All Questions
5 questions
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 ...
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 $k$?
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 ...
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 $R \subseteq V \subsetneq Q(R)$ ...