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main.tex
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main.tex
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\documentclass[a4paper]{amsart}
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%\documentclass[a4paper,twoside]{article}
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\usepackage{color}
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%\usepackage{authblk}
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%\usepackage{makeidx} % allows index generatio
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\makeindex
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\usepackage{graphicx}
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\usepackage[all]{xy}
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\usepackage{amssymb}
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\usepackage{hhline}
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\usepackage{tikz-cd}
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\usepackage{longtable}
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\usepackage{ulem}
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%\usepackage{amsfont}
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\usepackage[a4paper, left=2cm, right=2cm, top=2cm,bottom=2cm]{geometry}
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\usepackage{hyperref}
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\usepackage[draft=true]{minted}
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\usepackage{xspace}
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\usepackage[backend=bibtex, bibencoding=utf8, doi=false, url=false, isbn=false]{biblatex}
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\addbibresource{./sources.bib}
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%\usepackage{draftwatermark}
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%\SetWatermarkText{Early draft -- Do not circulate!}
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%\SetWatermarkScale{1.2}
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%\usepackage{cleveref}
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\theoremstyle{definition}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{definition}[theorem]{Definition}
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\newtheorem{remark}[theorem]{Remark}
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\newtheorem{result}[theorem]{Result}
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\newtheorem{proposition}[theorem]{Proposition}
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\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem{example}[theorem]{Example}%
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\newtheorem{conjecture}[theorem]{Conjecture}%
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\newtheorem{corollary}[theorem]{Corollary}
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\newtheorem{construction}[theorem]{Construction}%
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\newtheorem{observation}[theorem]{Observation}
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%\newenvironment{proof}{\noindent
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%{\bf Proof:}}{
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%\begin{flushright}$\Box$\end{flushright}}
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\usepackage[all]{xy}
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\usepackage{pdflscape}
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%\usepackage{nccmath}
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%\newenvironment{mpmatrix}{\begin{smallsize}\begin{pmatrix}}{\end{\pmatrix}\end{smallsize}}%
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\setcounter{MaxMatrixCols}{66}
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\newcommand{\Grass}{\mathrm{Grass}}
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\newcommand{\im}{\mathrm{im}\,}
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\newcommand{\RR}{\mathbb{R}}
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\newcommand{\Id}{\mathrm{Id}}
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\newcommand{\PP}{\mathbb{P}}
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\newcommand{\IA}{\mathbb{A}}
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\newcommand{\FF}{\mathbb{F}}
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\newcommand{\Inf}{\mathrm{Inf}}
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\newcommand{\NN}{\mathbb{N}}
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\newcommand{\Spec}{\mathrm{Spec}\,}
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\newcommand{\CC}{\mathbb C}
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\newcommand{\sk}{\mathrm{sk}}
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\newcommand{\QQ}{\mathbb Q}
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\newcommand{\bbk}{\mathbf k}
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\newcommand{\ZZ}{\mathbb Z}
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\renewcommand{\PP}{\mathbb P}
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\newcommand{\rank}{\mathrm{rank}\,}
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\newcommand{\Oscar}{OSCAR\xspace}
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%\newcommand{\D}{d}
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%CHANGE FOR WORKING PHASE -- DELETE SUBSEQUENT TWO LINES LATER
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\newcommand{\red}[1]{{\textcolor{red}{#1}}}
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\newcommand{\blue}[1]{{\textcolor{blue}{#1}}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{document}
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\title{Some thoughts on finite extensions of rings}
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\author{Others}
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\address{
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Other authors\dots
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}
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%\email{brandhorst@math.uni-sb.de}
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%
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\author{Matthias Zach}
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\address{
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Matthias Zach:
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RPTU Kaiserslautern-Landau,
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67663 Kaiserslautern,
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Germany
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}
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%\email{zach@mathematik.uni-kl.de}
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\begin{abstract}
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Dah.
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\end{abstract}
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\maketitle
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We are considering complex analytic maps $f \colon (\CC^n, 0) \to (\CC^p, 0)$ which
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are \textit{finite}. In particular, we may assume $n \geq p$. By definition, $f$ is
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finite if and only if its pullback $\varphi \colon \OO_p \to \OO_n$ is a finite
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extension of rings, i.e. $\OO_n = \CC\{x_1,\dots,x_n\}$ is a finite
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$\OO_p = \CC\{y_1,\dots,y_p\}$-module via $\varphi$.
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\begin{definition}
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Let $\varphi \colon R \to S$ be a morphism of commutative rings with unit. An element
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$x \in S$ is called \textit{integral} over $R$ if there exists a \textit{monic}
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polynomial
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\[
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P = X^r + c_{r-1} \cdot X^{r-1} + \dots + c_1 \cdot X + c_0 \in R[X]
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\]
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such that $P(x) = 0$ in $S$.
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\end{definition}
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\begin{proposition}
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\label{prp:Finiteness}
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Let $\varphi \colon R \to S$ be a morphism of commutative, Noetherian rings with unit.
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The set of elements in $S$ which are integral over $R$ is a subalgebra of $S$.
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Moreover, $S$ is finite over $R$ if and only if $S$ is generated
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as an $R$-algebra by finitely many integral elements.
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\end{proposition}
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\noindent
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For a proof see \cite[Theorem 4.2 and Corollary 4.5]{Eisenbud95}.
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\medskip
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For complex analytic mappings $f$ as above, we are working with rings of
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convergent power series in several variables. In practice, however, we will
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mostly encounter maps which admit a \textit{polynomial representative}
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\[
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f \colon \CC^n \to \CC^p, \quad (x_1,\dots,x_n) \mapsto (y_1,\dots, y_p) =
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(f_1(x), \dots, f_p(x)), \quad f_j(x) \in \CC[x_1,\dots,x_n].
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\]
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It seems to be no significant restriction for practical applications to
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assume that also this polynomial representative is actually finite.
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For finite extensions of polynomial rings we do have algorithmic methods
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to compute with the structure of $S$ as an $R$-module.
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\begin{proposition}
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Let $\varphi \colon \CC[y] \to \CC[x]$ be a mapping of polynomial
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rings given by $y_j \mapsto f_j(x_1,\dots,x_n)$. Let
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\[
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J = \langle y_j - f_j(x) : j = 1,\dots, p\rangle \subset \CC[y, x]
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\]
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and $LJ$ the \textit{leading ideal} of $J$ with respect to a monomial
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ordering which is an elimination ordering on the $x$-variables.
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Then $\varphi$ is finite if and only if $\langle x_1,\dots, x_n\rangle^N \subset LJ$
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for some $N \in \NN$. In this case, the set of monomials
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$\{ x^\alpha : \alpha \in \NN^n, \, x^\alpha \notin LI\}$
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are a generating set for $\CC[x]$ as a $\CC[y]$-module.
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\end{proposition}
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\noindent
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TODO: Find a reference for the proof.
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\medskip
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Based on this result we can implement seamless transitions for elements $h \in S$
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between their representation as polynomials $h \in \CC[x]$
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and module elements $v \in R \cdot e_1 + \dots + R \cdot e_m$ with respect to a
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chosen system of generators over $R$.
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\medskip
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Working with power series in computer algebra is an almost impossible undertaking
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beyond simple arithmetic. Recall that in general it must be deemed impossible to even decide
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equality of two given power series. Thus, the naive hope at this point would be
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that \textit{localizing} at the origins of the domain and codomain of $f$,
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respectively, would bring us sufficiently close to working with power
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series by restricting ourselves to those which are in fact \textit{rational functions}.
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\begin{proposition}
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\label{prp:NonFiniteness}
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Let $\varphi \colon R \to S$ be a finite extension of commutative Noetherian rings with
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unit. Let $P \subset R$ be a prime ideal and $Q \subset S$ prime with
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$\varphi^{-1}(Q) = P$ (i.e. $Q$ is a \textit{prime over} $P$).
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Then the naturally induced morphism on the localizations
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\[
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\phi \colon R_P \to S_Q
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\]
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is finite if and only if $Q$ is the only prime in $S$ with that property.
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\end{proposition}
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\begin{proof}
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According to Proposition
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\ref{prp:Finiteness}, we have finiteness of $\phi$ if and only if an arbitrary element
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$\frac{a}{b} \in S_Q$ is integral over $R_P$. Now by assumption both $a$ and $b$ are
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integral over $R$ and hence over $R_P$ and since $\frac{a}{b} = a \cdot \frac{1}{b}$
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it suffices to consider elements of the form $\frac{1}{b}$ for $b \notin Q$.
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Suppose we found a monic polynomial
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\[
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F = X^r + c_{r-1} \cdot X^{r-1} + \dots + c_1 \cdot X^1 + c_0 \in R_P[X]
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\]
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with $F\left(\frac{1}{b}\right) = 0$ in $S_Q$. Then we can clear denominators to
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obtain
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\[
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F' = c'_r \cdot X^r + c'_{r-1} \cdot X^{r-1} + \dots + c'_1 \cdot X^1 + c'_0 \in R[X]
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\]
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with $F'\left(\frac{1}{b}\right) = 0$ and $c'_r \notin P$.
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But then, by definition of localization at $Q$,
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\[
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u \cdot \left( c'_r + c'_{r-1} \cdot b + \dots + c'_1 \cdot b^{r-1} + c'_0 \cdot b^r
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\right) = 0
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\]
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in $S$ for some $u \notin Q$. This expression can be split as
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\begin{equation}
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\label{eqn:Splitting}
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-u \cdot c'_r = u \cdot b \cdot \left(c'_{r-1} + c'_{r-2} \cdot b + \dots + c'_1 \cdot b^{r-2} + c'_0 \cdot b^{r-1} \right).
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\end{equation}
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Now suppose $Q'\neq Q$ was another prime with $\varphi^{-1}(Q') = P = \varphi^{-1}(Q)$.
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Choosing $b \in Q' \setminus Q$ leads to a contradiction in (\ref{eqn:Splitting}):
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Clearly, the right hand side, containing $b$ as a factor, is an element in $Q'$.
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Comparing with the left hand side, we find that
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$c'_r \notin P = \varphi^{-1}(Q) = \varphi^{-1}(Q')$ and, hence, $\varphi(c'_r) \in S$ is
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neither an element of $Q$, nor of $Q'$. This leaves us with $u \in Q'$.
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To arrive at a contradiction, let
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\[
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\mathrm{ord}_{Q'} \colon S \to \NN_0, \quad a \mapsto \mathrm{max}
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\left\{k \in \NN_0 : a \in Q'^{(k)}\right\}
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\]
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where $Q'^{(k)}$ denotes the $k$-th \textit{symbolic power}, i.e.
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the preimage of $\left(Q'_{Q'}\right)^k$ in $S$ under the localization
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map $S \to S_{Q'}$. The map $\mathrm{ord}_{Q'}$ is a \textit{valuation}
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and it is now easy to see that the valuation of the left hand side of
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(\ref{eqn:Splitting}) is strictly less than the valuation of the
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right hand side.
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TODO: Find an argument to prove the converse.
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% On the contrary, if $Q$ was the only prime over $P$, we can argue as follows.
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% Let $b \in S \setminus Q$ be arbitrary and
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% \[
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% G = X^r + a_{r-1} \cdot X^{r-1} + a_{r-2} \cdot X^{r-2} + \dots + a_1 \cdot X + a_0 \in
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% R[X]
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% \]
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% a monic polynomial with $G(b) = 0$ in $S$. We may choose $G$ to be of minimal
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% degree with this property. Then
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% \[
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% G^\vee = 1 + a_{r-1} \cdot X + a_{r-2} \cdot X^2 + \dots + a_1 \cdot X^{r-1} + a_0 \cdot X^r
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% \]
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% satisfies $G^\vee\left(\frac{1}{b}\right) = 0$ in $S_Q$. Provided that we are allowed
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% to divide by $a_0$ in $R_P$, we would thus get a monic polynomial
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% in $R_P[X]$ annihilating $\frac{1}{b}$.
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%
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% Now suppose $a_0 \in P$ so that $\varphi(a_0) \in Q$. Expanding $G(b) = 0$
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% then yields
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% is a unit in $
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% Now suppose $\varphi(a_0) \in Q$, so
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% $a_0 \in P = \varphi^{1}(Q)$.
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\end{proof}
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\begin{example}
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Consider the map
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\[
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f \colon \CC \to \CC, \quad x \mapsto y = x - x^3.
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\]
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Clearly, $\varphi \colon \CC[y] \to \CC[x]$ is finite and, in fact,
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$\CC[x]$ is a free $\CC[y]$-module with generating set $1, x, x^2$.
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For the analytification of this map $f \colon (\CC, 0) \to (\CC, 0)$
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the implicit function theorem yields that there exists a local inverse
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\[
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y = x - x^3, \quad x = y + \sum_{k = 2}^\infty c_k \cdot y^k \in \CC\{y\}
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\]
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as a convergent power series in $y$. Therefore $\varphi^h \colon \CC\{y\} \to \CC\{x\}$
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is in fact an \textit{isomorphism} and $\CC\{x\}$ is generated as a $\CC\{y\}$-module
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by the unit element $1 \in \CC\{x\}$.
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Thus we find a discrepancy of $3 - 1 = 2$ in the ranks of the two modules over the
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respective rings. One could hope that the localization
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\[
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\phi \colon \CC[y]_{\langle y\rangle} \to \CC[x]_{\langle x \rangle}
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\]
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provides us with a reasonable approximation to $\varphi^h$, but
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Proposition \ref{prp:NonFiniteness} tells us that the elements
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\[
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M_p = \left\langle\left(\frac{1}{1 - x}\right)^k : k \leq p\right\rangle_{\CC[y]_{\langle y\rangle}}
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\]
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form an infinite, ever increasing sequence of
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$\CC[y]_{\langle y \rangle}$-submodules of $\CC[x]_{\langle x \rangle}$ and
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that, hence, $\CC[x]_{\langle x \rangle}$ does not have a finite
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$\CC[y]_{\langle y\rangle}$-module structure.
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\end{example}
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\printbibliography
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\end{document}
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