Euclidean, PID, UFD hierarchy

In one sentence

Euclidean $\subsetneq$ PID $\subsetneq$ UFD $\subsetneq$ integral domain, every inclusion is strict, and the examinable content is the witness in each gap plus the fact that irreducible and prime coincide only from UFD onwards.

Why the exam asks it

Because four classes with three strict inclusions generate a great many true-looking false statements, and because "irreducible" and "prime" are synonyms in $\mathbb{Z}$, which is where everyone's intuition was formed.

The idea

All rings here are commutative domains with $1$.

• Euclidean domain: has a division algorithm — a size function $d$ with $a = qb + r$ where $r = 0$ or $d(r) < d(b)$.
• PID: every ideal is principal.
• UFD: every non-zero non-unit factors into irreducibles, uniquely up to order and units.

$$\text{Euclidean} \subsetneq \text{PID} \subsetneq \text{UFD} \subsetneq \text{domain}.$$

The witnesses, each worth knowing by name:

• PID but not Euclidean: $\mathbb{Z}\!\left[\frac{1+\sqrt{-19}}{2}\right]$.
• UFD but not PID: $\mathbb{Z}[x]$, where $(2,x)$ is not principal. Also $K[x,y]$, where $(x,y)$ is not.
• Domain but not UFD: $\mathbb{Z}[\sqrt{-5}]$.

Irreducible against prime

$p$ is irreducible if $p = ab$ forces $a$ or $b$ to be a unit. $p$ is prime if $p \mid ab \Rightarrow p\mid a$ or $p \mid b$.

• Prime $\Rightarrow$ irreducible in any domain.
• Irreducible $\Rightarrow$ prime only in a UFD (and hence in PIDs and Euclidean domains).

The failure is the heart of the topic. In $\mathbb{Z}[\sqrt{-5}]$,

$$6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5}),$$

and all four factors are irreducible. Use the norm $N(a+b\sqrt{-5}) = a^2+5b^2$, which is multiplicative. $N(2) = 4$, so a proper factorisation of $2$ would need a factor of norm $2$; but $a^2+5b^2 = 2$ has no integer solution. Same for $3$ (norm $9$, would need norm $3$) and for $1\pm\sqrt{-5}$ (norm $6$, would need norm $2$ or $3$).

So $2$ is irreducible. It is not prime: it divides $(1+\sqrt{-5})(1-\sqrt{-5}) = 6$ and divides neither factor, since $\frac{1\pm\sqrt{-5}}{2}$ is not in the ring. Two genuinely different factorisations, so unique factorisation fails.

What each level buys

In a PID: gcds exist and are $\mathbb{Z}$-combinations (Bézout); irreducible $\iff$ prime; nonzero primes are maximal; every ascending chain of ideals stabilises.

In a UFD: gcds exist but need not be combinations; Gauss's lemma holds, so $R$ a UFD makes $R[x]$ a UFD. That is why $\mathbb{Z}[x]$ and $K[x_1,\dots,x_n]$ are UFDs.

Note the asymmetry: $R$ a PID does not make $R[x]$ a PID. $\mathbb{Z}$ is a PID and $\mathbb{Z}[x]$ is not. $K[x]$ is a PID and $K[x,y]$ is not. Adjoining a variable costs you principality every time and costs you nothing in unique factorisation.

Where intuition breaks

"Irreducible means prime." $2$ in $\mathbb{Z}[\sqrt{-5}]$. True in a UFD, false in a general domain, and $\mathbb{Z}$ being a UFD is why it feels like a definition.

"$\mathbb{Z}[x]$ is a PID because $\mathbb{Z}$ is." $(2,x)$ is not principal. $R[x]$ is a PID only when $R$ is a field.

"$K[x,y]$ is a PID." $(x,y)$ is not principal. It is a UFD, by Gauss.

"UFD implies PID." $\mathbb{Z}[x]$. The implication runs the other way.

"$\mathbb{Z}[\sqrt{-5}]$ is not a domain." It is a domain — it sits inside $\mathbb{C}$. It fails unique factorisation, not the absence of zero divisors.

"Every PID is Euclidean." $\mathbb{Z}[(1+\sqrt{-19})/2]$. The gap is real, though the witness is the only one you need.

"Gcds do not exist outside PIDs." They exist in any UFD; what fails is Bézout, the expression of the gcd as a combination.

The exam's angle

1. Use norms in quadratic rings. $N(a+b\sqrt{d}) = a^2 - db^2$ is multiplicative, so irreducibility becomes a small Diophantine check.
2. To disprove a UFD, exhibit two factorisations and show all factors irreducible by norm.
3. To disprove a PID, find a two-generator ideal and show no single generator works — $(2,x)$ and $(x,y)$ are the models.
4. Remember which direction each implication runs. Prime $\Rightarrow$ irreducible always; the converse needs a UFD.
5. Adjoining a variable: preserves UFD, destroys PID unless the base is a field.

The night before

• Euclidean $\subsetneq$ PID $\subsetneq$ UFD $\subsetneq$ domain, all strict.
• Witnesses: $\mathbb{Z}[(1+\sqrt{-19})/2]$ (PID not Euclidean), $\mathbb{Z}[x]$ and $K[x,y]$ (UFD not PID), $\mathbb{Z}[\sqrt{-5}]$ (domain not UFD).
• Prime $\Rightarrow$ irreducible in any domain. Irreducible $\Rightarrow$ prime only in a UFD.
• $\mathbb{Z}[\sqrt{-5}]$: $6 = 2\cdot3 = (1+\sqrt{-5})(1-\sqrt{-5})$, norm $a^2+5b^2$, and $2$ is irreducible but not prime.
• PID gives Bézout, irreducible $=$ prime, nonzero primes maximal.
• Gauss: $R$ UFD $\Rightarrow R[x]$ UFD. But $R$ PID does not give $R[x]$ PID.
• $R[x]$ is a PID $\iff R$ is a field.