Let $M$ be a finitely generated flat module over a commutative ring with unity $R$ . Then is $M[[x]]$ also flat over $R[[x]]$ ? If this is not always true , then what if we also assume $M$ is finitely presented and flat over $R$ ; is $M[[x]]$ flat over $R[[x]]$ then ?

  • $\begingroup$ You have to make clear what denote by $M[[X]]$, but since this question is a follow up of math.stackexchange.com/questions/2440035/… I suppose that $M[[X]]$ is the module of power series with coefficients in $M$. $\endgroup$ – user26857 Oct 7 '17 at 19:44
  • $\begingroup$ @user26857 : yes that's what I meant $\endgroup$ – user456828 Oct 8 '17 at 7:17
  • $\begingroup$ As I already told you the question follows if ones knows that $M[[X]]\simeq R[[X]]\otimes_RM$. As $R$-module, $M[[X]]$ is isomorphic to a direct product of countable many copies of $M$, and similarly for $R[[X]]$. If $M$ is finitely presented $M[[X]]$ and $R[[X]]\otimes_RM$ are canonically isomorphic $R$-modules. (This is a classical result.) Now it's easy to show that the canonical isomorphism is in fact an isomorphism of $R[[X]]$-modules. $\endgroup$ – user26857 Oct 17 '17 at 20:16
  • $\begingroup$ @user26857: can you please give me a reference for that classical result ? $\endgroup$ – user456828 Oct 18 '17 at 4:19
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    $\begingroup$ This is probably in Bourbaki, and also in the paper of Webb, Tensor and direct products. $\endgroup$ – user26857 Oct 18 '17 at 14:30

I think this is true. Let $0\to K\to N$ be an inclusion of $R[[x]]$ modules. Then, $0\to K\otimes_R M\to N\otimes_R M$ is exact, since $M$ is flat over $R$. Now, $K\otimes_R M=K\otimes_{R[[x]]} R[[x]]\otimes_R M=K\otimes_{R[[x]]} M[[x]]$ and similarly for $N$, proving flatness of $M[[x]]$ over $R[[x]]$.

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  • $\begingroup$ For the OP $M[[X]]$ is the module of power series with coefficients in $M$, and not $M\otimes_RR[[X]]$. They are isomorphic when $M$ is finitely presented, if I'm not mistaken. $\endgroup$ – user26857 Oct 7 '17 at 19:40
  • $\begingroup$ @user26857 My answer is clearly wrong unless $M[[x]]=M\otimes_R R[[x]]$. Thanks. I will be happy to delete the answer if that is the case. $\endgroup$ – Mohan Oct 7 '17 at 20:00
  • $\begingroup$ @user26857 : could you please provide an argument as to why $M[[x]]$ and $M \otimes_R R[[x]]$ are isomorphic if $M $ is finitely presented over $R$ ? $\endgroup$ – user456828 Oct 9 '17 at 6:25

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