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.

References

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