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:

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 $$m' \geq m.$$
 * 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 $$\mathrm{ker}(f) \subseteq \mathrm{ker}(f^2) \subseteq \mathrm{ker}(f^3) \subseteq \ldots$$

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.