Recovering the center of a ring

Question

Suppose $R$ is a ring and $M$ is a projective left $R$-module. Suppose also that $M$ is a generator, meaning that every left $R$-module is a quotient of some possibly infinite direct sum of copies of $M$. Is it true that the center of the ring of endomorphisms of $M$ is isomorphic to the center of $R$?

Or do we maybe need more hypotheses, like that $M$ is finitely generated?

(I would prefer conditions on $M$ that can be stated using just the structure of the category of left $R$-modules as an additive category. 'Projective' and 'generator' are conditions of this sort. 'Finitely generated' is not such a condition, as far as I can tell, since to state it we need to know which object in the category of modules is $R$.)

Answer 1

At this moment I can't say either way when $M$ is infinitely generated, but when $M$ is finitely generated, you have a progenerator. At that point, $R$ is Morita equivalent to $\mathrm{End}(M_R)$ and Morita equivalent rings indeed have isomorphic centers.

Also,"finitely generated" is usually considered a categorical property, if that helps ease your reservation. I think it is not necessary to know which object is $R$, since in some sense the generators of $\mathrm{Mod}_R$ are interchangeable, and they have a categorical description.

Big fan, by the way.

Answer 2

I think it's true that $Z(\operatorname{End}(G))\cong Z(R)$ for any generator $G$ (projective or not) of $\operatorname{Mod} R$.

I've not been able to find an explicit reference, but I think the following sketch of a proof works, although I may have missed some details. Or (equally likely) I may have made it more complicated than necessary.

What I shall actually show is that $Z(\operatorname{End}(G))$ is isomorphic to the centre $Z(\operatorname{Mod}R)$ of the module category (i.e., the endomorphism ring of the identity functor on $\operatorname{Mod}R$). It is well known that $Z(R)$ is isomorphic to $Z(\operatorname{Mod}R)$, but in any case the following argument will prove that when you apply it to $G=R$.

If $\alpha:\operatorname{id}_{\operatorname{Mod}R}\to\operatorname{id}_{\operatorname{Mod}R}$ is a natural transformation, then $\alpha_G\in Z(\operatorname{End}(G))$ as by naturality it commutes with any other endomorphism. So $S(\alpha)=\alpha_{G}$ defines a homomorphism $S:Z(\operatorname{Mod}R)\to Z(\operatorname{End}(G))$.

Conversely, suppose $\beta:G\to G$ is in the centre of $\operatorname{End}(G)$. For any $R$-module $M$, we can associate to it an exact sequence $$\bigoplus_JG\xrightarrow{\varphi}\bigoplus_IG\xrightarrow{\theta} M\to 0$$ in a functorial way. For example, let $I=\operatorname{Hom}(G,M)$ with $\theta$ acting on each summand of $\bigoplus_IG$ according to the element of $\operatorname{Hom}(G,M)=I$ labelling that summand, and similarly take $J=\operatorname{Hom}(G,K)$ where $K$ is the kernel of $\theta$.

Now $\beta$ induces endomorphisms $\beta_J$ and $\beta_I$ of $\bigoplus_JG$ and $\bigoplus_IG$. By the fact that $\beta$ is central in $\operatorname{End}(G)$, $\beta_I\varphi=\varphi\beta_J$, and so there is an induced map $\beta_M:M\to M$: i.e., a unique map so that $\beta_{M}\theta=\theta\beta_{I}$.

This is all functorial and additive, so we have produced a homomorphism $T:Z(\operatorname{End}(G))\to Z(\operatorname{Mod}R)$, given by $T(\beta)=(\beta_M)_{M\in\operatorname{Mod}R}$.

If $M=G$, then $\beta\theta=\theta\beta_{I}$, so $\beta_{G}=\beta$, and so $S\circ T$ is the identity on $Z(\operatorname{End}(G))$.

Also, $S$ is injective. If $\alpha\in Z(\operatorname{Mod}R)$ with $\alpha_{G}=0$, then by applying it to the exact sequences used in the construction of $T$, we get $\alpha_{M}=0$ for all $M$, so $\alpha=0$.

So $S$ and $T$ are mutually inverse isomorphisms.