# algebraic

Let $K$ be an extension field^{} of $F$ and let $a\in K$.

If there is a nonzero polynomial^{} $f\in F[x]$ such that
$f(a)=0$ (in $K$) we say that $a$ is *algebraic over $F$*.

For example, $\sqrt{2}\in \mathbb{R}$ is algebraic over $\mathbb{Q}$ since there is a nonzero polynomial with rational coefficients, namely $f(x)={x}^{2}-2$, such that $f(\sqrt{2})=0$.

If all elements of $K$ are algebraic over $F$, one says that
the field extension $K/F$ is algebraic^{}.

Title | algebraic |
---|---|

Canonical name | Algebraic |

Date of creation | 2013-11-05 18:32:06 |

Last modified on | 2013-11-05 18:32:06 |

Owner | drini (3) |

Last modified by | pahio (2872) |

Numerical id | 8 |

Author | drini (2872) |

Entry type | Definition |

Classification | msc 13B05 |

Classification | msc 11R04 |

Classification | msc 11R32 |

Related topic | AlgebraicNumber |

Related topic | FiniteExtension |

Related topic | ProofOfTranscendentalRootTheorem |