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3 votes
0 answers
174 views

What are non-archimedean norms on $\mathbb{R}$, whose restriction to $\mathbb{Q}$ is trivial?

I wonder if there is any classification result on non-archimedean norms on $\mathbb{R}$, with trivial restriction to $\mathbb{Q}$? Any references or examples would be welcomed! Some examples of such ...
16 votes
2 answers
943 views

Higher-rank Archimedean valuations of $\mathbb{Q}$, does it exist?

I was reading the proof of Ostrowski's theorem, with an eye toward the Zariski-Riemann space (as well as adic space, Berkovich space, etc.) In the proof, the value group is always assumed to be in $\...
10 votes
1 answer
854 views

Is it a valuation ring?

It is known that a one-dimensional Noetherian local ring is a discrete valuation ring if it is integrally closed. Then, even if it is not Noetherian, would a one-dimensional local ring become a ...
5 votes
0 answers
144 views

Is there a good notion of higher-rank archimedean norm?

Let $K$ be a field. I think I know what a norm (archimedean or not) $|-| : K \to \mathbb R_{\geq 0}$ is. In the case where the norm is nonarchimedean, it's equivalent to the data of a valuation of ...
2 votes
0 answers
104 views

References on topological ringed spaces

This is a follow up to this question of mine. First of all, let me fix some terminologies, which may or may not be standard: Definition: A topological ringed space is a pair $X := (|X|, \mathcal{O}_X)...
9 votes
1 answer
782 views

Is every field the residue field of a discretely valued field of characteristic 0?

Let $k$ be a field of positive characteristic $p$. Is there necessarily a discrete valuation ring of characteristic $0$ with maximal ideal $(p)$ and residue field isomorphic to $k$?
12 votes
1 answer
2k views

Extension of 2-adic valuation to the real numbers

I just want to know what properties of valuations extend to $\mathbb R$... Denote an extension of the 2-adic valuation from $\mathbb Q$ to $\mathbb R$ by $\nu$. Suppose $\nu(x)=\nu(y)=0$. Is it true ...
4 votes
1 answer
799 views

Is the integral closure of a valuation ring in a finite separable extension of its fraction field étale?

Let $K$ be a field endowed with a rank (height) one valuation with completion $\hat{K}$, which is not discrete. Let $R$ be the valuation ring of $K$. Let $L \subset \hat{K}$ be a separable finite ...
4 votes
1 answer
348 views

Given a non-field local domain $R$, finding a dominating Valuation ring whose residue field is algebraic/finite extension of the residue field of $R$

Let $(R, \mathfrak m)$ be a non-field local domain with fraction field $Q(R)$ . Let $k_{R}:=R/m$. We know that there is a Valuation ring $(V,\mathfrak m_V)$ such that $R \subseteq V \subsetneq Q(R)$ ...