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Given a commutative ring $R$ with unit and $a_1=(r_1,\ldots,r_n)^T \in R^n$ with coprime entries (i.e. $\sum_i Rr_i=R)$. Are there $a_2,\ldots,a_n \in R^n$ such that the matrix $A = (a_1,\ldots,a_n) \in R^{n \times n}$ is invertible?

The answer is "yes" if $R$ is a PID by "Is a vector of coprime integers column of a regular matrix?".

However none of the solutions there generalizes because the first answer uses division by a gcd and the second one (to my understanding) the elementary divisors theorem.

But because the answer is always "yes" in case $n=2$ (write $r_1s_1+r_2s_2=1$ and take $a_2=(-s_2,s_1)^T)$ I wonder if this is also true for $n > 2$ ?

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tj_
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  • Such vectors are called unimodular, and your question asks whether any unimodular vector in a commutative ring is equivalent to $(1,0,\dots,0)$ which is a well known problem with negative answer. (Btw, I've never seen an invertible matrix called "regular" and Wikipedia says "this usage is rare".) – user26857 Nov 24 '14 at 16:22

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No, this is not true, but it is quite non-trivial to find examples. At least at the top of my head I only know the one below - does anyone know a simpler one?

The entries of the vector $a_1 := (r_1,...,r_n)$ being coprime means that the map $$\alpha: R^n\xrightarrow{(r_1,...,r_n)} R$$ is surjective, so $P := \text{ker}(\alpha)$ is a rank projective $R$-module with $P\oplus R\cong R^{n}$. Conversely, any such module arises in this way.

The existence of the vectors $a_2,...,a_n$, however, is equivalent to the existence of a commutative diagram $$\begin{array}{ccc} R^n\ \ & \xrightarrow{\alpha} & R \\ \downarrow{\small\cong} && \ \ \downarrow{\small 1} \\ R^n \ \ & \xrightarrow{p_1} & R\end{array}$$ which is equivalent to $\text{ker}(\alpha)$ being free on $(n-1)$-generators.

Over a PID, any projective module is free, confirming your observation that any vector of coprime elements can be extended to a regular matrix. In general, however, there are non-free stably free modules: Taking $R:={\mathbb R}[x,y,z]/(x^2+y^2+z^2-1)$ and $a_1 := (x,y,z)$ gives such an example. See Is a stably free module always free? for details.

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Hanno
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  • I think there is a problem with the argument: Let $R$ be a Dedekind domain with class number $> 1$ and let $P\subseteq R$ be a non-principal ideal. Then $R^2 \cong P \oplus R$. Take $n=2$. By the part "Conversely, ..." there is $a_1$ such that $P=\ker(\alpha)$. Since $a_2$ exists (n=2), you say $\ker(\alpha)$ is free on one generator, i.e. $P$ is principal, in contrast to the choice of $P$. – tj_ Nov 24 '14 at 09:09
  • @tj_: Why do we have $R^2 \cong P\oplus R$? – Hanno Nov 24 '14 at 09:12
  • This is a property of Dedekind domains and holds for any ideal $P\neq 0$. See for instance Lemma 1.4.11 in Rosenberg: Algebraic K-theory and Applications. – tj_ Nov 24 '14 at 09:16
  • @tj_: Sorry, you might well be right in that I made a mistake, but at the moment I can't follow your objection: Lemma 1.4.11 states that $R\oplus I_1 I_2\cong I_1\oplus I_2$ for fractional ideals $I_1,I_2$. How do you apply this here? – Hanno Nov 24 '14 at 09:21
  • You are right, the Lemma doesn't show $R^2 \cong P \oplus R$. Could please give me a hint, how $\alpha$ is constructed for a given $R$-module $P$ with $R^n\cong P \oplus R$ ? – tj_ Nov 24 '14 at 10:12
  • Given $P$ and an isomorphism $P\oplus R\cong R^n$, take $\alpha: R^n \to P\oplus R\xrightarrow{p_2} R$. – Hanno Nov 24 '14 at 10:15
  • OK, and then take $r_i := \alpha(e_i)$ where $e_1,...,e_n$ is the standard basis of $R^n$. But if your answer is correct this would imply in a comm. ring with unit there is no non-principal ideal $P$ s.t. $R^2 \cong R \oplus P$. That's interesting becuse there are examples for such $P$ in the non-comm. case. I have to think about it in more detail. But it may ake a few days. So long! – tj_ Nov 24 '14 at 10:37
  • @tj_: Alright, let me know if there are other troubles with the answer. – Hanno Nov 24 '14 at 10:39
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    I don't know if this example is simpler, but it's definitely the most natural possible. – user26857 Nov 24 '14 at 16:14
  • @user26857: Interesting reference, thank you! – Hanno Nov 24 '14 at 16:40
  • @Hanno: Back again. Your answer is perfectly right. Thank you very much for the ingenious argument and for the counter-example! – tj_ Nov 28 '14 at 00:36
  • @user2685: Thanks for the correct terminology and the counter-example. – tj_ Nov 28 '14 at 00:38