On ${\Bbb Z}/m{\Bbb Z}$-torsors. I would like to know the explicit construction of ${\Bbb Z}/m{\Bbb Z}$-torsor $Y$'s over a scheme $X$. It is explained that $X$ are classified by $H_{et}^1(X, {\Bbb Z}/m{\Bbb Z})$, which is far from easy for me to understand. Moreover very recently, I found that I got confused about the difference of a torsor and a locally constant sheaf on a scheme $X$. It seems that a torsor is not necessarily a sheaf.
For instance, choose ${\Bbb Z}/2{\Bbb Z}$-torsor $Y$. As long as we consider the canonical isomorphism ${\Bbb Z}/2{\Bbb Z} \cong \mu_2$, we have $H_{et}^1(X, {\Bbb Z}/2{\Bbb Z}) \cong H_{et}^1(X, \mu_2)$. Especially for $X = {\mathrm{Spec}}\,k$ with a field $k$, $H_{et}^1(X, \mu_2) \cong H_{et}^1(\pi_1({\mathrm{Spec}} \,k), \mu_2) \cong k^{*}/{k^{*}}^2$. So for an element $a \in k^{*}/{k^{*}}^2$, the corresponding torsor $Y$ is defined as
\begin{equation}
Y \colon= {\mathrm{Spec}}\,k[X]/(X^2 - a) \phantom{A} \cdots \cdots \phantom{A}  (\lozenge),
\end{equation}
which turns out to be isomorphic to ${\mathrm{Spec}}\,k[X]/(X^2 - 1) \cong \mu_2$ over $k(\sqrt{a})$.
On the other hand, constant group scheme ${\Bbb Z}/3{\Bbb Z}$ over ${\mathrm{Spec}}\,k$ is defined as follows$\colon$
\begin{equation*}
{\Bbb Z}/3{\Bbb Z} \,\colon= {\mathrm{Spec}}\,(k \epsilon_0 \oplus k \epsilon_1 \oplus k \epsilon_2)/(\epsilon_i^2 = \epsilon_i,{\phantom{i}} \epsilon_i \epsilon_j = 0 \phantom{i} {\mathrm{for}}\,i \not= j) \phantom{A} \cdots \cdots \phantom{A}  (\lozenge),
\end{equation*}
where the comultiplication is given as $* \colon \epsilon_i \mapsto \underset{k + l = i}{\Sigma} \epsilon_k \otimes \epsilon_l$.
A ${\Bbb Z}/3{\Bbb Z}$-torsor $Y$ over ${\mathrm{Spec}}\,k$ is clasisfied by $H_{et}^1({\mathrm{Spec}}\,k, {\Bbb Z}/3{\Bbb Z}) \cong H_{et}^1(\pi_1({\mathrm{Spec}} \,k), {\Bbb Z}/3{\Bbb Z}) \cong {\mathrm{Hom}}(\pi_1({\mathrm{Spec}}\,k), {\Bbb Z}/3{\Bbb Z})$. Let us choose a character $\chi \colon \pi_1({\mathrm{Spec}}\,k) \to {\Bbb Z}/3{\Bbb Z}$.
Q. What is the explicit structure ring ${\cal O}_{\chi}$ which realise ${\Bbb Z}/3{\Bbb Z}$-torsor $Y \colon= {\mathrm{Spec}}\,{\cal O}_{\chi}$ corresponding to $\chi$ like $(\lozenge)$?
 A: There are two cases :

*

*either $\chi$ is trivial, in this case $\mathcal{O}_{\chi}=k\times k\times k$ whose spectrum is $\mathbb{Z/3Z}$,

*or $\chi$ is non trivial. In this case, it will necessarily be onto and its kernel is an open subgroup of $\operatorname{Gal}(\overline{k}/k)$ of index 3. By the fundamental theorem of Galois theory, it corresponds to an extension $L/k$ of degree 3. You then have $\mathcal{O}_\chi=L$.

Note that the extension $L/k$ do come with a co-action of $\mathbb{Z/3Z}$, in other word, $\operatorname{Spec}L$ has an action of $\mathbb{Z/3Z}$. This is given by the following :
first note that $\chi$ induces an isomorphism $$\overline{\chi}:\operatorname{Gal}(L/k)\simeq\operatorname{Gal}(\overline{k}/k)/\operatorname{Gal}(L/k)\to\mathbb{Z/3Z}$$ Write $\overline{\chi}^{-1}:\mathbb{Z/3Z}\to\operatorname{Gal}(L/k)$ for its inverse. Now consider the map $L\to \operatorname{Map}(\mathbb{Z/3Z},L)$ such that $l\mapsto (x\mapsto \overline{\chi}^{-1}(x)(l))$. I claim (but leave you the details) that this is indeed a co-action which makes $\operatorname{Spec} L$ a $\mathbb{Z/3Z}$-torsor.

Finally, I will give a construction which treats the two above cases in a uniform way. It is also necessary for $\mathbb{Z}/n\mathbb{Z}$-torsors where the character $\chi:\operatorname{Gal}(\overline{k}/k)\to\mathbb{Z}/n\mathbb{Z}$ is neither trivial nor surjective. Well, we can even consider the case of a $G$-torsor for a non-necessarily abelian but finite group. So we start with a character $\chi\in H^1_{et}(\operatorname{Spec}k,G)=\operatorname{Hom}(\operatorname{Gal}(\overline{k}/k),G)$.
Consider the following left $\operatorname{Gal}(\overline{k}/k)$-action on $\overline{k}[G]$ :

*

*the first one is the standard one, given by $\operatorname{Gal}(\overline{k}/k)\times \overline{k}[G]\to \overline{k}[G], \sigma.\lambda g= \sigma(\lambda) g$

*the second one comes from $\chi$, it is given by $\operatorname{Gal}(\overline{k}/k)\times\overline{k}[G]\to \overline{k}[G], \sigma.\lambda g=\lambda\chi(\sigma) g$.

We define $\mathcal{O}_\chi$ to be the invariants for the action.
$$\mathcal{O}_\chi=\operatorname{Eq}(\overline{k}[G]\rightrightarrows\operatorname{Map}(\operatorname{Gal}(\overline{k}/k),\overline{k}[G]) )$$
Finally, $\overline{k}[G]$ have a right co-action by $G$ which is compatible with the two $\operatorname{Gal}(\overline{k}/k)$-action. Hence it induces a right $\overline{k}[G]$-co-action on $\mathcal{O}_\chi$ making $\operatorname{Spec}\mathcal{O}_\chi$ a left $G$-torsor.
For instance, when $\chi$ is trivial, the second action is $\sigma.\lambda g=\lambda g$ so the two actions coincide on $\sum_g \lambda_g g$ iff for all $g\in G$ and all $\sigma, \lambda_g=\sigma(\lambda_g)$, in other words the two actions coincide iff for all $g\in G, \lambda_g\in k$.  Hence $\mathcal{O}_\chi=k[G]$.
You can also check that this construction gives the same thing as the first part when $\chi$ is a non-trivial character $\operatorname{Gal}(\overline{k}/k)\to\mathbb{Z/3Z}$.
