Let $X$ be an algebraic variety defined over a perfect field $k$, let $\bar k$ be the algebraic closure of $k$, let $X_{\bar k}$ be the base change of $X$ over $\bar k$ and let $G ={\rm Gal}(\bar k/k)$ be the absolute Galois group. Is it true that the equality $$k(X) = \bar k(X_{\bar k})^G$$ holds. That is any $G$-invariant rational function is a rational function defined over $k$?
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1$\begingroup$ Yes. See p7 and exercise 1.12 of "Silverman - The arithmetic of elliptic curves". $\endgroup$– Daniel LoughranCommented Aug 9, 2016 at 16:19
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