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free abelian group

Idea

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 $S$.

Definition

Definition

Let

$$U \colon Ab \longrightarrow Set$$

be the forgetful functor from the category Ab of abelian groups, to the category Set of sets. This has a left adjoint free construction:

$$\mathbb{Z}[-] \colon Set \longrightarrow Ab \,.$$

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.

Properties

In terms of formal linear combinations

Definition

A formal linear combination of elements of a set $S$ is a function

$$a : S \to \mathbb{Z}$$

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

$$a = \sum_{s \in S} a_s \cdot s \,.$$

In this expression one calls $a_s \in \mathbb{Z}$ the coefficient of $s$ in the formal linear combination.

Remark

Definition 2 of formal linear combinations makes sense with coefficients in any abelian group $A$, not necessarily the integers.

$$A[S] \coloneqq \mathbb{Z}[S] \otimes A \,.$$

Definition

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 $\mathbb{Z}$:

$$(\sum_{s \in S} a_s \cdot s) + (\sum_{s \in S} b_s \cdot s) = \sum_{s \in S} (a_s + b_s) \cdot s \,.$$

Proposition

The free abelian group on $S \in Set$ is, up to isomorphism, the group of formal linear combinations, def. 3, on $S$.

Proposition

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:

$$\mathbb{Z}[S] \simeq \oplus_{s \in S} \mathbb{Z} \,.$$

Subgroups

Proposition

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.

Remark

Prop. 2 implies that (assuming AC) every abelian group admits a free resolution of length 2, hence with trivial syzygies. See there.

Examples

• 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 $R$.

Related concepts

• free group

References

• Serge Lang Algebra, Graduate Texts in Mathematics 211 (Revised third ed.), Springer 2002