Let \(K=\QQ(\sqrt d)\) be a quadratic number field and let \(\OO_K\) be its ring of integers. What can you say about \(\OO_K/\mathfrak p\) when \(\mathfrak p\subset \OO_K\) is a nonzero prime ideal?
Solution
The quotient \(\OO_K/\mathfrak p\) is a finite field.
1. Every nonzero prime ideal of \(\OO_K\) is maximal.
The ring \(\OO_K\) is a Dedekind domain. In a Dedekind domain every nonzero prime ideal is maximal. Hence \(\OO_K/\mathfrak p\) is a field.
2. The field \(\OO_K/\mathfrak p\) is finite.
Because \(\mathfrak p\) is nonzero, choose \(0\ne\alpha\in\mathfrak p\). Since \(\alpha\) is an algebraic integer, its norm \[ N_{K/\QQ}(\alpha)\in\ZZ\setminus\{0\}. \] The principal ideal \((\alpha)\) has finite index in \(\OO_K\), with \[ |\OO_K/(\alpha)|=|N_{K/\QQ}(\alpha)|. \] Since \((\alpha)\subseteq\mathfrak p\), the quotient map \[ \OO_K/(\alpha)\twoheadrightarrow\OO_K/\mathfrak p \] is surjective. Therefore \(\OO_K/\mathfrak p\) is finite.
Thus every nonzero prime ideal of a quadratic integer ring has a finite residue field.