This is question 5 from section 3.9 in Jacobson's Basic Algebra I.
Suppose we have $A,B \in M_n(D)$ satisfy $\det AB \neq 0$, where D is a p.i.d. Let $\text{diag}\{a_1,a_2,...,a_n\}$ be the smith normal form for $A$, $\text{diag}\{b_1,b_2,...,b_n\}$ be the smith normal form for $B$, and $\text{diag}\{c_1,c_2,...,c_n\}$ be the smith normal form for $AB$. Prove that $a_i\mid c_i$ and $b_i \mid c_i$ for all $i$.
This question was asked before with no reply.
My current idea is as follows:
Let $A', B'$, and $C'$ be the smith normal forms of $A, B$, and $AB$ respectively. Consider that $A = UA'V$, $B = WB'X$, and $AB = YC'Z$, where $U,V,W,X,Y,Z$ are all invertible. Then, we have that $UA'VWB'X = YC'Z$ and so $Y^{-1}UA'VWB'XZ^{-1} = C'$. Then, the $i$-rowed minors of $C'$ are linear combinations of the $i$-rowed minors of $A'$, and so the g.c.d. of the $i$-rowed minors of $A'$ is a divisor of the g.c.d. of the $i$-rowed minors of $C'$. Let $\Delta_i^A$ be the g.c.d. of the $i$-rowed minors of $A'$ and similarly let $\Delta_i^C$ be the g.c.d. of the $i$-rowed minors of $C'$.
However, I get stuck after this point. We can show that $a_1 = \Delta_1^A \mid \Delta_1^C = c_1$. But after that, $a_1a_2 = \Delta_2^A \mid \Delta_2^C = c_1c_2$ does not necessarily imply that $a_2 \mid c_2$. The only other observation I have made is that we have a stronger condition than divisibility for the $n$-th minor, that being $\Delta_n^A = \Delta_n^B\Delta_n^C$. Is there any tips of what to do from here?
Perhaps my approach is entirely incorrect and there is a better way to use some combination of the invariance/structure theorems to get the desired result as this does fall under those sections of the text.
Any help/hints would be greatly appreciated!