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The ring of integers $\mathcal{O}_K$ of a number field $K$ is always isomorphic to some ring of the form $\mathbb{Z}[x_1, ..., x_r]/\mathfrak{p}$, where $\mathfrak{p} \subset \mathbb{Z}[x_1, ..., x_r]$ is a prime ideal. I would appreciate any reference to a source where any of the following questions were discussed:

i) Which quotients $\mathbb{Z}[x_1, ..., x_r]/\mathfrak{p}$ do occur as rings of algebraic integers of number fields? ($\mathfrak{p}$ cannot be maximal; what other restrictions are there for $r$ and $\mathfrak{p}$?)

ii) What can be said about the minimal $r$ such that a given $\mathcal{O}_K$ is isomorphic to $\mathbb{Z}[x_1, ..., x_r]/\mathfrak{p}$? (It is at most $[K : \mathbb{Q}]$ and is sometimes larger than $1$; given $d$, do all values $1, 2, ..., d$ occur for some $K$ of degree $[K : \mathbb{Q}] = d$?)

iii) What can be said about the minimal number of generators of $\mathfrak{p}$, when $\mathcal{O}_K \cong \mathbb{Z}[x_1, ..., x_r]/\mathfrak{p}$ and $r$ is minimal? (Does it sometimes have to be "very large"?)

Thank you!

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    $\begingroup$ For (ii), see mathoverflow.net/questions/21267/…, including the comments. $\endgroup$
    – KConrad
    Commented Oct 9, 2014 at 10:36
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    $\begingroup$ For (i): Obvious restrictions for $p$ are: $\text{codim}(p)=1$; $p \cap \mathbb{Z}=0$; there is some $f \in p$ such that $f(0)\neq 0$; the content of each $f \in p$ is $1$. $\endgroup$
    – tj_
    Commented Oct 9, 2014 at 11:55
  • $\begingroup$ Thank you; I was not aware of the result of P.A.B. Pleasants that mostly clarifies ii). For i), many quotients of one-variable polynomial ring would fail to be rings of integers e.g., for reasons other that already mentioned by tj_; since question i) is not well-defined, I am glad for any remarks that makes the setting clearer. $\endgroup$
    – Albertas
    Commented Oct 9, 2014 at 13:34
  • $\begingroup$ 2 tj_ : the remark about content seems imprecise ($2f$ must be also be in $\mathfrak{p}$, since $\mathfrak{p}$ is an ideal). $\endgroup$
    – Albertas
    Commented Oct 9, 2014 at 13:43
  • $\begingroup$ @Albertas: Yes, it's indeed imprecise. The correct statement is that $p$ is generated by polynomials of content $1$. $\endgroup$
    – tj_
    Commented Oct 9, 2014 at 23:01

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