TITLE: Projective module over $R[X]$
QUESTION [7 upvotes]: 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?

REPLY [6 votes]: 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.