $\mathfrak{m}$ is not annihilated by $\mathfrak{m}$ so cannot be identified with a vector space through the action of $R$. – metalspringpro Jun 06 '23 at 19:28
  • @Flyingpencil The set of maps you provided is a generating set, but it is not minimal. For example, in the $n=3$ case, we have the identity map $f:\mathfrak{m} \to \mathfrak{m}$ and the map $g:\mathfrak{m} \to \mathfrak{m}$ given by $g(x)=x^2$ and $g(y)=0$. But if we consider the map $h:\mathfrak{m} \to \mathfrak{m}$ given by $h(x)=0$ and $h(y)=xy$, then $h=xf-g$, so $h$ is not needed as a generator. If you trim out redundant generators, you will see that $n+2$ remain. – metalspringpro Jun 06 '23 at 19:36
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    @Flyingpencil Perhaps to be a bit more concrete, a minimal generating set for $\operatorname{Hom}_R(\mathfrak{m},\mathfrak{m})$ is the collection of maps $f$ admitting one of the following descriptions: 1. $f$ is the identity map, 2. $f(x)=0$ and $f(y)$ is a minimal monomial generator of $\mathfrak{m}^{n-1}$, or 3. $f(x)=y^{n-1}$ and $f(y)=0$. The argument I originally had in mind was homological, but I like this direct proof idea of yours better. – metalspringpro Jun 06 '23 at 20:05
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    Thanks! Now I got that the set of minimal generators should be ${id, g_{ij},h}$ where $g_{ij}(x)=x^iy^j, i+j=n-1, g_{ij}(y)=0$ and $h(x)=0, h(y)=x^{n-1}$. There are $n+2$ many. – Flying pencil Jun 06 '23 at 20:32