# Extensions of an algebraically closed field

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?

• I think every algebraically closed field is perfect automatically. So that extra restriction shouldn't matter. Commented May 6, 2021 at 7:18
• This is not the definition of algebraically closed, but you want to prove this is equivalent to it. No polynomial $\in k[x]$ of degree $\ge 2$ is irreducible so there are no finite extensions. Also $f$ splits completely: it has a root $a$ so $f=(x-a)g$ then $g=(x-b)h$ and so on, proving that $k$ is algebraically closed. Commented May 6, 2021 at 7:19

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, $$F= k$$