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 [math]\displaystyle{ \mathrm{im}(f) \supseteq \mathrm{im}(f^2) \supseteq \mathrm{im}(f^3) \supseteq \ldots }[/math],
• the second is the ascending chain [math]\displaystyle{ \mathrm{ker}(f) \subseteq \mathrm{ker}(f^2) \subseteq \mathrm{ker}(f^3) \subseteq \ldots }[/math]

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

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

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

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

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

References

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