In modern valuation theory, one studies not just absolute values on a field, but also Krull valuations. The motivation is easy enough:If $k$ is a field, a valuation ring of $k$ is a subring $R$ such that for every $x \in k^{\times}$, at least one of $x, x^{-1}$ is an element of $R$. (It follows of course that $k$ is the fraction field of $R$.) If $| \ |$ is a non-Archimedean norm on a field, then the set $\{x \in k \ | \ |x| \leq 1 \}$ is a valuation ring. However, the converse does not hold, since if $R$ is a valuation ring, then $k^{\tim...Read more

Let $F$ be a global function field with full constant field $\mathbb{F}_q$. We fix a place $\infty$ and let $A$ be the ring of elements of $F$ regular away from $\infty$. We denote by $v_\infty$ the normalized valuation associated to $\infty$. If $a\in A$ and $v_\infty(a)\geq 0$ then $a$ has no poles. Therefore (see for example Corollary 1.1.20 in Stichtenoth's Algebraic Function Fields and Codes) $a$ is algebraic over $\mathbb{F}_q$ and then $a\in \mathbb{F}_q$ since it is the full constant field. We then have $v_\infty(a) < 0$ for all $a\i...Read more

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 no totally ramified extensions except the trival ones".Let $K$ be Henselian field, $L/K$ be a finite, totally ramified extension. Let $\lambda$ and $\kappa$ be the residue field of $L$ and $K$ respectively. Because $L/K$ is totally ramified, $K$ is the maximal unramified subextension, so we have $\lambda=\kappa$. If $L\ne K$, let $a \in L-K$. Sinc...Read more

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 $v:K^{\times}\to\Gamma$ decomposes $K^{\times}$ as a disjoint union of nonempty open subsets, indexed by $\Gamma$. Each of these is homeomorphic to $R^{\times}$, which is in turn (using the reduction map to $k$) a disjoint union of nonempty open subsets, indexed by $k^{\times}$. We conclude that any basis for the topology of $K$ must have cardinality at least $\kappa:=\max(...Read more

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) it is enough to treat the case of a rank one valuation ("height one" in Bourbaki's terminology), with $F=k$ and $K/k$ finitely generated. (Instaed of the latter condition, we may assume $K$ complete).The answer is yes for a discrete valuation, because then (if $F=k$) the completion of $R$ is isomorphic to $k[[t]]$....Read more

I know at least one method of constructing a convex valuation ring of rank n (but it is rather complicated).. What are the easiest methods of doing this.. given a natural number n I want to have a valuation ring (preferably convex) whose rank is n. I have heard you can do this with polynomials and power series, but I am not really sure how this is done....Read more

I'm pretty sure trivial valuation over a field cannot be extended to a non-trivial one in a bigger field. Is there a simple way to show this without using the sledge hammer theorem on valuation extension over complete valued field?...Read more

Let $(K, \nu)$ be a valued field and $x$ is transcendental over $K$. Is there exist a henselian extension of $(K, \nu)$ in between $(K, \nu)$ and $(K(x), \nu^{'})$ where $\nu^{'}$ is an extension (arbitary) of $\nu$....Read more

This is a similar question from the book "Valued Fields by Antonio J. Engler and Alexander Prestel, Springer, 2005 " page 82, Exercise 3.5.4.(b).Let $(K_{1}, V_{1})\subseteq (K_{2}, V_{2})$ be finite extension of valued fields. Assume that $[K_{2} : K_{1}] = n$. Let $G_{1}$ and $G_{2}$ be value groups of $V_{1}$ and $V_{2}$ respectively. Prove that $[G_{2} : G_{1}] = n$. Thanks...Read more

I am looking for the original reference for Ostrowski's theorem of 1916 that the only valuations on the rational numbers are the trivial, Archimedean, and p-adic valuations.http://en.wikipedia.org/wiki/Ostrowski's_theoremWikipedia refers to Koblitz (p-adic Numbers, p-adic Analysis, and Zeta-Functions), but I can't find the reference in there....Read more

Usually absolute value $v$ on a field $K$ is defined as a map $x\to |x|_v$ such that $\forall x,y\in K$ one has$$\begin{align}&|x|_v=0\iff x=0\\&|xy|_v=|x|_v|y|_v\\&|x+y|_v\leq |x|_v+|y|_v\end{align}$$If one has $|x+y|_v\leq \max(|x|_v,|y|_v)$ which is stronger than the triangle inequality, the absolute value is ultra metric or non Archimedean.Now Marc Hindry in his “arithmétique” suggests a broader definition replacing the third inequality by $$\exists C_v\gt 0,\,\,|x+y|_v\leq C_ v\max(|x|_v,|y|_v)$$This definition covers the ultra...Read more

Does anyone know of an elementary proof that an algebraically maximal field is Henselian (ie one that does not assume knowledge of henselizations)? Definitions:We say a valued field $(K,v)$ is algebraically maximal if it has no proper algebraic intemediate extensions, that is any algebraic valued field extension of $(K,v)$ either increases the value group or the residue field.We say a valued field is Henselian if if satisfies Hensels lemma, for instance any polynomial $X^n+X^{n-1}+a_{n-2}X^{n-2}+...+a_0$ with $v(a_{n-2}),...,v(a_0)>0$ has a ...Read more

We define a valuation on the field of rational number $\mathbb Q$ as follows. For example if we choose a prime number $2$ then for $x \neq 0\in \mathbb Q$, $v(x) = v(2^{n}a/b)= n$ where $n$ is an integer and $a$ and $b$ are relatively prime integer. I have a question, how do we extend this valuation explicitly over the field $\mathbb Q(2^{1/2})$. Thanks...Read more

I'm currently learning the basic machinery of valuation theory from "ordered exponential fields" by Kuhlmann, and I have a question regarding an assertion made on page 1.She gives the standard axioms for a valuation $\nu:M\rightarrow\Gamma\cup\{+\infty\}$ with $M$ a left $R$-module and $\Gamma$ a totally ordered set order extended to $\Gamma\cup\{+\infty\}$. Namely, these axioms are$\nu$ is surjective,$\nu(x)=+\infty\iff x=0,$$\nu(rx)=\nu(x)$ for all $x\in M$ and $0\neq r\in R$,$\nu(x-y)\geq\min\{\nu(x),\nu(y)\}$ for all $x,y\in M$.She then goe...Read more

Let $F$ be an ordered field. Let $A$ be a convex subring of $F$. $A$ is a valuation ring on $F$, so a local ring. Let $m_A$ denote its maximal ideal.As a general result of valuation-theory, we know that for any subfield $C$ of $F$ contained in $A$ that is maximal for inclusion with respect to these conditions, the residue field $F_A:= A / m_A$ is algebraic over the image $C_A$ of $C$ under the residue map $A \rightarrow A /m_A$. Moreover $C$ is algebraically closed relatively to $F$.Now, in this more specific case, $F_A$ is ordered by $a + m_A ...Read more