# proof of transcendental root theorem

###### Proposition 1.

Let $F\mathrm{\subset}K$ be a field extension with $K$ an algebraically closed field. Let $x\mathrm{\in}K$ be transcendental over $F$. Then for any natural number^{} $n\mathrm{\ge}\mathrm{1}$, the element ${x}^{\mathrm{1}\mathrm{/}n}\mathrm{\in}K$ is also transcendental over $F$.

###### Proof.

Suppose $x$ is transcendental over a field $F$, and assume for a contradiction^{} that ${x}^{1/n}$ is algebraic over $F$. Thus, there is a polynomial^{} $P(y)\in F[y]$ such that $P({x}^{1/n})=0$ (note that the polynomial ${y}^{n}-x$ is not a polynomial with coefficients in $F$, so $P(y)$ might be more involved). Then the field $H=F({x}^{1/n})\subseteq K$ is a finite algebraic extension^{} of $F$, and every element of $H$ is algebraic over $K$. However $x\in H$, so $x$ is algebraic over $F$ which is a contradiction.
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Title | proof of transcendental root theorem |
---|---|

Canonical name | ProofOfTranscendentalRootTheorem |

Date of creation | 2013-03-22 14:11:41 |

Last modified on | 2013-03-22 14:11:41 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 6 |

Author | alozano (2414) |

Entry type | Proof |

Classification | msc 11R04 |

Related topic | AlgebraicElement |

Related topic | AlgebraicClosure |

Related topic | Algebraic |

Related topic | AlgebraicExtension |

Related topic | AFiniteExtensionOfFieldsIsAnAlgebraicExtension |