Existence of a invertible matrix with given first column 
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. 
 A: 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.
