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Let $(A_i)_{i\in I}$ be a family of right $R$-modules and $M$ be a left $R$-module, where $I$ is an index set. The natural homomorphism $$\varphi:(\prod_{i\in I}A_i)\otimes_RM\to \prod_{i\in I}(A_i\otimes_RM)$$ given by $(a_i)\otimes m\mapsto(a_i\otimes m)$ is not always a bijection. It is easy to obtain $\varphi$ is a surjection provided $M$ is finitely generated. If we assume that $M$ is finitely presented, can we prove $\varphi$ is a bijection?

Crostul
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liwolf
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1 Answers1

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$M$ being finitely presented means that there is an exact sequence $$R^m \to R^n \to M \to 0.$$ This gives a commutative diagram $$\require{AMScd}\begin{CD} \bigl(\prod_i A_i\bigr) \otimes_R R^m @>>>\bigl(\prod_i A_i\bigr) \otimes_R R^n @>>> \bigl(\prod_i A_i\bigr) \otimes_R M @>>> 0\\ @VVV @VVV @VVV \\ \prod_i (A_i \otimes_R R^m) @>>> \prod_i (A_i \otimes_R R^n) @>>> \prod_i (A_i \otimes_R M) @>>> 0 \end{CD}$$ with exact rows since tensoring is right exact. The first and second vertical arrows are easily seen to be isomorphisms. Then by the five lemma, the right vertical arrow is an isomorphism, too.

marlu
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  • How exactly do you apply the 5 Lemma here? – Flavius Aetius Nov 15 '22 at 22:32
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    If you extend both rows by another zero on the right, the five lemma applies. – marlu Nov 16 '22 at 19:12
  • Yes, I agree. Thanks. Do you use somewhere implicitly that $N$ is a flat $A$-module. I do not think but I want to be sure. Also, I adapted your proof (hopefully correctly) to inverse limits: See : https://math.stackexchange.com/questions/181004/inverse-limit-of-modules-and-tensor-product/4577954#4577954. I would be interested in your opinion on this generalization. . – Flavius Aetius Nov 16 '22 at 21:22
  • No, I'm not using flatness anywhere. I'll have a look at your proof. – marlu Nov 17 '22 at 01:36
  • Marlu, does your proof go through for modules over a non-commutative base ring $R$? The question is posed for a non-commutative base ring $R$. I think that it should work for a non-commutative ring $R$, too, as the Five Lemma works in the non-commutative setting, as well. Do you agree? – Flavius Aetius Sep 17 '23 at 21:38
  • I agree. The diagram in question is actually merely a diagram of abelian groups if $R$ is non-commutative. So we are just using the Five Lemma for abelian groups. – marlu Sep 18 '23 at 22:18
  • Marlu, thank you for the input! – Flavius Aetius Sep 25 '23 at 09:26