Over a principal ideal domain flatness is equivalent to torsion-freeness

Suppose $M$ is flat and $0\mathrm{\ne }r\in R$. Since $R$ is a domain, multiplication by $r$ is an injection $R\to R$. By [L2], tensoring with $M$ preserves this injection. Under [L3] the resulting map is multiplication by $r$ on $M$, so $rm=0$ implies $m=0$. Thus $M$ is torsion-free.

Conversely, suppose $M$ is torsion-free and fix an ideal $I$. If $I=0$, then $I{\otimes }_{R}M=0$, so its map to $M$ is injective. If $I\mathrm{\ne }0$, choose a generator $r\mathrm{\ne }0$ using [L1]. The map $f:R\to I$, $a\mapsto ar$, is bijective by principality and cancellation in the domain, and its inverse is $R$-linear.

By [L3], $f\otimes 1$ identifies $R{\otimes }_{R}M$ with $I{\otimes }_{R}M$. The composite with the ideal multiplication map sends $a\otimes m$ to $arm$, so under $R{\otimes }_{R}M\cong M$ it is multiplication by $r$. This is injective by torsion-freeness. Since $f\otimes 1$ is an isomorphism, the ideal multiplication map is injective.

The ideal test [L2] now proves $M$ flat. Only one generator of one fixed ideal was chosen at a time; no choice function on all ideals, and no AC, is used. The zero module is included in both directions.