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Hasse principle for Brauer groups of fields of transcendence degree 2

In his paper "A Hasse principle for function fields over PAC fields" (DOI link), Ido Efrat proves the following result: Let $F$ be an extension of a perfect PAC field $K$ of relative ...
aspear's user avatar
  • 41
16 votes
2 answers
943 views

Higher-rank Archimedean valuations of $\mathbb{Q}$, does it exist?

I was reading the proof of Ostrowski's theorem, with an eye toward the Zariski-Riemann space (as well as adic space, Berkovich space, etc.) In the proof, the value group is always assumed to be in $\...
zxx's user avatar
  • 343
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
2 votes
1 answer
183 views

Pair of recurrence relations with $a(2n+1)=a(2f(n))$

Let $f(n)$ be A053645, distance to largest power of $2$ less than or equal to $n$; write $n$ in binary, change the first digit to zero, and convert back to decimal. Let $g(n)$ be A007814, the ...
Notamathematician's user avatar
1 vote
1 answer
295 views

Formula from the recurrence relation

Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$. Then we have an integer sequence ...
Notamathematician's user avatar
3 votes
2 answers
458 views

Subsequence of the cubes

Let $p$ and $q$ be integers. Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$. Then ...
Notamathematician's user avatar
12 votes
1 answer
535 views

Open immersion of affinoid adic spaces

If $R$ and $S$ are complete Huber rings with $\varphi: R \to S$ a continuous map, then is it true in general that if $\mathrm{Spa}(S, S^\circ) \to \mathrm{Spa}(R, R^\circ)$ is an open immersion of ...
Ashwin Iyengar's user avatar
3 votes
2 answers
726 views

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

This question is inspired from the post linked below: Can an algebraic number on the unit circle have a conjugate with absolute value different from 1? What I am curious about is the following: let $\...
asrxiiviii's user avatar
3 votes
1 answer
289 views

Existence of algebraic integer with absolute value equal to reciprocal of maximum of $1$ and absolute value of a given algebraic number

Consider a number field $K$, and let $v_1, \cdots v_n$ ($n \in \mathbb N$) be some finite (i.e. non-archimedean) places of $K$. Is the following true? For every $\alpha \in K^\times$ there exists $\...
asrxiiviii'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
12 votes
1 answer
2k views

Extension of 2-adic valuation to the real numbers

I just want to know what properties of valuations extend to $\mathbb R$... Denote an extension of the 2-adic valuation from $\mathbb Q$ to $\mathbb R$ by $\nu$. Suppose $\nu(x)=\nu(y)=0$. Is it true ...
domotorp's user avatar
  • 19.1k
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
1 vote
1 answer
64 views

Valuation of congruent elements in a local division ring

Let $K$ be a complete local division ring (note $v$ its valuation). For $x,y\in K$ ($y\ne0$), one puts $x^y=yxy^{-1}$. Let $r\in\mathbb N$. Consider $x,y\in K$ and $a,b\in K^*$ such that $v(x-y)\ge r$ ...
joaopa's user avatar
  • 3,998
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
0 votes
1 answer
145 views

Understanding a valuation property of function fields

I came across this point in a paper recently and I'm having difficulty seeing why it's true. Any explanations or hints would be appreciated. For any prime $\mathfrak{p}$ of $\mathbb{F}_q [t]$ such ...
user221330's user avatar
0 votes
0 answers
205 views

Why is any non-archimedean field Huber?

Here a non-archimedean field means a field $k$ whose topology is induced from a non-archimedean norm $| \cdot |: k \to \mathbb{R}_{\geq 0}$. As a reminder, a ring $A$ is adic if there is an ideal $I \...
Aaron Johnson's user avatar
3 votes
0 answers
169 views

Rational power series and extensions

Let $F$ be a field, let $F(x)$ the field of rational functions, and let $F((x))$ the field of Laurent series (which contains $F(x)$). One may ask: which series $\sum_i a_i x^i$ lie in $F(x)$? The ...
Stefan Witzel's user avatar
5 votes
1 answer
966 views

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

(I apologize in advance for these simple questions, I am a beginner trying to go through Scholze's paper Perfectoid Spaces). Let $(R, R^+)$ be an affinoid $k$-algebra as defined in Scholze's paper ...
usr0192's user avatar
  • 785
9 votes
2 answers
2k views

Completion and algebraic closure

Following this question: Given a valued field $K$, denote with $\bar{K}$ its algebraic closure and with $\hat{K}$ the completion. Then both $\hat{\bar{K}}$ and $\hat{\bar{\hat{K}}}$ are complete and ...
Martin's user avatar
  • 1,101
12 votes
1 answer
778 views

Is every Polish ring topology on $\mathbb{C}$ defined by an absolute value?

There is a unique up to isomorphism algebraically closed field of characteristic 0 and cardinality of the continuum. Let's call it $K$. We usually call it $\mathbb{C}$, but by this we impose a ...
Uri Bader's user avatar
  • 11.6k
3 votes
1 answer
341 views

On conductors, levels and traces on quaternion algebras

I am currently working on level issues in the division central simple algebra case, say $D$ over a local non-archimedean field $F$ (e.g. $\mathbf{Q}_p$). Let say that $\mathcal{O}_D$ and $\mathcal{O}...
Desiderius Severus's user avatar
4 votes
0 answers
375 views

Extension of the product formula for valuations to a simultaneous completion

It is well known that $\mathbb{C}$ and $\mathbb{C}_p$ are "algebraically" isomorphic (that is, ignoring the topology), but an isomorphism depends on the axiom of choice and there is no canonical way ...
Maurizio Monge's user avatar
0 votes
2 answers
280 views

Goldbach-type problem: the valuation of irreducible elements of a subgroup of $\mathbb{Q}_{> 0}$

Let $\phi : \mathbb{Q}_{>0} \to \mathbb{Z}$ be the group morphism defined by $\phi(p) = p$ for $p$ a prime number. It follows that $\phi(\prod_i p_i^{n_i}) = \sum_i n_i p_i$, with $p_i$ a prime ...
Sebastien Palcoux's user avatar
2 votes
1 answer
1k views

Maximal unramified extension and inertia group for separable closure

I have a problem in understanding the inertia group of an infinite extension. I am studying it in this context. Let $K$ be a field, $v$ a discrete valuation on $K$, and $\mathcal{O}_v$ the discrete ...
Pgatti's user avatar
  • 147
1 vote
2 answers
354 views

Algebraic maximal extension and algebraic closure

Let $K$ be a valued field. We say that $K$ is algebraic maximal if any algebraic extension of $K$ has either a bigger value group or a bigger residue field. Under which condition is an algebraic ...
Richard's user avatar
  • 31
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
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
0 votes
1 answer
481 views

Relating $p$-adic Valuations of Elements in $\mathbb{C}$ and $\mathbb{C}_p$

Let $K = \mathbb{Q}(\theta)$, where $\theta$ is a root of an irreducible polynomial $g \in \mathbb{Z}[t]$. Fix a rational prime $p$. Let $\theta^{(1)}, \ldots, \theta^{(n)}$ be the roots of $g$ in $\...
OSHE's user avatar
  • 113
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
10 votes
1 answer
1k views

Existence of maximal totally ramified extensions of an arbitrary CDVF

Let $K$ be a complete, discretely valued field with (let's say) perfect residue field $k$. We have a unique maximal unramified extension $K^{unr}$ of $K$ and a unique maximal tamely ramified ...
Pete L. Clark's user avatar
2 votes
2 answers
671 views

Fiddling with p-adics

A paper I'm reading implicitly assumes the statement: Let $K_0$ be the completion of $\mathbb {Q}_ p^{un}$. Then any finite extension of $K_0$ is complete with residue field $\bar {\mathbb {F}} _p$. ...
Randy Brown's user avatar
  • 1,386