Generators for a free submodule of a free module
In this question, it can be seen that the basis of $2\mathbb{Z}\subset \mathbb{Z}$ as $\mathbb{Z}$ module is different. However, the basis of $\mathbb{2Z}$ which is $\left\{2\right\}$ is a scalar multiple of of the basis of $\mathbb{Z}$ which is $\left\{1\right\}$. It is indeed true that under the assumption of PID, if $N\subset M$ are free modules. Then there exists some basis $\mathcal{B}$ of $M$ for which there is some subset $S\in \mathcal{B}$ whose element-wise scalar multiples is a basis for $N$.
However, I wish to know whether there a possible generalization of this result? As a simple example:
Given two free Modules $N,M$ over a polynomial ring $R$ with more than one variable such that $N\subset M$, $\text{rank}(M)\leq n$. Is it possible to choose a $\mathcal{B}=\left\{b_1,b_2,\cdots,b_n\right\}$ a basis for $M$ and a subset $\mathcal{S}\subseteq\mathcal{B}$ consisting of $k\leq n$ elements such that $\mathcal{B'}=\left\{r_1b_1,r_2b_2,\cdots,r_kb_k\right\}$ (suitable reordering) is a basis for for $N$.
For $n=1$ and in two variable ring. I think the following example will hold.
Considering $R[x,y]$ as a module over itself and taking the $R[x,y]$-submodule as (say) the cyclic $R[x,y]$-submodule $\langle f(x,y) \rangle$ for some $f(x,y)\in R[x,y]$. In this case, we can take $\left\{f(x,y)\right\}$ to be a basis. Now, the module $R[x,y]$ will be generated by any unit in the polynomial ring. So my above question in this case translates to this simple query: can we find some $r(x,y) \in R[x,y]$ such that $r(x,y) u= f(x,y)$ where $u$ is a unit? This is obviously true by the choice of $r = f/u$.