值域的简单扩展的关键多项式

IF 0.4 Q4 MATHEMATICS Journal of Singularities Pub Date : 2014-06-03 DOI:10.5427/jsing.2022.25k
F. J. H. Govantes, W. Mahboub, M. Acosta, M. Spivakovsky
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引用次数: 22

摘要

设$\iota:K\hookrightarrow L\cong K(x)$为有值字段的简单超越扩展,其中$K$具有秩为1的估值$\nu$。也就是说,我们假设给定$K$的1级估值$\nu$及其扩展$\nu'$到$L$。设$(R_\nu,M_\nu,k_\nu)$表示$\nu$的估值环。本文的目的是提出MacLane的关键多项式理论的一个改进版本,类似于M. vaqui所考虑的那些,并使人想起Abhyankar和Moh(近似根)和T.C. Kuo研究的相关对象。也就是说,我们将$\iota$关联到一个可数的有序集合$$ \mathbf{Q}=\{Q_i\}_{i\in\Lambda}\subset K[x]; $$, $Q_i$被称为{\bf键多项式}。没有直接前导的键多项式$Q_i$称为{\bf极限键多项式}。让$\beta_i=\nu'(Q_i)$。我们给出了极限键多项式的显式描述(它可以看作是Artin- Schreier多项式的推广)。我们也给出了键多项式集合的阶型的上界。也就是说,我们证明,如果$\operatorname{char}\ k_\nu=0$,那么关键多项式的集合最多有阶型$\omega$,而在$\operatorname{char}\ k_\nu=p>0$的情况下,这个阶型由上面的$\omega\times\omega$限定,其中$\omega$代表第一个无限序数。
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Key polynomials for simple extensions of valued fields
Let $\iota:K\hookrightarrow L\cong K(x)$ be a simple transcendental extension of valued fields, where $K$ is equipped with a valuation $\nu$ of rank 1. That is, we assume given a rank 1 valuation $\nu$ of $K$ and its extension $\nu'$ to $L$. Let $(R_\nu,M_\nu,k_\nu)$ denote the valuation ring of $\nu$. The purpose of this paper is to present a refined version of MacLane's theory of key polynomials, similar to those considered by M. Vaqui\'e, and reminiscent of related objects studied by Abhyankar and Moh (approximate roots) and T.C. Kuo. Namely, we associate to $\iota$ a countable well ordered set $$ \mathbf{Q}=\{Q_i\}_{i\in\Lambda}\subset K[x]; $$ the $Q_i$ are called {\bf key polynomials}. Key polynomials $Q_i$ which have no immediate predecessor are called {\bf limit key polynomials}. Let $\beta_i=\nu'(Q_i)$. We give an explicit description of the limit key polynomials (which may be viewed as a generalization of the Artin--Schreier polynomials). We also give an upper bound on the order type of the set of key polynomials. Namely, we show that if $\operatorname{char}\ k_\nu=0$ then the set of key polynomials has order type at most $\omega$, while in the case $\operatorname{char}\ k_\nu=p>0$ this order type is bounded above by $\omega\times\omega$, where $\omega$ stands for the first infinite ordinal.
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CiteScore
0.90
自引率
0.00%
发文量
28
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