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I've come to think about this problem when reading a proof in Commutative Algebra by N. Bourbaki. Say, let $R$ be a commutative ring, given 3 $R-$modules $A$, $B$, $C$, and the $R$-homomorphism $f:B \to C$. Is the following equivalent?

  1. $f: B \to C$ is an isomorphism.

  2. $1_A \otimes f: A \otimes B \to A \otimes C$ is an isomorphism.

I think they are equivalent, as I see the author using this fact in the proof. $1 \Rightarrow 2$ is straight-forward. But I fail to see how to prove: $2 \Rightarrow 1$. Is it correct? Any hints would be appreciated.

Thank you guys very much,

And have a good day,

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doesn't this follow from functoriality of tensor product? note that a functor preserves isomorphisms. – adrido Jan 25 '14 at 7:53
Hi, I know that the functor $1_A \otimes -$ preserves isomorphism. However, I'm asking the other way round. – user49685 Jan 25 '14 at 7:57
On the other hand, $2 \implies 1$ will hold if $A$ is a faithfully flat $R$-module (e.g. a free $R$-module) – zcn Jan 25 '14 at 8:06
What is wrong with this question that would motivate you to delete it? Just because a question has an easy answer does not mean it won't be useful to others in the future. – PVAL Jan 25 '14 at 8:19
Even easier: Take $A=0$ to see that $2$ does not imply $1$. – Julian Kuelshammer Jan 25 '14 at 9:17

This question has been answered in comments:

The implication $2\implies 1$ does not hold. Take for instance $\mathbb{Z}$-modules $A=\mathbb{Q}$, $B=\mathbb{Z}/2\mathbb{Z}$, $C=\mathbb{Z}/3\mathbb{Z}$. Then $A\otimes B=A\otimes C=0$, but $B$ and $C$ are not isomorphic $\mathbb{Z}$-modules. – adrido Jan 25 at 8:04


Even easier: Take $A=0$ to see that 2 does not imply 1. – Julian Kuelshammer Jan 25 at 9:17


On the other hand, $2\implies 1$ will hold if $A$ is a faithfully flat $R$-module (e.g. a free $R$-module). – zcn Jan 25 at 8:06

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