In fact, ideal \( (x, y) \) is not principal. For suppose \( (x, y) = (f(x, y)) \), for some \( f(x, y) \in \mathbb{Z}[x, y] \). Then \( x = f(x, y) \cdot u(x, y) \), where \( \text{deg}(f) = 0, 1 \) and \( \text{deg}(u) = 1, 0 \). So \( f(x) = ax + by + c \) and \( u(x, y) = k\), with \( a, b, c, k \in \mathbb{Z} \), or similarly with the roles of \( f \) and \( u \) reversed. Either way, \( x = (ax + by + c) \cdot k \), so \( a = \pm 1 \), \( b = c = 0 \), and \( k = \pm 1 \).

There are two possibilities for \( f(x, y) \). One is that \( f(x, y) = \pm 1 \). This can't be, because in that case \( (x, y) = \mathbb{Z}[x, y] \), but constants are not in the former. The other possibility is that \( f(x, y) = \pm x \). Similarly by putting \( y = f(x, y) \cdot v(x, y) \), it follows that \( f(x, y) = \pm y \).

This contradiction proves that \( (x, y) \) is not a principal ideal, so \( \mathbb{Z}[x, y] \) is not a PID and therefore not a Euclidean domain. QED.

Note: This is similar to Ex. §9.1,#6 in Dummit & Foote, to prove that \( (x, y) \) is not a principal ideal in \( \mathbb{Q}[x, y] \).