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Let $R$ is commutative local ring, $M$ is $R$-module, $N$ is $R$-module. If $M,N$ are finitely generated, how to prove:

$M \otimes_R N=0$ if and only if $M=0$ or $N=0$.

If we delete the condition (local ring), then is the statement right?

A proof is existed by Nakayama lemma. Suppose $M \otimes_R N=0$, $J$ is maximal ideal in $R$, let $k=\frac{R}{J}$ so $k$ is field. $M_k=M\otimes k=M \otimes \frac{R}{J}=\frac{M}{JM}$.

If we delete the local condition the statement is not true. Example: $(m,n)=1 $ then $\Bbb Z_m \otimes \Bbb Z_n=0 (???)$, $\Bbb Z_m, \Bbb Z_n \not = 0$.

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  • $\begingroup$ @user26857 I wrote some proof by nakayama lemma . and Jacobson radical . I want the proof that is not used them , also counterexamples for deleted condition(local ring) I wrote that example but I don't understand that (???) $\endgroup$ – agustin Nov 12 '14 at 19:45
  • $\begingroup$ @user26857 in other words I don't understand the proof is used nakayama lemma completely. $\endgroup$ – agustin Nov 12 '14 at 19:48
  • $\begingroup$ For future references, here is a proof using Nakayama lemma. $\endgroup$ – Najib Idrissi Sep 9 '15 at 14:44
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About the proof: is it clear why $M\otimes_R R/J\simeq M/JM$? If yes, then consider $M\otimes_R N=0$ and tensor it (over $R$) with $R/J$ left and right. Then you get $M/JM\otimes_{R/J} N/JN=0$. Thus you have two vector spaces, namely $M/JM$ and $N/JN$, whose tensor product is $0$. But non-zero vector spaces have bases, so their tensor product, and then we have to admit that $M/JM=N/JN=0$. Now is time for Nakayama's lemma to enter the scene.

For the non-local case: $\mathbb Z_m=\mathbb Z/m\mathbb Z$, and using again that $M/IM\simeq M\otimes_R R/I$ we get $\mathbb Z_m\otimes_{\mathbb Z}\mathbb Z_n\simeq\mathbb Z_n/m\mathbb Z_n$. But the last one is isomorphic to $\mathbb Z/(m\mathbb Z+n\mathbb Z)$, and since $m\mathbb Z+n\mathbb Z=\gcd(m,n)\mathbb Z$ you are done.

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  • $\begingroup$ It's Ok . just two problems for me. P1:Why $M/JM\otimes_{R/J} N/JN=0$ ,,,,,,,,,,,,P2:Why $\mathbb Z/(m\mathbb Z+n\mathbb Z)\simeq\mathbb Z_n/m\mathbb Z_n$ $\endgroup$ – agustin Nov 12 '14 at 20:22
  • $\begingroup$ $M/JM\otimes_{R/J}N/JN=(M\otimes_RR/J)\otimes_{R/J}N/JN\simeq M\otimes_R N/JN$. (I've just realized that it's enough to tensor $M\otimes_RN=0$ with $R/J$ only on one side.) Here I've used a property of tensor product like $M\otimes_S S\simeq M$, where $S$ is a ring. $\endgroup$ – user26857 Nov 12 '14 at 20:26
  • $\begingroup$ $\mathbb Z_n=\mathbb Z/n\mathbb Z$ and $m(\mathbb Z/n\mathbb Z)=(m\mathbb Z+n\mathbb Z)/n\mathbb Z$. $\endgroup$ – user26857 Nov 12 '14 at 20:29
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    $\begingroup$ @user Yes, of course. Even more general, if $F$ is another ring then the tensor product is an $F$-module. $\endgroup$ – user26857 Oct 13 '15 at 14:14
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    $\begingroup$ @user $F$ can be any $R$-algebra. $\endgroup$ – user26857 Oct 13 '15 at 15:16

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