I have a principal ideal domain $A$, and am trying to show that a variety of $A$-modules are non-isomorphic. So I am looking for 'invariants' of $A$-modules, by which I mean mappings $M\mapsto f(M)\in\mathbb{Z}$ taking $A$-modules to the integers (or some other set if need be) where if $M$ and $N$ are isomorphic, then $f(M)=f(N)$. Then if $f(M)\neq f(N)$, we know that $M$ and $N$ are not isomorphic.
So far I have $f(M):=\dim_K(K\otimes_A M)$ and $g(M):=\dim_{A/I}(M/IM)$, where $K$ is the field of fractions of $A$ and $I$ is a maximal ideal. Composing with the contravariant hom-functor $D:=\text{Hom}(-,K/I_1)$ gives $f(D(M))$ and $g(D(M))$ as two others and finally $M\mapsto\text{ann}(M)$, mapping into the ideals in $A$.
These should be enough for what I am doing right now, but I anticipate needing to have a range of such invariants at my fingertips in the future... so what are some good examples?
There's going to be lots of invariants and describing all of them seems like a path that could easily lead astray. In the case of finitely generated modules, however, we can give a complete set of invariants (meaning that if all those invariants agree on two modules, they are isomorphic), which should be sufficient for most purposes. One such set can be obtained by pushing one of your ideas a little bit further, namely by considering $r(M)=\dim_K(K\otimes_AM)$ and $r_{\mathfrak{p},n}(M)=\dim_{A/\mathfrak{p}}(\mathfrak{p}^nM/\mathfrak{p}^{n+1}M)$ for all prime ideals $\mathfrak{p}$ and $n\ge0$. Then, if $M$ is a finitely generated $A$-module, we have $$M\cong A^{r(M)}\oplus\bigoplus_{\mathfrak{p}}\bigoplus_n(A/\mathfrak{p}^n)^{r_{\mathfrak{p},n-1}(M)-r_{\mathfrak{p},n}(M)}.$$ There are also other sets of complete invariants, e.g. it might be worthwhile to figure out how all the $r_{\mathfrak{p},n}$ determine all the $s_{\mathfrak{p},n}=\dim_{A/\mathfrak{p}}(M[\mathfrak{p}^n]/M[\mathfrak{p}^{n-1}])$, where $M[\mathfrak{p}^n]$ denotes the $\mathfrak{p}^n$-torsion of $M$, and vice versa (in the finitely generated case, that is).