# transcendental root theorem

Suppose a constant $x$ is transcendental over some field $F$. Then $\sqrt[n]{x}$ is also transcendental over $F$ for any $n\ge 1$.

###### Proof.

Let $\overline{F}$ denote an algebraic closure^{} of $F$. Assume for the sake of contradiction^{} that $\sqrt[n]{x}\in \overline{F}$. Then since algebraic numbers^{} are closed under multiplication^{} (and thus exponentiation by positive integers), we have ${(\sqrt[n]{x})}^{n}=x\in \overline{F}$, so that $x$ is algebraic over $F$, creating a contradiction.
∎

Title | transcendental root theorem |
---|---|

Canonical name | TranscendentalRootTheorem |

Date of creation | 2013-03-22 14:04:23 |

Last modified on | 2013-03-22 14:04:23 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 8 |

Author | mathcam (2727) |

Entry type | Theorem^{} |

Classification | msc 11R04 |