Extensions of an algebraically closed field

Question

First some definitions: A field $k$ is algebraically closed if every non-constant polynomial $f(x) \in k[x]$ has a zero in $k$.Then if I am adding another restriction that if the field is also perfect, then why does this field $k$ has no non-trivial finite separable extensions or no non-trivial finite Galois extensions?

Answer 1

Let $F$ be a finite extension of $k$ and $x\in F$. Since $F|k$ is finite, there is a non-zero $f\in k[X]$ such that $f(x)= 0$. Since $k$ is algebraically closed, $f$ has all its roots in $k$, so $x\in k$. Therefore, $F= k$