# algebraic

Let $B$ be a ring with a subring $A$. An element $x\in B$ is algebraic over $A$ if there exist elements ${a}_{1},\mathrm{\dots},{a}_{n}\in A$, with ${a}_{n}\ne 0$, such that

$${a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\mathrm{\cdots}+{a}_{1}x+{a}_{0}=0.$$ |

An element $x\in B$ is transcendental over $A$ if it is not algebraic.

The ring $B$ is algebraic over $A$ if every element of $B$ is algebraic over $A$.

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

Canonical name | Algebraic1 |

Date of creation | 2013-03-22 12:07:50 |

Last modified on | 2013-03-22 12:07:50 |

Owner | djao (24) |

Last modified by | djao (24) |

Numerical id | 8 |

Author | djao (24) |

Entry type | Definition |

Classification | msc 13B02 |

Related topic | AlgebraicExtension |

Defines | transcendental |