Article

Quasi-Valuations Extending a Valuation

Journal of Algebra (Impact Factor: 0.6). 09/2012; 372. DOI: 10.1016/j.jalgebra.2012.09.019
Source: arXiv

ABSTRACT Suppose $F$ is a field with valuation $v$ and valuation ring $O_{v}$, $E$ is
a finite field extension and $w$ is a quasi-valuation on $E$ extending $v$. We
study quasi-valuations on $E$ that extend $v$; in particular, their
corresponding rings and their prime spectrums. We prove that these ring
extensions satisfy INC (incomparability), LO (lying over), and GD (going down)
over $O_{v}$; in particular, they have the same Krull Dimension. We also prove
that every such quasi-valuation is dominated by some valuation extending $v$.
Under the assumption that the value monoid of the quasi-valuation is a group
we prove that these ring extensions satisfy GU (going up) over $O_{v}$, and a
bound on the size of the prime spectrum is given. In addition, a 1:1
correspondence is obtained between exponential quasi-valuations and integrally
closed quasi-valuation rings.
Given $R$, an algebra over $O_{v}$, we construct a quasi-valuation on $R$; we
also construct a quasi-valuation on $R \otimes_{O_{v}} F$ which helps us prove
our main Theorem. The main Theorem states that if $R \subseteq E$ satisfies $R
\cap F=O_{v}$ and $E$ is the field of fractions of $R$, then $R$ and $v$ induce
a quasi-valuation $w$ on $E$ such that $R=O_{w}$ and $w$ extends $v$; thus $R$
satisfies the properties of a quasi-valuation ring.

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    ABSTRACT: Suppose $F$ is a field with a nontrivial valuation $v$ and valuation ring $O_{v}$, $E$ is a finite field extension and $w$ is a quasi-valuation on $E$ extending $v$. We study the topology induced by $w$. We prove that the quasi-valuation ring determines the topology, independent of the choice of its quasi-valuation. Moreover, we prove the weak approximation theorem for quasi-valuations.
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    ABSTRACT: Let $T$ be a totally ordered set and let $D(T)$ denote the set of all cuts of $T$. We prove the existence of a discrete valuation domain $O_{v}$ such that $T$ is order isomorphic to two special subsets of Spec$(O_{v})$. We prove that if $A$ is a ring (not necessarily commutative) whose prime spectrum is totally ordered and satisfies (K2), then there exists a totally ordered set $U \subseteq \text{Spec}(A)$ such that the prime spectrum of $A$ is order isomorphic to $D(U)$. We also present equivalent conditions for a totally ordered set to be a Dedekind totally ordered set. At the end, we present an algebraic geometry point of view

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