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

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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. - [Show abstract] [Hide abstract]

**ABSTRACT:**Suppose $F$ is a field with valuation $v$ and valuation domain $O_{v}$, and $R$ is an $O_{v}-$algebra. We prove that $R$ satisfies SGB (strong going between) over $O_{v}$. We give a necessary and sufficient condition for $R$ to satisfy LO (lying over) over $O_{v}$. Using the filter \qv constructed in [Sa1], we show that if $R$ is torsion-free over $O_{v}$ then $R$ satisfies GD (going down) over $O_{v}$. In particular, if $R$ is torsion-free and $(R^{\times} \cap O_{v}) \subseteq O_{v}^{\times}$, then for any chain in $\text{Spec}(O_v)$ there exists a chain in $\text{Spec}(R)$ covering it. Assuming $R$ is torsion-free over $O_{v}$ and $[R \otimes_{O_{v}}F:F]< \infty$, we prove that $R$ satisfies INC (incomparabilty) over $O_{v}$. Assuming in addition that $(R^{\times} \cap O_{v}) \subseteq O_{v}^{\times}$, we deduce that $R$ and $O_{v}$ have the same Krull dimension and a bound on the size of the prime spectrum of $R$ is given. Under certain assumptions on $R$ and a \qv defined on it, we prove that the \qv ring satisfies GU (going up) over $O_{v}$. Combining these five properties together, we deduce that any maximal chain of prime ideals of the \qv ring is lying over $\text{Spec}(O_{v})$, in a one-to-one correspondence. - [Show abstract] [Hide abstract]

**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