Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 4 / 9E (Groups, Rings and Modules)

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.

Make $V$ a $k[x]$-module by $xv=\phi (v)$. The structure theorem over the Euclidean domain $k[x]$ decomposes its torsion module aswhere the $p_{\alpha }$ are monic irreducibles. Each summand is indecomposable: its submodules form a chain, so two nonzero submodules cannot form a direct sum. This is the desired decomposition into invariant indecomposable subspaces.

The characteristic polynomial iswhereas the minimal polynomial is the least common multiple

The prime ideals of $\mathbb{R}[x]$ are $(0)$ and the maximal ideals generated by the monic irreduciblesFor a nonzero prime $I=(p)$, use the residue classes of $1,x,\dots ,x^{N-1}$, where $N=n\mathrm{deg}p$. Multiplication by $x$ on $\mathbb{R}[x]/(p^n)$ is represented by the companion matrix of the monic polynomial $p^n$. In particular, for $p=x-a$ this may equivalently be written as one size-$n$ Jordan block with eigenvalue $a$. The ideal $(0)$ does not give a finite-dimensional quotient.

The upper-left block has characteristic and minimal polynomial $(x^2+1{)}^2$, while the lower-right block has characteristic and minimal polynomial $x^2+1$. Hence the corresponding module isIn the power bases from part (ii), an explicit normal form isthe direct sum of the companion matrices of $(x^2+1{)}^2$ and $x^2+1$.