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Consider the following situation. Let $k$ be a characteristic $0$ field, and consider an étale morphism of $k$ schemes $f:X\rightarrow Y$. Moreover, let $K$ and $L$ be two extension fields of $k$ such that $K\subseteq L$ is a finite Galois extension, and suppose we have an $L$-point of $X$, say $x\in X(L)$ such that its image under $f$ defines a point $y\in Y(K)$. Now consider the absolute Galois Group of $K$, say $G=\text{Gal}(\bar{K},K)$, and it's action on $X(\bar{K})$, sending a point $\bar{x}:X\rightarrow\text{Spec}(\bar{K})$ to $\sigma^{\ast}\circ\bar{x}$. Is it true that for each element $\sigma$ of the Galois group $G$, the image of $\sigma^{\star}\circ\bar{x}$ is again $y$? Thank you for any suggestion!

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I think I'm missing something as it seems the answer is yes to me: Since $f$ is defined over $k$ (and $\sigma$ leaves both $k \subseteq K$ invariant), $f(\sigma{\overline{x}})=\sigma (f(\overline{x}))$. And since $f(\overline{x})$ is a $K$-point, $\sigma$ leaves it invariant, i.e. $\sigma(f(\overline{x}))=f(\overline{x})=y$.

I didn't use $f$ is etale anywhere, or $k$ had characteristic $0$, so again, maybe I'm missing something.

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  • $\begingroup$ That's the point! Me too I argued as you did, but it sounded me strange! By the way everything seems ok. $\endgroup$ – Simone Jun 16 '16 at 21:27

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