Fitting lemma

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 bijective 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.

To prove Fitting's lemma, we take an endomorphism f of M and consider the following two sequences of submodules. The first sequence is the descending sequence im(f), im(f²), im(f³),..., the second sequence is the ascending sequence ker(f), ker(f²), ker(f³),.... Because M has finite length, the first sequence cannot be strictly decreasing forever, so there exists some n with im(fⁿ) = im(fⁿ⁺¹). Likewise (as M has finite length) the second sequence cannot be strictly increasing forever, so there exists some m with ker(f^m) = ker(f^m+1). It is easily seen that im(fⁿ) = im(fⁿ⁺¹) yields im(fⁿ) = im(fⁿ⁺¹) = im(fⁿ⁺²) = ..., and that ker(f^m) = ker(f^m+1) yields ker(f^m) = ker(f^m+1) = ker(f^m+2) = ... . Putting k = max(m,n ), it now follows that im(f^k) = im(f^2k) and ker(f^k) = ker(f^2k). Hence, $\mathrm{ker}\left(f^k\right) \cap \mathrm{im}\left(f^k\right) = 0$ (because every $x\in \mathrm{ker}\left(f^k\right) \cap \mathrm{im}\left(f^k\right)$ satisfies $x=f^k\left(y\right)$ for some $y\in M$ but also $f^k\left(x\right)=0$ , so that $0=f^k\left(x\right)=f^k\left(f^k\left(y\right)\right)=f^{2k}\left(y\right)$ , therefore $y\in\mathrm{ker}\left(f^{2k}\right)=\mathrm{ker}\left(f^k\right)$ and thus $0=f^k\left(y\right)=x$ ) and $\mathrm{ker}\left(f^k\right) + \mathrm{im}\left(f^k\right) = M$ (since for every $x\in M$ , there exists some $y\in M$ such that $f^k\left(x\right)=f^{2k}\left(y\right)$ (since $f^k\left(x\right)\in\mathrm{im}\left(f^k\right)=\mathrm{im}\left(f^{2k}\right)$ ), and thus $f^k\left(x-f^k\left(y\right)\right)=f^k\left(x\right)-f^k\left(f^k\left(y\right)\right)=f^k\left(x\right)-f^{2k}\left(y\right)=0$ , so that $x-f^k\left(y\right)\in\mathrm{ker}\left(f^k\right)$ and thus $x\in \mathrm{ker}\left(f^k\right)+f^k\left(y\right)\subseteq \mathrm{ker}\left(f^k\right) + \mathrm{im}\left(f^k\right)$ ). Consequently, M is the direct sum of im(f^k) and ker(f^k). Because M is indecomposable, one of those two summands must be equal to M, and the other must be equal to {0}. Depending on which of the two summands is zero, we find that f is bijective or nilpotent.^[1]

Notes

↑ Jacobson (2009), p. 113–114.

References

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

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