Let $R$ be a principal ideal ring, which is not a field, and $M$ be a torsion module, then there is an isomorphism due to the theorem for finitely generated modules over a principal domain
$ M \cong \bigoplus_{i=1}^s R/(c_i) $
where $c_i \in R\setminus (R^* \cup \{0\})$ and $c_1\mid\ldots\mid c_s$. We call $\chi(M):= (c_1\cdot\ldots\cdot c_s)\subset R$ the characteristic ideal of $M$. Now let $M'$ and $M''$ be torsion modules as well and $0\to M'\to M\to M''\to 0$ be an exact sequence of $R$-modules, then it holds that $\chi(M)=\chi(M')\cdot\chi(M'')$. Why?
I know that for $R=\mathbb{Z}$ this is in fact nothing but Lagrange's theorem, but why is it the case for any principal ideal domain?
Assume $M$ is generated by $x_1, \cdots, x_n$, define $\eta: R^n \to M, \sum_i a_i e_i \mapsto \sum_i a_i x_i$, where $e_1, \cdots, e_n$ is basis of $R^n$.
$N=$ker($\eta$) and $N''=$ker($\beta \circ \eta$) are free $R$-module of rank $n$.($\beta$ is the map $M \to M''$.) Then $N \subset N''$.
a set of generator of $N''$: $(f_1, \cdots, f_n)=(e_1, \cdots, e_n)A$. $A$ is a matrix with entry in $R$.
a set of generator of $N$: $(g_1, \cdots, g_n)=(f_1, \cdots, f_n)B=(e_1, \cdots, e_n)AB$.
isomorphism theorem says a matrix is equivalent to a diagonal matrix diag($d_1, \cdots, d_n)$ with $d_i | d_{i+1}$.
Since determinant of an invertible matrix is an invertible element in $R$. We have $\chi(M'')=(\text{det} A)$, $\chi(M)=(\text{det} AB)$, notice $M' \simeq N''/N$, $N$ is a submodule of rank $n$ of a free $R$-module $N''$, so $\chi(M')=(\text{det} B)$.