Existence of a invertible matrix with given first column

Question

Let $A$ be a PID and $a_1,a_2,...,a_n$ be elements of $A$ such that the ideal generated by these elements is $A$. Show that there exists a $M \in GL_n(A)$ such that the first column of $M$ is $(a_1,...,a_n)^t$.

I think this should be easy by using the fact that every ideal in $A$ is generated by one element but i cannot see it.

For $n=1$, this is trivial. For $n=2$, this is easy. $n=3$ is the crucial case and if you can figure it out, rest should be fine. The point, is for $n\geq 3$, you can find $M\in E_n(A)$, the group of elementary matrices. — Mohan, Apr 18 '17 at 14:16
Just to clarify, PID is a red herring. The fact you need is that $\dim A\leq 1$. — Mohan, Apr 18 '17 at 16:58
@Mohan: I was thinking about it but i really cannot see it.Any hints? Thanks for your continuous help! — Math Lover, Apr 19 '17 at 06:03

score 2 · Accepted Answer · answered Apr 19 '17 at 12:53

Too long to be a comment.

First, let $A$ be any commutative ring, $F=\oplus_{i=1}^n Ae_i$ a free module and $v_1=\sum a_ie_i\in F$ be unimodular. Then, there is a matrix $M\in GL_n(A)$ with its first row $(a_1,\ldots, a_n)$ if and only if $F/Av_1$ is free. The proof is easy, let me indicate it. For example, $F/Av_1$ is free just means you can find $v_2,\ldots, v_n\in F$ such that their images generate $F/Av_1$ and then, $v_1,\ldots, v_n$ is free basis for $F$. Two bases, $\{e_i\}, \{v_i\}$ are connected by an $M\in GL_n(A)$ and rest is clear. The converse is just following the steps backwards.

So, for the PID case, it boils down to proving that the projective module $P=F/Av_1$ as above is free. If $P=0$, there is nothing to prove, so assume $P\neq 0$. Then suffices to show that $P\cong Q\oplus A$ and by inductions on rank, you would be done. Since $P$ is torsion-free and non-zero, you can find a non-zero homomorphism $P\to A$. Since $A$ is a PID, image of this map is $Aa\cong A$. So, we get a surjection $P\to A$ and if $Q$ is the kernel, we get $P\cong Q\oplus A$, since the surjection splits.

For arbitrary one dimensional Noetherian rings, proof is a bit more complicated.

Existence of a invertible matrix with given first column

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