How to show that this function is an isomorphisn??

Question

Thank you all for helping me in the last question :)

the question is how to prove that the 'natural map' from k to K is an isomorphism? I checked that it is injective, but can't show it is surjective...Could anyone help me show it is surjective?

Answer 1

The (algebraic) extension $K/k$ is finite. Since $k$ is algebraically closed, $K$ must be $k$ itself.

Answer 2

$\pi$ is injective , hence $k$ is isomorphic to it's image $k'=\pi(k)$ which is a subfield of $K$, since $K$ is an algebraic over $k$, it is also algebraic over $k'$. Then $K=k'=\pi(k)$ since $k'$ is algebraically closed.