I already know that this is true for galois extensions of Q... how do you extend this result to any finite extension of Q?(adsbygoogle = window.adsbygoogle || []).push({});

I was thinking given a finite extension of Q, call it K... find a galois extension that includes the finite extension, call it L... then somehow use the fact that the ring of algebraic integers of L has an integral basis to show that the integers of K must have an integral basis.

By integral basis, I mean that there is a finite set of elements in the ring w1,w2,w3...wn such that an element in the ring can be written in the form a1w1+a2w2+a3w3+...anwn

where a1,a2,a3,...an belong to Z.

I'd appreciate any help. Thanks.

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Algebraic integers of a finite extension of Q has an integral basis

Loading...

Similar Threads - Algebraic integers finite | Date |
---|---|

I Semi-simple Lie algebra | Mar 10, 2018 |

I Can we construct a Lie algebra from the squares of SU(1,1) | Feb 24, 2018 |

I Splitting ring of polynomials - why is this result unfindable? | Feb 11, 2018 |

A≡b mod n true in ring of algebraic integers => true in ring of integers | Apr 10, 2012 |

Integer algebra homework | May 31, 2009 |

**Physics Forums - The Fusion of Science and Community**