Let $R$ be a discrete valuation ring (DVR) and let $M$ be a projective module of finite type over the polynomial ring $R[t]$. Is $M$ free over $R[t]$?

As far as I understand, this should be a consequence of the Bass-Quillen conjecture for $R$. Is it proven in this particular case?


2 Answers 2


The Bass-Quillen conjecture is known to be true for principal ideal domains (that is, if $A$ is a PID, all finitely generated projective modules over $A[T_1,\dots,T_n]$ are free). This was proven in theorem 4 of the paper

Quillen, Daniel, Projective modules over polynomial rings, Invent. Math. 36, 167-171 (1976). ZBL0337.13011.

  • $\begingroup$ This is precisely it! Thank you so much. $\endgroup$
    – Stabilo
    Jun 11, 2021 at 7:11
  • $\begingroup$ I think that the particular case in the question was already covered by the previous Suslin-Vaserstein paper (every projective module over $A[t]$ or $A[t,s]$ for $A$ PID, is free). The $A[t]$ might even have been established before. $\endgroup$
    – YCor
    Jun 11, 2021 at 8:01
  • 3
    $\begingroup$ As I understand , CS Seshadri wast the first one to prove Serre's conjecture over $A[x]$ where $A$ is a PID. ( doi.org/10.1073/pnas.44.5.456) $\endgroup$
    – skeptic
    Jun 11, 2021 at 8:46

By a result of Vasconcelos-Simis (Notices AMS, 1971), projective modules over R[t], R any valuation ring of finite rank, are free. A detailed proof appears in Lequain-Simis, J. Pure Appl. Algebra, 1980, which contains more general results.


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