# Basis of a free submodule of a free module

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 $$r = f/u$$.

• I would think the case $n=1$ would easily imply that $R$ is a PID. Feb 7, 2020 at 20:29
• Note that the problem linked at the top of the question is a much simpler one as the basis for $M$ Is fixed, whereas this question allows the basis for $M$ to be chosen separately for each $N$.
– Ben
Feb 8, 2020 at 10:31

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.