Are all cyclic groups $\mathbb{Z}/n\mathbb{Z}$ cotorsion?

Question

Is it correct that all cyclic groups $\mathbb{Z}/n\mathbb{Z}$ are cotorsion, i.e., $Ext^1(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})=0$?

I believe yes, since $Ext^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z}/n\mathbb{Z})\simeq\mathbb{Z}/n\mathbb{Z}$, so I suspect $\mathbb{Z}/n\mathbb{Z}$ is a reduced cotorsion group (it is pretty clear it is reduced.) Yet I couldn't find this result online, and it's resisting my standard tricks since $\mathbb{Q}$ is not projective, and $\mathbb{Z}/n\mathbb{Z}$ is not injective, etc.

Answer 1

An easier proof only uses the fact that $\operatorname{Ext}^1(-,-)$ is an additive functor in each argument.

Consider how multiplication by $n$ acts on $\operatorname{Ext}^1(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})$.

On the one hand, multiplication by $n$ acts as an isomorphism on the first argument $\mathbb{Q}$, and so as an isomorphism on $\operatorname{Ext}^1(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})$.

On the other hand, multiplication by $n$ acts as zero on the second argument $\mathbb{Z}/n\mathbb{Z}$, and so as zero on $\operatorname{Ext}^1(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})$.

Which is a contradiction unless $\operatorname{Ext}^1(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})=0$

Answer 2

I belive it is so. The following reasoning uses, however, the fact that $\mathrm{Ext}_\mathbb{Z}^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z}/n\mathbb{Z}) \simeq \mathbb{Z}/n\mathbb{Z}$, which I do not know how to obtain.

Observe that $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})=0$: Whenever $(p/q)$ is mapped to some $m$, then $(p/nq)$ is mapped to some $k$ and it holds that $0=nk=m$ (in $\mathbb{Z}/n\mathbb{Z}$).

Now, consider a short exact sequence

$$0\rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow 0 \;\; .$$

Aplying $\mathrm{Hom}(-,\mathbb{Z}/n\mathbb{Z}),$ one obtains a long exact sequence

$$0 \rightarrow\mathrm{Hom}(\mathbb{Q}/\mathbb{Z},\mathbb{Z}/n\mathbb{Z})  \rightarrow \mathrm{Hom}(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})  \stackrel{\alpha}{\rightarrow}\mathrm{Hom}(\mathbb{Z},\mathbb{Z}/n\mathbb{Z})  \stackrel{\beta}{\rightarrow} \mathrm{Ext}_\mathbb{Z}^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z}/n\mathbb{Z})  \stackrel{\gamma}{\rightarrow} \mathrm{Ext}_\mathbb{Z}^1(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})  \stackrel{\delta}{\rightarrow} \mathrm{Ext}_\mathbb{Z}^1(\mathbb{Z},\mathbb{Z}/n\mathbb{Z})  \rightarrow \dots, $$

where:

1) $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})=0$, hence $\alpha=0$.

2) From exactness, $\beta$ is injective. Since $\mathrm{Hom}(\mathbb{Z},\mathbb{Z}/n\mathbb{Z})\simeq \mathbb{Z}/n\mathbb{Z} \simeq \mathrm{Ext}_\mathbb{Z}^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z}/n\mathbb{Z})$, it follows that $\beta$ is an isomorphism. In particular, it is surjective.

3) From exactness it follows that $\gamma=0$.

4) Since $\mathbb{Z}$ is projective, we have $\mathrm{Ext}_\mathbb{Z}^1(\mathbb{Z},\mathbb{Z}/n\mathbb{Z})=0$. Hence $\delta=0$.

5) Now it follows by exactness that $\mathrm{Ext}_\mathbb{Z}^1(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})=0$.

Answer 3

Multiplication by $n$ is an automorphism of this group (by functoriality in the first component), and at the same time zero (by functoriality in the second component). Therefore the group is zero.

Remarks: Remark that $\mathbf Q = \varinjlim \frac{1}{n}\mathbf Z$, where the limit is taken over the integers partially ordered by divisibility. Thus for any abelian group $G$, $$\mathrm{Hom}(\mathbf Q, G) = \varprojlim \mathrm{Hom}(\frac{1}{n}\mathbf Z, G) = \varprojlim \mathrm{Hom}(\mathbf Z, G),$$

where the arrows in the last limit have been replaced by the appropriate multiplication maps. Therefore $\mathrm{Hom}(\mathbf Q, G)$ consists of collections $$(g_n)_{n\geq 1} \in \prod_{n=1}^\infty G$$

such that $n g_{nm} = g_m$ for any $(n,m)$. Let us write $\widetilde{G}$ for $\mathrm{Hom}(\mathbf Q, G)$. We can prove:

Let $f: G \to H$ be surjective. Then $\widetilde{G}\to \widetilde{H}$ is surjective if and only if for every $h\in H$ which is a divisible element (i.e. one which is contained in $nH$ for all $n\geq 1$), there exists a divisible $g\in G$ with $f(g) = h$.

One direction of the equivalence is easy, and the other uses a construction by Zorn's lemma.

In particular, this gives examples of nontrivial extensions of $\mathbf Q$. For instance, the extension $$0 \to K \to \bigoplus\frac{1}{n}\mathbf Z \to \mathbf Q \to 0$$ is nontrivial.

The group $\mathrm{Ext}^1(\mathbf Q, \mathbf Z)$ is even uncountable (in fact it's isomorphic to $\mathbf A_f/\mathbf Q$, where $\mathbf A_f$ is the ring of finite adèles).