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?

  • $\begingroup$ I think every algebraically closed field is perfect automatically. So that extra restriction shouldn't matter. $\endgroup$
    – Arthur
    Commented May 6, 2021 at 7:18
  • $\begingroup$ 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. $\endgroup$
    – reuns
    Commented May 6, 2021 at 7:19

1 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, $F= k$


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