Every Prime Element is Irreducible in an Integral Domain

Theorem

If \(R\) is an integral domain and \(p \in R\) is prime, then \(p\) is irreducible.

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:

\[ p \mid a \quad \text{or} \quad p \mid b.\]

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

\[ pn = a.\]

Substituting this into \(p = ab\) we have that:

\[\begin{align*} &p = pnb \\ \implies& p - pnb = 0 \\ \implies& p(1 - nb) = 0 \\ \end{align*}\]

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.