Let $A$ be a free abelian group of rank $n$, and let $\alpha_1 \in A\setminus \{0\}$ such that $\alpha_1 \not \in kA$ for all $k > 1$. Do there always exist $\alpha_2,\ldots, \alpha_n \in A$ such that $\alpha_1,\ldots, \alpha_n$ is a basis for $A$?
This question implies that the answer is yes, but it doesn't give any justification, and there's no proof in the accepted answer. Can anyone point me to a proof of the claim?
Attempting to generalise the example in the linked question, we might try letting $e_1,\ldots, e_n$ be a basis and writing $\alpha_1 = c_1e_1 + \ldots + c_ne_n$ for integers $c_i$. Then we would need a series of elementary column operations taking the vector $(c_1,\ldots, c_n)$ to a vector containing one $1$ and all other entries $0$. Since $k\nmid \alpha_1$ for all $k>1$, we have that the $c_i$ are mutually coprime, so there exist integers $x_1,\ldots, x_n$ with $c_1x_1 + \ldots c_nx_n = 1$. Maybe we can somehow use this combination to construct the desired $1$. Beyond that, I'm not too sure how to proceed.
Best Answer
Here is a different approach. Consider the quotient group $A/\mathbb{Z}\alpha_1$. The assumption that $\alpha_1$ is not divisible by any integer $k>1$ tells us that this quotient is torsion-free. Indeed, if it did have torsion, there would exist $a\in A\setminus \mathbb{Z}\alpha_1$ such that $ka\in\mathbb{Z}\alpha_1$ for some nonzero integer $k$. Say $ka=n\alpha_1$. Dividing both sides by $\gcd(k,n)$, we may assume $k$ and $n$ are relatively prime. But now this implies $\alpha_1$ is divisible by $k$, and $|k|>1$ since $a\not\in \mathbb{Z}\alpha_1$, contradicting our assumption about $\alpha_1$.
So, $A/\mathbb{Z}\alpha_1$ is a finitely generated torsion-free abelian group, so by the classification of finitely generated abelian groups, it is free. Picking a basis for it and lifting each basis element to an element of $A$, you get elements of $A$ which form a basis together with $\alpha_1$.