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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.

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    $\begingroup$ 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. $R=\mathcal O$ $\endgroup$
    – KConrad
    Commented Nov 13, 2012 at 7:44
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    $\begingroup$ 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. $\endgroup$
    – Tommaso Centeleghe
    Commented Nov 13, 2012 at 11:00

3 Answers 3

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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.

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    $\begingroup$ 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.) $\endgroup$
    – LMN
    Commented Nov 13, 2012 at 15:02
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    $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Commented Nov 13, 2012 at 15:39
  • $\begingroup$ 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 $\mathbb{Z}$. $\endgroup$
    – Pete L. Clark
    Commented Jun 29, 2013 at 4:49
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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).

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

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  • $\begingroup$ 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 $\endgroup$
    – Will Sawin
    Commented Nov 13, 2012 at 6:48

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