Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $K$ be a number field over $\mathbb{Q}$ of degree $n$, and $\mathcal{O} \subset \mathcal{O}_K$ an order.

$\textbf{Questions:}$ $\newcommand{\Spec}{\textrm{Spec }}$ $\newcommand{\cO}{\mathcal{O}}$

1.) Is the natural map $\phi: \Spec \mathcal{O}_K \rightarrow \Spec \mathcal{O}$ flat?

2.) How many distinct primes (can) lie under a given prime? All but finitely many local rings of $\cO$ are canonically identified with local rings of $\cO_K$. Since it is not obvious to me that $\phi$ is surjective, how many more primes are in $\cO$ than in $\cO_K$ (w.r.t. the canonical identification above).

3.) In $\mathcal{O}_K$, and dedekind domains, prime ideals can be generated by two elements. How many elements are required to generate prime ideals of $\mathcal{O}$ ? Is it possible to give an answer depending on the degree $n = [K:\mathbb{Q}]$ and the index $[\mathcal{O}_K: \mathcal{O}]$?

4.) Is every ideal $I$ of $\cO$ also a proper $\cO$-ideal as is the case for the maximal order? That is, has ring of multipliers $R$ exactly $\cO$. (The rings of multipliers $R\subset K$ is the subring of elements $\alpha$ so that $\alpha \cdot I \subset I$). Certainly $\cO \subset R$.

$\textbf{Note:}$ If need be, feel free to assume that $K$ is quadratic imaginary. I'm primarily interested in this case, but I would like to have a clearer picture of the general situation.

share|improve this question
1  
For (4): if $I$ is an inv. frac. ${\mathcal O}$-ideal then $I$ is a proper ${\mathcal O}$-ideal. To say that the inv. frac. ${\mathcal O}$-ideals are the proper ${\mathcal O}$-ideals is equivalent to saying the ${\mathbf Z}$-dual ${\mathcal O}^\vee$ is an invertible fractional ${\mathcal O}$-ideal. This condition on the ${\mathbf Z}$-dual is satisfied if ${\mathcal O} = {\mathbf Z}[\alpha]$ for an $\alpha$, which covers all quad. orders, but if $[K:\mathbf Q] > 2$ there are inf. many orders ${\mathcal O}$ in $K$ containing a noninv. fractional ${\mathcal O}$-ideal s.t. $R=\mathcal O$ –  KConrad Nov 13 '12 at 7:44
2  
Check out Cox's book "Primes of the form x^2+ny^2". He talks a lot about imaginary quadratic orders, their invertible ideals and their class groups. Plus, it is a really beautiful book. –  Tommaso Centeleghe Nov 13 '12 at 11:00
add comment

3 Answers

2) I'm not sure what you mean by "lie under". Any number of equal-characteristic primes of $\mathcal O_K$ can map to a single prime of $\mathcal O$, but that's lying over, not under. The map on primes is surjective - for any local ring of $\mathcal O$, its integral closure is a local ring of $\mathcal O_K$.

3) They can be at least $n-1$. Take $p$ a totally split prime, and consider the subring of $\mathcal O_K$ of elements that are in $\mathbb Z$, modulo $p$. Then the primes lying over $p$ glue together into a single prime ideal, whose local ring is the inverse image of the diagonal $\mathbb F_p$ in the natural map $\mathbb Z_p^n \to \mathbb F_p^n$. If $m$ is the maximal ideal of this local ring, then $m/m^2 = \mathbb F_p^n = (R/m)^n$, so the ideal requires at least $n$ generators.

4) No. Certainly some ideals have ring of multipliers $\mathcal O_K$. in $\mathbb Z[\sqrt{-3}]$, say, the ideal $(2)$ has this property.

share|improve this answer
1  
Thanks, Will. Why do you say that any number of equal-characteristic primes of $\mathcal{O}_K$ can map to a single prime? (Also, The map on primes is also surjective since this is an integral extension so has going up and 0 is a contracted prime.) –  LMN Nov 13 '12 at 15:02
1  
To glue two primes, $\pi_1$ and $\pi_2$ together, consider the subring of $\mathcal O_K$ where the residue mod $\pi_1$ equals the residue mod $\pi_2$. You can do this whenever you can embed $\mathcal O_K/\pi_1$ and $\mathcal O_K/\pi_2$ in a common field, which you can do when they have the same characteristic. Then repeat this process to glue any number. –  Will Sawin Nov 13 '12 at 15:39
    
In 3), I'm confused why you say $n-1$, since it seems you are constructing an ideal which requires $n$ generators. Note also that any ideal in any order in a degree $n$ number field needs at most $n$ generators, just because the additive group of $\mathcal{O}$ is isomorphic to $\mathbb{Z}^n$ and an ideal is in particular a subgroup: now apply structure theory over the PID $\mathbb{Z}$. –  Pete L. Clark Jun 29 '13 at 4:49
add comment

1) No. The normalization of a ring $R$ is never flat over $R$, unless $R$ was normal in the first place.

share|improve this answer
    
Ah of course. I guess an example for an order in an imaginary quadratic field of index $n$ would be $0 \to \mathcal O/n \to \mathcal O/n^2 \to (\mathcal O/n)^2 \to 0$, which tensored with $\mathcal O_K$ is no longer exact –  Will Sawin Nov 13 '12 at 6:48
add comment

I can't figure out how to just comment, perhaps because I am a new user.

It seems to me like the example cited in Flatness of normalization should apply to 1).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.