Fitting lemma

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.^[1]

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 sequences of submodules:

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

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

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

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

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

Consequently, ${\displaystyle M}$ is the direct sum of ${\displaystyle \mathrm {im} (f^{k})}$ and ${\displaystyle \mathrm {ker} (f^{k})}$. Because ${\displaystyle M}$ is indecomposable, one of those two summands must be equal to ${\displaystyle M}$, and the other must be the trivial submodule. Depending on which of the two summands is zero, we find that ${\displaystyle f}$ is either bijective or nilpotent.^[2]

Notes[]

References[]

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