A Euclidean domain is a principal ideal domain.

$R$ is a PID $\Rightarrow R$ is a UFD.

Definition. Let $R$ be an integral domain.

• A unit is an element $a\in R$ such that there exists an element $b\in R$ such that $ab=1$.
• Let $p,q\in R$. The element $p$ divides $q$ if $q=ap$ for some $a\in R$.
• Let $p,q\in R$. The element $p$ is a proper divisor of $q$ if $q=ap$ for some $a\in R$ and neither $a$ nor $p$ is a unit.
• Let $p,q\in R$. The elements $p$ and $q$ are associates if $p=aq$ for some unit $a\in R$.
• An element $p\in R$ is irreducible if
  1. $p\ne 0$,
  2. $p$ is not a unit,
  3. $p$ has no proper divisor.

The following proposition shows that every property of divisors can be written in terms of containments of ideals and vice versa.

Let $p,q\in R$ and $\left(p\right)$ and $\left(q\right)$ denote the ideals generated by the elements $p$ and $q$ respectively. Then

1. $p$ is a unit $\iff \left(p\right)=R$.
2. $p$ divides $q\iff \left(q\right)\subseteq \left(p\right)$.
3. $p$ is a proper divisor of $q\iff \left(q\right)⊊\left(p\right)⊊R$.
4. $p$ is an associate of $q\iff \left(q\right)=\left(p\right)$.
5. $p$ is irreducible $\iff$
   1. $\left(p\right)\ne 0$,
   2. $\left(p\right)\ne R$,
   3. If $q\in R$ and $\left(q\right)\supseteq \left(p\right)$ then either $\left(q\right)=\left(p\right)$ or $\left(q\right)=R$.

Definition.

• A unique factorization domain (or UFD) is an integral domain such that
• If $x\in R$ then $x=p_1{p}_2\cdots p_n$ for some $p_1,\dots ,p_n\in R$ which are all irreducible.
• If $x\in R$ and $x=p_1{p}_2\cdots p_n=u{q}_1{q}_2\cdots q_m$ where $u\in R$ is a unit and $p_1{p}_2\cdots p_n,q_1{q}_2\cdots q_m\in R$ are irreducible, then $m=n$ and there exists a permutation $\sigma :\left\{1,2,\dots n\right\}\to \left\{1,2,\dots n\right\}$ such that for each $1\le i\le n$, $q_i=u_i{p}_{\sigma \left(i\right)},$ for some unit $u_i\in R$.

A principal ideal domain is a unique factorization domain.

The proof of Theorem 1.3 will require the following lemmas.

If $R$ is a principal ideal domain and $p\in R$ is an irreducible element of $R$ then $\left(p\right)$ is a prime ideal.

Let $R$ be a principal ideal domain. There does not exist an infinite sequence of elements $a_1,a_2,\dots \in R$ such that $$\left(0\right)⊊\left(a_1\right)⊊\left(a_2\right)⊊\cdots$$

Greatest common divisors

Definition.

• Let $R$ be a unique factorization domain. Let $a_0,a_1,\dots ,a_n\in R.$ A greatest common divisor, $gcd\left(a_0,a_1,\dots ,a_n\right)$ is an element $d\in R$ such that
  1. $d$ divides $a_i$ for all $i=0,1,\dots n$
  2. If $d\text{'}$ divides $a_i$ for all $i=0,1,\dots n$, then $d\text{'}$ divides $d$.

Let $R$ be a unique factorization domain and let $a_0,a_1,\dots ,a_n\in R.$ Then

1. $gcd\left(a_0,a_1,\dots ,a_n\right)$ exists.
2. $gcd\left(a_0,a_1,\dots ,a_n\right)$ is unique up to multiplication by a unit.

References