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.

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. 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.

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