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2 votes
1 answer
151 views

For an element in the integral closure of an ideal $I$ - which power is in $I$?

Consider an ideal $I$ in a ring $R$. If $f \in R$ belongs to the integral closure of $I$, then there is $k_0 \geq 0$ such that $f^k \in I^{k-k_0}$ for all $k \geq k_0$. Are there any known upper ...
pinaki's user avatar
  • 5,359
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
1 vote
0 answers
104 views

Does maximally incompleteness cause nonvanishing of the extension of maximal ideal of a valuation ring by rank 1 free module?

In B. Bhatt's lecture notes[1], Remark 4.2.5 says ... $\operatorname{Ext}_R^2(k,R)$ is non-zero if $K$ is not spherically complete. which amounts to the following pure algebraic question. Statement ...
XYC's user avatar
  • 441
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
1 vote
0 answers
37 views

Valuations of coefficients of minimal polynomials for tuples

Suppose you are given two valued fields $(K,v) \subseteq (L,w)$ and a tuple $a \in L^n$. What kind of restrictions do we have on the valuation of the coefficients of polynomials $q \in K[x_1,\dots x_n]...
Simone Ramello's user avatar
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
1 vote
1 answer
197 views

Chain of closed irreducible sets on Zariski Riemann spaces

Let $A$ be a domain and $K=\mathrm{Frac}(A)$. The Zariski Riemann space $\mathrm{ZR}(K,A)$ is the set of all valuation rings of $K$ containing $A$. It comes with a natural center map \begin{align}...
nowhere dense's user avatar
3 votes
1 answer
136 views

Rank 1 valuations that are not discrete on finite transcendental extensions of the rationals

Suppose $K=\mathbb{Q}(X_1,\dots,X_n)$ is a purely transcendental extension of the rationals on finitely many indeterminates. Can anyone give an example of a rank $1$ valuation on $K$ that fails to be ...
Harry Gindi's user avatar
  • 19.6k
4 votes
0 answers
357 views

Completeness of the field of fractions of a ring of formal power series

Let $k$ be a field and let $k[[X,Y]]$ be the ring of formal power series with coefficients in $k$. Let $k((X,Y))$ be its field of fractions. For $F\in k[[X,Y]]$, $F\neq 0$ define $v(F)$ as the least ...
Carlos's user avatar
  • 1,688
1 vote
1 answer
142 views

Valuation theory on semisimple algebras used in the paper of Cohen-Martinet: reference request

I'm currently reading the paper of Henri Cohen & Jacques Martinet "Etude heuristique des groupes de classes des corps de nombres" On the 2nd section, they recall some facts on valuations, ...
gualterio's user avatar
  • 1,013
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
2 votes
0 answers
118 views

Valuation Rings and Ultrafilters II

See my post here: Valuation Rings and Ultrafilters Let $K$ be a field, and let $\mathcal{S}$ be the set of pairs $(R, \mathfrak{p})$ of subrings $R$ of $K$ with designated prime ideals $\mathfrak{p}$ ...
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0 votes
1 answer
269 views

Valuation ring satisfying either a.c.c. or d.c.c. on prime ideals

If a commutative ring with unity has finite Krull dimension, then it satisfies a.c.c. and d.c.c. on prime ideals. The converse is not true in general, as can be seen from here An infinite dimensional ...
user521337's user avatar
  • 1,209
19 votes
2 answers
566 views

Ostrowski's Theorem for topological rings?

Ostrowski's theorem classifies all absolute values on a number field $K$. Questions: More generally, can one classify all Hausdorff topologies on $K$ making $K$ into a topological field? In ...
Tim Campion's user avatar
1 vote
1 answer
465 views

Valuation ring whose maximal ideal and every ideal of finite height are principal

Let $(R, \mathfrak m)$ be a valuation ring such that $\mathfrak m$ and every ideal of finite height is principal. Then is $R$ Noetherian , i.e. a discrete valuation ring ?
user avatar
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
3 votes
0 answers
96 views

Luroth's theorem for Discrete valuation rings?

Luroth's theorem states that if $k$ is a field and $L$ is a field extension of $k$ such that $k \subset L \subseteq k(X)$, then $L=k(f(X))$ for some $f(X) \in k(X) $ . My question is ; is there any ...
user avatar
7 votes
1 answer
282 views

Uniquely ordered commutative rings

I am wondering whether there are reasonable necessary and/or sufficient conditions to dedice whether a commutative ring can be uniquely ordered (like for instance $\mathbb{Z}$) or not. In the field ...
Alice's user avatar
  • 131
5 votes
0 answers
209 views

Completions of $K(x)$

Let $K$ be a field. Are there books or articles discussing completions of $K(x)$ with respect to the metric induced by the $p$-adic valuation $|\;\;|_p$ where $p\in k[x]$ is irreducible and different ...
Chilote's user avatar
  • 596
2 votes
0 answers
120 views

Group of units of a valuation

Let K be a field. Then a subring R of K is called a valuation ring if for all $x \in K^*,$ either $x \in R$ or $x^{-1} \in R$ (or both). It can be shown that for any valuation $v$ on $K,$ the ring $\...
Alice's user avatar
  • 131
3 votes
0 answers
108 views

Structure of valuations on $\mathbb{F}_q(X,Y)$?

I'm looking to construct all valuations on $\mathbb{Q}(X,Y)$ extending the p-adic valuation on $\mathbb{Q}$ and understand their structural properties. In doing this, to obtain 3 dimensional valuation ...
rmg512's user avatar
  • 147
3 votes
1 answer
2k views

Completion of a finite field extension is also finite?

Let $(L,w)/(K,v)$ be a finite extension of valuation fields, and let $L_w$, $K_v$ be the respective completions of $(L,w)$, $(K,v)$. Is the field extension $L_w/K_v$ finite? For nonarchimedean ...
Fan Zheng's user avatar
  • 5,169
1 vote
0 answers
79 views

Saturation of a subalgebra over the Tate-algebra inside the power series ring

Let $A$ be a discrete valuation ring and $\pi$ a uniformizer. Over $A$ we consider the Tate-algebra $$A\langle t \rangle =\{ f=\sum_{n=0}^\infty a_nt^n \mid a_n\in A, \lim_{n\to \infty} \lvert a_n\...
Andreas Maurischat's user avatar
0 votes
1 answer
469 views

Finite extensions of residue fields of Henselian DVRs

Let $K$ be an Henselian discrete valuation field such that its completion is separable over $K$. Let $F$ be its infinite residue field. Is it true that a finite extension of $F$ is a simple extension ...
Jana's user avatar
  • 2,032
0 votes
0 answers
383 views

Pseudo-cauchy sequence and valuation

Let $k$ be a field and $x$ is transcendental over $k$. Can we construct a pseudo-cauchy sequence $(a_{i})$ convergent to $x$ with each $a_{i}$ is algebraic over $k$ and $k(a_{i})\subseteq k(a_{i + 1})$...
Rajnish's user avatar
  • 173
2 votes
1 answer
601 views

Henselization of valued field

What is the importance of henselization in valuation theory, when the rank of valuation is bigger than one? Thanks
Rajnish's user avatar
  • 173
11 votes
2 answers
1k views

Valuations on tensor products

Let $A$ be a commutative ring, $B$ (resp. $C$) be a commutative $A$-algebra endowed with a valuation $v$ (resp. $w$), not necessarily of rank 1. Assume that $v$ and $w$ induce equivalent valuations on ...
Weizhe Zheng's user avatar
1 vote
0 answers
196 views

Extending commuting endomorphims of a complete discrete valued field to the algebraic closure?

Is it true that any two commuting endomorphisms of a complete discrete valued field extend to commuting automorphisms of the algebraic closure?
Artem Chernikov's user avatar
8 votes
1 answer
351 views

When is a valued field second-countable?

Let $R$ be a valuation ring, with fraction field $K$ and residue field $k$. Denote by $\Gamma=K^{\times}/R^{\times}$ the valuation group (assumed nontrivial). The valuation $v:K^{\times}\to\Gamma$ ...
Laurent Moret-Bailly's user avatar
2 votes
1 answer
399 views

Quotient field extension for an incomplete DVR

Let $R$ be a DVR with maximal ideal $xR$, and assume that $R$ is not complete in the $xR$-adic topology. Let $\hat{R}$ be the completion of $R$ in the $xR$-adic topology. Set $K=Q(R)$, the fraction ...
Sean Sather-Wagstaff's user avatar
7 votes
1 answer
2k views

The space of valuations of a function field

Hello, I'm looking for someone who can help me to understand Zariski's theory of valuations. First I outline the theory: we take a field $K$ which is a finitely generated transcendent extension of ...
Xin Nie's user avatar
  • 1,804
6 votes
2 answers
563 views

If the discriminant of a binary quadratic form has high valuation, is the form "almost a square".

For a binary quadratic form $ax^2+bxy+cy^2$ over a field (characteristic not 2), the discriminant $b^2-4ac$ is 0 if and only if that form is the square of a linear form. I am curious about an ...
Daniel Erman's user avatar
  • 2,955
3 votes
1 answer
1k views

Does totally ramified extension really exist?

The answer is certainly "Yes", but this is the problem I met in Algebraic Number Theory by Neukirch. I guess that I must be doing something wrong, since otherwise I will get the statement "There are ...
Li Zhan's user avatar
  • 95
4 votes
3 answers
2k views

Chevalley's valuation extension theorem and the axiom of choice

Hello, Do we know if the axiom of choice is needed for Chevalley's valuation/place extension theorem (i.e. the theorem that states that for every valued field and a field extension, one can extend ...
Itay's user avatar
  • 43
11 votes
2 answers
863 views

Valuations and separable extensions

Let $R$ be a valuation ring containing a field $k$, with residue field $F$ and quotient field $K$. Assume $F/k$ is separable. Is $K/k$ separable? I have convinced myself that (for a positive answer) ...
Laurent Moret-Bailly's user avatar
2 votes
2 answers
1k views

Why are extensions so heavily emphasized in valuation theory?

Whenever I read anything about valuations or things related to them (such as local fields) extensions always occupy a prominent position and a huge amount of effort is expended to derive results about ...
teil's user avatar
  • 4,351
3 votes
1 answer
928 views

How exotic can DVRs be in the ring of rational functions over a local field?

Suppose that $R$ is a complete DVR with field of fractions $K$, uniformiser $\pi$ and residue field $k$. Let $B$ be a subring of the ring $K(t)$ of rational functions over $K$. Moreover assume that $...
Simon Wadsley's user avatar
3 votes
1 answer
463 views

Are valuation rings regular?

This question is short, and to the point: Valuation rings are certainly integrally closed, but are they regular? The motivation is that I'm trying to understand the resolution of singularities of ...
Makhalan Duff's user avatar
5 votes
2 answers
2k views

Is a valuation domain PID when its maximal ideal is principal?

It is known that a valuation domain is a principal ideal ring if and only if its prime ideals are principal. Is it a principal ideal ring when its unique maximal ideal is principal?
TmobiusX's user avatar
  • 1,207