Let $M$ be a finitely generated torsion module over the PID $R$. Suppose that $M$ has $s$ elementary divisors and $t$ invariant factors. Define $$S:=\{n\in\mathbb{N}\,|\,M\text{ is isomorphic to a direct sum of $n$ cyclic modules}\}.$$ (a) Show that $s=\max(S)$.
(b) Show that $t=\min(S)$.
I know that I am supposed to use the classification theorem of finitely generated modules over PID. Thus,
$M \cong \frac{R}{({p_1}^{e_1})}\bigoplus \frac{R}{({p_2}^{e_2})} \bigoplus \ldots \bigoplus \frac{R}{({p_s}^{e_s})}$ where $p_i$'s are primes (not necessarily distinct) and $e_i\geq 1$
The invariant form gives $M \cong \frac{R}{(a_1)}\bigoplus \frac{R}{(a_2)} \bigoplus \ldots \bigoplus \frac{R}{(a_t)}$ where $a_i$'s are non-zero, non-units and $a_i|a_{i+1}$. We know from this that $ann_{R}M=(a_t)$
Naturally both $t,s \in S$. Also, any cyclic R-module is of the form $R/I$ for some ideal $I$. Thus, for some $k \in S$, we have $M \cong R/I_1 \bigoplus R/I_2 \bigoplus+\ldots \bigoplus R/I_k$ for some ideals $I_1=(b_1),I_2=(b_2), \ldots I_k=(b_k)$ of $R$. Since $ann_R R/I=I$, this gives $ann_RM=\cap I_j$
Thus, $(a_t)=\cap I_j$. Hence, $a_t|b_i$ for $1\leq i \leq k$
I am not sure how to show that $t \leq k$ from this. Also, any hints for part $(a)$?
Thanks!