Why is $\text{Hom}_S(k\otimes_R S, M\otimes_R N) \cong \text{Hom}_R(k,M) \otimes_R N$?

Asked Oct 25 '17 at 01:02

Active Oct 28 '17 at 22:29

Viewed 224 times

Here is the situation.

$\varphi: (R,\mathfrak{m},k) \to (S, \mathfrak{n},\ell)$ is a ring homomorphism. $M$ is a finitely generated $R$-module. $N$ is a finitely generated $S$-module and is flat over $R$.

I know there is a map $\Phi: \text{Hom}_R(k,M) \otimes_R N \to \text{Hom}_S(k\otimes_R S, M\otimes_R N)$ given by $f \otimes n \mapsto f\otimes (1_S \mapsto n)$, but how can I use the fact that $N$ is flat over $R$ and finitely generated to show that this is an isomorphism?

edited Oct 28 '17 at 22:29

asked Oct 25 '17 at 01:02

JJC94

This could be helpful (at least to handle the left hand side). If you don't want to assume $R$ noetherian then notice that in the linked thread the isomorphism holds for $M$ finitely presented. This means that here we have to suppose $\mathfrak m$ finitely generated. – user26857 Oct 28 '17 at 20:05
Yes, you're right. I fixed it. – JJC94 Oct 28 '17 at 22:29
Also, that link is useful. Thank you. – JJC94 Oct 28 '17 at 22:55
Related: https://math.stackexchange.com/questions/50699 – Watson Nov 22 '18 at 19:15

Why is $\text{Hom}_S(k\otimes_R S, M\otimes_R N) \cong \text{Hom}_R(k,M) \otimes_R N$?

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