Let $(R,m)$ be commutative noetherian local ring with unity. Suppose $P$ is a finitely generated projective module over $R[X]$ of rank $n$ . Is $P$ free? If not, what is the counter example?

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    $\begingroup$ Well the answear is affirmative by obvious application of serre's conjecture if $(R,m)$ is zero dimensional. In general I expect a counter example. But I am not getting it. thanking you $\endgroup$
    – A.G
    Feb 27, 2011 at 20:11
  • $\begingroup$ According to Wikipedia a counterexample occurs with R equal to the local ring of the curve $y^2 = x^3$ at the origin. It doesn't state what the counterexample is or provide a reference. en.wikipedia.org/w/… $\endgroup$ Feb 27, 2011 at 21:00
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    $\begingroup$ @Anjan: the Quillen-Suslin solution of Serre's conjecture actually showed a bit more, and in particular gives an affirmative answer to your question when $R$ is a DVR. Moreover, it a famous conjecture of Bass and Quillen that your question should have an affirmative answer if you add the hypothesis of regularity (since a regular one-dimensional local ring is a DVR, this is exactly what we did above). So I think you should look for counterexamples among singular one-dimensional local rings. (I don't know of one off the top of my head, but I also suspect they should exist.) $\endgroup$ Feb 27, 2011 at 21:03
  • $\begingroup$ Another good place to look would be Lam's new(ish) book Serre's Problem on Projective Modules. (I do not yet have a copy, or I would tell you whether a counterexample can be found there.) $\endgroup$ Feb 27, 2011 at 21:05
  • $\begingroup$ @Pete: If $R$ is a DVR there's a much easier proof (this is from Lam's book you mentioned): Start from Kaplansky's theorem that every submodule of a free module over a left hereditary ring (every left ideal is projective) is a sum of left ideals (see Lam, Lectures on modules and rings, (2.24), p. 42). For a DVR $R$ every left ideal $I$ of $R[X]$ is of the form $R[X] \cdot f$ where $f$ is a polynomial in $I$ of minimal degree (by the division algorithm). Since $R[X]$ has no zero-divisors, we have $I \cong R[X]$. Thus Kaplansky's theorem shows that every projective module over $R[X]$ is free. $\endgroup$
    – t.b.
    Feb 27, 2011 at 21:55

1 Answer 1


Here is some elaboration on the wiki entry in George's comment.

Suppose $R$ is a domain. $R$ is called seminormal if whenever $b^2=c^3$ in $R$ one can find $t \in R$ such that $b=t^3, c=t^2$.

The relevant thing here is the following fact:

R is seminormal if and only if $Pic(R) \cong Pic(R[X])$

So if $R$ is local and not seminormal then there will be a projective, non-free $R[x]$-module of rank $1$.

As for an implicit example, take $R = k[t^2,t^3]_{(t^2,t^3)}$. One can check that $I = (1-tx, t^2x^2)$ is an invertible (fractional) ideal of $R[x]$ which is non-free.

UPDATE: by request, a reference is this survey, see page 16. I am sure you can find more by googling the relevant terms.

  • $\begingroup$ Is $1-tx$ even in $R[x]$? $\endgroup$ Feb 27, 2011 at 22:26
  • $\begingroup$ George, it is inside the quotient field. $\endgroup$
    – curious
    Feb 27, 2011 at 22:28
  • $\begingroup$ Ok, yes, its a fractional ideal (I was forgetting that invertible ideal means invertible fractional ideal by definition). $\endgroup$ Feb 27, 2011 at 22:32
  • $\begingroup$ Can you tell me a reference for the proof of the fact that $R$ is seminormal iff $Pic(R) \cong Pic(R[X])$? Thanks for this nice answer $\endgroup$
    – A.G
    Feb 28, 2011 at 4:51

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