Geometric Intuition

Let $K = \mathbf{Q}(\alpha)$ be a number field, and let $\O_K$ be the ring of integers of $K$. To employ our geometric intuition, as the Lenstras did on the cover of [LL93], it is helpful to view $\O_K$ as a 1-dimensional scheme

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

over

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

There is a natural map $\pi :X\rightarrow Y$ that sends a prime ideal $\mathfrak{p}\in X$ to $\mathfrak{p}\cap \mathbf{Z}\in Y$. For example, if

$$\mathfrak{p}= (65537, 2^8 + i)\subset\mathbf{Z}[i],$$

then $\mathfrak{p}\cap \mathbf{Z}= (65537)$. For more on this viewpoint, see [Har77] and [EH00, Ch. 2].

If $p\in\mathbf{Z}$ 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 $\O_K$. We may imagine the decomposition of $p\O_K$ into prime ideals geometrically as the fiber $\pi^{-1}(p\mathbf{Z})$, where the exponents $e_i$ are the multiplicities of the fibers. Notice that the elements of $\pi^{-1}(p\mathbf{Z})$ are the prime ideals of $\O_K$ that contain $p$, i.e., the primes that divide $p\O_K$. This chapter is about how to compute the $\mathfrak{p}_i$ and $e_i$.

Remark 4.1.1   More technically, in algebraic geometry one defines the inverse image of the point $p\mathbf{Z}$ to be the spectrum of the tensor product $\O_K \otimes _{\mathbf{Z}} \mathbf{Z}/\mathfrak{p}\mathbf{Z}$; by a generalization of the Chinese Remainder Theorem, we have

$$\O_K \otimes _{\mathbf{Z}} \left(\mathbf{Z}/\mathfrak{p}\mathbf{Z}\right) \cong \oplus \O_K/\mathfrak{p}_i^{e_i}.$$