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Let $R$ be a semi-local ring, and $M$ a finite projective $R$-module. If the localizations $M_m$ have the same rank for all maximal ideals $m$ of $R$ then $M$ is free.

flag	asked Jan 25 2011 at 1:10 John 21●1

Is it Noetherian? – Harry Gindi Jan 25 2011 at 1:26

Also, this sounds like homework. I'm voting to close based on the fact that there is technically no question here, just a statement. – Harry Gindi Jan 25 2011 at 1:28

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Hey, I only need help. However it is not Noetherian. – John Jan 25 2011 at 2:51

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This would be better asked at math.stackexchange.com, since it is not a research level question, and will surely be closed soon. – Emerton Jan 25 2011 at 4:31

For a proof in the commutative case see math.stackexchange.com/questions/150944/… – YACP Jan 27 at 10:36

Liu Hang · Answer 1 · 2011-01-25 02:17:26Z UTC

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If R is commutative(even not Noetherian), I think the answer is yes. Please see this paper.

answered Jan 25 2011 at 2:17

Liu Hang
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I saw that paper, but how can I relate the theorem "Over a commutative indecomposable semilocal ring, any projective module is free" to my question? – John Jan 25 2011 at 2:54

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A semi-local ring can be decomposed as sum of finite many indecomposable rings. M is free with the same rank (by your assumption) over any of these indecomposable rings, so M is free over R. – Liu Hang Jan 25 2011 at 5:07

Projective modules over semi-local rings

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