Torsion-free abelian groups are flat

First every subgroup $K\le {\mathbb{Z}}^r$ is finite free, by induction on the finite integer $r$. The case $r=0$ is immediate. Project $K$ to the last coordinate. If its image is zero, apply induction in ${\mathbb{Z}}^{r-1}$. Otherwise let $d$ be the least positive integer in the image, using The well-ordering principle. Division with remainder by $d$, from Division with remainder in $\mathbb{Z}$: for $a\in \mathbb{Z}$ and $b>0$ there are unique $q,r\in \mathbb{Z}$ with $a=qb+r$ and $0\le r<b$, shows that every image element is a multiple of $d$: its remainder also lies in the image and cannot be positive and smaller than $d$. Choose one $u\in K$ with last coordinate $d$. Subtracting integer multiples of $u$ gives $K=\mathbb{Z}u\oplus \ker (\pi {\mathrm{\mid }}_K)$, and the kernel is a subgroup of ${\mathbb{Z}}^{r-1}$, so induction completes the proof. These are finitely many existential choices, not a choice function on an infinite family.

Let $f:A\to B$ be injective and $z={\sum }_{i=1}^t{a}_i\otimes m_i\in A\otimes M$ have zero image in $B\otimes M$. By The tensor product $M{\otimes }_{R}N$ from the additive group underlying the free $\mathbb{Z}$-module on $M\times N$, elementary tensors, and finite tensor sums, this equality to zero has a finite witness: the formal sum ${\sum }_i{e}_{(f(a_i),m_i)}$ is a finite integer linear combination of defining additivity and balancing relations in the free abelian group on $B\times M$.

For any subgroup $H\le M$ generated by a finite list $x_1,\dots ,x_v$, argue as follows. Choose a maximal integer-linearly-independent sublist $b_1,\dots ,b_r$; there are only finitely many sublists. For each $x_i$, maximality supplies an equation $d_i{x}_i={\sum }_j{a}_{ij}{b}_j$ with $d_i$ a positive integer: if $x_i$ is not already on the sublist, a dependence among it and the $b_j$ has nonzero coefficient at $x_i$, since the $b_j$ are independent. For a member of the sublist use $d_i=1$. The finitely many equations give $dH\subseteq G={\sum }_j\mathbb{Z}{b}_j$ for $d={\prod }_i{d}_i>0$, with empty product $1$. Multiplication by $d$ is injective on the torsion-free $H$, and $G\cong {\mathbb{Z}}^r$. Thus $H\cong dH$ is finite free by step 1.1. This also covers $H=0$ and $r=0$.

For one finite basis $h_1,\dots ,h_s$ of $H$ and any abelian group $T$, the relations of The tensor product $M{\otimes }_{R}N$ from the additive group underlying the free $\mathbb{Z}$-module on $M\times N$, elementary tensors, and finite tensor sums identify $T\otimes H$ with $T^s$: $a\otimes {\sum }_j{n}_j{h}_j\mapsto (n_{j}a{)}_j$, with inverse $(a_j)\mapsto {\sum }_j{a}_j\otimes h_j$. The composites are identities on elementary tensors and tuples. Therefore tensoring an injection with $H$ is the corresponding coordinatewise injection. No simultaneous choice of bases for all finitely generated subgroups is made.

Let $H$ be generated by all the elements of $M$ appearing in the original sum and in this finite witness. Every relation in the witness is then a valid relation in the tensor presentation for $B\otimes H$. Consequently the same sum $z_H\in A\otimes H$ maps to zero in $B\otimes H$. Step 3.1 implies $z_H=0$, so its image $z$ in $A\otimes M$ is zero.

Thus $-\otimes M$ preserves injections. It is right exact by Tensoring is right exact, whose commutative-ring hypothesis holds for $\mathbb{Z}$. It therefore preserves every short exact sequence and is exact, so $M$ is flat by Left and right flat modules over an arbitrary ring. The finite-rank and finite-witness arguments require no choice axiom.