# 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 |