A set of generators is a $\mathbb{Z}-$basis for a free abelian group

Question

Question Let $A$ be a free abelian group of $n$ dimension and $\{a_1,\dots,a_n\}$ a set of generators of $A$.Show that $\{a_1,\dots,a_n\}$ is a $\mathbb{Z}-$basis of $A$.

Attempt It remains to show that $\{a_1,\dots,a_n\}$ is $\mathbb{Z}-$linear independent.So let $$\sum_{i=1}^nc_ia_i=0$$ for some $c_i\in\mathbb{Z}$.It would suffice to show that $c_i=0,\forall i$ but can't continue.

Is this correct so far?How could I continue?

Answer 1

Assume the contrary. Let $a_1, \ldots, a_n \in A$ and $c_1, \ldots, c_n \in \mathbb{Z}$ be such that (1) $a_1, \ldots, a_n$ generate $A$; (2) $\displaystyle\sum_{i=1}^{n}c_i a_i = 0$; (3) $\displaystyle\sum_{i=1}^{n}|c_i|$ is the smallest possible (and positive). Now if exactly one of the $c_i$'s, say $c_1$, is nonzero, then $a_1=0$ and $A$ is generated by $a_2, \ldots, a_n$ (contradicting the condition on its dimension). And if at least two are, then we may assume that $0 < c_1 \leqslant c_2$ (permuting and changing signs of $a_i$'s if necessary), but then $c_1(a_1 + a_2) + (c_2-c_1)a_2+\ldots+c_n a_n=0$, and $a_1+a_2, a_2, \ldots, a_n$ still generate $A$, contradicting (3).

Answer 2

We express each $a_j$ in terms of the standard generators $e_1, \ldots,e_n$ of $A$, as in $a_j=\sum_{i=1}^n u_{i,j}e_i$. Consider the square matrix $M=(u_{i,j})_{i,j=1}^n$ with the coordinate vectors. That the $a_i$ generate $A$ is equivalent to the row-rank of $M$ being $n$. But then the column-rank is $n$ as well, meaning that the columns are linearly independent (doesn't matter if we mean this over $\Bbb Q$ or over $\Bbb Z$). Hence for any vector $c=(c_i)_{i=1}^n$, $Mc=0$ implies $c=0$.