# Characterisation of UFDs

## Theorem

Let $A$ be an integral domain.

The following statements are equivalent:

- $(1): \quad A$ is a unique factorisation domain

- $(2): \quad A$ is a GCD domain satisfying the ascending chain condition on principal ideals.

- $(3): \quad A$ satisfies the ascending chain condition on principal ideals and every irreducible element of $A$ is a prime element of $A$.

## Proof

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We use the notation $a \sim b$ to mean $a \mid b$ and $b \mid a$, or equivalently that $a = bu$ for some unit $u \in A^\times$. We say $a$ and $b$ are "associates".

$\left( 1 \rightarrow 2 \right)$ Let $a \sim p_1^{\alpha_1} \cdots p_k ^ {\alpha_k}$, $b \sim p_1^{\beta_1} \cdots p_k ^ {\beta_k}$, where $\alpha_i, \beta_i \in \N \cup \{0\}$. Then we claim that $\gcd(a,b) \sim d = p_1^{\text{min}(\alpha_1, \beta_1)} \cdots p_k^{\text{min}(\alpha_k, \beta_k)}$.

Clearly $a \sim d \cdot p_1^{\alpha_1 - \text{min}(\alpha_1,\beta_1)} \cdots p_k^{\alpha_k - \text{min}(\alpha_k,\beta_k)}$ noting all exponents are non-negative. Similarly for $b$, it suffices to prove that if $e \mid a$ and $e \mid b$ then $e \mid d$. $e$ only has factors associated to each $p_i$, and none of the exponents in $e$'s irreducible factorization can be greater than both $\alpha_i$ and $\beta_i$, so $e = p_1^{\gamma_1} \cdots p_n^{\gamma_n}$ where $\gamma_i \leq \text{min}(\alpha_i, \beta_i)$, clearly now $e \mid d$ so we have a GCD domain.

Aiming for a contradiction, suppose we had a chain:

- $\sequence {a_1} \subsetneq \sequence {a_2} \subsetneq \dots$

that did not terminate.

Let $a_i = p_1^{\large \gamma_{1,i}} \cdots p_n^{\large \gamma_{n,i}}$. It is easy to that the sum of the exponents must be a strictly decreasing sequence of positive integers, which cannot occur.

$\left( 2 \rightarrow 3 \right)$ By assumption we have the ascending chain condition on principal ideals.

Let $p$ be irreducible and $p \mid ab$, let $d = \gcd(a,p)$. Since $d \mid p$, either $d \sim p$, in which case since $d \mid a$ we have that $p \mid a$.

Otherwise, $d \sim 1$, in which case $\gcd(ab,pb) \sim b$ so since $p \mid ab$ and $p \mid pb$ we have $p \mid b$.

$\left( 3 \rightarrow 1 \right)$ Suppose irreducible factorization was false, let $d \in A$ be an element that cannot be written as a product of irreducible elements. Then $d$ is not irreducible, so $d = a_1 x_1$ for some non-units $a_1, x_1 \in A \setminus A^{\times}$. Not both of $a_1, x_1$ can be irreducible so assume without loss of generality that $x_1$ is reducible, say $x_1 = a_2 x_2$ where neither are units, continuing onwards we have the following chain:

- $\sequence d \subsetneq \sequence {x_1} \subsetneq \sequence {x_2} \subsetneq \dots$

This contradicts the ascending chain condition on principal ideals. Therefore, every elements can be factored as a product of irreducible elements.

Now to prove uniqueness, suppose $p_1 p_2 \cdots p_i = q_1 q_2 \cdots q_j$ were two equal products of irreducibles. Then $p_1 \mid q_1 \cdots q_j$ and since $p_1$ is prime, (without loss of generality) $p_1 \sim q_1$, since $A$ is an integral domain, by Cancellation Law for Ring Product of Integral Domain, we have that $p_2 \cdots p_i \sim q_2 \cdots q_j$, continuing onwards shows that each $p_a \sim q_a$ and $i=j$ so the factorizations are unique up to unit.

$\blacksquare$