# Relation/difference between proof of constant rank of f.g. projective module over commutative domain

Let $$M$$ be a f.g projective module over a domain $$R$$. I am wondering how I can see openness and closedness via $$K_0$$ group proof.

Since $$M$$ is f.g. projective, it is clear localization at $$p\in Spec(R)$$ $$M$$ is free. Pick any $$f\in R$$ s.t. $$p\in D(f)$$. Then it is not hard to see $$M_f$$ is free via $$M_p$$ free. Hence it shows local constant rank is open condition. Then from $$R$$ being domain, one deduces constant rank being open and closed condition or one forms partition by disjointness of 2 open sets. Hence $$M$$ is constant rank at every point.

Here is the proof provided in Milnor's Introduction to Algebraic K-theory if I understood correctly. Pick any $$p\in Spec(R)$$. It is clear that talking about rank $$R_p$$ and $$k_p$$ of $$M$$ is talking about the same thing where $$k_p$$ is the residue field at $$p$$ and $$R_p$$ is localization away from $$p$$. So I will pick $$R_p$$ as my choice. Since $$R\to Frac(R)$$ factors through $$R\to R_p\to Frac(R)$$ and diagram commutes, one deduces $$K_0(R)\to K_0(Frac(R))$$ factoring through $$K_0(R)\to K_0(R_p)\to K_0(Frac(R))$$. Note that $$K_0(R_p)=<[R_p]>\cong K_0(Frac(R))=<[Frac(R)]>$$ where $$[R_p]\to [Frac(R)]$$ via ring homomoprhism induced by $$K_0$$ maps. Hence every projective module at point level hits the same image of the same rank. Hence it is of constant rank at every point.

$$\textbf{Q:}$$ Is there a way to see openness of constant rank condition from $$K_0$$ level map without going through the proof as in paragraph 2? It seems that one did not invoke any topological argument in $$K_0$$ proof. This seems very strange. Do I have a dictionary translating between $$K_0$$ proof and topological proof?

I have not figured out how to deal with $$k_p$$'s argument at the moment.