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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?

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    $\begingroup$ 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. $\endgroup$ – Ariyan Javanpeykar May 8 '18 at 13:34
  • $\begingroup$ @AriyanJavanpeykar I edited my question, to ask the one I am actually interested in. $\endgroup$ – user124171 May 8 '18 at 13:40
  • $\begingroup$ Is smooth base change really needed to see that $H^1(k,G) = \{*\}$ if $k$ is algebraically closed? $\endgroup$ – Ariyan Javanpeykar May 8 '18 at 14:48
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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.

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    $\begingroup$ By "$n$ is constant", do you mean $n=1$? And $(\mathbb{Z}/2)^n$ should probably be $(\mathbb{Z}/2)^{n-1}$. $\endgroup$ – Laurent Moret-Bailly May 8 '18 at 13:59
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    $\begingroup$ @LaurentMoret-Bailly I don't see why it should be $n-1$. Consider the case $L=k$. $\endgroup$ – Will Sawin May 8 '18 at 15:02

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