Flat, complete modules vs. the completion of a free module

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This is a relatively elementary question in commutative algebra. My motivation comes from trying to compare some results in the theory of deformation quantisation. Let $k$ be a field, let $k[h]$ be a polynomial ring in one variable and let $k[[h]]$ be the ring of formal power series. If $M$ is a free $k[h]$-module then the $h$-adic completion of $M$ is flat over $k[[h]]$, and obviously it's $h$-adically complete (see Lemma 15.27.5 https://stacks.math.columbia.edu/tag/06LD for example). Someone once told me that the converse is true but I've never been able to piece together the proof.




Question: If $M$ is a flat, $h$-adically complete $k[[h]]$-module then is it true that $M$ can be constructed as the $h$-adic completion of some free $k[h]$-module?




Any references, proofs or counterexamples would be greatly appreciated. Thanks in advance.




commutative-algebra representation-theory




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    This is a relatively elementary question in commutative algebra. My motivation comes from trying to compare some results in the theory of deformation quantisation. Let $k$ be a field, let $k[h]$ be a polynomial ring in one variable and let $k[[h]]$ be the ring of formal power series. If $M$ is a free $k[h]$-module then the $h$-adic completion of $M$ is flat over $k[[h]]$, and obviously it's $h$-adically complete (see Lemma 15.27.5 https://stacks.math.columbia.edu/tag/06LD for example). Someone once told me that the converse is true but I've never been able to piece together the proof.




    Question: If $M$ is a flat, $h$-adically complete $k[[h]]$-module then is it true that $M$ can be constructed as the $h$-adic completion of some free $k[h]$-module?




    Any references, proofs or counterexamples would be greatly appreciated. Thanks in advance.




    commutative-algebra representation-theory




    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      This is a relatively elementary question in commutative algebra. My motivation comes from trying to compare some results in the theory of deformation quantisation. Let $k$ be a field, let $k[h]$ be a polynomial ring in one variable and let $k[[h]]$ be the ring of formal power series. If $M$ is a free $k[h]$-module then the $h$-adic completion of $M$ is flat over $k[[h]]$, and obviously it's $h$-adically complete (see Lemma 15.27.5 https://stacks.math.columbia.edu/tag/06LD for example). Someone once told me that the converse is true but I've never been able to piece together the proof.




      Question: If $M$ is a flat, $h$-adically complete $k[[h]]$-module then is it true that $M$ can be constructed as the $h$-adic completion of some free $k[h]$-module?




      Any references, proofs or counterexamples would be greatly appreciated. Thanks in advance.




      commutative-algebra representation-theory




      share|cite|improve this question














      This is a relatively elementary question in commutative algebra. My motivation comes from trying to compare some results in the theory of deformation quantisation. Let $k$ be a field, let $k[h]$ be a polynomial ring in one variable and let $k[[h]]$ be the ring of formal power series. If $M$ is a free $k[h]$-module then the $h$-adic completion of $M$ is flat over $k[[h]]$, and obviously it's $h$-adically complete (see Lemma 15.27.5 https://stacks.math.columbia.edu/tag/06LD for example). Someone once told me that the converse is true but I've never been able to piece together the proof.




      Question: If $M$ is a flat, $h$-adically complete $k[[h]]$-module then is it true that $M$ can be constructed as the $h$-adic completion of some free $k[h]$-module?




      Any references, proofs or counterexamples would be greatly appreciated. Thanks in advance.





      commutative-algebra representation-theory





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      edited Aug 16 at 11:43

























      asked Aug 16 at 10:36









      Lewis Topley

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