Every Prime Element is Irreducible in an Integral Domain

The converse holds in a unique factorisation domain.

Proof

Let $p$ be prime in an integral domain $R$ and let $p=ab$, a factorisation that necessarily at least exists since $1\in R$ and $p=1p$.

Hence $p\mid ab$ and since $p$ is prime, either:

We assume without loss of generality that $p\mid a$. Therefore there exists some $n\in R$ such that

Substituting this into $p=ab$ we have that:

Now since $R$ is an integral domain, we have the zero product property, so either $p=0$ or $1-nb=0$, however $p$ is prime and hence is non-zero by definition, hence $1-nb=0$ and therefore $1=nb$. Since we are working in a commutative ring, we also have $1=bn$ and $b$ and $n$ are units.

The fact that $b$ is a unit proves this general factorisation has one term a unit, and hence $p$ is irreducible.