# finite field cannot be algebraically closed

###### Theorem.

###### Proof.

The proof proceeds by the method of contradiction^{}. Assume that a field
$F$ is both finite and algebraically closed^{}. Consider the polynomial^{}
$p(x)={x}^{2}-x$ as a function from $F$ to $F$. There are two elements
which any field (in particular, $F$) must have — the additive identity
$0$ and the multiplicative identity^{} $1$. The polynomial $p$ maps both of
these elements to $0$. Since $F$ is finite and the function $p:F\to F$ is not one-to-one, the function cannot map onto $F$ either, so
there must exist an element $a$ of $F$ such that ${x}^{2}-x\ne a$ for
all $x\in F$. In other words, the polynomial ${x}^{2}-x-a$ has no root
in $F$, so $F$ could not be algebraically closed.
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Title | finite field cannot be algebraically closed |
---|---|

Canonical name | FiniteFieldCannotBeAlgebraicallyClosed |

Date of creation | 2013-03-22 16:29:09 |

Last modified on | 2013-03-22 16:29:09 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 6 |

Author | rspuzio (6075) |

Entry type | Theorem |

Classification | msc 12F05 |

Related topic | AlgebraicClosureOfAFiniteField |