# Rank and restriction of scalars

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 $$R$$.

• what is your definition of rank in that case? and further assumptions for $R,S$? Commented Jun 23, 2022 at 15:49
• @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 $R_{\mathfrak{p}}$. Commented Jun 23, 2022 at 17:06
• 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? Commented Jun 25, 2022 at 11:31
• @Simonsays No I'm not aware - what can one say in that case? Commented Jun 26, 2022 at 21:22

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 $$M_R\cong (S^{\oplus m})_R \cong R^{\oplus mn}$$.
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 $$f_*\tilde{M}=\mathcal{O}^{nm}_{Spec(R)}$$