Meaning of "free on the basis"

Question

T.Y. Lam in his book "A First Course in Noncommutative Rings" gave an example of "Hurwitz ring of integral quaternions" which is $$R=\lbrace(a+bi+cj+dk)/2 \mid a,b,c,d\in\mathbb{Z} \text{ are either all even or all odd}\rbrace.$$ Now he wrote that as an abelian group, $R$ is free on the basis $\lbrace(1+i+j+k)/2,i,j,k\rbrace$. I want to know that what is the meaning of "free on the basis".

Answer 1

Recall that given a set $S$, a free Abelian group on $S$ consists of an Abelian group $G$, together with a function $\eta : S \to G$, which is initial for such data. That is, given any Abelian group $H$ and function $h : S \to H$, there exists a unique group homomorphism $f : G \to H$ such that $f \circ \eta = h$.

An Abelian group $G$ is said to be free on basis $B \subseteq G$ if and only if $G$, together with the inclusion function $B \to G$, is a free Abelian group on $B$.

Note that if $G$ is free on $S$ with inclusion $\eta : S \to G$, then $G$ is free on the basis $\eta(S)$. Moreover, $\eta$ is injective, so we can identify $S$ with $\eta(S)$.

Answer 2

There is an abstract, category theoretic meaning explained in the other answer, and an equivalent meaning which says the following (in the general case).

First let me establish the terminology. We are given an abelian group $R$ and a subset $S \subset R$. I'll take $S$ to be finite, $S = \{s_1,...,s_n\}$. We want to define what it means to say that $R$ is free on the basis $S$. This means: for every $r \in R$ there exists a unique $n$-tuple of integers $a_1,...,a_n$ such that $$r = a_1 s_1 \, + \, ... \, + \, a_n s_n $$ So, in the special case of your question, what it means is that for each $r \in R$ there exists a unique 4-tuple $(a,b,c,d) \in \mathbb Z^4$ such that $$r = a\left(\frac{1+i+j+k}{2}\right) \, + \, b i \, + \, c j \, + \, d k $$ Motivated by your comment, let me explain some variations in the terminology.

Sometimes the phrase $G$ is free on the basis $S$ is abbreviated. For example, the word free is left out altogether and one simply says that $S$ is a basis for $G$; the meaning, however, is unchanged.

Also, one says that $G$ is a free abelian group if there exists a subset $S \subset G$ such that $G$ is free on the basis $S$.

So, as you can see, the terminology is a little redundant; the term "free" and the term "basis" have tight logical connections.

Regarding your comment, there is no such thing as a basis which is not free; you could, though, talk about a subset which is not a free basis.