# Finitely generated projective module , of constant rank, over semi-local ring [duplicate]

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.

• 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. Dec 9, 2019 at 16:41
• Check out this paper to see an elementary proof of Forster's theorem hlombardi.free.fr/publis/forster.pdf. Dec 9, 2019 at 16:42
• @BadamBaplan: very cool result! Thanks for sharing it. Dec 9, 2019 at 23:45

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.
• 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. Dec 9, 2019 at 8:12