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I have posted this question on MS for 2 weeks however there is no answer up to now. So I post it here and I hope MOers can help me.

Let $\mathfrak{R}$ be a ring, then we knew that a free module over $\mathfrak{R}$ is projective. Moreover, if $\mathfrak{R}$ is a principal ideal domain then a module over $\mathfrak{R}$ is free if and only if it is projective or if $\mathfrak{R}$ is local then a projective module is free.

We also have a very big question on free properties of projective module over a ring of polynomial, that was Serre conjecture, and now is Quillen-Suslind's theorem.

I wonder that, do we have a general condition for a ring $\mathfrak{R}$, so that every projective module over $\mathfrak{R}$ is also free over $\mathfrak{R}$ which involve all of the cases that mentioned above ?

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for completenss: the link to the MSE question: math.stackexchange.com/q/115187/15416 –  Julian Kuelshammer Jun 23 '13 at 8:35
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If $R$ is a regular local ring that contains a field, then any projective module over $R[x_1,\ldots,x_n]$ is free. If $R$ is a regular local ring of Krull dimension at most 2, then any projective module over $R[x]$ is free.

Beyond that, I don't believe there are any further results in the direction of generalizing Quillen-Suslin, so the answer to your intended question would be no.

(Technically, of course, the answer is yes --- the general condition you're looking for is the condition that every projective module over $R$ is free. But you're looking for a non-trivial general condition.....)

Note also that the results for PIDs and local rings are quite elementary compared to Quillen-Suslin, so this is a little like asking for a generalization of the twin facts that there are no postive integer solutions to the equations $x^2=-1$ and $x^n+y^n=z^n$.

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@Steven: Some f.g. assumptions are missing. –  Martin Brandenburg Mar 16 '12 at 8:59
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