The Stacks project

Lemma 10.78.7. Let $R$ be a semi-local ring. Let $M$ be a finite locally free module. If $M$ has constant rank, then $M$ is free. In particular, if $R$ has connected spectrum, then $M$ is free.

Proof. Omitted. Hints: First show that $M/\mathfrak m_ iM$ has the same dimension $d$ for all maximal ideal $\mathfrak m_1, \ldots , \mathfrak m_ n$ of $R$ using the rank is constant. Next, show that there exist elements $x_1, \ldots , x_ d \in M$ which form a basis for each $M/\mathfrak m_ iM$ by the Chinese remainder theorem. Finally show that $x_1, \ldots , x_ d$ is a basis for $M$. $\square$


Comments (1)

Comment #10961 by Junyan Xu on

I recently formalized this lemmas in Lean, and was able to weaken the locally free assumption to flatness. The proof ends up like this: to show that is a basis amounts to showing the induced map is bijective, for which it suffices to show is bijective for each maximal . (This is completed using bijective_of_localized_maximal in Lean; I'm not sure whether the counterparts of the surrounding lemmas exist in Stacks.) Since is finite flat over the local ring , from the proof of 00NZ, it suffices to show is bijective, but this map is the base change of , which is bijective by choice of the .

(In Lean the argument is all carried out using tensor products: localizations and quotients of doesn't appear, only the tensor product of with localizations and quotients of . There doesn't seem to be a statement of CRT for modules in Stacks, but of course one can just tensor the surjection with .)

A direct consequence of this lemma is that the Picard group of a semi-local ring is trivial. Combined with the observation that a Noetherian total ring of fraction must be semi-local, one can deduce that invertible modules over Noetherian rings are isomorphic to ideals: This has also been formalized in Lean.

There are also:

  • 4 comment(s) on Section 10.78: Finite projective modules

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