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# Fitting’s lemma

###### Theorem 1 (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

$M=\ker(\phi^{n})\oplus\operatorname{im}(\phi^{n}).$ |

###### 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

$\left\{\begin{array}[]{l}\ker(\phi^{n})=\ker(\phi^{{n+1}})=\cdots,\mbox{ and}% \\ \operatorname{im}(\phi^{n})=\operatorname{im}(\phi^{{n+1}})=\cdots.\end{array}\right.$ |

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. ∎

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:

###### Corollary 1 (Fitting Lemma).

In the theorem above, $\phi$ is either nilpotent ($\phi^{n}=0$ for some $n$) or an automorphism iff $M$ is indecomposable.

###### 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. ∎

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).

## Mathematics Subject Classification

16D10*no label found*16S50

*no label found*13C15

*no label found*

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