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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.

Martin
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moahmmad
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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 $(N\otimes_R S)^n$. – froggie May 26 '12 at 12:45
  • ...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 sums – uncookedfalcon May 26 '12 at 13:52
  • @froggie Please consider posting an answer, so that the question does not remain without one. – ˈjuː.zɚ79365 Jun 18 '13 at 11:19
  • @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. – froggie Jun 18 '13 at 13:39
  • This is only homological in punland. – Loki Clock Jun 18 '13 at 13:43
  • See also https://math.stackexchange.com/questions/2167674/dual-commutes-with-base-change – Watson Jan 11 '18 at 10:14
  • Related: https://math.stackexchange.com/questions/50699 – Watson Nov 22 '18 at 19:14

1 Answers1

2

  1. Since $M$ is a finitely generated free $R$-module, one has $M\cong R^n$ for some $n$ (this statement is equivalent to choosing a basis $e_1,\ldots, e_n$ for $M$).

  2. Moreover, since $M$ is free, any homomorphism $\phi\colon M\to N$ can be specified completely by specifying the images $\phi(e_1),\ldots, \phi(e_n)$ of the basis vectors. This says exactly that $\operatorname{Hom}(M,N) \cong N^n$. Thus the left hand side of the desired isomorphism is $$\operatorname{Hom}(M,N)\otimes_RS\cong N^n\otimes_R S.$$

  3. Since $M\cong R^n$ and tensor product commutes with direct sums, we have that $M\otimes_RS\cong S^n$. In terms of our previously chosen basis $e_1,\ldots, e_n$ of $M$, this statement says simply that $M\otimes_R S$ is a free $S$-module with basis $e_1\otimes 1,\ldots, e_n\otimes 1$.

  4. Just as in point (2), since $M\otimes_R S$ is a free $S$-module, specifying a homomorphism $\phi\colon M\otimes_R S\to N\otimes_RS$ is equivalent to specifying the images $\phi(e_1\otimes 1),\ldots, \phi(e_n\otimes 1)$ of the basis vectors of $M\otimes S$. Thus $$\operatorname{Hom}_S(M\otimes_R S, N\otimes_R S)\cong (N\otimes_RS)^n.$$

Thus, combining points (2) and (4), in order to prove the desired isomorphism it suffices to show that $N^n\otimes_R S\cong (N\otimes_R S)^n$. But this is exactly the fact that tensor product commutes with direct sums. This completes the proof.

It is possible to give a simple description of the isomorphism $$f\colon \operatorname{Hom}(M,N)\otimes_R S\to \operatorname{Hom}_S(M\otimes_R S, N\otimes_R S)$$ we just constructed. It is given on simple tensors by $$f(\phi\otimes s)(m\otimes t) = \phi(m)\otimes st.$$ Despite the simplicity of this definition, it is not clear that $f$ is indeed an isomorphism. The above argument proves that $f$ is an isomorphism when $M$ is a finitely generated free $R$-module.

Adam Higgins
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froggie
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    "The above argument proves ..." No. You have just constructed some isomorphism. A priori it is not clear that it coincides with the natural homomorphism (which exists without any assumptions on $M$). A better proof starts with this natural homomorphism, and remarks that a) it is an isomorphism for $M=R$, b) the class of $M$ where it is an isomorphism is closed under finite direct sums (this is purely formal), and hence concludes that it holds for finitely generated free $M$. – Martin Brandenburg Jun 18 '13 at 14:00