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 an automorphism or nilpotent.[1]

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.

Proof

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, ${\displaystyle \mathrm {ker} \left(f^{k}\right)\cap \mathrm {im} \left(f^{k}\right)=0}$ (because every ${\displaystyle x\in \mathrm {ker} \left(f^{k}\right)\cap \mathrm {im} \left(f^{k}\right)}$ satisfies ${\displaystyle x=f^{k}\left(y\right)}$ for some ${\displaystyle y\in M}$ but also ${\displaystyle f^{k}\left(x\right)=0}$, so that ${\displaystyle 0=f^{k}\left(x\right)=f^{k}\left(f^{k}\left(y\right)\right)=f^{2k}\left(y\right)}$, therefore ${\displaystyle y\in \mathrm {ker} \left(f^{2k}\right)=\mathrm {ker} \left(f^{k}\right)}$ and thus ${\displaystyle 0=f^{k}\left(y\right)=x}$) and ${\displaystyle \mathrm {ker} \left(f^{k}\right)+\mathrm {im} \left(f^{k}\right)=M}$ (since for every ${\displaystyle x\in M}$, there exists some ${\displaystyle y\in M}$ such that ${\displaystyle f^{k}\left(x\right)=f^{2k}\left(y\right)}$ (since ${\displaystyle f^{k}\left(x\right)\in \mathrm {im} \left(f^{k}\right)=\mathrm {im} \left(f^{2k}\right)}$), and thus ${\displaystyle f^{k}\left(x-f^{k}\left(y\right)\right)=f^{k}\left(x\right)-f^{k}\left(f^{k}\left(y\right)\right)=f^{k}\left(x\right)-f^{2k}\left(y\right)=0}$, so that ${\displaystyle x-f^{k}\left(y\right)\in \mathrm {ker} \left(f^{k}\right)}$ and thus ${\displaystyle x\in \mathrm {ker} \left(f^{k}\right)+f^{k}\left(y\right)\subseteq \mathrm {ker} \left(f^{k}\right)+\mathrm {im} \left(f^{k}\right)}$). 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.[2]

Notes

1. Jacobson, A lemma before Theorem 3.7. harvnb error: no target: CITEREFJacobson (help)
2. Jacobson (2009), p. 113–114.

References

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