Basis of a free submodule of a free module

Question

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 $r = f/u$.

Answer 1

Take $R = k[x,y]$ with the submodule $R \to R^{\oplus 2}$ given by the inclusion $1\mapsto (x,y)$.

The vector $(x,y)$ is primitive so the basis for $R^{\oplus 2}$ would be of the form $\{ (x,y), (p,q)\}$ for some $p,q\in R$.

This means there are $a,b\in R$ such that $a(x,y) + b(p,q) = (1,0)$.

From $ay + bq =0$ you can see $y|pq$, $y|q$ cant be a basis (image of second projection contained in $(y)$). Writing $bp = cy$ then $ax + bp = ax + cy = 1$. But this is impossible - just plug $x=y=0$.

By the way your question is about the existence of Smith Normal Form - which only holds for PIDs. This is related to the existence of bezout coefficients in a PID: for any $x,y$ there are elements $r,s$ such that $$rx+sy = gcd(x,y).$$ Clearly this fails in higher dimensions (PID has krull dimension 1).

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More generally for a commutative ring $R$, if a vector $v=(v_i) \in R^{\oplus n}$ can be member of a basis, then the ideal generated by the coefficients is $R$.

To see this, note If it generated a proper ideal $I$ then we could find a maximal ideal $\mathfrak m \supset I$. For $k=R/\mathfrak m$, the map $R^{\oplus n} \to k^{\oplus n}$ sends a basis to a basis, but sends $v$ to 0.