Introduction

A prime ideal of a commutative ring $R$ is an ideal  $\mathfrak{p}$ such that $R/\mathfrak{p}$ is an integral domain, and an ideal  $\mathfrak{p}$ is maximal if $R/\mathfrak{p}$ is a field. Let $K=\mathbb{Q}(\alpha)$ be a number field, and let $\O _K$ be the ring of all elements of $K$ that are integral over  $\mathbb{Z}$. As you know, $\O _K$ is a Dedekind domain.

In order to employ our geometric intuition, we may view $\O _K$ as a one-dimensional ``scheme''

$$X = \Spec(\O _K) = \{$$ all prime ideals of $O_K$ $$\}$$

``over''

$$Y=\Spec(\mathbb{Z}) = \{ (0) \} \cup \{ p\mathbb{Z}: p \in\mathbb{Z}$$ is prime $$\}.$$

There is a natural map $f:X\rightarrow Y$ that sends a prime ideal $\mathfrak{p}\in X$ to $\mathfrak{p}\cap \mathbb{Z}\in Y$. For much more on this point of view, see [2, Ch. 2].

Ideals were originally introduced by Kummer because in rings of integers of number fields ideals factor uniquely as products of primes ideals, which is something that is not true for general algebraic integers. (The failure of unique factorization for algebraic integers was used by Liouville to ``destroy'' Lamé's purported 1847 ``proof'' of Fermat's Last Theorem.)

If $p\in\mathbb{Z}$ is a prime number, then the ideal $p\O _K$ of $\O _K$ factors uniquely as a product $\prod \mathfrak{p}_i^{e_i}$, where the $\mathfrak{p}_i$ are maximal ideals of $\O _K$. Viewed geometrically, the decomposition of $p\O _K$ into prime ideals in $\O _K$ is the same as the fiber $f^{-1}(p\mathbb{Z})$ (plus ramification data). We are concerned with how to compute $f^{-1}(p\mathbb{Z})$ in practice.

Example 1.1   The following MAGMA session shows the commands needed to compute the factorization of $p\O _K$ in MAGMA for $K$ the number field defined by a root of $x^5+7x^4+3x^2-x+1$.

> K<a> := NumberField(x^5+7*x^4+3*x^2-x+1);
> OK := MaximalOrder(K);
> I := 2*OK;
> Factorization(I);
[
<Principal Prime Ideal of OK
Generator:
[2, 0, 0, 0, 0], 1>
]
> Factorization(Discriminant(OK));
[ <5, 1>, <353, 1>, <1669, 1> ]
> J := 5*OK;
> Factorization(J);
[
<Prime Ideal of OK
Two element generators:
[5, 0, 0, 0, 0]
[2, 1, 0, 0, 0], 1>,
<Prime Ideal of OK
Two element generators:
[5, 0, 0, 0, 0]
[3, 1, 0, 0, 0], 2>,
<Prime Ideal of OK
Two element generators:
[5, 0, 0, 0, 0]
[2, 4, 1, 0, 0], 1>
]
> [K!OK.i : i in [1..5]];
[ 1, a, a^2, a^3, a^4 ]

Thus $2\O _K$ is already a prime ideal, and

$$5\O _K = (5,2+a)\cdot(5,3+a)^2\cdot(5,2+4a+a^2).$$

Notice that in this example $\O _K=\mathbb{Z}[a]$. But be warned--in general, one can not find an $a$ such that $\O _K=\mathbb{Z}[a]$ (see Example 4.2 below). When $\O _K=\mathbb{Z}[a]$ it is very easy to factor $p\O _K$, as we will see below. The following factorization gives a hint as to why:

$$x^5+7x^4+3x^2-x+1 \equiv (x+2)\cdot (x+3)^2 \cdot (x^2+4x+2)\pmod{5}.$$

The exponent $2$ of $(5,3+a)^2$ in the factorization of $5\O _K$ above suggests ``ramification'', in the sense that the cover $X\rightarrow Y$ has less points (counting their ``size'', i.e., their residue class degree) in its fiber over $5$ than it has generically. Here's a suggestive picture:

Diagram of $\Spec(\O _K) \rightarrow \Spec(\mathbb{Z})$