# every algebraically closed field is perfect

###### Proposition 1.

Every algebraically closed field is perfect

###### Proof.

Let $K$ be an algebraically closed field of prime characteristic $p$. Take $a\in K$. Then the polynomial $X^{p}-a$ admits a zero in $K$. It follows that $a$ admits a $p$th root in $K$. Since $a$ is arbitrary we have proved that the field $K$ is perfect.∎

Title every algebraically closed field is perfect EveryAlgebraicallyClosedFieldIsPerfect 2013-03-22 16:53:06 2013-03-22 16:53:06 polarbear (3475) polarbear (3475) 6 polarbear (3475) Result msc 12F05