Let $R$ be a commutative ring and $R[U]$ the polynomial ring in one variable. What is the relation between projective modules over $R$ and projective modules over $R[U]$? Is every projective module over $R[U]$ of the form $P[U]$ for a projective $R$-module $P$? If not what are the obstructions?
Edit: I realised that this question was to general for what I was actually looking for. Since the question in the current form seems to be interesting on its own, I refrained from editing it and opened a new question instead.
Here is the answer for finitely generated modules of rank one. Recall that the isomorphism classes of these modules form a group, the Picard group $Pic(R)$, with tensor product as multiplication.
And what does it mean that $R$ is semi-normal?
It means that if $x,y\in R$ satisfy $y^2=x^3$, then there exists $s\in R$ with $x=s^2$ and $y=s^3$ .
(Geometrically: you can parametrize the cusp over $R$ ).
Admittedly this condition is a little strange, but at least it is easy to see that a normal ring $R$ (= integrally closed domain) is semi-normal :
Take $s=\frac {y}{x}\in Frac(R)$. Of course we have $x=s^2$ and $y=s^3$.
The key point is that $s\in R $ : the fraction $s=\frac {y}{x}$ is integral over $R$ because it satisfies the monic equation $T^2-x=0$ and since $R$ is integrally closed we must have $s\in R$ .