# Can I assume $R$ is a local ring?

I should prove this statement:

Let $R$ be a ring, $M$ be a $R$-module and $P$ a projective $R$-module of finite type. If $x=\sum_i m_i\otimes p_i$ is an element in $M\otimes_R P$ such that $\sum_i m_i\phi(p_i)=0 \quad \forall\phi\in P^*$, then $x=0$.

Well, I've proved it in the case $R$ is a local ring. Can I assume $R$ is a local ring without loss of generality?

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The statement is true For some finitely generated $R$- module $Q$ you have $P \oplus Q \simeq R^n$. From this deduce: a) that there is an injective morphism of $R$-modules $M\otimes P \to M\otimes R^n$ and b) that every linear form on $P$ is the restriction of one on $R^n$. Since your result is true for $M\otimes R^n$ because of the isomorphism $M\otimes R^n\simeq M^n$, it will also be true for the original $M\otimes P$ by using a) and b).
Your actual question Since the result is true it is impossible to say that it can't be proved by reducing to the case where $R$ a local ring. However in that case $P$ will be free and I suspect that you'll have to do something close to the above. Moreover you'll have to check one or two little facts to ensure that you can go back and forth between the global case and the local case, in particular that your hypothesis on linear forms in $P^\ast$ can be localized. So personally, I wouldn't go for the reduction to local rings since the general case is so simple.