Fitting lemma
# Fitting lemma

In mathematics, the Fitting lemma – named after the mathematician Hans Fitting – is a basic statement in abstract algebra. Suppose M is a module over some ring. If M is indecomposable and has finite length, then every endomorphism of M is either an automorphism or nilpotent.

As an immediate consequence, we see that the endomorphism ring of every finite-length indecomposable module is local.

A version of Fitting's lemma is often used in the representation theory of groups. This is in fact a special case of the version above, since every K-linear representation of a group G can be viewed as a module over the group algebra KG.

## Proof

To prove Fitting's lemma, we take an endomorphism f of M and consider the following two chains of submodules:

* The first is the descending chain $\mathrm {im} (f)\supseteq \mathrm {im} (f^{2})\supseteq \mathrm {im} (f^{3})\supseteq \ldots$,
* the second is the ascending chain $\mathrm {ker} (f)\subseteq \mathrm {ker} (f^{2})\subseteq \mathrm {ker} (f^{3})\subseteq \ldots$

Because $M$ has finite length, both of these chains must eventually stabilize, so there is some $n$ with $\mathrm {im} (f^{n})=\mathrm {im} (f^{n'})$ for all $n'\geq n$, and some $m$ with $\mathrm {ker} (f^{m})=\mathrm {ker} (f^{m'})$ for all $m'\geq m.$

Let now $k=\max\{n,m\}$, and note that by construction $\mathrm {im} (f^{2k})=\mathrm {im} (f^{k})$ and $\mathrm {ker} (f^{2k})=\mathrm {ker} (f^{k}).$

We claim that $\mathrm {ker} (f^{k})\cap \mathrm {im} (f^{k})=0$. Indeed, every $x\in \mathrm {ker} (f^{k})\cap \mathrm {im} (f^{k})$ satisfies $x=f^{k}(y)$ for some $y\in M$ but also $f^{k}(x)=0$, so that $0=f^{k}(x)=f^{k}(f^{k}(y))=f^{2k}(y)$, therefore $y\in \mathrm {ker} (f^{2k})=\mathrm {ker} (f^{k})$ and thus $x=f^{k}(y)=0.$

Moreover, $\mathrm {ker} (f^{k})+\mathrm {im} (f^{k})=M$: for every $x\in M$, there exists some $y\in M$ such that $f^{k}(x)=f^{2k}(y)$ (since $f^{k}(x)\in \mathrm {im} (f^{k})=\mathrm {im} (f^{2k})$), and thus $f^{k}(x-f^{k}(y))=f^{k}(x)-f^{2k}(y)=0$, so that $x-f^{k}(y)\in \mathrm {ker} (f^{k})$ and thus $x\in \mathrm {ker} (f^{k})+f^{k}(y)\subseteq \mathrm {ker} (f^{k})+\mathrm {im} (f^{k}).$

Consequently, $M$ is the direct sum of $\mathrm {im} (f^{k})$ and $\mathrm {ker} (f^{k})$. (This statement is also known as the Fitting decomposition theorem.) Because $M$ is indecomposable, one of those two summands must be equal to $M$ and the other must be the zero submodule. Depending on which of the two summands is zero, we find that $f$ is either bijective or nilpotent.

