Fitting’s lemma

Theorem 1 (Fitting Decomposition Theorem).

Let $R$ be a ring, and $M$ a finite-length module over $R$. Then for any $\varphi \mathrm{\in }\mathrm{End}\mathit{}\mathrm{(}M\mathrm{)}$, the endomorphism ring of $M$, there is a positive integer $n$ such that

$$M=\ker ({\varphi }^n)\oplus \mathrm{im}({\varphi }^n).$$

Proof.

Given $\varphi \in \mathrm{End}(M)$, it is clear that $\ker ({\varphi }^i)\subseteq \ker ({\varphi }^{i+1})$ and $\mathrm{im}({\varphi }^i)\supseteq \mathrm{im}({\varphi }^{i+1})$ for any positive integer $i$. Therefore, we have an ascending chain of submodules

$$\ker (\varphi )\subseteq \mathrm{\cdots }\subseteq \ker ({\varphi }^i)\subseteq \ker ({\varphi }^{i+1})\subseteq \mathrm{\cdots },$$

and a descending chain of submodules

$$\mathrm{im}(\varphi )\supseteq \mathrm{\cdots }\supseteq \mathrm{im}({\varphi }^i)\supseteq \mathrm{im}({\varphi }^{i+1})\supseteq \mathrm{\cdots }.$$

Both chains must be finite, since $M$ has finite length. Therefore, we can find a positive integer $n$ such that

$$\{\begin{array}{c}\ker ({\varphi }^n)=\ker ({\varphi }^{n+1})=\mathrm{\cdots },\text{ and}\hfill \\ \mathrm{im}({\varphi }^n)=\mathrm{im}({\varphi }^{n+1})=\mathrm{\cdots }.\hfill \end{array}$$

If $u\in M$, then ${\varphi }^n(u)\in \mathrm{im}({\varphi }^n)=\mathrm{im}({\varphi }^{2n})$. Therefore, ${\varphi }^n(u)={\varphi }^{2n}(v)$ for some $v\in M$. Write $u=(u-{\varphi }^n(v))+{\varphi }^n(v)$. Applying the ${\varphi }^n$ to the first term, we get ${\varphi }^n(u-{\varphi }^n(v))={\varphi }^n(u)-{\varphi }^{2n}(v)=0$, so it is in $\ker ({\varphi }^n)$. The second term is clearly in $\mathrm{im}({\varphi }^n)$. So

$$M=\ker ({\varphi }^n)+\mathrm{im}({\varphi }^n).$$

Furthermore, if $u\in \ker ({\varphi }^n)\cap \mathrm{im}({\varphi }^n)$, then $u={\varphi }^n(v)$ for some $v\in M$. Since ${\varphi }^{2n}(v)={\varphi }^n(u)=0$, $v\in \ker ({\varphi }^{2n})=\ker ({\varphi }^n)$. Therefore, $u={\varphi }^n(v)=0$. This shows that we can replace $+$ in the equation above by $\oplus$, proving the theorem.∎

Stated differently, the theorem says that, given an endomorphism $\varphi$ on $M$, $M$ can be decomposed into two submodules $M_1$ and $M_2$, such that $\varphi$ restricted to $M_1$ is nilpotent, and $\varphi$ restricted to $M_2$ is an isomorphism.

A direct consequence of this decomposition property is the famous Fitting Lemma:

Corollary 1 (Fitting Lemma).

In the theorem above, $\varphi$ is either nilpotent (${\varphi }^n\mathrm{=}\mathrm{0}$ for some $n$) or an automorphism iff $M$ is indecomposable.

Proof.

Suppose first that $M$ is indecomposable. Then either $\ker ({\varphi }^n)=0$ or $\mathrm{im}({\varphi }^n)=0$. If $n=1$, then the lemma is proved. Suppose $n>1$. In the former case, any $u\in M$ is the image of some $v$ under ${\varphi }^n$, so $u=\varphi ({\varphi }^{n-1}(v))$ and therefore $\varphi$ is onto. If $\varphi (u)=0$, then ${\varphi }^n(u)={\varphi }^{n-1}(\varphi (u))=0$, so $u=0$. This means $u$ is an automorphism. In the latter case, ${\varphi }^n(u)=0$ for any $u\in M$, so $\varphi$ is nilpotent.

Now suppose $M$ is not indecomposable. Then writing $M=M_1\oplus M_2$, where $M_1$ and $M_2$ as proper submodules of $M$, we can define $\varphi \in \mathrm{End}(M)$ such that $\varphi$ is the identity on $M_1$ and $0$ on $M_2$ ($\varphi$ is a projection of $M$ onto $M_1$). Since both $M_1$ and $M_2$ are proper, $\varphi$ is neither an automorphism nor nilpotent.∎

Remark. Another way of stating Fitting Lemma is to say that $\mathrm{End}(M)$ is a local ring iff the finite-length module $M$ is indecomposable. The (unique) maximal ideal in $\mathrm{End}(M)$ consists of all nilpotent endomorphisms (and its complement consists of, of course, the automorphisms).