PlanetMath: Fitting's lemma

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

$\begin{displaymath} \left\{ \begin{array}{l} \ker(\phi^n)=\ker(\phi^{n+1})=\cdot... ...^n)= \operatorname{im}(\phi^{n+1}) =\cdots. \end{array}\right. \end{displaymath}$

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. $\qedsymbol$

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:

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. $\qedsymbol$

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).