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.


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