Higher Tor over the integers vanishes

The canonical free cover $\epsilon :{\mathbb{Z}}^{(A)}\twoheadrightarrow A$ exists by Every module is a quotient of a free module. The integers are a PID: a nonzero ideal has a least positive integer $d$ by The well-ordering principle, and 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 element is a multiple of $d$; the zero ideal is principal as well.

For any supplied length-one projective resolution $P_{\bullet }\to A$, the tensor complex $P_{\bullet }\otimes B$ has zero terms in every degree $i\ge 2$. Its homology in those degrees is therefore zero. This proves the conditional choice-free clause and the corresponding right-resolution construction of Tor from a projective resolution of the right module.

Write $F={\mathbb{Z}}^{(A)}$ and $K=\ker \epsilon$. Under The Axiom of Choice, well-order its standard basis index set $A$ by The well-ordering theorem. For an index $a$ write $F_{\le a}$ and $F_{<a}$ for the spans of the indicated initial segments. Let $I_a\le \mathbb{Z}$ be the image of $K\cap F_{\le a}$ under the $a$-coordinate map. The integer calculation of step 1.1 gives $I_a=d_a\mathbb{Z}$ with $d_a>0$ if it is nonzero.

For every $a$ with $I_a\mathrm{\ne }0$, the set of vectors $x\in K\cap F_{\le a}$ having $a$-coordinate $d_a$ is nonempty. AC chooses one such $x_a$ for every such index at once. Each vector has finite support by The free module on a set and its standard basis. These simultaneous lifts and the preceding well-order are the two uses of AC in kernel freeness.

Use Transfinite induction on the basis order to prove that $K\cap F_{\le a}$ is spanned by the $x_b$ with $b\le a$. For $y$ in that subgroup, if $I_a=0$ its $a$-coordinate is zero; otherwise write that coordinate as $c{d}_a$ and subtract $c{x}_a$. The resulting vector lies in $K\cap F_{<a}$. If it is nonzero, its finite support has a greatest index $b<a$, and the induction hypothesis at $b$ applies. A zero remainder needs no terms. This proves the assertion at least, successor, and limit positions alike. Every nonzero member of $K$ has a greatest support index, so the chosen family spans $K$.

In a nonzero finite integer relation among the $x_a$, take the greatest index $a$ with nonzero coefficient $c_a$. Its $a$-coordinate is $c_a{d}_a\mathrm{\ne }0$, since every lower-index vector has zero coordinate there. This contradicts the relation. Thus the family is independent and is a basis of $K$; the empty family correctly handles $K=0$.

By Free modules are projective, with the exact choice boundary, AC makes both $K$ and $F$ projective. Hence $0\to K\to F\to A\to 0$ is a projective resolution of length at most one. This last use of AC selects preimages of arbitrary basis families in the free-projectivity theorem; the canonical free cover itself used no choice.

To obtain DC locally, given an entire relation $E$ on a nonempty set $X$ and $x_0\in X$, AC chooses $s(x)\in \{y:xEy\}$ for every $x\in X$; The recursion theorem then gives $x_{n+1}=s(x_n)$ from the prescribed $x_0$. This proves the exact DC hypothesis of The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain. The short resolutions from step 6.1 (also applied to $B$) and that DC hypothesis permit The balanced Tor bifunctor to identify the specified-resolution homology with the balanced groups ${\mathrm{Tor}}_i^{\mathbb{Z}}(A,B)$. Step 1.2 gives their vanishing for all $i\ge 2$. This accounts separately for the choice needed by the change-of-resolution interface.