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Let $R$ be a commutative Noetherian ring (with unity) and $M,N,P$ be finitely generated projective modules over $R$ such that for some $n\ge 1$, we have $M\otimes_R N \cong M \otimes_R P \cong R^n$. Then is it true that $N \cong P$ ?

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In general this is false. For a simple example, take $R=\mathbb{R}[x,y,z]/(x^2+y^2+z^2-1)$, the co-ordinate ring of the real sphere and let $P$ the tangent bundle, given by the presentation, $0\to R\stackrel{(x,y,z)}{\longrightarrow} R^3\to P\to 0$. Then, it is well known that $P$ is not free, but $P\otimes P=P\otimes R^2=R^4$.

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  • $\begingroup$ I see, thanks ... I think I can prove it is true when $n=1$, and you give a counter-example with $n=4$, so that makes me wonder, do we have counterexamples for $n=2,3$ ? $\endgroup$ – user102248 May 23 at 22:28
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    $\begingroup$ @user102248 If $n=1$, by rank considerations, rank of $M=1$ and then you can tensor by $M^{-1}$ to get what you want. If $n>1$, take a one dimensional domain $R$ which has a projective module of rank 1, say $P$ such that $P^{\otimes n}=R$, with $n>1$ the smallest such. (These exist for any $n$, for example over $R=\mathbb{C}[x,y]/(y^2-x^3-x)$). Then, $R^n\cong R^n\otimes P$, but $R$ is not isomorphic to $P$. $\endgroup$ – Mohan May 24 at 3:00

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