Consider the integers. You learned early on that every integer greater than $1$ can be broken into prime factors, and that this decomposition is essentially unique: $12 = 2^2 \cdot 3$, not $12 = 2 \cdot 6$ in any fundamentally different sense. This is so familiar it feels like a fact about numbers themselves rather than a structural property worth isolating. But when we pass to more general [rings](/page/Ring), the story changes — sometimes dramatically.

Take the ring $\mathbb{Z}[\sqrt{-5}] = \{a + b\sqrt{-5} : a, b \in \mathbb{Z}\}$. Inside this ring, we have the factorisation

\begin{align*} 6 &= 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}). \end{align*}

One can verify that $2$, $3$, $1 + \sqrt{-5}$, and $1 - \sqrt{-5}$ are all irreducible in $\mathbb{Z}[\sqrt{-5}]$ — none of them factor further — yet $6$ has two genuinely different factorisations into irreducibles. The elegant arithmetic of $\mathbb{Z}$ has broken down. Understanding precisely what property of $\mathbb{Z}$ fails in $\mathbb{Z}[\sqrt{-5}]$ — and when it holds in more general rings — is the central question of this chapter.

[example: Failure of Unique Factorisation in Z[sqrt(-5)]] We verify that $2$, $3$, $1 + \sqrt{-5}$, and $1 - \sqrt{-5}$ are all irreducible in $R = \mathbb{Z}[\sqrt{-5}]$. The key tool is the **norm map** \begin{align*} N: \mathbb{Z}[\sqrt{-5}] &\to \mathbb{Z}_{\geq 0} \\ a + b\sqrt{-5} &\mapsto a^2 + 5b^2. \end{align*} This satisfies $N(\alpha\beta) = N(\alpha)N(\beta)$ for all $\alpha, \beta \in R$. Indeed, \begin{align*} N((a + b\sqrt{-5})(c + d\sqrt{-5})) &= N((ac - 5bd) + (ad + bc)\sqrt{-5}) \\ &= (ac - 5bd)^2 + 5(ad + bc)^2 \\ &= a^2c^2 - 10abcd + 25b^2d^2 + 5a^2d^2 + 10abcd + 5b^2c^2 \\ &= (a^2 + 5b^2)(c^2 + 5d^2) = N(\alpha)N(\beta). \end{align*} The units of $R$ are exactly the elements of norm $1$: if $\alpha\beta = 1$ then $N(\alpha)N(\beta) = 1$, so $N(\alpha) = 1$ (since both are non-negative integers). The only elements with $N(\alpha) = a^2 + 5b^2 = 1$ are $\alpha = \pm 1$. Now we check irreducibility. $N(2) = 4$. If $2 = \alpha\beta$ then $N(\alpha)N(\beta) = 4$, so the norms are $(1,4)$, $(2,2)$, or $(4,1)$. Could $N(\alpha) = 2$? That would require $a^2 + 5b^2 = 2$, which has no integer solutions (since $b \neq 0$ forces $5b^2 \geq 5 > 2$, and $b = 0$ gives $a^2 = 2$, impossible). So the only factorisation of $2$ has one factor with norm $1$, hence a unit. Thus $2$ is irreducible. $N(3) = 9$. If $3 = \alpha\beta$ then $N(\alpha)N(\beta) = 9$. The only possibility for a non-unit factor is $N(\alpha) = 3$, requiring $a^2 + 5b^2 = 3$. Again $b \neq 0$ forces $5b^2 \geq 5 > 3$, and $b = 0$ gives $a^2 = 3$, impossible. So $3$ is irreducible. $N(1 + \sqrt{-5}) = 1 + 5 = 6$. If $1 + \sqrt{-5} = \alpha\beta$ then $N(\alpha)N(\beta) = 6$, so we need $N(\alpha) \in \{2, 3\}$. But as shown above, neither $2$ nor $3$ is a norm in $R$. So $1 + \sqrt{-5}$ is irreducible. By the same argument, so is $1 - \sqrt{-5}$. Finally, these four elements are genuinely different up to units: the only units are $\pm 1$, and $\pm 2 \neq \pm 3$, $\pm 2 \neq \pm(1 \pm \sqrt{-5})$, and $\pm 3 \neq \pm(1\pm \sqrt{-5})$. The two factorisations $6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$ are genuinely distinct. [/example]

The failure in $\mathbb{Z}[\sqrt{-5}]$ is not accidental. It reflects a deep structural deficiency: the irreducibles there are not prime in the ring-theoretic sense. A **Unique Factorisation Domain** is precisely a [ring](/page/Ring) where these two notions — irreducible and prime — coincide, and where unique factorisation is guaranteed as a consequence.

## Irreducibles and Primes

The $\mathbb{Z}[\sqrt{-5}]$ example forces us to distinguish two notions that coincide in $\mathbb{Z}$ but diverge in more general rings. An element can fail to factor further (irreducible) yet still fail to have the divisibility power we associate with primes in ordinary arithmetic. Making this distinction precise is the first step toward understanding when unique factorisation holds.

[definition: Irreducible Element] Let $R$ be an [integral domain](/page/Integral%20Domain). A non-zero non-unit element $a \in R$ is called **irreducible** if whenever $a = bc$ for $b, c \in R$, at least one of $b$ or $c$ is a unit. [/definition]

[definition: Prime Element] Let $R$ be an [integral domain](/page/Integral%20Domain). A non-zero non-unit element $p \in R$ is called **prime** if whenever $p \mid bc$ for $b, c \in R$, then $p \mid b$ or $p \mid c$. [/definition]

[remark: Prime Implies Irreducible] In any integral domain, every prime element is irreducible. To see this, suppose $p$ is prime and $p = ab$. Then $p \mid ab$, so $p \mid a$ or $p \mid b$. Say $p \mid a$, so $a = pc$ for some $c \in R$. Then $p = ab = pcb$, and since $R$ has no zero divisors, $1 = cb$. So $b$ is a unit. The converse fails in general: as the example above shows, the irreducible element $2 \in \mathbb{Z}[\sqrt{-5}]$ is not prime, since $2 \mid 6 = (1+\sqrt{-5})(1-\sqrt{-5})$ but $2 \nmid (1 + \sqrt{-5})$ and $2 \nmid (1 - \sqrt{-5})$. [/remark]

## Definition

With irreducibles and primes distinguished, we can now state what it means for a ring to have the arithmetic structure of $\mathbb{Z}$: every element factors, and the factorisation is unique up to the two unavoidable ambiguities of ordering and units.

With these pieces in place, we can give the central definition.

[definition: Unique Factorisation Domain] An [integral domain](/page/Integral%20Domain) $R$ is called a **unique factorisation domain** (UFD) if: 1. Every non-zero non-unit element $a \in R$ can be written as a finite product of irreducibles: \begin{align*} a &= p_1 p_2 \cdots p_n \end{align*} for irreducible elements $p_1, \ldots, p_n \in R$. 2. This factorisation is unique up to order and units: if $a = p_1 \cdots p_n = q_1 \cdots q_m$ are two factorisations into irreducibles, then $n = m$ and after reordering, each $p_i$ is an associate of $q_i$ (meaning $p_i = u_i q_i$ for some unit $u_i \in R^\times$). [/definition]

[explanation: Why Uniqueness Up to Units and Order] The two qualifications in the uniqueness statement are both essential and cannot be omitted.

**Up to order** is necessary because multiplication is commutative: $6 = 2 \cdot 3 = 3 \cdot 2$ should not count as two different factorisations.

**Up to units** is necessary because multiplying a factor by a unit and its inverse by that unit's inverse gives a formally different but arithmetically identical factorisation. In $\mathbb{Z}$, for instance, $6 = 2 \cdot 3 = (-2) \cdot (-3)$. If we required literal equality rather than associate equality, even $\mathbb{Z}$ would fail to be a UFD. Two elements $a, b \in R$ are called **associates** if $a = ub$ for some unit $u \in R^\times$, equivalently if $(a) = (b)$ as principal ideals.

The moral is: factorisation into irreducibles lives in the monoid $R \setminus \{0\}$ modulo the equivalence relation of being associates, and uniqueness is a statement in this quotient monoid. [/explanation]

[example: Z is a UFD] The integers $\mathbb{Z}$ form a UFD. While this is the motivating example, the proof already reveals the structural backbone that underpins the general theory. Existence of factorisation into primes follows by strong induction: if $n > 1$, either $n$ is prime (and we are done) or $n = ab$ with $1 < a, b < n$, and we apply induction to both factors. Uniqueness is the Fundamental Theorem of Arithmetic. The key step is that in $\mathbb{Z}$, every irreducible is prime: if $p$ is an irreducible integer and $p \mid ab$, then either $p \mid a$ (and we are done) or $\gcd(p, a) = 1$. By [Bézout's theorem](/page/Bezout%27s%20Theorem), there exist $s, t \in \mathbb{Z}$ with $sp + ta = 1$. Multiplying by $b$: $spb + tab = b$. Since $p \mid ab$, both terms on the left are divisible by $p$, so $p \mid b$. This argument uses the Euclidean algorithm, and it is precisely this algorithmic structure that makes $\mathbb{Z}$ special. What makes $\mathbb{Z}$ non-trivial as an example is the following: the primes $2, 3, 5, 7, \ldots$ are not just irreducible elements but generators of the height-one prime ideals $(2), (3), (5), (7), \ldots$, and every nonzero non-unit ideal in $\mathbb{Z}$ factors uniquely as a product of these: $(n) = (p_1)^{e_1} \cdots (p_k)^{e_k}$. In $\mathbb{Z}[\sqrt{-5}]$, the analogous statement for element ideals fails — but Dedekind showed that it is restored for ideals in any ring of integers, at the cost of introducing non-principal ideals. [/example]