17 votes
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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 ...
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16 votes
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How did height in algeb. number theory/elliptic curves started?

I think that Keith Conrad is correct and heights via maxs of valuations are in Weil's "Mordell-Weil" paper. But note that Weil's paper generalized Mordell's work in two ways. First, he extended from $\...
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16 votes
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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 ...
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15 votes
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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 ...
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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. ...
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10 votes
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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, ...
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  • 2,177
10 votes
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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, ...
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  • 7,271
8 votes
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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. $\frac{4^m-1}{3} = 1 + 2^2 + 2^4 + \cdots + 2^{2m-2}$ ...
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  • 4,582
7 votes
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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. ...
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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 ...
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  • 1,525
6 votes

Completion of a finite field extension is also finite?

Since $L/K$ is finite, we can write $L = K(a_1,\ldots,a_r)$ for some $a_i$. These same $a_i$ will generate $L_w$ over $K_v$. You can't hope for anything simpler than that! Well, you could hope that a $...
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  • 42.5k
6 votes

Finite extension of local fields

Yes. For an example, let $k$ be an uncountable algebraically closed field of characteristic $p>0$, and let $K = k((x))$, the field of fractions of the ring of formal power series $\mathcal{O}_K = k[...
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6 votes
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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 ...
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  • 12.5k
6 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 $\mathbb Q$. ...
6 votes
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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 ...
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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. (...
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5 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 ...
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4 votes
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Finite extension of local fields

Let $k$ be any field and $K=k((x,y))$ the fraction field of the ring $k[[x,y]]$ of formal Laurent series in two variables. Then Harbater and Stevenson proved that the absolute Galois group of $K=k((x,...
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4 votes
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Subgroup of Projective general linear group on complete discrete valuation ring

I don't know if the following qualifies as "elementary", but here is how I would prove this (for $2$ replaced by some $d\in \mathbb{N}$): $\DeclareMathOperator{\gl}{GL}$ First, choose some finitely ...
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4 votes
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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 ...
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4 votes
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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 ...
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4 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-...
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4 votes
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Rank 1 valuations that are not discrete on finite transcendental extensions of the rationals

Example: $val(X_1^{e_1}X_2^{e_2})=e_1 + e_2 \sqrt{2}$.
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4 votes
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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 ...
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  • 2,015
4 votes
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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 ...
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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 $\hat{\bar{K}}$.
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  • 17.1k
3 votes
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'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 ...
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  • 2,015
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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3 votes
Accepted

Completion of discrete valuation ring

The transcendence degree of $\hat{R}$ over $R$ can be large. If $R$ is the Henselization of $k[t]_{(t)}$ then the transcendence degree is max(${2^{\aleph_0}}$,$|R|$), and there are many ways to ...
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