$\DeclareMathOperator\Hom{Hom}$Fix a commutative ring $R$. For $R$-modules $M$ and $N$, there is an inclusion of abelian groups $\Hom\_R(M,N) \to \Hom\_{\mathbb{Z}}(M,N).$ Are there conditions on $R$ that will ensure this is an inclusion of a direct summand (preferably with a section natural in $M$ and $N$)?


(I have tried googling this question, but all the words are too common. Pointers to the right terminology would also be appreciated!)



R-module hom a direct summand of Z-module hom?
There exists a natural section if and only if $R$ is *separable* over $\mathbb Z$ (more generally over $k$, if you have a $k$-algebra $R$ and are asking about the morphism $\hom\_R(M,N)\to \hom\_k(M,N)$). If you're only asking for a section, no naturality, then I don't know any reasonable condition, but maybe someone can chime in and say something about it.


Here's a proof of my claim about separability (I'll do it over a base commutative ring $k$, because $\mathbb Z$ adds nothing specific here): $\hom\_k(M,N)\cong \hom\_R(R\otimes\_k M, N)$ and the forgetful map $\hom\_R(M,N)\to \hom\_k(M,N)$ corresponds to precomposition by the action map $R\otimes\_k M\to M$ under this equivalence.


So by the Yoneda lemma, a natural section $\hom\_R(R\otimes\_k M, {-}) \to \hom\_R(M,{-})$ is the same thing as a natural section $M\to R\otimes\_k M$ to the action map $R\otimes\_k M\to M$ (note that we use naturality in $N$ to get a section $M\to R\otimes\_k M$, and naturality in $M$ to get that this section is natural).


We now apply this to $M = R$, we get a section $R\to R\otimes\_k R$ to the multiplication. Now, this is very weak as is, but this section is natural in endomorphisms of $R$ as a left $R$-module, i.e. in multiplications on the right by elements of $R$. If you unravel what this naturality means, it will imply that $R\to R\otimes\_k R$ is a section of $R\otimes\_k R^\text{op}$-modules ($R$-bimodules over $k$), which is what is called a "separability idempotent" for $R$. Recall:



> 
> **Definition** : A $k$-algebra $R$ is called separable (over $k$) if the multiplication map $R\otimes\_k R^\text{op}\to R$ admits a $R\otimes\_k R^\text{op}$-module section.
> 
> 
> 


The converse is not hard to show: if $R$ is separable, then the action map has a natural section $M = R\otimes\_R M \to (R\otimes\_k R)\otimes\_R M = R\otimes\_k M$, and so $\hom\_R(M,N)\to \hom\_R(R\otimes\_k M, N) = \hom\_k(M,N)$ has a natural section too.


**Examples of separable algebras**:


* Finite étale extensions of $k$ are separable.
* Matrix rings over commutative rings are separable over their base (they all "look like this" étale-locally).
* A group ring $k[G]$ is separable if and only if $\lvert G\rvert\in k^\times$.
* Finally, separable algebras are closed under many familiar operations: taking the center, finite products, taking tensor products (over the same base or over a different base).


