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.^[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 chains of submodules:

• The first is the descending chain i m ( f ) ⊇ i m ( f 2 ) ⊇ i m ( f 3 ) ⊇ … {\displaystyle \mathrm {im} (f)\supseteq \mathrm {im} (f^{2})\supseteq \mathrm {im} (f^{3})\supseteq \ldots } ,
• the second is the ascending chain k e r ( f ) ⊆ k e r ( f 2 ) ⊆ k e r ( f 3 ) ⊆ … {\displaystyle \mathrm {ker} (f)\subseteq \mathrm {ker} (f^{2})\subseteq \mathrm {ker} (f^{3})\subseteq \ldots }

Because M {\displaystyle M} has finite length, both of these chains must eventually stabilize, so there is some n {\displaystyle n} with i m ( f n ) = i m ( f n ′ ) {\displaystyle \mathrm {im} (f^{n})=\mathrm {im} (f^{n'})} for all n ′ ≥ n {\displaystyle n'\geq n} , and some m {\displaystyle m} with k e r ( f m ) = k e r ( f m ′ ) {\displaystyle \mathrm {ker} (f^{m})=\mathrm {ker} (f^{m'})} for all m ′ ≥ m . {\displaystyle m'\geq m.}

Let now k = max { n , m } {\displaystyle k=\max\{n,m\}} , and note that by construction i m ( f 2 k ) = i m ( f k ) {\displaystyle \mathrm {im} (f^{2k})=\mathrm {im} (f^{k})} and k e r ( f 2 k ) = k e r ( f k ) . {\displaystyle \mathrm {ker} (f^{2k})=\mathrm {ker} (f^{k}).}

We claim that k e r ( f k ) ∩ i m ( f k ) = 0 {\displaystyle \mathrm {ker} (f^{k})\cap \mathrm {im} (f^{k})=0} . Indeed, every x ∈ k e r ( f k ) ∩ i m ( f k ) {\displaystyle x\in \mathrm {ker} (f^{k})\cap \mathrm {im} (f^{k})} satisfies x = f k ( y ) {\displaystyle x=f^{k}(y)} for some y ∈ M {\displaystyle y\in M} but also f k ( x ) = 0 {\displaystyle f^{k}(x)=0} , so that 0 = f k ( x ) = f k ( f k ( y ) ) = f 2 k ( y ) {\displaystyle 0=f^{k}(x)=f^{k}(f^{k}(y))=f^{2k}(y)} , therefore y ∈ k e r ( f 2 k ) = k e r ( f k ) {\displaystyle y\in \mathrm {ker} (f^{2k})=\mathrm {ker} (f^{k})} and thus x = f k ( y ) = 0. {\displaystyle x=f^{k}(y)=0.}

Moreover, k e r ( f k ) + i m ( f k ) = M {\displaystyle \mathrm {ker} (f^{k})+\mathrm {im} (f^{k})=M} : for every x ∈ M {\displaystyle x\in M} , there exists some y ∈ M {\displaystyle y\in M} such that f k ( x ) = f 2 k ( y ) {\displaystyle f^{k}(x)=f^{2k}(y)} (since f k ( x ) ∈ i m ( f k ) = i m ( f 2 k ) {\displaystyle f^{k}(x)\in \mathrm {im} (f^{k})=\mathrm {im} (f^{2k})} ), and thus f k ( x − f k ( y ) ) = f k ( x ) − f 2 k ( y ) = 0 {\displaystyle f^{k}(x-f^{k}(y))=f^{k}(x)-f^{2k}(y)=0} , so that x − f k ( y ) ∈ k e r ( f k ) {\displaystyle x-f^{k}(y)\in \mathrm {ker} (f^{k})} and thus x ∈ k e r ( f k ) + f k ( y ) ⊆ k e r ( f k ) + i m ( f k ) . {\displaystyle x\in \mathrm {ker} (f^{k})+f^{k}(y)\subseteq \mathrm {ker} (f^{k})+\mathrm {im} (f^{k}).}

Consequently, M {\displaystyle M} is the direct sum of i m ( f k ) {\displaystyle \mathrm {im} (f^{k})} and k e r ( f k ) {\displaystyle \mathrm {ker} (f^{k})} . (This statement is also known as the Fitting decomposition theorem.) Because M {\displaystyle M} is indecomposable, one of those two summands must be equal to M {\displaystyle M} and the other must be the zero submodule. Depending on which of the two summands is zero, we find that f {\displaystyle f} is either bijective or nilpotent.^[2]

Notes

References

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