Irreducible Modules Over PIDs (12.1.14)

Dummit and Foote Abstract Algebra, section 12.1, exercise 14:

Letting $M$ be a torsion module over $R$ a P.I.D., prove $M$ is irreducible if and only if $M=Rm$ for some nonzero element $m∈R$ with prime annihilator $(p)$.

Proof: ($⇒$) $M$ is clearly generated by a single element $m$, as a partial generator would imply a nontrivial submodule. Assume $\text{Ann }(m)=(r)$ for composite, nonzero $r$, so write $r = ab$ for nonzero nonunits $a$ and $b$. We show $Rbm ⊂ Rm = M$. Assuming otherwise, we can write $xbm = m$ for some $x∈R$, and now $axbm = am ⇒ am = 0$ despite $(r) ⊂ (a)$, a contradiction. ($⇐$) Take nonzero $rm ∈ M$; we thus have $p \not \mid r$ and there are elements $a,b∈R$ providing $ar+bp = 1$. We have $arm = (ar+bp)m=m$ so that $Rrm = M$ and $M$ has no nontrivial submodules.$~\square$