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 $\mathrm{i}\mathrm{m}(f)\supseteq \mathrm{i}\mathrm{m}(f^2)\supseteq \mathrm{i}\mathrm{m}(f^3)\supseteq \dots$,
• the second is the ascending chain $\mathrm{k}\mathrm{e}\mathrm{r}(f)\subseteq \mathrm{k}\mathrm{e}\mathrm{r}(f^2)\subseteq \mathrm{k}\mathrm{e}\mathrm{r}(f^3)\subseteq \dots$

Because $M$ has finite length, both of these chains must eventually stabilize, so there is some $n$ with $\mathrm{i}\mathrm{m}(f^n)=\mathrm{i}\mathrm{m}(f^{n^{\prime }})$ for all $n^{\prime }\ge n$, and some $m$ with $\mathrm{k}\mathrm{e}\mathrm{r}(f^m)=\mathrm{k}\mathrm{e}\mathrm{r}(f^{m^{\prime }})$ for all $m^{\prime }\ge m.$

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

We claim that $\mathrm{k}\mathrm{e}\mathrm{r}(f^k)\cap \mathrm{i}\mathrm{m}(f^k)=0$. Indeed, every $x\in \mathrm{k}\mathrm{e}\mathrm{r}(f^k)\cap \mathrm{i}\mathrm{m}(f^k)$ 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 $y\in \mathrm{k}\mathrm{e}\mathrm{r}(f^{2k})=\mathrm{k}\mathrm{e}\mathrm{r}(f^k)$ and thus $x=f^k(y)=0.$

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

Consequently, $M$ is the direct sum of $\mathrm{i}\mathrm{m}(f^k)$ and $\mathrm{k}\mathrm{e}\mathrm{r}(f^k)$. (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.

References

• Jacobson, Nathan (2009), Basic algebra, 2 (2nd ed.), Dover, ISBN 978-0-486-47187-7