Skip to main content

All Questions

Filter by
Sorted by
Tagged with
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 ...
George's user avatar
  • 328
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 $k$?
Alex Mennen's user avatar
  • 2,130
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 ...
Tim Campion's user avatar
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 ...
userm's user avatar
  • 41
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 $R \subseteq V \subsetneq Q(R)$ ...
user avatar