If $R$ is a commutative ring, $S$ is a commutative $R$ algebra, and $M$ is a finite free $R$ module, I can show $$S \mathop{\otimes}\limits_R \mathsf{End}_R(M) \cong \mathsf{End}_S(S \mathop{\otimes}\limits_R M) \rlap{\quad \text{as $S$ algebras.}}$$ Can I broaden this at all past just finite free $M$? Do you know a counterexample when $M$ is merely finitely generated, for instance?
Base extension commutes with taking endomorphism rings
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abstract-algebra
commutative-algebra
tensor-products
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1Bourbaki (Algebra II par. 5 no.3) says you have an iso in either of the cases 1) $M$ f.g. projective or 2) $S$ f.g. projective as $R$-module. – 2017-02-18
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0Actually, Bourbaki gives a stronger statement about $Hom(N, M)$, not just $End(M)$, which might help, since only $N$ needs to be f.g. projective. -- A (not finitely generated) counterexample would be $M = \mathbb{Q}_p/\mathbb{Z}_p$, $R =\mathbb{Z}_p$, $S= \mathbb{Q}_p$ (p-adics); if I'm not mistaken, we have $End_R(M) = R$ here and hence the left hand side $\mathbb{Q}_p$, the right hand side $0$. – 2017-02-19