Euclidean, PID, UFD hierarchy

The chain

Field ⊂ Euclidean domain ⊂ PID ⊂ UFD ⊂ integral domain

Inclusion	Strict because
Euclidean ⊊ PID	$\mathbb{Z}[(1 + \surd-19)/2]$ is a PID, not Euclidean
PID ⊊ UFD	$\mathbb{Z}[X]$ and 𝔽[X, Y]: (2, X) and (X, Y) are not principal
UFD ⊊ domain	$\mathbb{Z}[\surd-5]: 6 = 2\cdot3 = (1+\surd-5)(1-\surd-5)$

Facts that follow

• R UFD ⇒ R[X] UFD (Gauss). So $\mathbb{Z}[X_{1}, \dots, X_{n}]$ is a UFD.
• R PID ⇏ R[X] PID: $\mathbb{Z}$ is a PID, $\mathbb{Z}[X]$ is not.
• 𝔽 field ⇒ 𝔽[X] Euclidean (degree), hence a PID; 𝔽[X, Y] is not. $- \mathbb{Z}[i]$ and $\mathbb{Z}[\omega]$ are Euclidean (norm), hence PIDs and UFDs.
• In a UFD: irreducible ⇔ prime. In a general domain only prime ⇒ irreducible; in $\mathbb{Z}[\surd-5], 2$ is irreducible but not prime.
• Every PID is Noetherian; $\mathbb{Z}[X_{1}, X_{2}, \dots] ($infinitely many variables) is a UFD that is not Noetherian.

Norms for quadratic rings

$N(a + b\sqrt{d}) =$ |$a^{2} -$ db$^{2}$| is multiplicative; units are the elements of norm 1. In $\mathbb{Z}[\surd-5], N(2) = 4$ and no element has norm 2, so 2 is irreducible; yet 2 | $(1+\surd-5)(1-\surd-5)$ without dividing either factor.

Key takeaways

• Learn one counterexample per strict inclusion: $\mathbb{Z}[(1+\surd-19)/2], \mathbb{Z}[X], \mathbb{Z}[\surd-5]$.
• UFD is preserved by adjoining variables; PID is not.
• Norm arguments settle irreducibility in quadratic rings.