Flat modules need not have projective dimension zero

A nonzero integer cannot annihilate a nonzero rational number, so $\mathbb{Q}$ is torsion-free. The choice-free Torsion-free abelian groups are flat makes it flat over $\mathbb{Z}$.

Every homomorphism $f:\mathbb{Q}\to \mathbb{Z}$ is zero. For $q\in \mathbb{Q}$ and every positive integer $m$, additivity gives $f(q)=mf(q\mathrm{/}m)$. Thus the integer $f(q)$ is divisible by every positive integer. A nonzero integer $z$ cannot have this property: choose $m>\mathrm{\mid }z\mathrm{\mid }$; if $z=mk$ with integer $k\mathrm{\ne }0$, then $\mathrm{\mid }z\mathrm{\mid }=m\mathrm{\mid }k\mathrm{\mid }\ge m$, a contradiction. Hence $f(q)=0$ for every $q$.

The canonical cover $\epsilon :{\mathbb{Z}}^{(\mathbb{Q})}\twoheadrightarrow \mathbb{Q}$ sends $e_q$ to $q$, by Every module is a quotient of a free module. If $\mathbb{Q}$ were projective, its lifting property in Projective modules and the lifting property would give a section $s:\mathbb{Q}\to {\mathbb{Z}}^{(\mathbb{Q})}$ with $\epsilon s={\mathrm{id}}_{\mathbb{Q}}$. For each index $q$, coordinate projection ${\pi }_q$ is a homomorphism to $\mathbb{Z}$ by The free module on a set and its standard basis, so ${\pi }_{q}s=0$ by step 1.2. All coordinates of every $s(x)$ are zero; hence $s=0$, contradicting $\epsilon s(1)=1$. Thus $\mathbb{Q}$ is not projective, without assuming projective abelian groups are free.

By Projective dimension of an object, a length-zero projective resolution exists exactly for a projective module. Step 2.1 rules it out for the flat module in step 1.1, refuting the claim. No arbitrary projective-resolution existence is asserted in the choice-free conclusion.