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.<ref>Template:Harvnb</ref>
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.
ProofEdit
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>\mathrm{im}(f) \supseteq \mathrm{im}(f^2) \supseteq \mathrm{im}(f^3) \supseteq \ldots</math>,
- the second is the ascending chain <math>\mathrm{ker}(f) \subseteq \mathrm{ker}(f^2) \subseteq \mathrm{ker}(f^3) \subseteq \ldots</math>
Because <math>M</math> has finite length, both of these chains must eventually stabilize, so there is some <math>n</math> with <math>\mathrm{im}(f^n) = \mathrm{im}(f^{n'})</math> for all <math>n' \geq n</math>, and some <math>m</math> with <math>\mathrm{ker}(f^m) = \mathrm{ker}(f^{m'})</math> for all <math>m' \geq m.</math>
Let now <math>k = \max\{n, m\}</math>, and note that by construction <math>\mathrm{im}(f^{2k}) = \mathrm{im}(f^{k})</math> and <math>\mathrm{ker}(f^{2k}) = \mathrm{ker}(f^{k}).</math>
We claim that <math>\mathrm{ker}(f^k) \cap \mathrm{im}(f^k) = 0</math>. Indeed, every <math>x \in \mathrm{ker}(f^k) \cap \mathrm{im}(f^k)</math> satisfies <math>x=f^k(y)</math> for some <math>y \in M</math> but also <math>f^k(x)=0</math>, so that <math>0=f^k(x)=f^k(f^k(y))=f^{2k}(y)</math>, therefore <math>y \in \mathrm{ker}(f^{2k}) = \mathrm{ker}(f^k)</math> and thus <math>x=f^k(y)=0.</math>
Moreover, <math>\mathrm{ker}(f^k) + \mathrm{im}(f^k) = M</math>: for every <math>x \in M</math>, there exists some <math>y \in M</math> such that <math>f^k(x)=f^{2k}(y)</math> (since <math>f^k(x) \in \mathrm{im}(f^k) = \mathrm{im}(f^{2k})</math>), and thus <math>f^k(x-f^k(y)) = f^k(x)-f^{2k}(y)=0</math>, so that <math>x-f^k(y) \in \mathrm{ker}(f^k)</math> and thus <math>x \in \mathrm{ker}(f^k)+f^k(y) \subseteq \mathrm{ker}(f^k) + \mathrm{im}(f^k).</math>
Consequently, <math>M</math> is the direct sum of <math>\mathrm{im}(f^k)</math> and <math>\mathrm{ker}(f^k)</math>. (This statement is also known as the Fitting decomposition theorem.) Because <math>M</math> is indecomposable, one of those two summands must be equal to <math>M</math> and the other must be the zero submodule. Depending on which of the two summands is zero, we find that <math>f</math> is either bijective or nilpotent.<ref>Jacobson (2009), p. 113–114.</ref>
NotesEdit
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