# Cancelling finitely generated projective modules from a tensor product of finitely generated projective modules

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 $$N \cong P$$ ?

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 $$P\otimes P=P\otimes R^2=R^4$$.
• 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 $n=2,3$ ? – user102248 May 23 at 22:28
• @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 $P$. – Mohan May 24 at 3:00