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

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


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

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

$\displaystyle \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

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

William Stein 2012-09-24