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

1. Oct 13, 2007

### learningphysics

I already know that this is true for galois extensions of Q... how do you extend this result to any finite extension of Q?

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.

2. Oct 13, 2007

### Hurkyl

Staff Emeritus
Written uniquely, you mean.

3. Oct 13, 2007

### learningphysics

yes, you're right. it must be unique.

4. Oct 19, 2007

### mathwonk

this is not my forte, but doesn't finitely generated as an algebra plus integral imply finite as a module?

and a finite module over Z, is a direct sum of free ones and finite cyclic ones, but your module is torsion free hen ce free.

i do not see what galois has to do with it, unkless i am missing some hypotheses here.

i.e. take a finite field extension of Q, then take the elements which are integral over Z, i guess you need to know that is a finitely generated Z algebra to use my argument. is that the missing step?

dummitt and foote page 697 use your idea of embedding in a galois extension and taking traces to prove finite generation.

Last edited: Oct 19, 2007
5. Oct 20, 2007

### learningphysics

Hi mathwonk. This is an exercise from "Elements of Abstract Algebra" by Allan Clark. Almost at the end of it. :)

He gives a proof that a Galois extension of Q has a finite basis where the coefficients can be taken over Z... so this is what a Z-module is correct? Module is analogous to a vector space, but taking coefficients from a ring?

Then as an exercise he asks to prove that any finite extension of Q is a Z-module (he doesn't use the term module though). That's the reason I'm trying to take this approach.

He hasn't introduced modules in the text. He's about to introduce trace and norm in the next part...