Irreducible module

An irreducible, or simple, module is nonzero and has no submodules other than zero and itself.

For a simple module $M$ and a central element $z$ of an algebra $A$, the image $zM$ is a submodule. If it is nonzero, it equals $M$, and iterating gives $z^{r}M=M$ for every $r$. This is impossible when $z$ is nilpotent. Hence $zM=0$. It follows, for example, that the only simple module for $k[X]/(X^r)$ is the one-dimensional quotient on which $X$ acts by zero.

If $A$ is a finite module over a commutative central subring $R$ and $M$ is a simple left $A$-module, then $M=Am$ is finite over $R$. A central element not annihilating $M$ acts bijectively, since its kernel and image are $A$-submodules. Put $B=R/{\mathrm{Ann}}_R(M)$. For $0\ne r\in B$, write generators as $m_i=r{\sum }_j{b}_{ij}{m}_j$. The adjugate matrix identity shows $\det (I-r(b_{ij}))$ annihilates $M$, hence is zero in $B$. Since this determinant has form $1-rq$, $r$ is invertible. Thus $B$ is a field. If $R$ is a finite-type algebra over an algebraically closed field $k$, the residue field is $k$, and every such simple module is finite-dimensional over $k$.

Every simple module over a commutative ring $R$ is isomorphic to $R/\mathfrak{m}$ for a maximal ideal $\mathfrak{m}$; its annihilator is therefore maximal.

For a finite-dimensional commutative complex algebra, a simple module is a cyclic quotient of the algebra and hence finite-dimensional. Every action map is a module endomorphism, since the algebra is commutative. The complex Schur's lemma makes each action scalar, so every vector subspace is a submodule; simplicity forces dimension one. The finite-dimensional hypothesis cannot simply be deleted: $\mathbb{C}(t)$ acting on itself is a simple module of infinite complex dimension.