Is every finitely generated projective $\mathbf{Z}[x]$module free?

17$\begingroup$ QuillenSuslin theorem amathew.wordpress.com/2012/01/16/thequillensuslintheorem $\endgroup$ – Fernando Muro Mar 2 '13 at 21:03

1$\begingroup$ "If k is a principal ideal domain then any projective module over k[x1,...,xn] is free." Thanks Fernando! $\endgroup$ – yatayr Mar 2 '13 at 21:11

1$\begingroup$ You're welcome. $\endgroup$ – Fernando Muro Mar 2 '13 at 21:51

5$\begingroup$ And again the comment box is misused for answers. $\endgroup$ – Martin Brandenburg Mar 3 '13 at 0:15

13$\begingroup$ I do not understand the votes to close. The QuillenSuslin theorem is wellknown to experts, but I am sure there are a great many research mathematicians who are unfamiliar with it. $\endgroup$ – Steven Landsburg Mar 3 '13 at 1:41
While it's certainly true (per Fernando's comment) that this is a special case of the QuillenSuslin theorem, it was certainly known long before Quillen and Suslin came along.
There's a paper of Murthy from the mid1960s which shows that every projective $R[x]$module is extended whenever $R$ is a regular ring of dimension at most 2. ("Extended" here means "of the form $P[x]$ where $P$ is a projective $R$module". Since all projective ${\mathbb Z}$modules are free, extended is equivalent to free in this case.)
But there's an even earlier paper of Bass which covers the case where $R$ is regular of dimension 1, which is all you need. The paper is called "Torsion Free and Projective Modules".
Edited to add: And the case of a PID predates even Bass; I think it's due to Seshadri in the 1950s.

6$\begingroup$ Glad this informative answer got posted before the question gets closed... $\endgroup$ – Yemon Choi Mar 3 '13 at 1:29

4$\begingroup$ This is a great answer. I was pretty sure when Fernando gave his answer in the comments he was nuking flies and that this special case must be much easier. Also I should add that I am pretty sure Serre's original conjecture was over a field and many people might not realize QuillenSuslin works over PIDs. $\endgroup$ – Benjamin Steinberg Mar 3 '13 at 2:01

1$\begingroup$ Well it was known as the Serre conjecture although he himself apparently never owned the name. en.m.wikipedia.org/wiki/Quillen–Suslin_theorem $\endgroup$ – Benjamin Steinberg Mar 3 '13 at 2:45

1$\begingroup$ Steve Thanks for this, and for convincing me not to recycle! $\endgroup$ – yatayr Mar 3 '13 at 2:51

1$\begingroup$ Todd: Serre surely knew that the unimodular extension property is equivalent to Quillen/Suslin (this is basically a restatement of Serre's computation of $K_0$). I think the issue Fernando has in mind is not whether Serre was interested in this question but whether he expected a particular answer. $\endgroup$ – Steven Landsburg Mar 3 '13 at 17:38
When $R$ is a PID, then every finitely generated projective $R[x]$module is free. As Steven already said, this is due to Seshadri. Here is the reference:
Seshadri, C.S., Triviality of vector bundles over the affine space $K^2$, Proc. Nat. Acad. Sci. USA 44 (1958), 456458.