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Elements $a$ and $b$ from an $R$-module $M$ are linearly dependent if there are scalars $x$, $y$ in $R$, $x \neq 0$ or $y \neq 0$, such that $xa = yb$.

Let $M$ be a torsion-free module over an integral domain $R$.
In this case the binary relation has the following properties:
1. It is an equivalence relation on nonzero elements of $M$;
2. Each equivalence class combined with the zero element is a submodule of $M$.

If $M$ is not a free module, then the submodules may not be isomorphic to each other (e.g. $\mathbb Z \times \mathbb Q$), or may not be free: $\langle 2,x \rangle$ over $\mathbb{Z}[x]$ (Show that $\langle 2,x \rangle$ is not a principal ideal in $\mathbb Z [x]$).

If $M$ is a free module over a PID, all the submodules are free modules isomorphic to $R$ over itself.

But what if $M$ is a free module, and $R$ is not a PID?

Questions:
1. Are the described submodules of a free module over an integral domain free?
2. If such submodule $N$ of a free module over an integral domain is free, is it possible to extend a basis of $N$ to a basis of $M$?

Alex C
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  • What do you call a "line" in the second question? – lisyarus Feb 10 '19 at 23:06
  • @lisyarus: By line I mean a free module of rank 1. – Alex C Feb 10 '19 at 23:10
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    It is easy to find a non-free submodule when $R$ fails the "pre-pre-Schreier condition" (which it does whenever $R$ is a Noetherian non-UFD). Indeed, pick any four elements $a, b, c, d$ of $R$ satisfying $ab = cd$ for which we can find no four elements $x, y, z, w$ in $R$ such that $a = xy$, $b = zw$, $c = xz$ and $d = yw$. It is easy to see that none of $a, b, c, d$ is $0$. Now, the two vectors $\left(a,c\right)$ and $\left(d,b\right)$ in the free $R$-module $R^2$ are ... – darij grinberg Aug 15 '19 at 18:07
  • ... linearly dependent (since $b\left(a,c\right) = c\left(d,b\right)$). Consider the $R$-submodule of $R^2$ formed by all vectors linearly dependent of them (and the zero vector). If this $R$-submodule was free, then it would be free of rank $1$ (since tensoring it with the quotient field of $R$ gives a $1$-dimensional vector space over the latter), and thus it would be generated by a single vector $\left(y,z\right)$. Thus, there would be $x, w \in R$ such that $\left(a,c\right) = x\left(y,z\right)$ and $\left(d,b\right) = w\left(y,z\right)$. These would yield ... – darij grinberg Aug 15 '19 at 18:10
  • ... the precise four equalities we have assumed impossible. – darij grinberg Aug 15 '19 at 18:10
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    I am quite interested to know what happens when $R$ is a UFD but not a PID! I think it has to do with the existence of a "least common denominator" for a finite list of elements of $\operatorname{Quot}\left(R\right)$. That's equivalent to being an lcm-domain, or isn't it? – darij grinberg Aug 15 '19 at 18:11

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