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Let $M$ be a finitely generated projective module over a commutative ring $R$ with finitely maximal ideals. Also assume that $M$ has constant rank i.e. rank of $M_P$ as $R_P$-module remains constant as $P$ varies over all prime ideals of $R$. How to show that $M$ is free ?

I'm allowed to assume that finitely generated projective modules over local rings are free.

My try: Let $J$ be the Jacobson radical of $R$. Then $R/J$ is a finite direct product of fields. Now $M/JM$ is a projective $R/J$-module. Now I'll be done if I can show $M/JM$ is free over $R/J$. Unfortunately I'm unable to show this last part.

Please help.

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  • $\begingroup$ Personally, I think the best way to understand this is by way of Forster's theorem, which says that when an f.g. module in a $d$-dimensional ring is locally generated by $n$ elements, it is in fact generated by $d + n$ elements. For a projective module $M$ of rank $n$ over a $0$-dimensional ring, one easily sees that $M$ will be $n$-generated hence free. Since, as you mention, isomorphisms of projective modules lift modulo the Jacobson radical, you immediately have the generalization of your result whenever $R/J$ is $0$-dimensional. $\endgroup$ Dec 9, 2019 at 16:41
  • $\begingroup$ Check out this paper to see an elementary proof of Forster's theorem hlombardi.free.fr/publis/forster.pdf. $\endgroup$ Dec 9, 2019 at 16:42
  • $\begingroup$ @BadamBaplan: very cool result! Thanks for sharing it. $\endgroup$ Dec 9, 2019 at 23:45

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As you say, one may reduce to the case when $R$ is a finite direct product of fields. Say $R = \prod_{i=1}^{n} F_{i}$, and $M$ has constant rank $r \in \mathbb{N}$. Let $e_{1}, \ldots, e_{n}$ be the corresponding standard idempotents of $R$, and put $J_{i} = e_{i}R$; it is both an ideal of $R$ and a ring (not a subring of $R$, though!) which is isomorphic to $F_{i}$. The maximal ideals $P_{1}, \ldots, P_{n}$ of $R$ are given by $P_{i} = \bigoplus_{j \neq i} e_{i}R$. Here is the outline of an approach, whose details I leave to you.

$(1)$ Show that the localization $R_{P_{i}} \cong F_{i}$, so that $M_{P_{i}}$ is a free $F_{i}$-module of rank $r$. Observe that the isomorphism $M_{P_{i}} \cong F_{i}^{r}$ is also an isomorphism of $R$-modules, suitably interpreted.

$(2)$ Show that $M \cong \prod_{i=1}^{n} J_{i}M$ as $R$-modules.

$(3)$ Show that $M_{P_{i}} \cong J_{i}M$ as $R$-modules.

$(4)$ Use $(1)$-$(3)$ to conclude that $M \cong R^{r}$ as $R$-modules.

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  • $\begingroup$ Maybe a "better" proof, though it's more of less the same idea: by (a specific proof of) $(1)$, one shows that the map $R \to \prod_{i = 1}^{n} R_{P_{i}}$ induced by each of the canonical localization maps is an isomorphism. This gives a decomposition of $M$ as a product $\prod_{i=1}^{n} M_{i}$, where $M_{i} \cong M_{P_{i}}$ (as both $R$-modules and $R_{P_{i}} \cong F_{i}$-modules!). Writing this out, it really is just the same proof, but who knows? Maybe someone will like this phrasing better. $\endgroup$ Dec 9, 2019 at 8:12

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