Euclidean Domains, PIDs, and UFDs

11.1 Euclidean Domains

An integral domain $R$ is a Euclidean domain if there exists a function $\delta : R \setminus \{0\} \to \mathbb{N}_0$ such that for all $a, b \in R$ with $b \neq 0$ :

There exist $q, r \in R$ with $a = bq + r$ and either $r = 0$ or $\delta(r) \lt \delta(b)$ .

Example. $\mathbb{Z}$ is a Euclidean domain with $\delta(a) = |a|$ .

Example. $F[x]$ is a Euclidean domain with $\delta(f) = \deg(f)$ .

Example. $\mathbb{Z}[i]$ is a Euclidean domain with $\delta(a + bi) = a^2 + b^2$ .

11.2 Principal Ideal Domains

An integral domain $R$ is a principal ideal domain (PID) if every ideal of $R$ is principal (generated by a single element).

Theorem 11.1. Every Euclidean domain is a PID.

Proof. Let $I$ be a non-zero ideal of the Euclidean domain $R$ . Choose $d \in I \setminus \{0\}$ Minimising $\delta(d)$ . We claim $I = (d)$ . For any $a \in I$ Write $a = qd + r$ with $r = 0$ or $\delta(r) \lt \delta(d)$ . Since $r = a - qd \in I$ Minimality of $\delta(d)$ Forces $r = 0$ So $a = qd \in (d)$ . $\blacksquare$

Corollary 11.2. $\mathbb{Z}$ , $F[x]$ And $\mathbb{Z}[i]$ are PIDs.

11.3 Unique Factorization Domains

An integral domain $R$ is a unique factorization domain (UFD) if:

Every non-zero, non-unit element factors into irreducibles.
The factorization is unique up to ordering and associates.

Theorem 11.3. Every PID is a UFD.

The chain of implications is:

$\mathrm{Euclidean\ domain} \Rightarrow \mathrm{PID} \Rightarrow \mathrm{UFD}$

None of the reverse implications hold .

Example. $\mathbb{Z}[x]$ is a UFD but not a PID. (The ideal $(2, x)$ is not principal.)

Example. $\mathbb{Z}[\sqrt{-5}]$ is not a UFD: $6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$ Gives two distinct factorizations into irreducibles.

Problem. Show that $\mathbb{Z}[\sqrt{-5}]$ is not a UFD.

Solution

Solution. We show that $2, 3, 1 + \sqrt{-5}, 1 - \sqrt{-5}$ are all irreducible in $\mathbb{Z}[\sqrt{-5}]$ and that $2, 3$ are not associates of $1 \pm \sqrt{-5}$ .

The norm is $N(a + b\sqrt{-5}) = a^2 + 5b^2$ . Note $N(\alpha\beta) = N(\alpha)N(\beta)$ and $\alpha$ is a unit iff $N(\alpha) = 1$ .

$N(2) = 4$ . If $2 = \alpha\beta$ with neither a unit, then $N(\alpha)N(\beta) = 4$ So $N(\alpha) = N(\beta) = 2$ . But $a^2 + 5b^2 = 2$ has no integer solutions. So $2$ is irreducible.

Similarly, $N(3) = 9$ And $a^2 + 5b^2 = 3$ has no solutions, so $3$ is irreducible.

$N(1 \pm \sqrt{-5}) = 6$ . If $1 + \sqrt{-5} = \alpha\beta$ with neither a unit, then $N(\alpha), N(\beta) \in \{2, 3\}$ . But $a^2 + 5b^2 = 2$ and $a^2 + 5b^2 = 3$ have no solutions. So $1 \pm \sqrt{-5}$ are irreducible.

Now $2 \cdot 3 = 6 = (1 + \sqrt{-5})(1 - \sqrt{-5})$ gives two distinct factorizations into Irreducibles (the factors are not associates since their norms are different: $4, 9$ vs $6, 6$ ). Therefore $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. $\blacksquare$

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