The free abelian group $\mathbb{Z}[S]$ on a set $S$ is the abelian group whose elements are formal $\mathbb{Z}$-linear combinations of elements of $S$.
Let
be the forgetful functor from the category Ab of abelian groups, to the category Set of sets. This has a left adjoint free construction:
This is the free abelian group functor. For $S \in$ Set, the free abelian group $\mathbb{Z}[S] \in$ Ab is the free object on $S$ with respect to this free-forgetful adjunction.
Explicit descriptions of free abelian groups are discussed below.
A formal linear combination of elements of a set $S$ is a function
such that only finitely many of the values $a_s \in \mathbb{Z}$ are non-zero.
Identifying an element $s \in S$ with the function $S \to \mathbb{Z}$ which sends $s$ to $1 \in \mathbb{Z}$ and all other elements to 0, this is written as
In this expression one calls $a_s \in \mathbb{Z}$ the coefficient of $s$ in the formal linear combination.
Definition 2 of formal linear combinations makes sense with coefficients in any abelian group $A$, not necessarily the integers.
For $S \in$ Set, the group of formal linear combinations $\mathbb{Z}[S]$ is the group whose underlying set is that of formal linear combinations, def. 2, and whose group operation is the pointwise addition in $\mathbb{Z}$:
The free abelian group on $S \in Set$ is, up to isomorphism, the group of formal linear combinations, def. 3, on $S$.
For $S$ a set, the free abelian group $\mathbb{Z}[S]$ is the direct sum in Ab of ${|S|}$-copies of $\mathbb{Z}$ with itself:
Assuming the axiom of choice, then every subgroup of a free abelian group (def. 1) is itself a free abelian group.
(e.g. Lang 02, Appendix 2 §2, page 880) For a full proof see at principal ideal domain this theorem.
Prop. 2 implies that (assuming AC) every abelian group admits a free resolution of length 2, hence with trivial syzygies. See there.
The free abelian group on the singular simplicial complex of a topological space $X$ consists of the singular chains on $X$.
For $R$ a ring and $S$ a set, the tensor product of abelian groups $\mathbb{Z}[S] \otimes R$ is the free module over $R$ on the basis $S$. If $R = k$ is a field, then this is the vector space over $k$ with basis $S$.
For $R$ a ring, the tensor product of abelian groups $\mathbb{Z}[\mathbb{N}]\otimes R$ is the abelian group underlying the ring of polynomials over $R$.