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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.

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2 Answers 2

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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 $\alpha_1$.

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You've made a good start. The method continues by applying the Euclidean algorithm.

Notice that $\text{g.c.f.}(c_1,\ldots,c_n) = 1$.

If the vector $(c_1,\ldots,c_n)$ is already equal to one of the standard unit coordinate vectors $\pm \, \vec e_1 \, , \, \ldots \, , \, \pm \, \vec e_n$, we're done.

Otherwise, we may choose $i \ne j \in \{1,\ldots,n\}$ so that $0 \ne |c_j| < |c_i|$.

Next, depending on the signs of $c_i$ and $c_j$, apply the column operator that adds $\pm c_j$ to $c_i$, with the effect of reducing $|c_i|$, without altering the fact that $\text{g.c.f.}(c_1,\ldots,c_n) = 1$.

Continuing by induction on the quantity $\sum_i |c_i|$, by a sequence of column operations one eventually reaches one of the standard unit coordinate vectors. By appropriately multiplying the elementary matrices corresponding to the sequence of column operations, one obtains an invertible integer matrix taking $(c_1,\ldots,c_n)$ to a standard unit coordinate vector. By applying the inverse of that matrix to a basis of standard unit coordinate vectors, one obtains a basis containing $(c_1,\ldots,c_n)$.

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  • $\begingroup$ Thank you so much! Sorry I accepted the other answer - this one is really helpful too $\endgroup$ Mar 11, 2022 at 12:00

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