Let $(R,\mathfrak{m})$ be a commutative noetherian local ring. Let $\Lambda$ be an $R$-algebra which is finitely generated as an $R$-module. Let $M$ be a finitely generated $\Lambda$-module. If $\widehat{M}$(the $\mathfrak{m}$-adic completion of $M$) is a projective $\widehat{‎‎\Lambda}$-module, then can we prove that $M$ is a projective $\Lambda$-module?

In fact, this problem can be generalized to the following question: Let $R$ ans $S$ be two commutative noetherian local rings and let $S$ be a faithfully falt $R$-algebra. Let $\Lambda$ be a finitely generated $R$-module. Let $M$ be a finite $\Lambda$-module. If $M\otimes_RS$ is a projective $\Lambda\otimes_RS$-module, then can we deduce that $M$ is a projective $\Lambda$-module? How can we prove it?

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    $\begingroup$ Is $\Gamma$ finitely generated as a module or an algebra? If an algebra, what if $R= k[x], \Gamma = k[x,x^{-1}]$? $\endgroup$ – Will Sawin Jun 14 '17 at 18:06
  • $\begingroup$ $\Lambda$ is finitely genreated as an $R$-module. In fact, $\Lambda$ is a noether $R$-algebra. $\endgroup$ – Homa81 Jun 14 '17 at 18:20
  • $\begingroup$ If $\Lambda$ is a finitely generated $R$-module, this is correct. If you merely assume that it is just a finitely generated $R$-algebra (as you state in the question), this is false in general. $\endgroup$ – Mohan Jun 14 '17 at 20:30
  • $\begingroup$ Thank you very much. But how can I prove this? $\endgroup$ – Homa81 Jun 15 '17 at 4:26
  • $\begingroup$ Bourbaki, Commutative Algebra, ch. I, §3, Proposition 12. $\endgroup$ – abx Jul 28 '17 at 4:28

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