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
Canonical name	EveryAlgebraicallyClosedFieldIsPerfect
Date of creation	2013-03-22 16:53:06
Last modified on	2013-03-22 16:53:06
Owner	polarbear (3475)
Last modified by	polarbear (3475)
Numerical id	6
Author	polarbear (3475)
Entry type	Result
Classification	msc 12F05