Fitting Decomposition

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.

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 $\mathrm(f) \supseteq \mathrm(f^2) \supseteq \mathrm(f^3) \ldots$,
* the second sequence is the ascending sequence $\mathrm(f) \subseteq \mathrm(f^2) \subseteq \mathrm(f^3) \ldots$
Because $M$ has finite length, both of these sequences must eventually stabilize, so there is some $n$ with $\mathrm(f^n) = \mathrm(f^)$ for all $n^\prime \geq n$, and some $m$ with $\mathrm(f^m) = \mathrm(f^)$ for all $m^\prime \geq m$.

Let now $k = \max\$, and note that by construction $\mathrm (f^) = \mathrm (f^)$ and $\mathrm (f^) = \mathrm (f^)$.

We claim that $\mathrm\left(f^k\right) \cap \mathrm\left(f^k\right) = 0$. Indeed, every $x\in \mathrm\left(f^k\right) \cap \mathrm\left(f^k\right)$ satisfies $x=f^k\left(y\right)$ for some $y\in M$ but also $f^k\left(x\right)=0$, so that $0=f^k\left(x\right)=f^k\left(f^k\left(y\right)\right)=f^\left(y\right)$, therefore $y\in\mathrm\left(f^\right)=\mathrm\left(f^k\right)$ and thus $x=f^k\left(y\right)=0$.

Moreover, $\mathrm\left(f^k\right) + \mathrm\left(f^k\right) = M$: for every $x\in M$, there exists some $y\in M$ such that $f^k\left(x\right)=f^\left(y\right)$ (since $f^k\left(x\right)\in\mathrm\left(f^k\right)=\mathrm\left(f^\right)$), and thus $f^k\left(x-f^k\left(y\right)\right) = f^k\left(x\right)-f^\left(y\right)=0$, so that $x-f^k\left(y\right)\in\mathrm\left(f^k\right)$ and thus $x\in \mathrm\left(f^k\right)+f^k\left(y\right)\subseteq \mathrm\left(f^k\right) + \mathrm\left(f^k\right)$.

Consequently, $M$ is the direct sum of $\mathrm(f^k)$ and $\mathrm(f^k)$. (This statement is also known as the ''Fitting decomposition theorem''.) Because $M$ is indecomposable, one of those two summands must be equal to $M$, and the other must be the trivial submodule. Depending on which of the two summands is zero, we find that $f$ is either bijective or nilpotent.Jacobson (2009), p. 113–114.

Notes

References

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{{DEFAULTSORT:Fitting Lemma
Module theory
Lemmas in algebra