17 votes
Accepted

Completion and algebraic closure

First you have to observe that since all extensions of the valuation to $\bar{K}$ are conjugate, $\hat{\bar{K}}$ is well-defined up to (non-unique) isomorphism. Now, since $\hat{\bar{K}}$ is complete ...
Laurent Moret-Bailly's user avatar
17 votes
Accepted

Is it a valuation ring?

This is (essentially) a conjecture of Krull; a counter-example was given by P. Ribenboim: Sur une note de Nagata relative à un problème de Krull, Math. Zeit. 64, 159-168 (1956). Note that "...
abx's user avatar
  • 37.3k
16 votes
Accepted

Extension of 2-adic valuation to the real numbers

No. The important thing to know is that, if $K \subseteq L$ is a field extension and $v: K \to \mathbb{R}$ is a valuation, then $v$ can be extended to $L$. So I can answer all of your questions by ...
David E Speyer's user avatar
15 votes
Accepted

simple questions on topological rings arising in the context of Perfectoid Spaces

Firstly, I don't think you should start to learn about adic spaces by thinking about perfectoid spaces. The sensible examples of adic spaces for a beginner to think about are sane Noetherian things ...
Kevin Buzzard's user avatar
12 votes
Accepted

In a CM field, must all conjugates of an algebraic integer lying outside the unit circle lie outside the same?

Zero is the only algebraic integer which has all its conjugates strictly inside the complex unit circle. (Look at the norm.) For explicit examples with conjugates on either side of the unit circle, ...
GNiklasch's user avatar
  • 2,391
11 votes

Uniquely ordered commutative rings

In a field, an element can be made negative in some order iff it is not a sum of squares. Thus, $F$ is uniquely ordered iff for every $a\in F^\times$, $a$ or $-a$, but not both, is a sum of squares. ...
Emil Jeřábek's user avatar
11 votes
Accepted

Is every field the residue field of a discretely valued field of characteristic 0?

Yes, by Hasse-Schmidt ("Die Struktur diskret bewerteter Koerper", Crelle's Journal, 1934) for any field $k$ of characteristic $p$ there exists a strict Cohen ring $A$, which is a Noetherian, ...
John Rognes's user avatar
  • 8,837
8 votes

Ostrowski's Theorem for topological rings?

Thanks to YCor's examples in the comments, I decided this question was worth a deep dive. It turns out on the one hand that There are lots of exotic (Hausdorff) field topologies on $\mathbb Q$. ...
8 votes
Accepted

Subsequence of the cubes

Experimenting with a CAS suggests an induction. In order to handle the induction, we need to consider the forms of the numbers involved. $\frac{4^m-1}{3} = 1 + 2^2 + 2^4 + \cdots + 2^{2m-2}$ ...
Peter Taylor's user avatar
  • 6,571
7 votes
Accepted

Literature on non-Archimedean analogues of basic complex analysis results

Benedetto has a textbook that discusses basic $p$-adic analysis, although his aim is to study $p$-adic dynamics. And it's for a single variable. But might be a good place to get some information. ...
Joe Silverman's user avatar
6 votes

Is the integral closure of a valuation ring in a finite separable extension of its fraction field étale?

EDIT I may have overlooked the assumption that $L$ is contained in $\hat{K}$. In general, if $v$ is a valuation of $K$ and $L/K$ is finite separable then $L \otimes_K \hat{K}$ is reduced and thus ...
François Brunault's user avatar
6 votes

Ostrowski's Theorem for topological rings?

The following relevant classification result, due to Kowalsky and Dürbaum [2], appears in Appendix B of Engler and Prestel [1]. Let $(K,\tau)$ be a topological field. Then $\tau$ is called a V-...
Emil Jeřábek's user avatar
6 votes

Examples of NIP fields of characteristic $p$

For any field $K$ and ordered group $\Gamma$, the Malc'ev-Neumann field $K((\Gamma))$ is maximal by a Theorem of Krull, hence algebraically maximal. Taking a perfect infinite NIP field $K$, for ...
Drike's user avatar
  • 1,555
6 votes
Accepted

The notions of $H^0(\widehat{ D})$ and $h^0(\widehat{D})$ are not satisfactory

The Arakelov-theoretical analogue of Riemann's inequality is Minkowski's theorem. As Felipe Voloch's answer indicates, Tate's approach makes the Riemann-Roch theorem a consequence of the Poisson ...
ACL's user avatar
  • 12.8k
6 votes
Accepted

Open immersion of affinoid adic spaces

This is not correct in general. There are in fact two examples in Bosch's Lectures on Formal and Rigid Geometry p.61-63. Let me sketch the first one. While it uses rigid-analytic spaces, it can be ...
Louis Jaburi's user avatar
5 votes

Birational Group Law

Thanks to nfdc23 for explaining the definition. There are counterexamples. For instance, begin with $S=\text{Spec}(R)$ for a discrete valuation ring $(R,\mathfrak{m})$. Let $\nu:\text{Spec}(\...
5 votes

The notions of $H^0(\widehat{ D})$ and $h^0(\widehat{D})$ are not satisfactory

This issue already comes up in Tate's proof of the functional equation for zeta functions. The functional equation should come out of some version of Riemann-Roch and, for function fields, it does. (...
Felipe Voloch's user avatar
4 votes
Accepted

Understanding a valuation property of function fields

Let $m_0$ be the degree of $\mathfrak p$, so the splitting field of $\mathfrak p$ is the extension of $\mathbb F_q$ of degree $m_0$. The set of roots of $t^{p^m}-t$ contains this field extension if ...
Peter Mueller's user avatar
4 votes
Accepted

Examples of NIP fields of characteristic $p$

Understanding concretely which fields (in the "pure" language of rings, $\mathcal{L} = \{0, 1, +, \cdot\}$) are NIP is a topic of current interest in model theory. The 2015 paper "Dp-minimal valued ...
John Goodrick's user avatar
4 votes
Accepted

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

Example: $val(X_1^{e_1}X_2^{e_2})=e_1 + e_2 \sqrt{2}$.
David Lampert's user avatar
4 votes
Accepted

Henselian valued fields for characteristic $0$: a characterization

In general, i.e. for any valued field $(K,v)$, the implication $K = K_v \cap \overline{K}$ (or more precisely that $(K,v)$ have no immediate algebraic extension, i.e. that $(K,v)$ be algebraically ...
nombre's user avatar
  • 2,367
4 votes

In a CM field, must all conjugates of an algebraic integer lying outside the unit circle lie outside the same?

To expand on GNiklasch's answer, and analyse what you write as well: we always have (when complex conjugation is central in the Galois group) have $\overline{\alpha^{\sigma}} = {\bar \alpha}^{\sigma}$ ...
Geoff Robinson's user avatar
4 votes
Accepted

Recurrence for the sum

Let $n=m^tk$ where $m\nmid k$. Then $f(n)=m^t$.  Furthermore, if $t>0$, then $f(n/m)=m^{t-1}$ and $n-f(n/m)=m^{t-1}(mk-1)$. It follows that $a(n)=a(m^{t-1}k)+a(m^{t-1}(mk-1))$ and further by ...
Max Alekseyev's user avatar
3 votes
Accepted

Definition of model functions and their density in $C^0(X^\text{an})$

As alluded to in the question, the space $\mathcal{D}(X)_{\mathbb{Z}}$ is indeed the space of model functions arising from integral divisors. Now, in order to deduce that $\mathcal{D}(X)_{\mathbb{Q}}$ ...
msteve's user avatar
  • 572
3 votes
Accepted

Valuation of congruent elements in a local division ring

$\newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}}$ Not necessarily. Take $K = \Q_3 + \Q_3 i + \Q_3 j + \Q_3 ij$ with $i^2=-1$ and $j^2=3$, $r=2$, $x=y=j$, $a=3i$ and $b=3(1+i)$. Then $v(j)=1$, ...
Aurel's user avatar
  • 4,933
3 votes
Accepted

What is the definable functor associated to an algebraic scheme (model theory of valued fields)

This follows by elimination of imaginaries in algebraically closed fields. Given a finite affine cover of your scheme, one obtains a functor to Set by gluing the affine charts. This is indeed not ...
Cubikova's user avatar
  • 315
3 votes
Accepted

Is $\mathbb{F}_{p}(t)^{h}$ an elementary substructure of/existentially closed in $\mathbb{F}_{p}((t))$?

There is apparently not a very short answer to this. It is, I think, really not known whether this extension is elementary and it would be very surprising if it was known, since we do not even know if ...
Florian Felix's user avatar
3 votes

Completion and algebraic closure

The completion of the algebraic closure of a valued field is algebraically closed and complete. So any further operation of closure or completion gives you a field isomorphic to $\hat{\bar{K}}$.
Myshkin's user avatar
  • 17.4k
3 votes
Accepted

'Smallest' subfield of the Surreals which is isomorphic to the Surreals as an ordered group

$\DeclareMathOperator{\Noo}{\mathbf{No}}$This might actually be a dead end. This is because if $F$ is isomorphic as an ordered group to $\Noo$, then their value classes under natural ordered group ...
nombre's user avatar
  • 2,367
3 votes

Given a non-field local domain $R$, finding a dominating Valuation ring whose residue field is algebraic/finite extension of the residue field of $R$

Yes, an algebraic extension is always possible. If one applies Zorn's lemma not to $$\{R\subseteq V\subseteq Q(R) \mid V\,\text{local}\}$$ but to $$\{R\subseteq V\subseteq Q(R) \mid V\,\text{local} \...
Johannes Hahn's user avatar

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