0
$\begingroup$

Let $M$ be a finitely generated projective module over a ring $R$, and let $X = \operatorname{Spec}R$. For each prime $\mathfrak p$ of $R$, $M_{\mathfrak p}$ is a finite rank free $R_{\mathfrak p}$-module, because projective modules over local rings are free. Define the rank function $r: |X| \rightarrow \mathbf N \cup \{0\}$ by $r(\mathfrak p) = \operatorname{Rank}_{R_{\mathfrak p}} M_{\mathfrak p}$. I am trying to understand why this function is locally constant.

If I consider neighborhoods $D(f)$ of $\mathfrak p$ (for $f \in R, \not\in \mathfrak p$), it would be enough to show that for $D(f)$ sufficiently small, $M_f$ is a free $A_f$-module. I don't know whether this is too much to hope for.

$\endgroup$
2
$\begingroup$

This is correct certainly for Noetherian rings. If $M_P$ is free, you can find a free module and a map $R^n\to M$ such that it is an isomorphism when you localize at $P$. Both kernel and cokernel are finitely generated and when localize at $P$, they are zero so you can find an $f\not\in P$ such that when you localize at $f$, both are zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.