Unique Factorization Domains

Unique factorization domains are the algebraic backbone of several major areas. In algebraic number theory, the failure of unique factorization in rings of integers of number fields is measured by the class group, and Dedekind's theorem that ideals do factor uniquely into prime ideals (even when elements do not) is the modern repair of Kummer's original insight — this is one of the deepest classical results building directly on the distinction between prime and irreducible elements developed here. In polynomial algebra and computer algebra systems, the fact that $F[x]$ is a UFD (as a Euclidean domain, hence a PID, hence a UFD by this theorem) underlies polynomial factorization algorithms used in symbolic computation software, and Gauss's Lemma extends unique factorization from $R[x]$ down from $\mathrm{Frac}(R)[x]$ back to $R[x]$ whenever $R$ itself is a UFD — this is exactly how one proves $\mathbb{Z}[x]$ is a UFD despite not being a PID. In cryptography, factorization-based hardness assumptions (RSA, and Gaussian-integer variants of primality testing) rely on unique factorization in $\mathbb{Z}$ and $\mathbb{Z}[i]$ to guarantee well-defined notions of "prime" moduli. In coding theory, factorizations of polynomials over finite fields $\mathbb{F}_q[x]$ (a UFD) determine cyclic code structure. In algebraic geometry, whether a coordinate ring is a UFD governs whether every codimension-one subvariety is cut out by a single equation — a geometric shadow of the algebraic uniqueness property proved here.

:::quiz[In $\mathbb{Z}[\sqrt{-5}]$, why does $2$ fail to be prime even though it is irreducible?]

• $2$ is not irreducible in $\mathbb{Z}[\sqrt{-5}]$, so the question is moot
• $2$ divides $6=(1+\sqrt{-5})(1-\sqrt{-5})$ but divides neither factor ✓
• $\mathbb{Z}[\sqrt{-5}]$ has no units other than $1$, blocking primality
• $2$ has norm $6$, matching neither factor's norm

:::faq[Why doesn't the existence-of-factorization argument work verbatim in $\mathbb{Z}[\sqrt{-5}]$?] It actually does work: $\mathbb{Z}[\sqrt{-5}]$ does satisfy ACC on principal ideals (it is Noetherian, being a finitely generated $\mathbb{Z}$-module as a ring), so every nonzero nonunit does factor into irreducibles there. What fails is uniqueness, because Part I of the PID proof — "irreducible implies prime" — depended crucially on writing $(p,a)$ as a single principal ideal $(d)$, which is only guaranteed when every ideal of the ring is principal. In $\mathbb{Z}[\sqrt{-5}]$ some ideals, like $(2, 1+\sqrt{-5})$, are not principal, so that step of the argument breaks down, and indeed irreducibles like $2$ turn out not to be prime.

Ask the tutor: Can you show me why $\mathbb{Z}[x]$ is a UFD even though it isn't a PID, using Gauss's Lemma?

Think of an integral domain as a universe of "numbers" where multiplication behaves, but the notion of "divides evenly" may be looser than in $\mathbb{Z}$. Irreducibility is a syntactic condition: an element resists being split into two nontrivial factors. Primality is a semantic condition about how the element interacts with products: it can only divide a product by dividing one of the pieces. In $\mathbb{Z}$ these two ideas trace out the same set of elements — that fusion is so automatic we rarely distinguish them. The whole subtlety of unique factorization theory is that this fusion is a theorem, true in nice rings like PIDs, and false in general. The picture to keep in mind: primes are the "well-behaved" atoms that never cause ambiguity when they divide a product, while irreducibles are merely atoms with respect to the multiplication operation, with no promise about how they interact with divisibility of products. The diagram below shows the containment of classes of rings, and where the logical implication proven in this article — PID $\Rightarrow$ UFD — sits.

Example 1: Factoring in $\mathbb{Z}[i]$, the Gaussian integers, a PID. Let $R = \mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{Z}\}$, which is known to be a Euclidean domain via the norm $N(a+bi) = a^2+b^2$, hence a PID, hence by our theorem a UFD. Consider the element

We factor $a$ into irreducibles two different-looking ways and confirm they agree up to units and reordering — units in $\mathbb{Z}[i]$ are $\{1, -1, i, -i\}$. Route A. Factor out the obvious real integer content: $50 + 50i = 50(1+i) = 2 \cdot 25 (1+i) = 2 \cdot 5^2 (1+i)$. Now $2 = -i(1+i)^2$ (check: $(1+i)^2 = 2i$, and $-i \cdot 2i = 2$), and $5 = (2+i)(2-i)$ since $(2+i)(2-i) = 4+1 = 5$. Both $1+i$ and $2\pm i$ are irreducible in $\mathbb{Z}[i]$ because their norms ($2$ and $5$ respectively) are primes in $\mathbb{Z}$, and a Gaussian integer of prime norm must be irreducible (if it factored as a product of two nonunits, the norms — which are multiplicative — would give a nontrivial factorization of the prime norm in $\mathbb{Z}$, impossible). So:

Route B. Directly test divisibility: $50+50i = (1+i)(a+bi)$ requires $(a+bi) = \frac{50+50i}{1+i} = \frac{(50+50i)(1-i)}{2} = \frac{50 - 50i + 50i +50}{2}= \frac{100}{2}=50$. So $50+50i = (1+i)\cdot 50 = (1+i)\cdot 2\cdot 5^2$. Continue factoring $2 = -i(1+i)^2$ and $5 = (2+i)(2-i)$ exactly as before, arriving at the same multiset of irreducibles:

up to the unit $-i$, confirming uniqueness as guaranteed by the theorem, regardless of the order in which factors were extracted.

Example 2: The non-example $\mathbb{Z}[\sqrt{-5}]$ is not a UFD — sanity-checking why the PID hypothesis matters. Let $R = \mathbb{Z}[\sqrt{-5}] = \{a + b\sqrt{-5} : a, b \in \mathbb{Z}\}$, with norm $N(a+b\sqrt{-5}) = a^2 + 5b^2$, multiplicative: $N(xy) = N(x)N(y)$. Consider $6 \in R$. We exhibit two genuinely different factorizations into irreducibles:

Check: $(1+\sqrt{-5})(1-\sqrt{-5}) = 1 - (-5) = 6$. Good. We must verify each of $2, 3, 1+\sqrt{-5}, 1-\sqrt{-5}$ is irreducible, and no two of them are associates. Compute norms: $N(2) = 4$, $N(3) = 9$, $N(1+\sqrt{-5}) = 1 + 5 = 6$, $N(1-\sqrt{-5}) = 6$. Irreducibility of $2$: if $2 = xy$ with $x,y$ nonunits, then $N(x)N(y) = 4$ with $N(x), N(y) > 1$ (norm $1$ forces a unit, since $a^2+5b^2=1$ only for $(a,b)=(\pm1,0)$, giving units $\pm 1$), so $N(x) = N(y) = 2$. But $a^2+5b^2 = 2$ has no integer solutions (checking $b=0$ gives $a^2=2$, impossible; $|b|\geq 1$ already exceeds $2$). So $2$ is irreducible. Irreducibility of $3$: similarly $N(x)N(y)=9$ forces $N(x)=N(y)=3$, and $a^2+5b^2=3$ has no solution. So $3$ is irreducible. Irreducibility of $1\pm\sqrt{-5}$: a nontrivial factorization would force norm factors $N(x)N(y)=6$ with both $>1$, i.e. $\{2,3\}$; but we just showed no element has norm $2$ or norm $3$. So $1\pm\sqrt{-5}$ are irreducible. No two are associates: the only units are $\pm1$ (norm $1$), so associates of $2$ are $\pm 2$, of $3$ are $\pm3$, and $1\pm\sqrt{-5}$ have norm $6\neq 4, 9$, so they cannot be associates of $2$ or $3$; and $1+\sqrt{-5}$ is not $\pm(1-\sqrt{-5})$. Hence $6$ has two factorizations into irreducibles, $2\cdot 3$ and $(1+\sqrt{-5})(1-\sqrt{-5})$, that are not related by reordering and units. So $\mathbb{Z}[\sqrt{-5}]$ is not a UFD — and by our theorem (contrapositive), it cannot be a PID either. Indeed, one can show the ideal $(2, 1+\sqrt{-5})$ is not principal, consistent with this.