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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 ...