Tor Flatness and Global Dimension — Examples

Tensoring with $\mathbb{Z}\mathrm{/}18$ gives the degree-one kernel of multiplication by $12$ on $\mathbb{Z}\mathrm{/}18$.

The congruence $12x\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}18)$ has six solutions, namely the subgroup generated by $3$.

That subgroup is cyclic of order $6$, so the Tor group is $\mathbb{Z}\mathrm{/}6$.

Its kernel consists of residues $x$ with $8x\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}12)$.

These are $0,3,6,9$, the subgroup generated by $3$.

The subgroup has order $4$, which exhibits exactly the $8$-torsion detected by Tor.

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

Torsion-free modules over a PID are flat, hence $\mathbb{Q}$ is flat.

A projective abelian group is free, but a nonzero free abelian group has a nonzero homomorphism to $\mathbb{Z}$, whereas every homomorphism $\mathbb{Q}\to \mathbb{Z}$ is zero; therefore $\mathbb{Q}$ is not projective.

Localization is exact because an equality $x\mathrm{/}1=0$ is witnessed by some $2^{k}x=0$, and the same witness lifts exactness through a short exact sequence.

The natural map $N{\otimes }_{\mathbb{Z}}\mathbb{Z}[1\mathrm{/}2]\to S^{-1}N$, $n\otimes a\mathrm{/}{2}^k\mapsto an\mathrm{/}{2}^k$, is an isomorphism.

Thus $-\otimes \mathbb{Z}[1\mathrm{/}2]$ is exact, so the module is flat and its positive Tor groups vanish.

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 $n=1$.

The first group is $\mathbb{Z}\mathrm{/}\gcd (4,6)=\mathbb{Z}\mathrm{/}2$.

Interchanging the two integers leaves their gcd unchanged and realizes the tensor-factor swap.

Thus this concrete calculation agrees with the natural symmetry over a commutative ring.

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.

Every vector space is free and hence projective, so every $k$-module has projective and flat dimension $0$.

Every abelian group has a length-one free resolution, giving both dimensions of $\mathbb{Z}$ at most $1$.

The nonzero group ${\mathrm{Tor}}_1^{\mathbb{Z}}(\mathbb{Z}\mathrm{/}2,\mathbb{Z}\mathrm{/}2)$ gives the lower bound $1$ for weak dimension, and nonzero ${\mathrm{Ext}}^1$ gives the global lower bound.