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

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    $\begingroup$ Quillen-Suslin theorem amathew.wordpress.com/2012/01/16/the-quillen-suslin-theorem $\endgroup$ – Fernando Muro Mar 2 '13 at 21:03
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    $\begingroup$ "If k is a principal ideal domain then any projective module over k[x1,...,xn] is free." Thanks Fernando! $\endgroup$ – ya-tayr Mar 2 '13 at 21:11
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    $\begingroup$ You're welcome. $\endgroup$ – Fernando Muro Mar 2 '13 at 21:51
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    $\begingroup$ And again the comment box is misused for answers. $\endgroup$ – Martin Brandenburg Mar 3 '13 at 0:15
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    $\begingroup$ I do not understand the votes to close. The Quillen-Suslin theorem is well-known 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 Quillen-Suslin theorem, it was certainly known long before Quillen and Suslin came along.

There's a paper of Murthy from the mid-1960s 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.

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    $\begingroup$ Glad this informative answer got posted before the question gets closed... $\endgroup$ – Yemon Choi Mar 3 '13 at 1:29
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    $\begingroup$ 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 Quillen-Suslin works over PIDs. $\endgroup$ – Benjamin Steinberg Mar 3 '13 at 2:01
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    $\begingroup$ 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
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    $\begingroup$ Steve Thanks for this, and for convincing me not to recycle! $\endgroup$ – ya-tayr Mar 3 '13 at 2:51
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    $\begingroup$ 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), 456-458.


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