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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.

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If your $R[U]$-module is finitely generated, then the Quillen-Suslin theorem says that it must be free. This actually holds for $R[U_1, \ldots, U_n]$-modules. I don't know about the general case. –  student May 28 '12 at 18:41
    
@Leandro: surely one needs some hypothesis such as that the $R[U]$-module is free over $R$. –  Qiaochu Yuan May 28 '12 at 18:43
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Dear Leandro, what you say is true if $R$ is a field and false for general commutative rings. –  Georges Elencwajg May 28 '12 at 18:44
    
@Georges,@Qiaochu you're absolutely right. I had the field case in mind. –  student May 28 '12 at 18:48
    
Just to clarify, the theorem is that a finitely generated projective $k[x_1,\ldots,x_n]$-module for $k$ a field is free, right? –  Keenan Kidwell May 28 '12 at 20:56

2 Answers 2

up vote 6 down vote accepted

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.

Theorem (Traverso, Swan)
For a commutative ring $R$ the following are equivalent:
a) The reduced ring $R_{red}=R/Nil(R)$ is semi-normal
b) The natural group morphism $Pic(R)\stackrel {\cong}{\to} Pic(R[U])$ is an isomorphism ($U=$ indeterminate)

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 $s\in R$ .

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That is a nice answer. Unfortunately it is hardly helpful for me (since my question wasn't precise enough, not since the answer wasn't accurate). I'll just wait for a while whether there will be other answers and accept yours if nothing new comes up. –  Simon Markett May 29 '12 at 9:14

If $R$ is a left regular ring, then the canonical map $K_0(R) \to K_0(R[t])$ is an isomorphism. This result is due to Grothendieck at least when $R$ is commutative. The general case can be found in the paper "The Whitehead group of a polynomial extension" (Bass, Heller, Swan) or in Rosenberg's book on Algebraic K-Theory.

Of course, this does not imply that every f.g. projective $R[t]$-module has the form $P[t]$ for some f.g. projective $R$-module (but we cannot expect that!); but this turns out to be true "up to exact sequences".

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Thx, I really should have seen this myself. In fact I am trying to prove a generalised form of the fundamental theorem. And somehow I got lost between what I may assume and what I have to prove... –  Simon Markett May 29 '12 at 9:09
    
Perhaps you can open a new question which is more specific and addresses your generalized form of the fundamental theorem? –  Martin Brandenburg May 29 '12 at 10:03

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