- Show that every Euclidean domain is a principal ideal domain.
In parts (b) and (c) assume that \(\mathbf{Z}[i]\) is a Euclidean domain with respect to the norm function \(N\) defined by \[N(a+bi)=a^2+b^2\text{ for all }a+bi\in\mathbf{Z}[i].\]
Let \(I\) be a nonzero ideal of \(\mathbf{Z}[i]\) generated by \(\alpha\). Show that every coset of \(I\) in \(\mathbf{Z}[i]\) contains an element of norm less than \(N(\alpha)\).
Show that \(\mathbf{Z}[i]/I\) is finite.
Solution
1. In part (a), every ideal of a Euclidean domain \(R\) with Euclidean function \(N\) is principal.
Let \(I\) be an ideal of \(R\). If \(I=0\), then \(I=(0)\). Otherwise choose \(d\in I\setminus\{0\}\) with \(N(d)\) minimal; then \((d)\subseteq I\). For \(a\in I\), division gives \(q,r\in R\) with \(a=qd+r\) and either \(r=0\) or \(N(r)<N(d)\). Since \(r=a-qd\in I\), minimality of \(N(d)\) excludes \(r\ne0\), so \(a=qd\in(d)\). Hence \(I=(d)\), and \(R\) is a principal ideal domain.
2. In part (b), every coset of \(I=(\alpha)\) contains an element of norm less than \(N(\alpha)\).
A coset has the form \(\beta+I\) with \(\beta\in\ZZ[i]\). Division by \(\alpha\ne0\) gives \(q,r\in\ZZ[i]\) with \(\beta=q\alpha+r\) and either \(r=0\) or \(N(r)<N(\alpha)\). Then \(r=\beta-q\alpha\in\beta+I\), and in both cases \(N(r)<N(\alpha)\), because \(N(0)=0<N(\alpha)\).
3. In part (c), \(\ZZ[i]/I\) is finite.
Let \(n=N(\alpha)\). The Gaussian integers \(a+bi\) with \(a^2+b^2<n\) satisfy \(\abs{a},\abs{b}<\sqrt n\), so there are finitely many of them. By step 2 every coset of \(I\) contains one of them, so there are finitely many cosets.
4. Q.E.D.
Steps 1, 2, and 3 answer parts (a), (b), and (c).