# Is the "basis" of a free module linearly independent?

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.

• If the $r_i$ were not all $0$ then $0$ would have two distinct representations, namely $0=r\cdot 0$ choosing any $r\in R$ different from the $r_i$ and $0 = r_1 a_1 + r_2 a_2 + \cdots + r_n a_n$ Oct 12, 2018 at 1:27
• @Matematleta In the definition, it say "for every non-zero $x \in F ...$ so if we take a module which is free according to this definition, we are not guaranteed that $0$ has a unique representation.
– Ovi
Oct 12, 2018 at 1:29
• $x$ is not assumed to be nonzero, but the scalars are. Oops nevermind after edit Oct 12, 2018 at 1:30
• @AndresMejia I'm sorry I see I forgot to write that in the definition, I edited the question; D&F's definition does say non-zero $x$. Do you think the definition is slightly wrong?
– Ovi
Oct 12, 2018 at 1:31
• @Matematleta Please see my comment above.
– Ovi
Oct 12, 2018 at 1:31

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.