Tor Flatness and Global Dimension — Examples

By [L1], tensoring with $\mathbb{Z}\mathrm{/}18$ gives the degree-one kernel of multiplication by $12$ on $\mathbb{Z}\mathrm{/}18$. The comparison with balanced Tor uses the stated AC hypothesis; no additional resolution-existence assertion is needed.

The congruence is equivalent to $3\mid 2x$, hence to $3\mid x$, since $x=2(2x)-3x$. The kernel residues modulo $18$ are therefore $0,3,6,9,12,15$.

The additive map $\mathbb{Z}\mathrm{/}6\to \mathbb{Z}\mathrm{/}18$, $[t{]}_6\mapsto [3t{]}_{18}$, is well defined because $3\cdot 6=18$, has image the listed kernel, and is injective because $18\mid 3t$ iff $6\mid t$. This proves the isomorphism and the choice-free kernel calculation.

Apply [L1] to the explicit right resolution of $\mathbb{Z}\mathrm{/}8$ and $M=\mathbb{Z}\mathrm{/}12$. Its tensor homology is the kernel of multiplication by $8$; AC is used only for the comparison giving balanced Tor. A left resolution, if used, is also explicitly supplied by the same multiplication construction for $12$.

The kernel residues satisfy $12\mid 8x$, equivalently $3\mid 2x$. Since $x=2(2x)-3x$, this is equivalent to $3\mid x$. Modulo $12$ the solutions are exactly $0,3,6,9$.

The map $\mathbb{Z}\mathrm{/}4\to \mathbb{Z}\mathrm{/}12$, $[t{]}_4\mapsto [3t{]}_{12}$, is a well-defined homomorphism, with image those four residues and kernel zero since $12\mid 3t$ iff $4\mid t$. It identifies $\mathbb{Z}\mathrm{/}4$ with the kernel in step 2.1, proving the claimed Tor calculation. The arithmetic uses no choice.

No nonzero integer annihilates a nonzero rational number, so $\mathbb{Q}$ is torsion-free.

The specific choice-free counterinstance in Flat modules need not have projective dimension zero proves that $\mathbb{Q}$ is not projective: every homomorphism $\mathbb{Q}\to \mathbb{Z}$ is zero, so a section of the canonical free cover would have every coordinate zero and could not be a section. Its proof does not assume arbitrary projective abelian groups are free.

By Torsion-free abelian groups are flat, step 1.1 makes $\mathbb{Q}$ flat. Together with step 1.2 this verifies the stated example with both premises supplied choice-free.

In the fraction definition Multiplicative subsets and the localisation $S^{-1}R$ as equivalence classes of fractions, $(a,2^k)$ and $(b,2^l)$ are equivalent precisely when some $2^j$ annihilates $2^{l}a-2^{k}b$. Since no nonzero integer is annihilated by $2^j$, this is equivalent to $2^{l}a=2^{k}b$. Therefore sending the class of $(a,2^k)$ to the rational $a\mathrm{/}{2}^k$ is a well-defined injective homomorphism with image $\mathbb{Z}[1\mathrm{/}2]$. This proves the asserted identification and makes $L$ torsion-free.

By Torsion-free abelian groups are flat, $L$ is flat. Thus tensoring with $L$ preserves short exact sequences, as in Left and right flat modules over an arbitrary ring. This proof of flatness uses no general localization-exactness or unproved fraction-tensor isomorphism.

Let $Q_{\bullet }\to N$ be a supplied projective resolution, and fix $i>0$. Write $Z_i=\ker d_i$ and $B_{i-1}=\mathrm{im}{d}_i$. Tensor the short exact sequence $0\to Z_i\to Q_i\to B_{i-1}\to 0$ with $L$. Tensoring the inclusion $B_{i-1}\to Q_{i-1}$ is also injective by flatness. Since $d_i$ factors through these two maps, $\ker (d_i\otimes 1)$ is the image of $Z_i\otimes L$ in $Q_i\otimes L$. Exactness of the resolution gives a surjection $Q_{i+1}\to Z_i$; tensoring preserves this surjection, so the same image is $\mathrm{im}(d_{i+1}\otimes 1)$. Thus $H_i(Q_{\bullet }\otimes L)=0$ in every positive degree. This proves the specified-resolution clause of Tor from a projective resolution of the right module, without dimension shifting or choice.

Under The Axiom of Choice, Higher Tor over the integers vanishes supplies projective resolutions of $N$ and $L$. 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 interface The balanced Tor bifunctor therefore identifies the homology computed in step 3.1 with the usual balanced Tor. Its vanishing holds for all $i>0$, not just for $i=1$.

In bidegrees $(p,q)$ with $p,q\in \{0,1\}$, the tensor of the two free rank-one groups is $\mathbb{Z}$.

The horizontal differential is multiplication by $m$. Under the supplied double-complex convention the vertical differential already includes the factor $(-1{)}^p$, so it is $(-1{)}^p$ times multiplication by $n$. The total differential is therefore $d_h+d_v$, with no second sign inserted.

Hence ${\mathrm{Tot}}_0=\mathbb{Z}$, ${\mathrm{Tot}}_1=\mathbb{Z}\oplus \mathbb{Z}$, and ${\mathrm{Tot}}_2=\mathbb{Z}$, where the degree-one summands are ordered $K_{1,0},K_{0,1}$. The total complex is $0⟶\mathbb{Z}\stackrel{c\mapsto (-nc,mc)}{\to }{\mathbb{Z}}^2\stackrel{(a,b)\mapsto ma+nb}{\to }\mathbb{Z}⟶0.$ The composite is $-mnc+nmc=0$, checking the sign and both axis labels. This also covers $m=1$ or $n=1$.

AC implies the DC required by [L2]: for an entire relation on a nonempty set, choose one successor at each point using AC and iterate that self-map from the prescribed start using The recursion theorem. This gives the required dependent-choice sequence. The two cyclic modules have the explicit resolutions in [L1], so no unspecified resolution-existence theorem is used.

By [L1], the first Tor group is the kernel of multiplication by $4$ on $\mathbb{Z}\mathrm{/}6$, namely $\{0,3\}$, and the reversed group is the kernel of multiplication by $6$ on $\mathbb{Z}\mathrm{/}4$, namely $\{0,2\}$. Each is cyclic of order two, in agreement with $\gcd (4,6)=2$.

Apply [L2] under step 1.1. Its termwise tensor swap and coherent comparison give an isomorphism between the two groups in step 2.1. An isomorphism between groups of order two must send the unique nonzero element to the unique nonzero element. Thus under the explicit kernel identifications it sends $[3{]}_6$ to $[2{]}_4$, so the concrete calculation agrees with the natural symmetry, not merely with equality of the two gcds.

A balanced tensor relation needs a right action on the first factor: $(xr)\otimes y=x\otimes (ry)$.

The notation ${}_{R}V$ specifies only $r\cdot v$ for the first copy; it supplies no value for $v\cdot r$. One could ask for additional right-module or bimodule data, but it is not part of the two given left modules.

Therefore the proposed tensor and Tor expressions have missing type data; this is a concrete handedness counterexample.

Under The Axiom of Choice, Every vector space has a basis supplies a basis for any $k$-module $V$, identifying it with the free module on that basis. By Free modules are projective, with the exact choice boundary, $V$ is projective: this second use of AC chooses preimages of basis values in each lifting problem. It is flat by Projective left and right modules are flat over an arbitrary ring. Thus $0\to V\stackrel{\mathrm{id}}{\to }V\to 0$ is both a length-zero projective and a length-zero flat resolution. For a supplied finite basis, the same free-projectivity theorem uses only finite choice, so these conclusions hold without AC.

Under AC, The integers have weak and global dimension one proves all four integer dimensions equal one, using actual length-one resolutions and the nonflat module $\mathbb{Z}\mathrm{/}2\mathbb{Z}$. No independent undeclared Tor or Ext calculation is needed here.

The definitions Projective dimension of an object and The flat dimension of a module assign dimension zero to the modules in step 1.1. Their suprema, as in Left and right global dimension of a ring and Left and right weak global dimension, are therefore zero. Because $k$ is commutative, the right-module assertion is the same statement with the scalar action written on the other side. Together with step 1.2 this gives all the claimed values.