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Let $K$ be an algebraically closed field and $F$ be a subfield of $K$ with ${tr.}\;{deg}(K/F)$ finite. If $\phi: K \to K$ be an $F$-embedding show that $\phi$ is surjective.

I know the result when ${tr.}\;{deg}(K/F)=0.$ I need some help to prove it in the more general case. Thanks

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Consider the sequence of extensions $K/\phi(K)/F$. Since $\newcommand\trdeg{\operatorname{tr.deg}}\trdeg K/F=n<\infty$, we have that $\trdeg\phi(K)/F=n$ as well, so $\trdeg K/\phi(K) =0$. However $\phi(K)\cong K$ is algebraically closed, so since $K$ is an algebraic extension of $\phi(K)$, it must be trivial. I.e., we have $K=\phi(K)$.

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