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Let $G$ be a profinite group and let $A$ be a $G$-group. A torsor over $A$ is a non-empty $G$-set $P$ endowed with a simply transitive right action $\star$ of $A$ which is compatible with the left action of $G$, i.e.

$$\sigma \cdot (x \star a) = (\sigma \cdot x) \star (\sigma \cdot a), \, \textrm{for all} \, \sigma \in G, a \in A, x \in P.$$

A homomorphism between torsors is a map that is equivariant with respect to the left action of $G$ and the right action of $A$. Any homomorphism between torsors is automatically an isomorphism.

What I want to show is that the isomorphism classes of torsors over $A$ are in bijective correspondence with $H^{1}(G,A)$.

I am able to fill in most of the details in the proof given by Serre in his book on Galois Cohomology but there is one thing I'm having trouble with.

Let $\alpha \in \mathrm{Z}^{1}(G,A)$ be be a 1-cocycle. We construct a torsor $P_{\alpha}$ over $A$ by endowing $A$ with a left action of $G$ given by $(\sigma,x) \mapsto \alpha_{\sigma} (\sigma \cdot x)$ and a right action of $A$ given by $(x,a) \mapsto xa$.

I then want to show that if two 1-cocycles $\alpha$ and $\alpha'$ are cohomologous, then $P_{\alpha} \simeq P_{\alpha'}$. I tried to do so by constructing a homomorphism of torsors $f: P_{\alpha} \rightarrow P_{\alpha'}$ (since any homomorphism of torsors is an isomorphism). I feel like the crucial part here is the fact that this homomorphism needs to be equivariant with respect to the left action $(\sigma,x) \mapsto \alpha_{\sigma} (\sigma \cdot x)$ of $G$. Writing down the equivariant condition unfortunately didn't give me any insight on how to define my homomorphism and I am stuck.

Any help and comments would be highly appreciated.

Edit: what I find is that we should have $f(\alpha_{\sigma})= \alpha'_{\sigma}$ (equivariance with respect to the left action) but this doesn't help me. If I am not mistaken, equivariance with respect to the right action would yield the identity which seems completely off.

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    $\begingroup$ The first 2 sections of Chapter X of Silverman's Arithmetic of Elliptic Curves I gives quite a detailed and (I think) accessible introduction to this, filling in most of the details $\endgroup$ Sep 23, 2021 at 22:17
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    $\begingroup$ I find the story much clearer in terms of homotopy theory in the case that $G$ is a discrete group, which could be generalized to profinite groups by passing to condensed mathematics. For the discrete case, see the nLab page: ncatlab.org/nlab/show/… $\endgroup$
    – Yai0Phah
    Sep 23, 2021 at 22:26

1 Answer 1

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I was working too late last night. Here is the solution to my problem.

Since $\alpha$ and $\alpha'$ are cohomologous, there exists $a \in A$ such that $\alpha'_{\sigma} = a \alpha_{\sigma} (\sigma \cdot a^{-1})$ for all $\sigma \in G$.

For $f: P_{\alpha} \rightarrow P_{\alpha'}$ to be a morphism of $G$-torsors, we need to have $f(\sigma x) = \sigma f(x)$, where $\sigma x$ is the left action of $G$ on $A$ defined by $\sigma x= \alpha_{\sigma} (\sigma \cdot x)$.

Computing both sides gives

$f(\sigma x) = f(\alpha_{\sigma} (\sigma \cdot x))$ and $\sigma f(x) = \alpha'_{\sigma} (\sigma \cdot f(x)) = (a \alpha_{\sigma} (\sigma \cdot a^{-1}))(\sigma \cdot f(x))$.

We thus want to construct a map $f$ such that $f(\alpha_{\sigma} (\sigma \cdot x))=(a \alpha_{\sigma} (\sigma \cdot a^{-1}))(\sigma \cdot f(x))$. It is now easy to see that setting $f(x) = ax$ gives the desired equality. It is also easy to check that $f$ is invariant with regard to the right action $(x,a) \mapsto xa$.

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