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\documentclass{article} |
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\usepackage[utf8]{inputenc} |
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\usepackage{amsfonts} |
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% \usepackage{yfonts} % WIP |
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\usepackage{amsthm} |
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\usepackage{amsmath} |
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\usepackage{enumerate} |
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\usepackage{hyperref} |
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\usepackage{amssymb} |
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\usepackage{tikz} % diagram |
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|
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\begin{filecontents}[overwrite]{commutative-algebra-notes.bib} |
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@misc{am, |
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author = {M. F. Atiyah and I. G. MacDonald}, |
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title = {{Introduction to Commutative Algebra}}, |
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year = {1969} |
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} |
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@misc{reid, |
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author = {Miles Reid}, |
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title = {{Undergraduate Commutative Algebra}}, |
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year = {1995} |
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} |
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@misc{mit-course, |
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author = {Steven Kleiman}, |
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title = {{Commutative Algebra - MIT OpenCourseWare}}, |
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year = {2008}, |
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note = {\url{https://ocw.mit.edu/courses/18-705-commutative-algebra-fall-2008/}}, |
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url = {https://ocw.mit.edu/courses/18-705-commutative-algebra-fall-2008/} |
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} |
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\end{filecontents} |
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\nocite{*} |
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\theoremstyle{definition} |
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\newtheorem{innerdefn}{Definition} |
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\newenvironment{defn}[1] |
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{\renewcommand\theinnerdefn{#1}\innerdefn} |
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{\endinnerdefn} |
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\newtheorem{innerthm}{Theorem} |
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\newenvironment{thm}[1] |
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{\renewcommand\theinnerthm{#1}\innerthm} |
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{\endinnerthm} |
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\newtheorem{innerlemma}{Lemma} |
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\newenvironment{lemma}[1] |
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{\renewcommand\theinnerlemma{#1}\innerlemma} |
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{\endinnerlemma} |
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\newtheorem{innerprop}{Proposition} |
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\newenvironment{prop}[1] |
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{\renewcommand\theinnerprop{#1}\innerprop} |
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{\endinnerprop} |
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\newtheorem{innercor}{Corollary} |
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\newenvironment{cor}[1] |
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{\renewcommand\theinnercor{#1}\innercor} |
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{\endinnercor} |
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\newtheorem{innereg}{Example} |
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\newenvironment{eg}[1] |
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{\renewcommand\theinnereg{#1}\innereg} |
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{\endinnereg} |
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\newtheorem{innerex}{Exercise} |
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\newenvironment{ex}[1] |
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{\renewcommand\theinnerex{#1}\innerex} |
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{\endinnerex} |
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\newcommand{\aA}{\mathfrak{a}} % TODO: use goth font |
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\newcommand{\mM}{\mathfrak{m}} |
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\title{Commutative Algebra notes} |
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\author{arnaucube} |
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\date{} |
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\begin{document} |
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\maketitle |
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\begin{abstract} |
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Notes taken while studying Commutative Algebra, mostly from Atiyah \& MacDonald book \cite{am} and Reid's book \cite{reid}. |
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Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$. |
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The proofs may slightly differ from the ones from the books, since I try to extend them for a deeper understanding. |
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\end{abstract} |
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\tableofcontents |
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\section{Ideals} |
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\subsection{Definitions} |
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% WIP |
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\subsection{Lemmas, propositions and corollaries} |
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\begin{thm}{1.X}{Zorn's lemma} \label{zorn} |
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TODO |
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\end{thm} |
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\begin{thm}{1.3} \label{1.3} |
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Every ring $A \neq 0$ has at lleast one maximal ideal. |
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\end{thm} |
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\begin{proof} |
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By Zorn's lemma \ref{zorn}. |
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\end{proof} |
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\begin{cor}{1.4} \label{1.4} |
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if $I \neq (1)$ an ideal of $A$, $\exists$ a maximal ideal of $A$ containing $I$. |
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\end{cor} |
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\begin{cor}{1.5} \label{1.5} |
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Every non-unit of $A$ is contained in a maximal ideal. |
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\end{cor} |
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\begin{defn}{Jacobson radical} |
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The \emph{Jacobson radical} of a ring $A$ is the intersection of all the maximal ideals of $A$. |
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Denoted $Jac(A)$. |
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$Jac(A)$ is an ideal of $A$. |
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\end{defn} |
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\begin{prop}{1.9} \label{1.9} |
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$x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$. |
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\end{prop} |
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\begin{proof} |
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Suppose $1-xy$ not a unit. |
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By \ref{1.5}, $1-xy \in \mM$ for $\mM$ some maximal ideal. |
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But $x \in Jac(A) \subseteq \mM$, since $Jac(A)$ is the intersection of all maximal ideals of $A$. |
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Hence $xy \in \mM$, and therefore $1 \in \mM$, which is absurd, thus $1-xy$ is a unit. |
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Conversely:\\ |
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Suppose $x \not\in \mM$ for some maximal ideal $\mM$. |
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Then $\mM$ and $x$ generte the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$. |
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Hence $1 -xy \in \mM$, and is therefore not a unit. |
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\end{proof} |
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\section{Modules} |
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\subsection{Modules} |
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\subsection{Cayley-Hamilton theorem, Nakayama lemma, and corollaries} |
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\begin{prop}{2.4}(Cayley-Hamilton Theorem) \label{2.4} |
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Let $M$ a fingen $A$-module. Let $\aA$ an ideal of $A$, let $\psi$ an |
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$A$-module endomorphism of $M$ such that $\psi(M) \subseteq \aA M$. |
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Then $\psi$ satisfies |
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$$\psi^n + a_1 \psi^{n-1} + \ldots + a_{n-1} \psi + a_n = 0$$ |
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with $a_i \in \aA$. |
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\end{prop} |
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\begin{proof} |
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Since $M$ fingen, let $\{ x_1, \ldots, x_n \}$ be generators of $M$.\\ |
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By hypothesis, $\psi(M) \subseteq \aA M$; so for any generator $x_i$, it's image $\psi(x_i) \in \aA M$. |
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Any element in $\aA M$ is a linear combination of the generators with coefficients in the ideal $\aA$, thus |
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$$\psi(x_i)= \sum_{j=1}^n a_{ij} x_j$$ |
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with $a_{ij} \in \aA$. |
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Thus, for a module with $n$ generators, we have $n$ different $\psi(x_i)$ equations: |
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$$ |
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\left. |
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\begin{aligned} |
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\psi(x_1) &= a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n\\ |
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\psi(x_2) &= a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n\\ |
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\ldots\\ |
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\psi(x_n) &= a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n |
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\end{aligned} |
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\right\} |
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\begin{aligned} |
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&\text{n elements $\psi(x_i) \in \aA M$ which}\\ |
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&\text{are linear combinations of the}\\ |
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&\text{$n$ generators of $M$} |
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\end{aligned} |
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$$ |
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Next step: rearrange in order to use matrix algebra. |
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Observe that each row equals $0$, and rearranging the elements at each row we get |
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\begin{align*} |
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&\psi(x_1) - (a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n) = 0\\ |
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&\psi(x_2) - (a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n) = 0\\ |
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&\ldots\\ |
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&\psi(x_n) - (a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n) = 0 |
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\end{align*} |
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Then, group the $x_i$ terms together; as example, take the row $i=1$: |
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$$(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0$$ |
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for $i=2$: |
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$$-a_{2,1} x_1 + (\psi - a_{2,2}) x_2 - \ldots - a_{2,n} x_n = 0$$ |
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So, $\forall i \in [n]$, as a matrix: |
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$$ |
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\begin{pmatrix} |
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\psi - a_{1,1} & -a_{1,2} & \ldots & -a_{1,n}\\ |
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-a_{2,1} & \psi-a_{2,2} & \ldots & -a_{2,n}\\ |
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\vdots\\ |
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-a_{n,1} & -a_{n,2} & \ldots & \psi-a_{n,n}\\ |
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\end{pmatrix} |
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\begin{pmatrix} |
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x_1\\ x_2\\ \vdots\\ x_n |
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\end{pmatrix} |
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= |
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\begin{pmatrix} |
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0\\ 0\\ \vdots\\ 0 |
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\end{pmatrix} |
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$$ |
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Kronecker delta: |
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$\delta_{ij} = |
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\begin{cases} |
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1 & \text{if } i = j,\\ |
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0 & \text{otherwise} |
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\end{cases}$ |
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With the Kronecker delta, $\psi(x_i)$ can be expressed as |
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$$\psi(x_i) = \sum_{j=1}^n \delta_{ij} \psi(x_j)$$ |
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so the previous matrix ccan be characterized as |
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$$\sum_{j=1}^n (\delta_{ij} \psi - a_{ij}) x_j = 0$$ |
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The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$) |
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\begin{itemize} |
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\item the term $(\psi - a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$ |
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\item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12} \cdot x_2)$ |
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\end{itemize} |
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Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$, |
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the ring of operators $A[\psi]$ is a commutative ring. |
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$\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant. |
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We're interested in the determinant because it is the only way to turn a system of multiple equations in a single scalar-like equation that describes the endomorphism $\psi$.\\ |
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$\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\ |
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$\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$ |
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$$det(M) \cdot x_i = 0~~ \forall i$$ |
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where $x_i$ are the generators of $M$. |
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Since $det(M)$ kills every generator, it must kill every element in $M$\\ |
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$\Longrightarrow~~ det(M)$ is the zero map. |
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Leibniz formula of the determinant of an $n \times n$ matrix: |
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$$ |
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det(M) = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n M_{i, \sigma(i)} |
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$$ |
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so, |
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$$(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})$$ |
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expanding it, |
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\begin{itemize} |
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\item highest power is $1 \cdot \psi^n$ |
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\item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\ |
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where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$ |
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\item the rest of coefficients of $\psi^k$ are also elements in $\aA$ |
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\end{itemize} |
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So we have |
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$$p(\psi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$ |
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with $a_i \in \aA$. |
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Since this determinant annihilates the generators (ie. $det(M)x_i=0$), the resulting enddomorphism $p(\psi)$ is the zero map on the entire module $M$, so: |
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$$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$ |
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with $a_i \in \aA$, as stated in the Cayley-Hamilton theorem. |
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\end{proof} |
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\begin{cor}{2.5} \label{2.5} |
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Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA M = M$. |
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Then, $\exists~ x \equiv 1 \pmod \aA$ such that $xM = 0$. |
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\end{cor} |
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\begin{proof} |
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take $\psi = \text{identity}$. Then in Cayley-Hamilton (\ref{2.4}): |
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\begin{align*} |
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&\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0\\ |
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\Longrightarrow~ &id_M + a_1 id_M + a_2 id_M + \ldots + a_{n-1} id_M + a_n = 0\\ |
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\Longrightarrow~ &(1 + a_1 + \ldots + a_n) id_M = 0 |
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\end{align*} |
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apply it to $m \in M$, where since $id_M(m)=m$ (by definition of the identity), we then have |
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$$(1 + a_1 + \ldots + a_n) \cdot m = 0$$ |
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with $a_i \in \aA$. |
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\begin{enumerate}[\text{part} i.] |
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\item $xM=0$:\\ |
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Thus the scalar $x = (1 + a_1 + \ldots + a_n)$ annihilates every $m \in M$, ie. the entire module $M$. |
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\item $x \equiv 1 \pmod \aA$:\\ |
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$x \equiv 1 \pmod \aA ~~ \Longleftrightarrow (x-1) \in \aA$\\ |
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then from $x = (1 + \underbrace{a_1 + \ldots + a_n}_b) \in \aA$, set $b=a_1 + \ldots + a_n$,\\ |
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so that $x=(1+b) \in \aA$.\\ |
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Then $x-1 = (1+b)-1 = b \in \aA$\\ |
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so $x-1 \in \aA$, thus $x \equiv 1 \pmod \aA$ as stated. |
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\end{enumerate} |
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\end{proof} |
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\begin{prop}{2.6}{Nakayama's lemma} \label{2.6} |
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Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA \subseteq Jac(A)$. |
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Then $\aA M = M$ implies $M=0$. |
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\end{prop} |
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\begin{proof} |
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By \ref{2.5}: since $\aA M = M$, we have $x M =0$ for some $x \equiv 1 \pmod {Jac(A)}$. (notice that at \ref{2.5} is $\pmod \aA$ but here we use $\pmod {Jac(A)}$, since we have $\aA \subseteq Jac(A)$). |
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By \ref{1.9}, $x$ is a unit in $A$. |
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Hence $M = x^{-1} \cdot \underbrace{x \cdot M}_{=0~ \text{(by \ref{2.5})}} = 0$. |
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Thus, if $\aA M = M$ then $M=0$. |
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\end{proof} |
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\begin{cor}{2.7} \label{2.7} |
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Let $M$ a fingen $A$-module, let $N \subseteq M$ a submodule of $M$, let $\aA \subseteq Jac(A)$ an ideal. |
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Then $M = \aA M + N \stackrel{\text{implies}}{\Longrightarrow} M=N$. |
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\end{cor} |
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\begin{proof} |
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The idea is to apply Nakayama (\ref{2.6}) to $M/N$. |
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Since $M$ fingen $\Longrightarrow~~ M/N$ is fingen and an $A$-module. |
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Since $\aA \subseteq Jac(A) ~\Longrightarrow~$ Nakayama applies to $M/N$ too. |
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By definition, |
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$$\aA M = \left\{ \sum a_i \cdot m_i ~~|~~ a_i \in \aA, m_i \in M \right\}$$ |
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where $m_i$ are the generators of $M$. |
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Then, for $M/N$, |
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$$\aA (\frac{M}{N}) = \left\{ \sum a_i \cdot (m_i + N) ~~|~~ a_i \in \aA, m_i \in M \right\}$$ |
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observe that $a_i(m_i+N)= a_i m_i +N$, thus |
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$$\sum_i a_i \cdot (m_i + N) = \underbrace{(\sum_i a_i \cdot m_i)}_{\in \aA M} + N \in \aA M + N$$ |
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Hence, |
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\begin{equation} |
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\aA (\frac{M}{N}) = \left\{ x + N ~~|~~ x \in \aA M \right\} = \aA M + N |
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\label{eq:2.7.1} |
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\end{equation} |
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By definition, if we take $\frac{\aA M + N}{N}$, then |
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$$\frac{\aA M + N}{N} = \left\{ y + N ~~|~~ y \in \aA M +N \right\} = \aA M + N$$ |
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thus every $y \in \aA M +N$ can be written as |
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$$y=x+n,~~ \text{with} x \in \aA M,~ n\in M$$ |
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which comes from \eqref{eq:2.7.1}. |
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Thus, $y + N = (x+n)+N = x+N$, since $n \in N$ is zero in the quotient. |
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Hence, every element of $\frac{\aA M +N}{N}$ has the form |
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$$\frac{\aA M + N}{N} = \left\{ x + N ~~|~~ x \in \aA M \right\}$$ |
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as in \eqref{eq:2.7.1}. |
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Thus |
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\begin{equation} |
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\aA (\frac{M}{N}) = \aA M + N = \frac{\aA M +N}{N} |
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\label{eq:2.7.2} |
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\end{equation} |
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By the Collorary assumption, $M = \aA M + N$; quotient it by $N$: |
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\begin{equation} |
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\frac{M}{N} = \frac{\aA M +N}{N} |
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\label{eq:2.7.3} |
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\end{equation} |
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So, from \eqref{eq:2.7.2} and \eqref{eq:2.7.3}: |
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$$\aA (\frac{M}{N}) = \aA M +N = \frac{\aA M +N}{N} = \frac{M}{N}$$ |
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thus, $\aA (\frac{M}{N}) = \frac{M}{N}$. |
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By Nakayama's lemma \ref{2.6}, if $\aA (\frac{M}{N}) = \frac{M}{N} ~\stackrel{implies}{\Longrightarrow}~ \frac{M}{N}=0$ |
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Note that |
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$$\frac{M}{N} = \{ m + N ~|~ m \in M \}$$ |
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(the zero element in $\frac{M}{N}$ is the coset $N=0+N$) |
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Then, $\frac{M}{N}=0$ means that the quotient has exactly one element, the zero coset $N$. |
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Thus, every coset $m + N$ equals the zero coset $N$, so $m-0 \in N ~\Longrightarrow~ m \in N$. |
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Hence every $m \in M$ lies in $N$, ie. $\forall m \in M,~ m \in N$. |
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So $M \subseteq N$. But notice that by the Corollary, we had $N \subseteq M$, therefore $M = N$. |
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Thus, if $M = \aA M + N \stackrel{implies}{\Longrightarrow} M = N$. |
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\end{proof} |
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\begin{prop}{2.8} \label{2.8} |
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Let $x_i \forall i \in [n]$ be elements of $M$ whose images $\frac{M}{m M}$ from a basis of this vecctor space. Then the $x_i$ generate $M$. |
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\end{prop} |
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\begin{proof} |
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Let $N$ submodule $M$, generated by the $x_i$. |
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|
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Then the composite map $N \longrightarrow M \longrightarrow \frac{M}{m M}$ maps $N$ onto $\frac{M}{m M}$, hence $N + \aA M = M$, which by \ref{2.7} implies $N = M$. |
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\end{proof} |
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\subsection{Noetherean rings} |
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\section{Exercises} |
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For the exercises, I follow the assignements listed at \cite{mit-course}. |
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The exercises that start with \textbf{R} are the ones from the book \cite{reid}, and the ones starting with \textbf{AM} are the ones from the book \cite{am}. |
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\subsection{Exercises Chapter 1} |
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\begin{ex}{R.1.1} |
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Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$? |
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\end{ex} |
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\begin{proof} |
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Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$. |
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$I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\ |
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Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$. |
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Thus $I \cup J$ is not closed under addition; thus is not an ideal. |
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Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum: |
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$$I+J = \{ a+b | a \in I, b \in J \}$$ |
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$gcd(2,3)=1$, so $I+J = \mathbb{Z}$. |
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Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$. |
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\end{proof} |
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\begin{ex}{R.1.5} |
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let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\ |
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In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$. |
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\end{ex} |
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\begin{proof} |
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(Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$). |
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Let |
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$$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$ |
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The claim is that $\psi^{-1}(P)$ is prime iddeal of $A$. |
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\begin{enumerate}[i.] |
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\item show that $\psi^{-1}(P)$ is an ideal of $A$:\\ |
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$0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$). |
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If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so |
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$$\psi(a-b)= \psi(a) - \psi(b) \in P$$ |
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hence $a-b \in \psi^{-1}(P)$. |
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If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\ |
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Thus $ra \in \psi^{-1}(P)$. |
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$\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$. |
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\item show that $\psi^{-1}(P)$ is prime:\\ |
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$\psi^{-1}(P) \neq A$, since if $\psi^{-1}(P)=A$, then $1_A \in \psi^{-1}(P)$, so $\psi(1_A)=1_B \in P$, which would mean that $P=B$, a contradiction since $P$ is prime ideal of $B$. |
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Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$. |
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Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\ |
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Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$. |
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Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$. |
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\end{enumerate} |
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\end{proof} |
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\begin{ex}{R.1.6} |
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prove or give a counter example: |
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\begin{enumerate}[a.] |
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\item the intersection of two prime ideals is prime |
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\item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime |
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\item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$ |
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\end{enumerate} |
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\end{ex} |
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\begin{proof} |
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\begin{enumerate}[a.] |
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\item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime. |
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Then $I \cap J = (2) \cap (3) = (6)$. |
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The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$. |
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Thus the intersection of two primes can not be prime. |
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\item $P_1=(2),~ P_2=(3)$, both prime. |
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Then, |
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$$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$ |
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$\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$. |
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(recall: principal= generated by a single element) |
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So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$. |
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The whole ring is not a prime ideal (by the definition of the prime ideal), so $P_1+P_2$ is not a prime ideal. |
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Henceforth, the sum of two prime ideals is not necessarily prime. |
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\item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$. |
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Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$. |
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Then |
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\begin{align*} |
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\psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\ |
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\text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A |
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\end{align*} |
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But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field. |
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Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal. |
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\end{enumerate} |
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\end{proof} |
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\subsection{Exercises Chapter 2} |
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\bibliographystyle{unsrt} |
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\bibliography{commutative-algebra-notes.bib} |
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\end{document} |
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