Endomorphism ring

The endomorphism ring ${\mathrm{End}}_R(M)$ consists of the $R$-module homomorphisms $M\to M$, with pointwise addition and composition as multiplication.

An endomorphism ring is local when its nonunits form a proper two-sided ideal; commutativity is not required. The Fitting lemma gives this property for an indecomposable finite-length module. If two nonunits had an invertible sum, multiplication by its inverse would give $I=f+g$ with noninvertible $f,g$. Fitting makes $f$ nilpotent, so $I-f=g$ would be invertible, a contradiction. Products with a nonunit are nonunits by finite-length injectivity/surjectivity. The quotient by the nonunit ideal is a division ring.

For a finite-dimensional module over an algebraically closed field, write a Krull-Schmidt decomposition $X={⨁}_a{M}_a^{\oplus m_a}$ with distinct indecomposable types. Each endomorphism ring of $M_a$ is a local endomorphism ring with residue division algebra $k$, by the finite-dimensional division algebra over an algebraically closed field result. Modulo the Jacobson radical of $\mathrm{End}(X)$, the blocks of one type become $M_{m_a}(k)$ and all maps between different types vanish. A composite through a different indecomposable type cannot be invertible, since that would make one type a direct summand of the other. This yields the displayed product and the Levi decomposition of a quiver automorphism group.

A nonzero module is a brick when every endomorphism is either zero or an isomorphism. Equivalently, its endomorphism ring is a division ring.

An indecomposable finite-dimensional quiver representation that is not a brick module contains a brick with nonzero self-extensions. The Ringel form then gives $q_Q(\dim B)=1-\dim {\mathrm{Ext}}_Q^1(B,B)\le 0$, impossible for a positive definite Tits form of a quiver.

The Fitting lemma supplies a nonzero nilpotent endomorphism $f$ of a non-brick indecomposable $X$. Minimize its nonzero rank; nilpotence and minimality give $f^2=0$. Set $I=\mathrm{im}f\subset K=\ker f={⨁}_j{K}_j$. Choose a nonzero component $u:I\to K_j$. The square-zero endomorphism $X\stackrel{f}{\to }I\stackrel{u}{\to }{K}_j\hookrightarrow X$ has rank at least $\dim I$, so $u$ is injective.

If ${\mathrm{Ext}}^1(I,K_j)=0$, the projection $K\to K_j$ extends to $X\to K_j$ by the long exact sequence of Ext groups, giving a module retraction and contradicting indecomposability. Thus this extension group is nonzero. Since path algebras are hereditary, $u$ induces a surjection ${\mathrm{Ext}}^1(K_j,K_j)\twoheadrightarrow {\mathrm{Ext}}^1(I,K_j)$. Hence $K_j$ is a proper indecomposable submodule with self-extensions. Iterate until a brick module is reached; dimensions strictly decrease.

This minimal-rank argument is given in section 2 of William Crawley-Boevey's quiver lectures.