I am reading Dummit and Foote (Section 10.3):
Definition: An $R-$ module $F$ is said to be free on the subset $A$ of $F$ if for every non-zero $x$ of $F$, there exist unique non-zero elements $r_1, r_2, ..., r_n$ of $R$ and unique $a_1, a_2, ..., a_n$ in $A$ such that $x = r_1 a_1 + r_2 a_2 + \cdots + r_n a_n$ for some $n \in \mathbb{Z}^+$. In this situation we say that $A$ is a basis or a set of free generators for $F$.
The definition is silent about the element $0 \in F$. I would like to know if $0 = r_1 a_1 + r_2 a_2 + \cdots + r_n a_n$ with $r_i \in R$ and $a_i \in A$ forces $r_1 = r_2 = \cdots = r_n = 0$. I don't see really how to go about proving it; if I try contradiction and assume that some $r_j \not = 0$, then all I can think to do is subtract the other terms to the other side, which doesn't help. If $r_j a_j$ is not-zero (which isn't even guaranteed) then we have the unique linear cobination for the term $r_j a_j \in F$, but that' about it.
First of all, let me remark that this is not the standard definition of a basis and free module. The standard definition does not include the restriction that $x$ must be nonzero, and allows $n\in\mathbb{Z}_{\geq 0}$ instead of just $n\in\mathbb{Z}_+$ (when $n=0$, the "empty sum" $r_1 a_1 + r_2 a_2 + \cdots + r_n a_n$ is by definition $0$). With the standard definition, it is obvious that $A$ must be linearly independent, since that's exactly what the uniqueness of the representation says for $x=0$ (one representation is the empty sum with $n=0$, and any linear relation would give another representation).
Second, Dummit and Foote's definition is not equivalent to the standard definition, and does not imply that $A$ is linearly independent. Indeed, if $F=\{0\}$, then Dummit and Foote's definition vacuously holds for any subset of $F$, since it has no nonzero elements. In particular, $A=\{0\}$ is a "basis" for $F$ but is not linearly independent.
However, Dummit and Foote's definition implies $A$ is linearly independent (and the two definitions are equivalent) as long as $F$ is not the trivial module. Indeed, suppose $x\in F$ is any nonzero element. Then given any representation of $x$ as a linear combination of elements of $A$, and any linear relation between elements of $A$, we can add the two to get another representation of $x$, contradicting uniqueness.