FittingsLemma 2007-08-20 14:05:52 2008-04-30 23:27:51 Theorem length of a module Fitting's decomposition theorem \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{mathrsfs} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics \usepackage{xypic} \usepackage{pst-plot} \usepackage{psfrag} % define commands here \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{cor}{Corollary} \newtheorem{ex}{Example} \newcommand{\real}{\mathbb{R}} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}} \PMlinkescapeword{decomposition} \PMlinkescapeword{theorem} \PMlinkescapeword{lemma} \begin{thm}[Fitting Decomposition Theorem] Let $R$ be a ring, and $M$ a finite-length module over $R$. Then for any $\phi \in \operatorname{End}(M)$, the endomorphism ring of $M$, there is a positive integer $n$ such that $$M=\ker(\phi^n)\oplus \operatorname{im}(\phi^n).$$ \end{thm} \begin{proof} Given $\phi\in \operatorname{End}(M)$, it is clear that $\ker(\phi^i)\subseteq \ker(\phi^{i+1})$ and $\operatorname{im}(\phi^i)\supseteq \operatorname{im}(\phi^{i+1})$ for any positive integer $i$. Therefore, we have an ascending chain of submodules $$\ker(\phi)\subseteq \cdots \subseteq \ker(\phi^i)\subseteq \ker(\phi^{i+1}) \subseteq \cdots,$$ and a descending chain of submodules $$\operatorname{im}(\phi)\supseteq \cdots \supseteq \operatorname{im}(\phi^i)\supseteq \operatorname{im}(\phi^{i+1}) \supseteq \cdots.$$ Both chains must be finite, since $M$ has finite length. Therefore, we can find a positive integer $n$ such that \begin{displaymath} \left\{ \begin{array}{l} \ker(\phi^n)=\ker(\phi^{n+1})=\cdots, \mbox{ and} \\ \operatorname{im}(\phi^n)= \operatorname{im}(\phi^{n+1}) =\cdots. \end{array} \right. \end{displaymath} If $u\in M$, then $\phi^n(u)\in \operatorname{im}(\phi^n)=\operatorname{im}(\phi^{2n})$. Therefore, $\phi^n(u)=\phi^{2n}(v)$ for some $v\in M$. Write $u=(u-\phi^n(v))+\phi^n(v)$. Applying the $\phi^n$ to the first term, we get $\phi^n(u-\phi^n(v))=\phi^n(u)-\phi^{2n}(v)=0$, so it is in $\ker(\phi^n)$. The second term is clearly in $\operatorname{im}(\phi^n)$. So $$M=\ker(\phi^n)+\operatorname{im}(\phi^n).$$ Furthermore, if $u\in \ker(\phi^n)\cap \operatorname{im}(\phi^n)$, then $u=\phi^n(v)$ for some $v\in M$. Since $\phi^{2n}(v)=\phi^n(u)=0$, $v\in \ker(\phi^{2n})=\ker(\phi^n)$. Therefore, $u=\phi^n(v)=0$. This shows that we can replace $+$ in the equation above by $\oplus$, proving the theorem. \end{proof} Stated differently, the theorem says that, given an endomorphism $\phi$ on $M$, $M$ can be decomposed into two submodules $M_1$ and $M_2$, such that $\phi$ restricted to $M_1$ is nilpotent, and $\phi$ restricted to $M_2$ is an isomorphism. A direct consequence of this decomposition property is the famous Fitting Lemma: \begin{cor}[Fitting Lemma] In the theorem above, $\phi$ is either nilpotent ($\phi^n=0$ for some $n$) or an automorphism iff $M$ is indecomposable. \end{cor} \begin{proof} Suppose first that $M$ is indecomposable. Then either $\ker(\phi^n)=0$ or $\operatorname{im}(\phi^n)=0$. If $n=1$, then the lemma is proved. Suppose $n>1$. In the former case, any $u\in M$ is the image of some $v$ under $\phi^n$, so $u=\phi(\phi^{n-1}(v))$ and therefore $\phi$ is onto. If $\phi(u)=0$, then $\phi^n(u)=\phi^{n-1}(\phi(u))=0$, so $u=0$. This means $u$ is an automorphism. In the latter case, $\phi^n(u)=0$ for any $u\in M$, so $\phi$ is nilpotent. Now suppose $M$ is not indecomposable. Then writing $M=M_1\oplus M_2$, where $M_1$ and $M_2$ as proper submodules of $M$, we can define $\phi\in \operatorname{End}(M)$ such that $\phi$ is the identity on $M_1$ and $0$ on $M_2$ ($\phi$ is a projection of $M$ onto $M_1$). Since both $M_1$ and $M_2$ are proper, $\phi$ is neither an automorphism nor nilpotent. \end{proof} \textbf{Remark}. Another way of stating Fitting Lemma is to say that $\operatorname{End}(M)$ is a local ring iff the finite-length module $M$ is indecomposable. The (unique) maximal ideal in $\operatorname{End}(M)$ consists of all nilpotent endomorphisms (and its complement consists of, of course, the automorphisms).