how ${\{1, -1}\}$ is the basis for $\mathbb{Z}$?

Question

In Adkin's textbook of Algebra, a/the basis is defined as:

$S \subset M$ is a basis of $M \ne {\{0}\}$ if and only if every $x \in M$ can be written uniquely as $$x = a_1 x_1 + \dots + a_n x_n$$ for $a_1, \dots, a_n \in R$ and $x_1, \dots, x_n \in S$. This definition implies that whenever there is an equation $$a_1 x_1 + \dots + a_n x_n = 0$$ where $x_1, \dots, x_n$ are distinct elements of $S$ and $a_1, \dots ,a_n$ are in $R$ then $$a_1 = \dots = a_n =0.$$

If so, how ${\{1, -1}\}$ is the basis for $\mathbb{Z}$ ? Of course, $0=a(1)+a(-1)$ for all $a\in R$.

Answer 1

$\mathbb Z$ is the linear span of $\{1,-1\}$ but $\{1,-1\}$ is linearly dependent subset of $\mathbb Z$. Hence it can't be called a basis of $\mathbb Z$ however $\{1\}$ is already a basis for $\mathbb Z$.

Answer 2

It is not a basis for the $\mathbb Z$-module $\mathbb Z$ for exactly the reason you gave. $\{1\}$ is already a basis.

If you provide more context, maybe we can determine that the problem you are looking at actually says something slightly different.