From: hale@mailhost.tcs.tulane.edu (William Hale)
Subject: Re: Number Fields (elementary level qn)
Date: Wed, 03 May 2000 16:46:35 -0500
Newsgroups: sci.math
Summary: Recognizing rings of integers in number fields

In article <8eq1tq$fl1$1@pegasus.csx.cam.ac.uk>, "Pete Bartlett"
<pcb21@cam.ac.uk> wrote:

> Could some kind soul please help me get a bit further with this homework
> problem.
> 
> I have shown O_k = Z[sqrt(2)] for k=Q(sqrt(2)) and that { 1, sqrt(2) } is an
> integral basis for k.
> 
> Now let K be the quartic field Q(alpha) where alpha = 4th-root(2).
> I have been asked to consider Trace_K/k (theta) and Trace_K/k (alpha*theta)
> where
> 
> theta = a + b.alpha + c.alpha^2 + d.alpha^3
> 
> Using the basis { 1 , alpha } of K/k I compute these to be
> 
> trace_K/k (theta) = 2a + 2b.sqrt(2)
> trace_K/k (theta*alpha) = 4d + 2c.sqrt(2)
> 
> Now I need to deduce that the algebraic integers in L are of form: theta = a
> + b.alpha + c.alpha^2 + d.alpha^3
> where 2a, 2b, 2c and 4d are (rational) integers.

I have the following diagram:

               K
              / \
             /   \
            k     O_K
           / \    /
          /   \  /
         Q     O_k
          \    /
           \  /
            Z

Note that k is the field of fractions of O_k.
Note that K is an extension of finite degree 2 of k.
Note that O_k is integrally closed.
Note that k has characteristic zero.
Then, by well known theorem, for any element x in O_K,
trace_K/k (x) lies in O_k.

In particular, let theta of yours be an element of O_K.
Then, trace_K/k (theta) lies in O_k.
But you calculated above that trace_K/k (theta) = 2a + 2b.sqrt(2).
Thus, 2a + 2b.sqrt(2) lies in O_k, which is spanned by { 1, sqrt(2) } over Z.
This implies that 2a and 2b are integers.

Similarly, trace_K/k (theta*alpha) = 4d + 2c.sqrt(2) implies 4d and 2c in Z.

>(in fact later in the
> problem I should be able to show that (using norms) 1,alpha,alpha^2,alpha^3
> form an integral basis for k but I haven't got that far yet!)
> 
> How do I do this? Certainly if theta is of this form then the traces I have
> calculated are in O_k

That's the key.

> and so the traces
> 
> trace_K/Q (theta)= trace_k/Q (Trace_K/k(theta)
> trace_K/Q (theta.alpha)= trace_k/Q (Trace_K/k(theta.alpha))
> 
> are both integers.

What do theses traces calculate out to be? I suspect that they provide
even less information about a, b, c, and d. Thus, this leads to a dead end.

> However this is a necessary for theta to be an algebraic
> integer. Is it also sufficient?

As you pointed out above, this is not sufficient. You are concluding
that 2*a is necessarily an integer. But, in fact, a itself must be
an integer, thereby ruling out a = 1/2 as being ok.

> Moreover if theta is an algebraic integer do
> I necessarily have to have that Trace_K/k (theta) is in O_k?

Yes, using a theorem that should have been proved by now.

--
Bill Hale