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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.

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  • $\begingroup$ Is it Noetherian? $\endgroup$ Jan 25, 2011 at 1:26
  • $\begingroup$ Also, this sounds like homework. I'm voting to close based on the fact that there is technically no question here, just a statement. $\endgroup$ Jan 25, 2011 at 1:28
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    $\begingroup$ Hey, I only need help. However it is not Noetherian. $\endgroup$
    – John
    Jan 25, 2011 at 2:51
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    $\begingroup$ This would be better asked at math.stackexchange.com, since it is not a research level question, and will surely be closed soon. $\endgroup$
    – Emerton
    Jan 25, 2011 at 4:31
  • $\begingroup$ For a proof in the commutative case see math.stackexchange.com/questions/150944/… $\endgroup$
    – user26857
    Jan 27, 2013 at 10:36

2 Answers 2

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If R is commutative(even not Noetherian), I think the answer is yes. Please see the paper of Hinohara, Projective modules over semilocal rings.

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    $\begingroup$ 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? $\endgroup$
    – John
    Jan 25, 2011 at 2:54
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    $\begingroup$ 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. $\endgroup$
    – Liu Hang
    Jan 25, 2011 at 5:07
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This follows both from Forster's Theorem and from Serre's Splitting Off Theorem.

If $R$ is a commutative ring, Noetherian or not, and $M$ is an $R$-module, let $\mu(M,R)$ be its minimal number of generators ($0\leq \mu(M,R)\leq\infty$).

A version of Forster's Theorem states that $\mu(M,R) \leq \max(\mu(M_{\mathfrak m},R_{\mathfrak m}))+\dim(R/\mathrm{rad}(R))$ if $M$ has finite presentation. The maximum is taken over the maximal ideals $\mathfrak m$ of $R$, and $\dim$ is Krull dimension.

If $M$ is finitely generated and projective, then $\mu(M_{\mathfrak m},R_{\mathfrak m})$ is just the $\mathfrak m$-rank of $M$, and $M$ admits finite presention. And if $R$ is semi-local, we have $\dim(R/\mathrm{rad}(R)) = 0$. It follows that when $n$ is the largest of the $\mu(M_{\mathfrak m},R_{\mathfrak m})$, there is a projection $R^{n}\twoheadrightarrow M$, giving $R^{n}=M\oplus P$ by projectivity. And when all the $\mu(M_{\mathfrak m},R_{\mathfrak m})$ are equal to $n$, $P$ must be the zero module, and $M$ is free.

Serre's Splitting Off Theorem says that if $m:=\min(\mu(M_{\mathfrak m},R_{\mathfrak m}))- \dim(R/\mathrm{rad}(R))>0$ for a finitely generated projective module $M$, then $M=R^{m}\oplus P$ with $\min(\mu(P_{\mathfrak m},R_{\mathfrak m}))=\dim(R/\mathrm{rad}(R))$, for a suitable submodule $P$. So $\mu(M_{\mathfrak m},R_{\mathfrak m})=m+\mu(P_{\mathfrak m},R_{\mathfrak m})$, and if $R$ is semi-local, $\mu(P_{\mathfrak m},R_{\mathfrak m})=0$ for some maximal ideal $\mathfrak m$. But if the $\mu(M_{\mathfrak m},R_{\mathfrak m})$ are all equal, so are the $\mu(P_{\mathfrak m},R_{\mathfrak m})$, and necessarily $P=0$.

These observations are the generalizations to semi-local rings of the well-known corresponding results for local rings.

The proofs of the above theorems are quite elementary, based on the elementary characterization of Krull dimension given by Coquand, Lombardi, Quitté et alii. Cf. http://hlombardi.free.fr/publis/KroBasSer.pdf, where these results are shown in slightly stronger form using $Hdim(R)$, the Heitmann dimension, a number less than or equal to $\dim(R/\mathrm{rad}(R))$.

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