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

flag	asked Mar 2 at 21:00 ya-tayr 268●7

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Quillen-Suslin theorem amathew.wordpress.com/2012/01/16/… – Fernando Muro Mar 2 at 21:03

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"If k is a principal ideal domain then any projective module over k[x1,...,xn] is free." Thanks Fernando! – ya-tayr Mar 2 at 21:11

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You're welcome. – Fernando Muro Mar 2 at 21:51

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And again the comment box is misused for answers. – Martin Brandenburg Mar 3 at 0:15

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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. – Steven Landsburg Mar 3 at 1:41

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Steven Landsburg · Accepted Answer · 2013-03-03 00:39:44Z UTC

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.

Martin Brandenburg · Answer 2 · 2013-03-03 03:37:32Z UTC

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.

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

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