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[parent] every algebraically closed field is perfect (Result)
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 $p$ root in $K$ Since $a$ is arbitrary we have proved that the field $K$ is perfect. $ \qedsymbol$




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Cross-references: perfect, root, polynomial, characteristic, prime, field, algebraically closed

This is version 3 of every algebraically closed field is perfect, born on 2007-04-01, modified 2007-04-02.
Object id is 9139, canonical name is DerivationOfAlgebraicallyClosed.
Accessed 880 times total.

Classification:
AMS MSC12F05 (Field theory and polynomials :: Field extensions :: Algebraic extensions)

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