# Descent of étale torsors

Let $$X$$ be a scheme over a field $$k$$, $$G$$ a finite abelian group of size invertible on $$X$$. Suppose $$K/k$$ is a Galois field extension and let $$Y\to X_K$$ be an étale $$G$$-torsor.

For what field extensions $$K/k$$ does any such $$Y$$ descend to $$k$$? I.e.: for what field extensions $$K/k$$ does there exist $$Y_0\to X$$ an étale $$G$$-torsor, such that $$(Y_0)_K\cong Y$$ and $$(Y_0)_K\to X_K$$ is the map $$Y\to X_K$$?

Equivalently, for what field extensions $$K/k$$ is the map $$H^1(X,G)\to H^1(X_K, G)$$ surjective?

For example, if $$k$$ is algebraically closed, for any extension $$K/k$$ of algebraically closed fields the map is an isomorphism, by the smooth base change theorem.

Example Let $$k$$ be the maximal unramified extension of $$\mathbf{Q}_p$$. Does any finite field extension $$K/k$$ satisfy this property?

• You are asking about the surjectivity of the map $H^1(X,G)\to H^1(X_K, G)$. Consider the case that $G=\mathbb{Z}/2\mathbb{Z}$, $X=$ Spec $\mathbb{Q}$, and $K$ is a quadratic number field. – Ariyan Javanpeykar May 8 '18 at 13:34
• @AriyanJavanpeykar I edited my question, to ask the one I am actually interested in. – user124171 May 8 '18 at 13:40
• Is smooth base change really needed to see that $H^1(k,G) = \{*\}$ if $k$ is algebraically closed? – Ariyan Javanpeykar May 8 '18 at 14:48

Let $L$ be a finite separable extension of $k$. Let $X$ be the Weil restriction from $L$ to $k$ of $\mathbb G_m$. Then for any field extension $K$ of $k$ (including $k$ itself),if $L \otimes_k K$ is a product of $n$ distinct fields, then $H^1(X_K, \mathbb Z/2) / H^1(K, \mathbb Z/2) = (\mathbb Z/2)^n$. This can be calculated by examining the Galois action on $H^1(X_{\overline{K}}, \mathbb Z/2)$.
Hence this descent only holds if $L \otimes_k K$ is a field, so that $n(K)= n(k)=1$. In particular, we must have this for all extensions $L$.
So the extension $K/k$ must be entirely inseparable and transcendental.
• By "$n$ is constant", do you mean $n=1$? And $(\mathbb{Z}/2)^n$ should probably be $(\mathbb{Z}/2)^{n-1}$. – Laurent Moret-Bailly May 8 '18 at 13:59
• @LaurentMoret-Bailly I don't see why it should be $n-1$. Consider the case $L=k$. – Will Sawin May 8 '18 at 15:02