The integers have global dimension one

Take the canonical surjection $\epsilon :{\mathbb{Z}}^{(A)}\to A$ and put $F={\mathbb{Z}}^{(A)}$, $K=\ker \epsilon$. The sequence $0\to K\to F\to A\to 0$ is exact. Under the stated AC hypothesis, [L2] makes $K$ free and both $K$ and $F$ projective. Hence this is a projective resolution of length at most one, and ${\mathrm{pd}}_{\mathbb{Z}}A\le 1$. No family of resolutions of all groups is selected: this construction applies separately to each $A$.

The group $\mathbb{Z}\mathrm{/}n\mathbb{Z}$ is not projective. Otherwise its identity map would lift through the quotient surjection $\pi :\mathbb{Z}\to \mathbb{Z}\mathrm{/}n\mathbb{Z}$ to a homomorphism $s$ with $\pi s=\mathrm{i}\mathrm{d}$. But $ns([1])=s(n[1])=s(0)=0$, and an integer annihilated by the positive integer $n$ is zero. Thus $s([1])=0$, contradicting $\pi s([1])=[1]\mathrm{\ne }0$ for $n>1$. Consequently [L4] and step 1.1 give ${\mathrm{pd}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}n\mathbb{Z})=1$, while step 1.1 bounds every projective dimension by one. Therefore the left global dimension is exactly one.

Left and right $\mathbb{Z}$-modules are both precisely abelian groups with the integer action given by repeated addition and additive inverses. Passing from $ma$ to $am=ma$ preserves the underlying homomorphisms, exact sequences, and lifting diagrams; thus it preserves projectivity and lengths of projective resolutions in both directions. Their two suprema are therefore equal, so the right global dimension is also one.