Divisibility, Euclidean Domains, Principal Ideal Domains and Unique Factorisation

Let $R$ be an integral domain. For $a,b\in R$, say that $a$ divides $b$, written $a\mid b$, if $b=ac$ for some $c\in R$.

Elements $a,b\in R$ are associates if $a=ub$ for a unit $u\in R^\times$. Thus associates differ only by multiplication by an invertible factor. This relation is used to state uniqueness of factorisations without artificially choosing representatives of unit classes.

Let $R$ be an integral domain and let $p\in R$ be nonzero and not a unit. The element $p$ is irreducible if every factorisation $p=ab$ has $a$ or $b$ a unit. It is prime if

$p\mid ab\quad\Longrightarrow\quad p\mid a\text{ or }p\mid b$

for all $a,b\in R$.

An integral domain $R$ is a Euclidean domain if it has a function

$\delta:R\setminus\{0\}\longrightarrow\mathbb N$

such that for every $a\in R$ and nonzero $b\in R$, there are $q,r\in R$ with

$a=bq+r,\qquad r=0\ \text{or}\ \delta(r)<\delta(b).$

Such a function $\delta$ is a Euclidean function. No multiplicative monotonicity condition on $\delta$ is included in this convention.

An integral domain $R$ is a principal ideal domain (PID) if every ideal $I\mathrel{\trianglelefteq}R$ is principal: there is an $a\in R$ with $I=(a)$.

The domain hypothesis is part of the term. A ring in which all ideals are principal but which has zero divisors is not called a PID here.

An integral domain $R$ is a unique factorisation domain (UFD) when every nonzero nonunit is a finite product of irreducible elements, and whenever

$a=p_1\cdots p_m=q_1\cdots q_n$

are products of irreducibles, then $m=n$ and, after a permutation, each $p_i$ is associate to the corresponding $q_i$. Thus uniqueness is only up to order and multiplication of factors by units.

Every Euclidean domain is a principal ideal domain.

Every prime element of an integral domain is irreducible.