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# 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.∎

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## Recent Activity

Jul 5

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo