Proof of Unique Factorization in PIDs

We prove that every principal ideal domain is a unique factorization domain, establishing the key link in the hierarchy ED $\subset$ PID $\subset$ UFD.

Statement

Theorem6.9PID implies UFD

Every principal ideal domain is a unique factorization domain. That is, every nonzero non-unit element of a PID can be written as a product of irreducible elements, uniquely up to order and associates.

Proof

Let $R$ be a PID. We prove existence and uniqueness of factorization separately.

Part 1: Existence of factorization. We show every nonzero non-unit has a factorization into irreducibles using the ascending chain condition.

Claim: $R$ satisfies the ACC on principal ideals. If $(a_1) \subseteq (a_2) \subseteq \cdots$ , then $I = \bigcup (a_n)$ is an ideal. Since $R$ is a PID, $I = (d)$ for some $d$ . Then $d \in (a_N)$ for some $N$ , so $I = (d) \subseteq (a_N)$ , giving $(a_n) = (a_N)$ for all $n \geq N$ .

Now suppose some nonzero non-unit $a$ has no factorization into irreducibles. Then $a$ is not irreducible, so $a = b_1 c_1$ with neither $b_1$ nor $c_1$ a unit. Then $(a) \subsetneq (b_1)$ and $(a) \subsetneq (c_1)$ . At least one of $b_1, c_1$ (say $b_1$ ) also has no irreducible factorization. Repeat: $b_1 = b_2 c_2$ with $(b_1) \subsetneq (b_2)$ , and $b_2$ has no factorization. This produces a strictly ascending chain $(a) \subsetneq (b_1) \subsetneq (b_2) \subsetneq \cdots$ , contradicting the ACC. So every nonzero non-unit factors into irreducibles.

Part 2: Irreducible implies prime in a PID.

Let $p$ be irreducible. Since $R$ is a PID, $(p)$ is an ideal. We show $(p)$ is prime.

Suppose $p \mid ab$ , i.e., $ab \in (p)$ . Consider $d = \gcd(p, a)$ . In a PID, $(d) = (p, a) = (p) + (a)$ . Since $d \mid p$ and $p$ is irreducible: either $d$ is a unit or $d$ is an associate of $p$ .

If $d$ is an associate of $p$ : $p \mid a$ , done. If $d$ is a unit: then $(p, a) = R$ , so $1 = sp + ta$ for some $s, t \in R$ . Then $b = spb + tab$ . Since $p \mid ab$ : $p \mid tab$ . And $p \mid spb$ . So $p \mid b$ .

Part 3: Uniqueness of factorization.

Suppose $p_1 \cdots p_m = q_1 \cdots q_n$ with all factors irreducible (hence prime). Since $p_1 \mid q_1 \cdots q_n$ and $p_1$ is prime, $p_1 \mid q_j$ for some $j$ . Since $q_j$ is irreducible, $p_1$ and $q_j$ are associates. After reordering, $p_1 \sim q_1$ . Cancel (valid in an integral domain): $p_2 \cdots p_m \sim q_2 \cdots q_n$ . By induction, $m = n$ and $p_i \sim q_i$ after reordering. $\blacksquare$

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Consequences

ExampleFactorization in specific PIDs

In $\mathbb{Z}[i]$ : $5 = (2+i)(2-i)$ and $N(2+i) = 5$ is prime in $\mathbb{Z}$ , so $2+i$ and $2-i$ are irreducible (and prime) in $\mathbb{Z}[i]$ .

The rational prime $p$ factors in $\mathbb{Z}[i]$ as:

$p = \pi \bar{\pi}$ with $\pi \not\sim \bar{\pi}$ if $p \equiv 1 \pmod{4}$ (e.g., $5 = (2+i)(2-i)$ ).
$p$ remains prime if $p \equiv 3 \pmod{4}$ (e.g., $3$ is prime in $\mathbb{Z}[i]$ ).
$2 = -i(1+i)^2$ (ramified).

RemarkBeyond PIDs

The proof uses two PID-specific properties: ACC on principal ideals (for existence) and the Bezout property $\gcd(a,b) \in (a,b)$ (for irreducible $\implies$ prime). In general UFDs, existence needs the ACC, while irreducible $\implies$ prime is given as an axiom.