Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 4 / 9E / a

If $M$ is irreducible and $m\ne 0$, the image $Rm$ is a nonzero submodule, hence $Rm=M$ and the map is onto. Conversely, if every nonzero $m$ is cyclic and $N\le M$ is nonzero, choose $m\in N\setminus \{0\}$. Then $M=Rm\subseteq N$, so $N=M$ and $M$ is irreducible.

The annihilator $I={\mathrm{Ann}}_R(M)$ is an ideal. For $r\notin I$, choose $m$ with $rm\ne 0$. Irreducibility and part (i) give $s\in R$ such that $srm=m$. The element $(sr-1)$ therefore kills the generator $m$, and hence all of $M$; thus $sr-1\in I$. Every nonzero class in $R/I$ consequently has an inverse, so $R/I$ is a field.