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Let $R\to S$ be a faithfully flat ring map between commutative rings and $M$ any projective $S$-module. Is it true that $M$ is projective $R$-module?

Here, is my attempt: Let $M\oplus X\cong S^{I}$, for some $S$-module $X$ and indexing set $I$. Since it also an isomorphism as $R$-modules. Then it is enough to prove that $S$ is projective over $R$.

Note that the map $h:S\to S\otimes_R R$, $s\mapsto s\otimes 1_R$ is an $R$-module isomorphism. Then $h$ is also an $S$-module homomoprphism as follows: $h(ts)=ts\otimes 1_R=t(s\otimes 1_R)=th(s)$. Hence, $S\cong S\otimes_R R$ as $S$-modules. So, $S\otimes_R R$ is projective over $S$. Then the result follows from Theorem 9.6 in the following link: https://arxiv.org/pdf/1011.0038v1.pdf

My above attempt is correct or there is some mistake. Please guide me.

waqas mahmood
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  • Have you shown that there is an $R$-module $Y$ such that $S\oplus Y\equiv R^{J}$ as $R$-modules for some indexing set $J$? – Mohan Apr 29 '19 at 20:23
  • In general $S$ is not projective over $R$. Example: $R=\mathbb Z$ and $S=\mathbb Z[[X]]$. In this case $R\to S$ is faithfully flat, but not projective (that is, free). – user26857 Apr 29 '19 at 22:43
  • Thanks for the comments. Could you please tell me in which case a projective module over $S$ becomes projective $R$-modules. I know it is true if $S$ is finitely presented (because any finitely presented flat module is projective ). Is there some other possibility which is weaker than finitely presented condition on $S$? Aslo, another possibility is given in Theorem 9.6 in the following link: https://arxiv.org/pdf/1011.0038v1.pdf. – waqas mahmood Apr 30 '19 at 05:48

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