All Questions
4 questions
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 $\...
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 ...
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 ...