What is Euclidean domain?

Example 2.15

$\Z$ (with $\nu = \abs\cdot$) and $K[X]$ (with $\nu = \deg$) are Euclidean — the Year 2 volume proved both divisions. So is $\Z[\iu]$, with $\nu = N$ the square norm (Exercise 2.4); the geometry of the proof is in the figure below. A PID that is not Euclidean exists but is delicate to certify (the standard example is $\Z\bigl[\frac{1+\iu\sqrt{19}}2\bigr]$); a UFD that is not a PID is easy: $K[X, Y]$ (Exercise 2.6), or $\Z[X]$.

Example 2.20 (A ring without unique factorization)

None of the implications Euclidean $\Rightarrow$ PID $\Rightarrow$ UFD is an equivalence, and the failure of the last is worth seeing once in complete detail. In

the norm is multiplicative and $N(z) = 1$ iff $z \in A^\times = \{\pm1\}$. Consider

All four factors are irreducible: their norms are $4, 9, 6, 6$, and a proper factorization $z = z_1z_2$ would force $N(z_1) \in \{2, 3\}$ — but $a^2 + 5b^2$ never equals $2$ or $3$ ($b = 0$ leaves the non-squares $2, 3$; $\abs b \geq 1$ gives $\geq 5$). Yet $2$ is associate to neither $1 \pm \iu\sqrt5$ (norms $4 \neq 6$): two genuinely different factorizations of $6$ into irreducibles. Equivalently, irreducible $\neq$ prime here: $2$ divides the product $(1 + \iu\sqrt5)(1 - \iu\sqrt5) = 6$ but neither factor (norms again). The ideal-theoretic repair of this failure — factorizing ideals rather than elements — is the birth of algebraic number theory; at our level, the example calibrates how special the Euclidean rings $\Z$, $K[X]$, $\Z[\iu]$ of this chapter really are.