I exemplify the previous question On ${\Bbb Z}/m{\Bbb Z}$-torsors. Let $k$ be a field of charactersitic $p$ and suppose that one has a character $\chi \colon {\mathrm{Gal}}(\overline{k}/k) \to {\Bbb Z}/p{\Bbb Z}$. Concretely, the character $\chi$ is determined by the value of Frobenius transformation ${\mathrm{Frob}}_p$, so we may assume ${\chi}({\mathrm{Frob}}_p) = a \in {\Bbb Z}/p{\Bbb Z}$.
On the other hand, we may reinterpret this $\chi$ by choosing some $x \in {\overline{k}}$ so that $$ \chi({\mathrm{Frob}}_p) = {\mathrm{Frob}}_p(x) - x = x^p - x = a. $$ Once we have this interpretation for $\chi$, we can immediately obtain the ${\Bbb Z}/p{\Bbb Z}$-torsor $({\Bbb Z}/p{\Bbb Z})_{\chi} = {\mathrm{Spec}}\,{\cal O}_{\chi}$ over ${\mathrm{Spec}}\,k$ as follows$\colon$
$$ {\cal O}_{\chi} \colon= k[X]/(X^p - X - a) \,\,\cdots \cdots \,(\lozenge). $$ Actually after base change over the field $k[\alpha]$ with $\alpha^p -\alpha = a$, we have $$ ({\Bbb Z}/p{\Bbb Z})_{\chi} \otimes_{k} k[\alpha] = {\mathrm{Spec}}\,k[X]/(X^p - X - a) \otimes_{k} k[\alpha] \cong {\mathrm{Spec}}\,k[X]/(X^p - X) = ({\Bbb Z}/p{\Bbb Z})_{{\mathrm{Spec}}\,k[\alpha]}. $$
I received the following explanation in the loc.cit. site:
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 $\mathcal{O}_\chi=k[G]$.