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If I extend scalars from a ring $R$ to a ring $S$ by a homomorphism $f:R \to S$, then starting with an $R$ module $M$, I get an $S$ module $S \otimes_R M$. Given $\sigma \in \text{End}_R(M)$, I know that $1 \otimes \sigma \in \text{End}_S(S \otimes _R M)$. But is there any other way to build elements of $\text{End}_S(S \otimes_R N)$?

I ask because I'm thinking about extending scalars from $k$ to $K$ on a vector space, and I'm trying to get a hold on $\text{End}_K(V \otimes_k K)$, besides just elements of the form $\sigma \otimes 1$, for $\sigma \in \text{End}_k(V)$.

Eric Auld
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We have $$\text{End}_S(S \otimes _R M)=\text{Hom}_S(S \otimes _R M,S \otimes _R M)\simeq\text{Hom}_R(M,S \otimes _R M)\simeq\text{Hom}_R(M,M)\otimes _R S=\text{End}_R(M)\otimes_RS$$ when $S$ is $R$-flat and $M$ finitely presented; see Hom and tensor with a flat module.

In general it's not true, but for the case of finitely dimensional vector spaces we can't have other endomorphisms.

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user26857
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  • Where is the isomorphism $\text{Hom}_S(S \otimes_R M, S \otimes_R M) \simeq \text{Hom}_R(M, S \otimes_R M)$ coming from? I don't see anything like that at the link you mention. – Eric Auld Mar 13 '15 at 21:08
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    @EricAuld Well, I've thought you know the basic properties of tensor product, and this is why I didn't mention http://en.wikipedia.org/wiki/Tensor-hom_adjunction (The link refers to the second isomorphism which is not well known.) – user26857 Mar 13 '15 at 21:10
  • It seems we're doing $\text{Hom}_S(S \otimes_R M, S \otimes_R M) \simeq \text{Hom}_R(M, \text{Hom}_S(S,S \otimes_R M)) \simeq \text{Hom}_R(M, S \otimes_R M)$. In doing so, we consider an R isomorphism $\text{Hom}_S(S,S \otimes_R M) \xrightarrow{\sim} S \otimes_R M$. To regard $S \otimes_R M$ as a left $R$ module, do we need to assume that $f(R)$ is in the center of $S$? – Eric Auld Mar 13 '15 at 21:34
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    @EricAuld I've considered all rings as being commutative, because I wanted to show that the case you are interested in works. If you want to take a broader setting, then yes. (Anyway, the general case fails for commutative rings unless $S$ is $R$-flat.) – user26857 Mar 13 '15 at 21:36
  • We need that $M$ is finitely presented over $R$ to get $\text{Hom}_R(M,S \otimes _R M)\simeq\text{Hom}_R(M,M)\otimes _R S$, right? So we'd need to be talking about a finite-dimensional vector space? – Eric Auld Mar 14 '15 at 05:08
  • You may want to include in your answer that the same holds if $S$ is not necessarily $R$-flat, but $M$ is finite free. I'll provide a reference for anyone looking at this page: page 15 http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod2.pdf – Eric Auld Mar 21 '15 at 10:32