Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be a PID. Is every finitely generated projective $R[T]$-module free? In other words, is every vector bundle on $\mathbb{A}^1_R$ trivial?

For $R=k[X]$ this is true by the Theorem of Quillen-Suslin. If it fails in general, what happens for $k[X,X^{-1}]$?

share|improve this question
    
Just one variable (it should stay a PID). –  Martin Brandenburg Feb 5 '13 at 19:26
    
Quillen-Suslin is for any (PID)$[x_1, \dots, x_n]$. –  user18119 Feb 5 '13 at 21:13
    
@QiL: Oh, really? Do you have a reference for that? –  Martin Brandenburg Feb 10 '13 at 14:33
1  
See Quillen's Inventiones paper in 1976, Or Ferrand's Bourbaki talk. –  user18119 Feb 11 '13 at 21:57
add comment

1 Answer

up vote 3 down vote accepted

Quote from Lam's Serre's problem on projective modules:

In 1958, Seshadri showed that Serre's conjecture is true for two variables (i.e. for $A = k[x_1,x_2]$). In fact, Seshadri proved that f.g. projectives over R[t] are free if $R$ is any commutative PID.

Reference: Seshadri, C.S., Triviality of vector bundles over the affine space $K^2$, Proc. Nat. Acad. Sci. USA 44, 456-458.

share|improve this answer
    
As the answer is entirely included in that Seshadri's article, I realise I should delete any reference to Lam's book. But this book is so wonderful (as is anything written by Lam) I won't do so. –  PseudoNeo Feb 5 '13 at 10:25
    
Great! Thank you. –  Martin Brandenburg Feb 5 '13 at 11:22
    
Why is it enough to prove the case n=2 in Proposition 1? –  Martin Brandenburg Feb 5 '13 at 19:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.