2
$\begingroup$

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.

  • 0
    Related: https://math.stackexchange.com/questions/506992018-11-22

1 Answers 1

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 $Hom(M,N) \cong N^n$. Thus the left hand side of the desired isomorphism is $Hom(M,N)\otimes_RS\cong N^n\otimes_R S.$

  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 $Hom_S(M\otimes_R S, N\otimes_R S)\cong (N\otimes_RS)^n.$

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 $f\colon Hom(M,N)\otimes_R S\to Hom_S(M\otimes_R S, N\otimes_R S)$ 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.

  • 0
    "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 $M$.2013-06-18