# $\phi: K \to K$ be an $F$-embedding with ${tr.}\;{deg}(K/F)$ finite then $\phi$ is surjective.

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

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 $$K=\phi(K)$$.