# Alternative definition on free modules.

I 'm trying to compare Free Modules and Free Abelian Groups.

We know that,

Definition. An abelian group $$G$$ is called free abelian group with rank $$n\in \Bbb N^*$$, if $$G$$ is the direct sum of $$n$$ infinite cyclic groups. That is, $$G=\langle x_1 \rangle \oplus ... \oplus \langle x_n \rangle$$ where $$U_i:=\langle x_i \rangle, \ i=1,...,n$$ are infinite cyclic groups. Then, $$x_1,...,x_n$$ are called free generators of $$G$$, $$X:=\{x_1,...,x_n\}$$ is called basis (or free generators set) of $$G$$ and $$n$$ is called rank of $$G$$.

We already know the following definition that seems equivalent.

Definition. Let $$R$$ be a ring with $$1_R$$. An $$R$$-module $$F$$ is called free on a subset $$A$$ of $$F$$, if for every nonzero element $$x\in F$$, there exist unique nonzero elements $$r_1,...,r_n\in R$$ and unique $$a_1,...a_n \in A$$ such that $$x=r_1a_1+...+r_na_n$$ for some $$n\in \mathbb{Z}^+$$ In this situation we say $$A$$ is a basis or set of free generators for $$F$$. If $$R$$ is commutative ring, the cardinality of $$A$$ is called the rank of $$F$$.

Questions.

1. Can we define similarly free modules (with the use of direct sum and with infinite cyclic modules)?

2. Is it useful to make such comparisons?

Abelian groups are precisely $$\mathbb{Z}$$-modules: the data of a $$\mathbb{Z}$$-module $$M$$ consists of an abelian group $$M$$ together with the specification of a ring homomorphism $$\eta: \mathbb{Z} \to \text{End}(M)$$, where the latter denotes the ring of group endomorphisms of $$M$$. But there is only one possible ring homomorphism from $$\mathbb{Z}$$ to any ring $$S$$, given by sending $$1 \in \mathbb{Z}$$ to the multiplicative unit in $$S$$, so $$\eta$$ above is uniquely specified hence adds no additional data to the abelian group structure of $$M$$.
Hence, free abelian groups are the same thing as free $$\mathbb{Z}$$-modules. The translation between the definition you gave for free $$R$$-modules and a formulation given in terms of direct sums can be obtained by observing that having a free $$R$$-module $$F$$ on a set of generators $$A \subset F$$ is equivalent to being able to write down an isomorphism of $$R$$-modules:
$$\bigoplus_{a \in A} Ra \to F, \\ (r_a)_{a \in A} \mapsto \sum_{a \in A} r_a a.$$
• Thank you for your answer, and sorry for the delay. So, an R-module F is called a free $R$-module if $F$ is isomorphic to a direct sum of copies of $R$: that is $$F\cong R_1 \oplus ... \oplus R_n \oplus ...$$ where $R_i=\langle x_i \rangle \cong R,\ \forall i\in \{1,2,...,n,...\}$. And $X:=\{x_1,...,x_n\}$ is called a basis of $F$. Correct? Commented Nov 27, 2018 at 1:34