$p$-power torsion dimensions recover the elementary divisors of a PID module

Since $p$ is prime, $(p)$ is a prime ideal; in a PID it is maximal by the same divisibility argument as for irreducibles, so $R\mathrm{/}(p)$ is a field. Every element of $p^{k-1}M[p^k]$ is killed by $p$, making this set an $R\mathrm{/}(p)$-vector space.

On one summand $C=R\mathrm{/}(p^e)$, if $e\ge k$ then $C[p^k]=p^{e-k}C$ and $p^{k-1}C[p^k]=p^{e-1}C$, a one-dimensional $R\mathrm{/}(p)$-space. If $e<k$, then $p^{k-1}C=0$, so the contribution is zero. For a $q$-primary summand with $q$ not associated to $p$, unique factorisation [L2] makes $p^k$ and $q^e$ coprime, so choose $u,v$ with $u{p}^k+v{q}^e=1$; multiplication by $u$ is inverse to multiplication by $p^k$ on $R\mathrm{/}(q^e)$, and its $p^k$-torsion is zero. Free summands likewise contribute no $p$-power torsion.

Direct sums commute with $M[p^k]$ and multiplication by $p^{k-1}$, so dimensions add. By step 2.1, $d_k$ counts exactly the exponents $e\ge k$; for $k=1$ it counts every $p$-primary cyclic summand, and beyond the largest exponent it is zero.

The summands counted by $d_k$ but not by $d_{k+1}$ are exactly those with exponent $e=k$, so $d_k-d_{k+1}$ recovers their multiplicity. Empty primary data and the zero module give the zero sequence.