Grassmannian as a functor and the rank of a projective module.

109 Views Asked by At

Consider the following definition, taken from Introduction to Algebraic Geometry and Algebraic Groups by Demazure and Gabriel:

3.4 Example: Let $n,r$ be two integers $\geq 0$; the Grassmannian is the functor $\underline{G}_{n,r}$ which assigns to each $R\in \mathop M\limits_ \sim $ the set of direct factors $P$ of rank $n$ of the $R$-module $R^{n+r}$. [...]

Here we are interested in functors

$$ \mathop M\limits_ \sim \to \mathop E\limits_ \sim, $$ where $\mathop E\limits_ \sim $ is the category of sets and $\mathop M\limits_ \sim $ is the category of models, i.e., unital, commutative rings belonging to some fixed universe $\mathop U\limits_ \sim $ – I don't think it's too important here and for the sake of this question we can just treat $\mathop M\limits_ \sim $ as the category of rings, but please correct me if I'm wrong.

What is meant by "direct factors $P$ of rank $n$"? I understand that direct factors of a free module are just projective modules, but what is a projective module of rank $n$ in this case? Is it a projective module $P$ such that the fibres of $\tilde{P}$ are of rank $n$ at all points of $\operatorname{Spec} R$? By rank of the fibre at $p\in\operatorname{Spec}R$ I mean $\dim_{\operatorname{Frac} R/p}(P \otimes_R \operatorname{Frac} R/p)$.