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 ...

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 "...

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 ...

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 ...

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, ...

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.
...

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, ...

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 $\mathbb Q$.
...

Community wiki

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. $\frac{4^m-1}{3} = 1 + 2^2 + 2^4 + \cdots + 2^{2m-2}$ ...

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.
...

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 ...

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-...

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 ...

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 ...

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 ...

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}(\...

Community wiki

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. (...

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 ...

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 ...

4
votes

Accepted

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

Example: $val(X_1^{e_1}X_2^{e_2})=e_1 + e_2 \sqrt{2}$.

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 ...

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
$\overline{\alpha^{\sigma}} = {\bar \alpha}^{\sigma}$ ...

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 ...

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 $\mathcal{D}(X)_{\mathbb{Q}}$ ...

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 $v(j)=1$, ...

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 ...

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 ...

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 $\hat{\bar{K}}$.

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 ...

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} \...

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