Let $R$ be a principal ideal domain and $F$ be a free, finitely generated $R$-module of rank $n$. The structure theorem for this scenario tells us that
For every submodule $N ⊆ F$ there exists a basis $v_1, …, v_n$ of $F$ and a unique sequence of ideals $\mathfrak a_1 \supseteq … \supseteq \mathfrak a_n$ in $R$ such that $N = \mathfrak a_1 v_1 + … + \mathfrak a_n v_n$.
The base $v_1, …, v_n$ itself need not be unique for a given submodule $N$. If $a_1, …, a_n$ are generators of $\mathfrak a_1, …, \mathfrak a_n$ respectively, then (some subsystem of) $a_1v_1, …, a_nv_n$ yields a $R$-basis for $N$.
Is any basis of $N$ of this form? That is – can one refine the statement to the following?
Every submodule of $F$ is free and for every submodule $N ⊆ F$ of rank $m ≤ n$ there exists a unique sequence of ideals $\mathfrak a_1 \supseteq … \supseteq \mathfrak a_n$ in $R$ such that for any $R$-basis $w_1, …, w_m$ of $N$ there is a $R$-basis $v_1, …, v_n$ of $F$ and some generators $a_1, …, a_n$ of $\mathfrak a_1, …, \mathfrak a_n$ respectively such that $$w_1 = a_1v_1,~ …,~ w_m = a_mv_m$$ and possibly $0 = a_{m+1}v_{m+1},~ …,~ 0 = a_nv_n$ if $m < n$.
Nevermind, $N = ℤ \oplus 2ℤ$ in $F = ℤ \oplus ℤ$ has $(1,0), (1,2)$ as $ℤ$-basis, but clearly, both of these elements are no $ℤ$-multiples of elements in $ℤ \oplus ℤ$ (other than $\pm 1$-multiples).