Assume that $$ 0 \to A \to \Bbb Z^n \overset f\to \Bbb Z^m \to 0 \tag{1} $$ is a short exact sequence of $\Bbb Z$-modules. What is $A$?
If instead of $\Bbb Z^n$ and $\Bbb Z^m$ we had the vector spaces $\Bbb R^n$ and $\Bbb R^m$, then it would be easy to compute $A$: In this case we would have $A = \Bbb R^k$ for some $k \leq n$. Further, the dimension formula yields \begin{align} n &= \dim(\ker(f)) + \dim(\operatorname{im}(f)) \\ &= \dim(A) + m \\ &= k + m, \end{align} and hence $k = n - m$.
Is there any suitable substitute for the dimension theorem in the more general situation (1)?
Since $\mathbb{Z}^m$ is free as a $\mathbb{Z}$-modul this sequence splits by the splitting lemma i.e. we have an isomorphism of $\mathbb{Z}$-modules $A\oplus \mathbb{Z}^m\cong\mathbb{Z}^n$. Notice that this is only possible if $A$ is finitely generated torsion-free and $n\geq m$ so let's assume these conditions are true. Then the classification theorem for finitely genrated $\mathbb{Z}$-modules gives $A\cong\mathbb{Z}^r$ for some $r$ since $A$ and for degree reasons $r=n-m$.