# algebraic sum and product

Let $\alpha ,\beta $ be two elements of an extension field^{} of a given field $K$. Both these elements are algebraic over $K$ if and only if both $\alpha +\beta $ and $\alpha \beta $ are algebraic over $K$.

Proof. Assume first that $\alpha $ and $\beta $ are algebraic^{}. Because

$$[K(\alpha ,\beta ):K]=[K(\alpha ,\beta ):K(\alpha )][K(\alpha ):K]$$ |

and both here are finite (http://planetmath.org/ExtendedRealNumbers), then $[K(\alpha ,\beta ):K]$ is finite. So we have a finite field extension $K(\alpha ,\beta )/K$ which thus is also algebraic, and therefore the elements $\alpha +\beta $ and $\alpha \beta $ of $K(\alpha ,\beta )$ are algebraic over $K$. Secondly suppose that $\alpha +\beta $ and $\alpha \beta $ are algebraic over $K$. The elements $\alpha $ and $\beta $ are the roots of the quadratic equation ${x}^{2}-(\alpha +\beta )x+\alpha \beta =0$ (cf. properties of quadratic equation) with the coefficients in $K(\alpha +\beta ,\alpha \beta )$. Thus

$$[K(\alpha ,\beta ):K]=[K(\alpha ,\beta ):K(\alpha +\beta ,\alpha \beta )][K(\alpha +\beta ,\alpha \beta ):K]\leqq 2[K(\alpha +\beta ,\alpha \beta ):K].$$ |

Since $[K(\alpha +\beta ,\alpha \beta ):K]$ is finite, then also $[K(\alpha ,\beta ):K]$ is, and in the finite extension (http://planetmath.org/FiniteExtension) $K(\alpha ,\beta )/K$ the elements $\alpha $ and $\beta $ must be algebraic over $K$.

Title | algebraic sum and product |

Canonical name | AlgebraicSumAndProduct |

Date of creation | 2013-03-22 15:28:03 |

Last modified on | 2013-03-22 15:28:03 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 8 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 11R32 |

Classification | msc 11R04 |

Classification | msc 13B05 |

Synonym | sum and product algebraic |

Related topic | FiniteExtension |

Related topic | TheoryOfAlgebraicNumbers |

Related topic | FieldOfAlgebraicNumbers |