Let $R$ be a principal ideal domain and let $M$ be a finitely-generated (not necessarily free) $R$-module. If I choose $m$ large enough, I can find a homomorphism $\phi: R^m \xrightarrow{} M $. For example if $(z_1,...,z_m)$ generate $M$ then let $(e_1,...,e_m)$ be the canonical basis for $R^m$. Let $\phi: R^m \xrightarrow{} M $ be a module homomorphism with $\phi(e_i) = z_i$. Now $\ker\phi$ is a submodule of $R^m$ and thus I can find a homomorphism $\psi: R^n \xrightarrow{} \ker\phi $, $R^n$ a free, finitely generated $R$-module. This yield an exact sequence, or a finite presentation of $M$:
$$R^n \xrightarrow{\psi} R^m \xrightarrow{\phi} M \rightarrow0.$$ My question is this: what happens when I choose $m$ larger? Specifically, how do (or don't) the the elementary divisors (units and nonunits) change? (Will there be more? Only additional zeroes? Only additional nonunits?)
Edit: clarification and specification concerning "when I choose $m$ larger": what I mean is that $M$ remains the same. And as we don't know how large a generating set of $M$ has to be (or even if there is a minimal generating set (minimal with respect to cardinality)) I can choose $m$ as large as I want (and arbitrarily larger than necessary) in order to make $\phi: R^m \xrightarrow{} M $ a surjective homomorphism. My question is how the elementary divisors of the finite presentation $$R^n \xrightarrow{\psi} R^m \xrightarrow{\phi} M \rightarrow 0.$$ change with larger $m$.