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I want a proof for

$$\operatorname{Hom}_R(M,N) \otimes_RS \cong \operatorname{Hom}_S(M\otimes_R S,N\otimes_R S)$$

where $\phi\colon R \to S$ is a homomorphism and $M$ is finitely generated free $R$-module and $N$ is an $R$-module.

or if anyone can say me about the isomorphism map between two side.

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    Hint: If $M$ is a finitely generated free $R$-module, then $M\cong R^n$ for some $n$. You then have $Hom_R(M,N)\cong N^n$ (why?). Similarly, $M\otimes_R S\cong S^n$, so the right hand side is isomorphic to $(N\otimes_R S)^n$.2012-05-26
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    ...so the LHS is $N^n \otimes_R S$, and the RHS $(N \otimes_R S)^n$, which are isomorphic cuz tensor product commutes with direct sums2012-05-26
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    @froggie Please consider posting an answer, so that the question does not remain without one.2013-06-18
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    @ˈjuː.zɚ79365: answer added. Also, if ever in the future you see a question left unanswered by me that you can answer, please feel free to add an answer yourself. I won't mind, seriously.2013-06-18
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    This is only homological in punland.2013-06-18
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    See also https://math.stackexchange.com/questions/2167674/dual-commutes-with-base-change2018-01-11
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    Related: https://math.stackexchange.com/questions/506992018-11-22

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