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Suppose $K=\mathbf{Q}(\sqrt{d})$; it's well known that $\mathcal{O}_K$ is $\mathbf{Z}+\frac{D+\sqrt{D}}{2}\mathbf{Z}$, where $D$ is the discriminant.

What is the analogue of this for a CM extension $F(\sqrt{-\alpha})/F$, where $F$ is totally real (of class number one) and $\alpha \in \mathcal{O}_F$ is totally positive?

Is there a canonical element $\xi$ of $F(\sqrt{-\alpha})$ such that $\mathcal{O}_{F(\sqrt{-\alpha})}=\mathcal{O}_F+\xi \mathcal{O}_F$?

I have looked in the literature, but all I can find are various theorems guaranteeing the nonexistence of relative integral bases in various situations. However a theorem in Chapter 7 of Narkiewicz's book guarantees that some $\xi$ does exist in the above situation.

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What would make such an element $\xi$ canonical? Even in the case of a quadratic extension of $Q$, I'm aware that you can take $\xi = (D + \sqrt{D}) / 2$, but what is so special about that choice? Certainly $\xi = (-D + \sqrt{D}) / 2$ would work just as well. One possible answer to my question -- find an element $\xi$ whose trace is a totally negative generator of the discriminant ideal? Is this what you had in mind perhaps? –  Marty May 13 '10 at 16:11
    
@Marty: Yes, that is exactly the kind of thing I had in mind, stupid of me not to say it. –  David Hansen May 13 '10 at 16:31
    
David, can you settle the simpler question of finding a "nice" basis at all, before you try to pin down more properties that you would like? That is, can you give a basis in terms of some properties of alpha (chosen to be squarefree, which can be done without wrecking its total positivity)? The way 2 splits in the base ring is presumably going to have an effect and I have doubts about whether you can really hope to get a clean uniform answer here. –  KConrad May 13 '10 at 19:34
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Is there a reason for wanting the trace of the mystery element to be totally negative other than it being one way Marty saw to recover D from (D + sqrt(D))/2? That is, David, would this trace property actually be useful for whatever you have in mind? –  KConrad May 13 '10 at 19:36
    
Keith, I am looking for an answer of the form "xi = a_i*sqrt(-alpha) + b_i works, where a_i and b_i depend in some explicit way on alpha and on how 2 splits in F". –  David Hansen May 13 '10 at 20:48
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I don't think that you will find an answer of the form $\xi = a_i \sqrt{-\alpha} + b_i$; not only must the coefficients depend on the way $2$ splits in the field, they also depend on the residue class of $\alpha$ mod $4$ (or some higher power of $2$ if $\alpha$ has even norm). In addition you probably should assume that $\alpha$ is squarefree. On the other hand I do not see where the condition that the base field be CM should play a role.

My suggestion where to look for an answer points in a different direction. I recall that Hasse once complained that $\frac{−D+\sqrt{D}}2$ is a rather arbitrary generator of an integral basis of a quadratic field, and that in order to get a canonical generator one should look at quadratic Gauss sums modulo the discriminant. For number fields, there are "Hecke Gauss sums", which are studied in the last chapter of Hecke's algebraic number theory.

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