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, k e r ( f k ) ∩ i m ( f k ) = 0 (because every x ∈ k e r ( f k ) ∩ i m ( f k ) satisfies x = f k ( y ) for some y ∈ M but also f k ( x ) = 0 , so that 0 = f k ( x ) = f k ( f k ( y ) ) = f 2 k ( y ) , therefore y ∈ k e r ( f 2 k ) = k e r ( f k ) and thus 0 = f k ( y ) = x ) and k e r ( f k ) + i m ( f k ) = M (since for every x ∈ M , there exists some y ∈ M such that f k ( x ) = f 2 k ( y ) (since f k ( x ) ∈ i m ( f k ) = i m ( f 2 k ) ), and thus f k ( x − f k ( y ) ) = f k ( x ) − f k ( f k ( y ) ) = f k ( x ) − f 2 k ( y ) = 0 , so that x − f k ( y ) ∈ k e r ( f k ) and thus x ∈ k e r ( f k ) + f k ( y ) ⊆ k e r ( f k ) + i m ( f k ) ). 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.