2
$\begingroup$

Let $R \hookrightarrow S$ be commutative rings. Suppose that $M$ is a finitely generated projective $S$-module. Let $f : \text{Spec}(S) \rightarrow \text{Spec}(R)$.

We have locally constant rank functions $r_R^M : \text{Spec}(R) \rightarrow \mathbb{N}$ and $r_S^M : \text{Spec}(S) \rightarrow \mathbb{N}$.

Suppose that $S$ is finitely generated and free over $R$, of rank $n$. In particular, $M$ is then finitely generated and projective over $R$.

My questions are:

(1) If $M$ is of constant rank $m$ over $S$, then is $M_R$ of constant rank $nm$ over $R$?

(2) Perhaps more generally: does $r_R^M \circ f = n \cdot r_S^M$?

There are well known relations relating these ranks when one base changes modules from $R$ to $S$, but I can't find any for restriction from $S$ to $R$.

$\endgroup$
4
  • $\begingroup$ what is your definition of rank in that case? and further assumptions for $R,S$? $\endgroup$
    – Simonsays
    Commented Jun 23, 2022 at 15:49
  • $\begingroup$ @Simonsays In which case? The definition of the rank function is that for $\mathfrak{p} \in \text{Spec}(R)$, $r_R^M(\mathfrak{p})$ is the natural number such that $M_{\mathfrak{p}}$ is free of rank $r_R^M(\mathfrak{p})$ over $R_{\mathfrak{p}}$. $\endgroup$
    – James
    Commented Jun 23, 2022 at 17:06
  • $\begingroup$ sorry for my later answer, without more assumptions for $R,S$, I'm afraid I don't know much. Though the case for $R,S$ being polynomial rings over a field is not difficult, you know about that? $\endgroup$
    – Simonsays
    Commented Jun 25, 2022 at 11:31
  • $\begingroup$ @Simonsays No I'm not aware - what can one say in that case? $\endgroup$
    – James
    Commented Jun 26, 2022 at 21:22

2 Answers 2

1
$\begingroup$

Yes, both your questions have affirmative answers.

The key observation is that we may reduce to the case when $M$ is in fact free: for a finite projective module, there are finitely many elements $s_1,\cdots,s_a\in S$ so that $M_{s_i}$ is free over $S_{s_i}$ (e.g. Stacks 00NX). Since $\operatorname{Spec} S\to\operatorname{Spec} R$ is finite, it is closed, and therefore each $f(V(s_i))$ is closed in $\operatorname{Spec} R$. So we can cover $\operatorname{Spec} R\setminus f(V(s_i))$ by open affines which have affine preimage in $\operatorname{Spec} S$ as $f$ is finite hence affine.

Now suppose $M\cong S^{\oplus m}$ is free. Then the pushforward of $\widetilde{M}$ to $\operatorname{Spec} R$ is just $M_R\cong (S^{\oplus m})_R \cong R^{\oplus mn}$.

$\endgroup$
1
$\begingroup$

sorry for the late answer: at least the special case $S$ being a polynomial ring over a field $k$ is an immediate consequence of the Quillen-Suslin theorem:
Every finitely generated, projective module $M$ over a polynomial ring $S/k$ is actually free.

As a consequence, if $R\subset S$ is an integral extension of rings such that $S$ is free of rank $n$ over $R$, the induced map $f: Spec(S)\rightarrow Spec(R)$ becomes a finite flat surjective morphism of degree $n$, in particular $f_*\mathcal{O}_{Spec(S)}\cong \mathcal{O}^n_{Spec(R)}$.
Now if $\tilde{M} \cong \mathcal{O}^m$ as an $\mathcal{O}_S$-module, you indeed get $f_*\tilde{M}=\mathcal{O}^{nm}_{Spec(R)}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .