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 $F$.


  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?

Thank you in advance.

Reference: Abstract Algebra, Dummit and Foote, 3rd ed.


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:

$$ \bigoplus_{a \in A} Ra \to F, \\ (r_a)_{a \in A} \mapsto \sum_{a \in A} r_a a. $$

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  • $\begingroup$ 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? $\endgroup$ – Chris Nov 27 '18 at 1:34

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