Fitting lemma

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.

Notes

References

References

1. {{harvnb. Jacobson. 2009
2. Jacobson (2009), p. 113–114.