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(f 2), im(f 3),..., the second sequence is the ascending sequence ker(f), ker(f 2), ker(f 3),.... Because M has finite length, the first sequence cannot be strictly decreasing forever, so there exists some n with im(f n) = im(f n+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(f n) = im(f n+1) yields im(f n) = im(f n+1) = im(f n+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

1. ^ Jacobson (2009), p. 113–114.