%%This is a standard LaTeX2e article document template. personal version 12/5/200%% \documentclass[11pt,twoside]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%packages%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \pagestyle{empty} \usepackage{latexsym} \usepackage{amssymb} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{amstext} \usepackage{multicol} \usepackage[all]{xy} \usepackage{amsxtra} \usepackage[pdftex, breaklinks, linktocpage=true, bookmarksopen=true,bookmarksopenlevel=0,bookmarksnumbered=true]{hyperref} \hypersetup{ pdftitle = {Math 371 Assignment 1}, pdfauthor = {\textcopyright\ Bryden Cais}, pdfcreator = {\LaTeX\ with package \flqq hyperref\frqq}, colorlinks = {true} } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%formatting%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \setlength{\topmargin}{0in} %%% This sets all the spacing stuff to use the page more \setlength{\oddsidemargin}{0in} %%% efficiently than the normal "article" setup would. \setlength{\evensidemargin}{0in} %%% It's OK to play with these some. \setlength{\textheight}{8.5in} %%% \setlength{\textwidth}{6.5in} %%% \setlength{\headsep}{0in} %%% \setlength{\headheight}{0in} %%% %\setlength{\footskip}{0in} %%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \DeclareMathOperator{\ann}{ann} \DeclareMathOperator{\Tor}{Tor} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\End}{End} \newcommand{\Q}{\mathbf{Q}} \newcommand{\R}{\mathbf{R}} \newcommand{\Z}{\mathbf{Z}} \newcommand{\C}{\mathbf{C}} \newcommand{\F}{\mathbf{F}} \newcommand{\p}{\mathfrak{p}} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}{Corollary} \begin{document} \begin{center} {\bf \Large \underline{Honors Algebra 4, MATH 371 Winter 2010}}\\ {\large Assignment 5}\\Due Wednesday, March 17 at 08:35 \end{center} For the problems 1--7, we fix a principal ideal domain $R$. \begin{enumerate} \item Let $M$ be any $R$-module. \begin{enumerate} \item For $m\in M$, the {\em annhilator of $m$ in $R$} is defined to be $$\ann_R(m):=\{r\in R\ :\ rm=0\}.$$ Prove that $\ann_R(m)$ is an ideal of $R$. \item We say that $m\in M$ is {\em torsion} if $\ann_R(m)\neq 0$ and we define the {\em torsion submodule of $M$} to be $$\Tor(M):=\{m\in M\ :\ \ann_R(m)\neq 0\}.$$ We say that an $R$-module $N$ is {\em torsion free} if $\Tor(N)=0$. Prove that $\Tor(M)$ really is a submodule of $M$ and that the quotient $M/\Tor(M)$ is torsion free. \end{enumerate} \item Let $M$ be any submodule of a free module $R^n$. Show that $M$ is itself a free module, of rank at most $n$ as follows: \begin{enumerate} \item Let $\pi_i:M\rightarrow R$ be the composition of the inclusion $M\hookrightarrow R^n$ with projection $R^n\rightarrow R$ on to the $i$th factor; it is an $R$-module homomorphism. If $\pi_1(M)=0$, show that $M$ is a submodule of $R^{n-1}$ in a natural way. \item If $\pi_1(M)\neq 0$ then it is an ideal of $R$, necessarily principal, say $\pi_1(M)=(d).$ For $m\in M$ with $\pi_1(m)=d$, show that $M\simeq Rm \oplus \ker \pi_1$ and that $\ker \pi_1$ is naturally a submodule of a free module of rank $n-1$ \item Conclude by induction on $n$. \end{enumerate} \item Prove that any finitely generated and torsion free $R$-module $M$ is a submodule of a free module, and hence free as follows: \begin{enumerate} \item Let $\{m_1,\ldots, m_s\}$ be a minimal set of generators of $M$, and let $M_i$ be the submodule of $M$ generated by $\{m_1,\ldots,m_i\}$. Show that $M_1$ a free $R$-module. \item Let $j\ge 1$ be the greatest integer such that $M_j$ is free. If $j=s$ we are done. Otherwise, $M_{j+1}$ is not free so there exists a relation $$xm_{j+1} + \sum_{1\le i\le j} r_i m_i=0$$ with $x\in R$ nonzero. Show that multiplication by $x$ on $M_{j+1}$ is an $R$-module homomorphism whose image is contained in a free $R$-module. \item Show that the kernel of multiplication by $x$ is zero, and deduce that $M_{j+1}$ is free after all. Conclude that $M$ is free. \end{enumerate} \item Prove that the short exact sequence $$\xymatrix{0\ar[r] & {\Tor(M)} \ar[r] & M \ar[r] & {M/\Tor(M)} \ar[r] & 0}$$ splits, so $M\simeq \left(M/\Tor(M)\right)\oplus \Tor(M)$ is the direct sum of its torsion submodule and a free module. Show that this decomposition of $M$ as a direct sum of a torsion module and a free module is unique. \item Let $T$ be a {\em torsion} $R$-module, i.e. the inclusion $\Tor(R)\hookrightarrow T$ is an isomorphism. For each nonzero (principal) ideal $(r)\subseteq R$, we define the {\em $(r)$-primary submodule of $T$} to be $$T_{(r)}:=\{t\in T\ :\ r^nt=0\ \text{for some}\ n\ge 0\}.$$ \begin{enumerate} \item Show that $T_{(r)}$ is a submodule of $T$ that depends only on the ideal $(r)$ (and not on a specific choice of generator for this ideal). \item If $r,s\in R$ have $gcd(r,s)=1$, show that $T_{(r)}\cap T_{(s)} = 0$. Conclude that for such $r,s$ we have $T_{(rs)}\simeq T_{(r)}\oplus T_{(s)}$. Hint: use the fact that for any nonnegative integers $n,m$, there exist $u,v\in R$ with $ur^n+vs^m = 1$. \item Prove that there is a canonical isomorphism of $R$-modules $$T\simeq \bigoplus_{p} T_p$$ where $p$ ranges over the distinct prime ideals of $R$. \end{enumerate} \item Let $p$ be a prime ideal of $R$ and $T_p$ a finitely generated, nonzero $p$-primary $R$-module. \begin{enumerate} \item Show that any quotient or sub-module of $T_p$ is again $p$-primary and finitely generated (be careful for finite generation of submodules!) \item Let $\ann(T_p):=\{ r\in R\ :\ rt=0\ \text{for all}\ t\in T_p\}$ be the {\em annihilator of $T_p$ in $R$}. Prove that $\ann(T_p)$ is a proper ideal of $R$, and is equal to $p^N$ for some positive integer $N$. \item Suppose $\{t_1,\ldots,t_s\}$ is a set of generators for $T_p$ as an $R$-module. Show that there exists a set of $s$ generators $\{y_1,\ldots,y_s\}$ of $T_p$ with $$\ann_R(y_1) = \ann_R(T_p)=p^N$$\label{mingenfix} \item Let $\langle y_1\rangle$ be the $R$-submodule of $T_p$ generated by $y_1$ and set $T_p':=T_p/\langle y_1\rangle$; it is again a finitely generated $p$-primary $R$-module. Suppose $y'\in T_p'$ has $\ann_R(y')=p^m$ for some positive integer $m$. Show that $m\le N$ and that there exists $y\in T_p$ projecting to $y'$ with $\ann_R(y)=p^m$.\label{lift} \item Prove that there exist integers positive $m_1\le m_2\le \ldots \le m_s$ with $\ann_R(T_p)=p^{m_s}$ and an isomorphism of $R$-modules \begin{equation} T_p\simeq R/p^{m_1} \oplus R/p^{m_2}\oplus \cdots \oplus R/p^{m_s} \label{primarydecomp} \end{equation} as follows: Fix a minimal set of generators $\{y_1,\ldots,y_s\}$ of $T_p$ and proceed by induction on $s$. \begin{enumerate} \item By (\ref{mingenfix}), we may assume $\ann_R(y_1)=\ann_R(T_p)$. If $s=1$ conclude. \item If $s > 1$, consider the short exact sequence of $R$-modules \begin{equation} \xymatrix{ 0\ar[r] & {\langle y_1 \rangle} \ar[r] & T_p \ar[r] & T_p' \ar[r] & 0 } \end{equation} Using the induction hypothesis and part (\ref{lift}) to appropriately lift generators of the direct summands occurring in the decomposition of $T_p'$ as in (\ref{primarydecomp}), show that this sequence is split, and conclude as desired. \end{enumerate} \item Show that the $m_i$ as in (\ref{primarydecomp}) are uniquely determined by $T_p$. Hint: for each $j\ge 0$, consider the $R$-module $p^jT_p/p^{j+1}T_p$. Show that this is a vector space over the field $R/p$ and compute its dimension. \end{enumerate} You have just proved: \begin{theorem}[Structure theorem for finitely generated modules over a PID: Primary decomposition] Let $R$ be a principal ideal domain and $M$ a finitely generated $R$-module. Then there exist direct sum decompositions $$R \simeq F\oplus T\quad\text{and}\quad T\simeq \bigoplus_p T_p$$ where $F$ is a free $R$-module of finite rank, $T$ is a torsion $R$-module and $T_p$ is a $p$-primary torsion $R$-module. Here, $F,T,T_p$ are each uniquely determined by $M$. Furthermore, for each $p$ there exist integers $0 < m_1 \le,\ldots, \le m_s$ and a direct sum decomposition $$T_p\simeq R/p^{m_1}\oplus \cdots \oplus R/p^{m_s}$$ with $s$ the minimal number of generators of $T_p$ and $p^m_s=\ann_R(T_p)$. The integers $\{m_i\}$ are uniquely determined by $M$. \end{theorem}\label{primary} \item Now prove: \begin{corollary}[Structure theorem for finitely generated modules over a PID: Invariant factor decomposition] Let $R$ be a principal ideal domain and $M$ a finitely generated $R$-module. Then there exists a direct sum decomposition $$M\simeq F \oplus T$$ with $F$ a free module and $T=\Tor(M)$ the torsion submodule of $M$. Moreover, there exists a nonnegative integer $d$ and nonzero elements $a_1,a_2,\ldots,a_m$ of $R$ which are not units and which satisfy the divisibility relations $$a_1 | a_2 | \cdots | a_m$$ such that there is a canonical isomorphism of $R$-modules $$T\simeq R/(a_1)\oplus R/(a_2) \oplus \cdots \oplus R/(a_d).$$ The ideals $(a_i)$ are uniquely determined by $T$ $($and hence by $M$$)$; they are called the {\em invariant factors} of $M$. \end{corollary} Hint: Use Theorem \ref{primary} and the Chinese Remainder Theorem. \item Now let $F$ be a field and $V$ a finite dimensional $F$-vector space equipped with a linear transformation $A:V\rightarrow V$. We consider $V$ as an $F[X]$-module via \begin{equation*} (a_0 + a_1 X + \cdots + a_n X^n)v := a_0v + a_1Av + \cdots + a_nA^nv, \end{equation*} where $A^i$ denotes the composition of $A$ with itself $i$-times. Since $F[X]$ is a PID, the structure theorems for $V$ as an $F[X]$-module that you proved above apply. \begin{enumerate} \item Let $(a_i)$ for $i=1,\ldots, d$ be the invariant factors of $V$. Show that each $a_i$ may be taken to be a monic polynomial, and that with this normalization the $a_i$'s are uniquely determined by $V$ and $A$ (not just up to units). \item Let $$g:=b_0 + b_1X + \cdots + X^k \in F[X]$$ be any monic polynomial and consider the finite-dimensional $F$-vector space $$V_g:=F[X]/(g).$$ Multiplication by $X$ gives a linear transformation of $V_g$, which we denote by $m_X$. Show that the matrix of $m_X$ with respect to the basis $1,X,X^2,\cdots, X^{k-1}$ of $V_g$ is the {\em companion matrix} of $g$, given by the $k\times k$ matrix over $F$\label{companion} \begin{equation*} C_g:=\left(\begin{array}{cccccc} 0 & 0 & \cdots & \cdots & \cdots & -b_0\\ 1 & 0 & \cdots & \cdots & \cdots & -b_1 \\ 0 & 1 & \cdots & \cdots & \cdots & -b_2 \\ 0 & 0 & \ddots & & & \vdots \\ \vdots & \vdots && \ddots & & \vdots \\ 0 & 0 & \cdots & \cdots & 1 & -b_{k-1} \end{array}\right) \end{equation*} \item Prove that there exists a basis of $V$ with respect to which the matrix of $A$ is in {\em rational canonical form}, i.e. has the block form \begin{equation} \left( \begin{array}{cccc} C_{a_1} & & &\\ & C_{a_2} & & \\ & & \ddots & \\ & & & C_{a_d} \end{array} \right)\label{rcan} \end{equation} Hint: First observe that $V$ is a torsion $F[X]$-module. Now use the invariant factor form of the structure theorem for modules over a PID and part (\ref{companion}). \item Prove that any square matrix $M$ over $F$ is similar to a matrix in rational canonical form; i.e. there exists an invertible matrix $P$ such that $P^{-1}MP$ has the form (\ref{rcan}), and show moreover that $P$ and this rational canonical form of $M$ are uniquely determined by $M$. Conclude that two $n\times n$ matrices over $F$ are similar if and only if they have the same rational canonical form. \end{enumerate} \end{enumerate} \end{document}