Why is $\text{Aut}(\overline{K}(x)\mid K(x))$ equal to $\text{Aut}(\overline{K}\mid K)$?

Question

Let $\overline{K}$ be an algebraic closure of $K$. I saw in a paper that $\text{Aut}(\overline{K}(x)\mid K(x))$ is equal to $\text{Aut}(\overline{K}\mid K)$, but there was no explanation why this happens. I think that it is enough to prove that if $\sigma\in \text{Aut}(\overline{K}(x)\mid K(x))$ then $\sigma(\overline{K}) = \overline{K}$. Is it true? How can I prove that $\text{Aut}(\overline{K}(x)\mid K(x))$ is equal to $\text{Aut}(\overline{K}\mid K)$ (or maybe isomorphic)?

Answer 1

Any field automorphism $ \sigma $ of $ \bar K(x) $ fixing $ K $ will send elements algebraic over $ K $ to elements algebraic over $ K $, since it will preserve the polynomial equation satisfied by those elements over $ K $. This is equivalent to the assertion $ \sigma(\bar K) = \bar K $, from which the claim follows trivially.