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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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    @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.2011-02-27

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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.

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    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 answer2011-02-28