By the Chinese remainder theorem, we know that $\mathbb{Z}_m \cong \prod_{i=1}^l \mathbb{Z}_{p_i^{k_i}}$, where $m=p_1^{k_1} ... p_l^{k_l}$.
Now, let $\Lambda = A(\mathbb{Z}^n) \subseteq \mathbb{Z}^n$, where $A$ is an integer valued matrix. Does $\mathbb{Z}^n/\Lambda$ decompose similarly as in the $n=1$-case?
EDIT: So, can I say something like: Let $\lambda_i$ be the $i$th diagonal entry of the Smith normal form of $A$. Then $\mathbb{Z}^n/\Lambda \cong \mathbb{Z}^n/\oplus_{i=1}^n \mathbb{Z} \cdot \lambda_i \cong \bigoplus_{i=1}^n \mathbb{Z}_{\lambda_i}$?
EDIT: For the record, the answer is yes: Structure theorem for finitely generated modules over a principal ideal domain