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Let $R$ be a ring an let $A$ and $B$ be two $R$-algebras. If $S$ is faithfully flat over $R$ then we say that $A$ is a $S$-twisted form of $B$ if $A\otimes_R S$ and $B\otimes_R S$ are isomorphic as $S$-algebras.

Now, if $F$ is a free $R$-module, $ rank R=n$, and $S$ is a commutative algebra which is faithfully flat over the commutative and unitary ring $R$, could you prove that a $R$-module $P$ is a $S$-twisted form of $F$ if and only if $P\otimes_R S$ is a free $S$-module of rank n?

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You define the notion for algebras but then your question is about modules. If the definition of $S$-twisted forms for modules is what I think it is (simply replacing "algebra" by "module" in the defintion above), then the answer is obviously yes. – Plop Jun 2 '11 at 21:02
Yes, it's exactly what I meant! well, I cannot formalize a proof... – user11428 Jun 2 '11 at 22:02
$F \otimes_R S \simeq S^n$, so it seems obvious to me. Are you sure you wrote the right statement that you want to prove? – Plop Jun 2 '11 at 22:20
Oh, yes! It was obvious! How could I be so silly!!!!! :) Sorry! – user11428 Jun 2 '11 at 22:31

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