Projective and Injective Resolutions — Examples

The composite of multiplication by $n$ with the quotient modulo $n$ is zero. The kernel of $\mathbb{Z}\to \mathbb{Z}\mathrm{/}n\mathbb{Z}$ is exactly $n\mathbb{Z}$, which is the image of the left map, and the left map is injective. So the sequence is exact.

Both copies of $\mathbb{Z}$ are free abelian groups of rank one and are therefore projective by [L1]. Hence the exact sequence of step 1.1 is a projective resolution of $\mathbb{Z}\mathrm{/}n\mathbb{Z}$.

A vector $a{e}_0+b{e}_1$ maps to $\bar{b}$, so it lies in $\ker ({\epsilon }_0)$ exactly when $b$ is even. Therefore $\ker ({\epsilon }_0)=\mathbb{Z}{e}_0\oplus \mathbb{Z}(2e_1)\mathrm{.}$

The next free object in the canonical construction is therefore $F_1={\mathbb{Z}}^{(\ker ({\epsilon }_0))}$ together with its canonical surjection $F_1\twoheadrightarrow \ker ({\epsilon }_0)$. This makes concrete what [L1] does: the construction simply repeats the free-on-the-underlying-set cover on the current kernel, with no arbitrary choices. The same pattern starts even for the zero module.

The inclusion $\mathbb{Z}\hookrightarrow \mathbb{Q}$ is injective, its cokernel is $\mathbb{Q}\mathrm{/}\mathbb{Z}$, and both $\mathbb{Q}$ and $\mathbb{Q}\mathrm{/}\mathbb{Z}$ are divisible. Therefore both are injective by [L3], and the displayed sequence is exact.

Thus the sequence is already an injective resolution of $\mathbb{Z}$. It starts with an embedding into the divisible group promised by [L2], and it is a concrete instance of the general existence statement [L1].

Multiplication by $m$ commutes with multiplication by $n$ on $\mathbb{Z}$, so the square with the two differentials commutes. It also commutes with the quotient maps to $\mathbb{Z}\mathrm{/}n\mathbb{Z}$, because both routes send $x$ to $\overline{mx}$.

Therefore the degreewise multiplication-by-$m$ maps form an augmentation-preserving chain map between the two resolutions. This is the explicit comparison map whose existence is asserted abstractly in [L1].

Because $1+n\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}n)$ and $(1+n)n=n(1+n)$, the degreewise maps $f_i=1$ and $g_i=1+n$ are both augmentation-preserving chain maps lifting $1_{\mathbb{Z}\mathrm{/}n\mathbb{Z}}$. They are distinct as maps on $\mathbb{Z}$ whenever $n\mathrm{\ne }0$.

Let $s_0:\mathbb{Z}\to \mathbb{Z}$ be the identity map. Then $g_0-f_0=n=d_1{s}_0,g_1-f_1=n=s_0{d}_1,$ so $s_0$ is a chain homotopy from $f_{\bullet }$ to $g_{\bullet }$. This is the explicit homotopy predicted by [L1].

Exactness is checked componentwise: the kernel of the quotient map onto $\mathbb{Z}\mathrm{/}m\mathbb{Z}\oplus \mathbb{Z}\mathrm{/}n\mathbb{Z}$ is $m\mathbb{Z}\oplus n\mathbb{Z}$, which is exactly the image of the displayed differential, and that differential is injective.

The resolution in step 1.1 is the direct sum of the two side resolutions from [L2], so [L1] identifies it as the horseshoe resolution for this split extension.

The kernel of $\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\mathbb{Z}$ is $2\mathbb{Z}$, and the kernel of $(a,b)\mapsto \bar{a}$ is $2\mathbb{Z}\oplus \mathbb{Z}$. Thus the two displayed rows are short exact in the sense of [L2].

Since $2\mathbb{Z}\cong \mathbb{Z}$, both sides of Schanuel's conclusion are free abelian groups of rank three. Hence they are isomorphic, exactly as [L1] predicts.

The standard resolution $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\mathbb{Z}\to 0$ has first syzygy $2\mathbb{Z}\cong \mathbb{Z}$. The stabilized resolution $0\to \mathbb{Z}\oplus \mathbb{Z}\stackrel{(a,b)\mapsto (2a,b)}{\to }\mathbb{Z}\oplus \mathbb{Z}\to \mathbb{Z}\mathrm{/}2\mathbb{Z}\to 0$ is again projective and exact, and its first syzygy is $2\mathbb{Z}\oplus \mathbb{Z}\cong \mathbb{Z}\oplus \mathbb{Z}$.

The groups $\mathbb{Z}$ and $\mathbb{Z}\oplus \mathbb{Z}$ are not isomorphic, so the first syzygies of these two projective resolutions are not isomorphic. This is why [L2] stops at stable isomorphism.

Kernels, images, and cokernels of homomorphisms between torsion abelian groups are torsion again, so $\mathbf{T}\mathbf{o}\mathbf{r}\mathbf{A}\mathbf{b}$ is an abelian full subcategory of the abelian category from [L1]. If $A$ is torsion, [L3] embeds it into a divisible group $D$; the torsion subgroup $t(D)$ is still divisible and still contains $A$. Hence [L2] makes $t(D)$ injective, so $\mathbf{T}\mathbf{o}\mathbf{r}\mathbf{A}\mathbf{b}$ has enough injectives.

Suppose $P$ were a nonzero projective torsion group. Let $P^{\times }=P\setminus \{0\}$, and for each $x\in P^{\times }$ let $n_x$ be the order of $x$. Form $G:={⨁}_{x\in P^{\times }}\mathbb{Z}\mathrm{/}{n}_x\mathbb{Z}$ with generators $e_x$, and define the surjection $\pi :G\to P$ by $\pi (e_x)=x$. Projectivity gives a section $s:P\to G$. Since $s(P)\mathrm{\ne }0$, some coordinate projection restricts to a nonzero map $P\to \mathbb{Z}\mathrm{/}{n}_x\mathbb{Z}$. Let $H$ be its nonzero cyclic image, write $\mathrm{\mid }H\mathrm{\mid }=d>1$, and regard the resulting map $u:P\twoheadrightarrow H\cong \mathbb{Z}\mathrm{/}d\mathbb{Z}$ as surjective.

Choose $x\in P$ with $u(x)=1\in \mathbb{Z}\mathrm{/}d\mathbb{Z}$. For each $n\ge 1$, projectivity lifts $u$ through the reduction $\mathbb{Z}\mathrm{/}{d}^n\mathbb{Z}\twoheadrightarrow \mathbb{Z}\mathrm{/}d\mathbb{Z}$ to a map $u_n:P\to \mathbb{Z}\mathrm{/}{d}^n\mathbb{Z}$. The element $u_n(x)$ is congruent to $1$ modulo $d$, hence is relatively prime to $d$ and has order $d^n$. But the order of $u_n(x)$ must divide the fixed finite order of $x$, impossible for arbitrarily large $n$.

Therefore $\mathbf{T}\mathbf{o}\mathbf{r}\mathbf{A}\mathbf{b}$ has enough injectives but no nonzero projective objects, so it does not have enough projectives. This refutes the statement.