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In linear algebra over a field $K$ as set of column vectors $v_1,\dots,v_k\in K^n$ is a basis of the subspace $V$ iff

  1. $v_1,\dots,v_k\in V$,
  2. and the matrix $[v_1,\dots,v_k]$is of rank $k$.

I wonder under what circumstances something similar can be said in linear algebra over a UFD $R$. More precisely, I am aiming for the following (wrong!) statement. Let $V$ be a free submodule of $R^n$ such that $v\in V\Leftrightarrow rv\in V$ for $r\in R$ (e.g. the kernel of a matrix). A set of column vectors $v_1,\dots,v_k\in R^n$ is a basis of $V$ if

  1. $v_1,\dots,v_k\in V$,
  2. the matrix $[v_1,\dots,v_k]$ is of rank $k$,
  3. and the entries of each $v_i$ are coprime.

If this would be true any free resolution of $V$ would have length $1$. It is easy to come up with counterexamples. My question is: Does it become true if we add $\dim R=1$?

HCH
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  • What do you mean by "free divisible submodule" here? If we take, say, $R = \mathbb{Z}$ then no nonzero submodule of $\mathbb{Z}^n$ is divisible. – Qiaochu Yuan Sep 10 '22 at 20:58
  • I want $rv\in V \Leftrightarrow v\in V$ for $r \in R$ (is this not what is meant by divisible?). Any kernel of a matrix is then divisible. $2\mathbb Z\subset \mathbb Z$ is free. $2\in 2\mathbb Z$, but $1\notin 2\mathbb Z$. – HCH Sep 10 '22 at 22:45
  • I don't know a name for that condition. Divisible means if $v \in V$ and $r \in R$ is not a zero divisor then there exists $w \in V$ such that $rw = v$ which is a much stronger condition, see https://mathworld.wolfram.com/DivisibleModule.html. – Qiaochu Yuan Sep 10 '22 at 22:49
  • I edited the post according to Qiaochu Yuan's comment on the correct definition of divisibility. – HCH Sep 13 '22 at 15:06

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