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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.
Of course, this notion is meant to be invariant under isomorphism: it doesn’t depend on the left adjoint chosen. Thus, if a functor of the form $\hom_{Set}(S, U-): Ab \to Set$ is representable by an abelian group $A$, then we may say $A$ is a free abelian group on $S$. A specific choice of isomorphism
corresponds, via the Yoneda lemma, to a function $S \to U A$ which exhibits $S$, or rather its image under this function, as a specific basis of $A$. If $A$ is so equipped with such a universal arrow $S \to U A$, then it is harmless to call $A$ “the” free abelian group on $S$.
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 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. , 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. , of elements of $S$.
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:
The free abelian group of a Cartesian product $S \times Z$ of sets $S, T \,\in\, Sets$ is naturally isomorphic to the tensor product of the free abelian groups of the factors:
In other words, as a functor from Set to Ab
this is a strong monoidal functor.
In homotopy theory/algebraic topology, the free abelian group construction is frequently applied degreewise to a simplicial set $S_\bullet$ which is (or is weakly equivalent to) the singular simplicial complex of a topological space — because the resulting chain complex (under the Dold-Kan correspondence) then computes the ordinary singular homology of that space. But since
the product of simplicial sets is degreewise that in Set,
the tensor product of simplicial abelian groups is degreewise the tensor product of abelian groups
the above Prop. implies that also the functor
from sSet to sAb is strong monoidal.
\bgein{proposition} \label{SubgroupsOfFreeAbelianGroupsAreFree} Assuming the axiom of choice, then every subgroup of a free abelian group (def. ) is itself a free abelian group. \end{proposition} (e.g. Lang 02, Appendix 2 §2, page 880) For a full proof see at principal ideal domain this theorem.
Prop. 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 $R$.
Textbook accounts:
Last revised on April 26, 2023 at 10:44:10. See the history of this page for a list of all contributions to it.