Every Irreducible Element is Prime in a Unique Factorisation Domain

Theorem

If \(R\) is a unique factorisation domain and \(a \in R\) is irreducible, then \(a\) is prime.

Proof

Let \(a\) be an irreducible element in a unique factorisation domain \(R\). Then consider some \(s, t \in R\) for which

\[ a \mid st.\]

Then, there exists some \(k\) such that

\[ ak = st.\]

Since we have uniqueness of factorisation, we can factorise each side into irreducibles in a unique way (up to reordering and associates), that is

\[ a \cdot k_1 \cdot \dots \cdot k_l = s_1 \cdot \dots \cdot s_n \cdot t_1 \cdot \dots \cdot t_m.\]

The fact that \(a\) is left on its own is a consequence of the irreducibility of \(a\) (it is already fully factored).

This means that each term in the left hand side of our factorisation is an associate with an element on the right hand side, and we can pair the elements as such.

Therefore

\[ a = u_i s_i \quad \text{or} \quad a = u_j t_j\]

for some unit \(u_i\) or \(u_j\) and some positive integer \(i \leq n\) or \(j \leq m\).

Hence \(a \mid s_i\) and \(a \mid s\) with the multiple being the product of all factors of \(s\) except \(s_i\), or \(a \mid t_j\) and therefore similarly \(a \mid t\).

This proves that \(a\) is prime.

Corollary

The set of prime elements is equal to the set of irreducible elements in a unique factorisation domain.

This follows simply from the above, and the fact that every unique factorisation domain is an integral domain with every prime being irreducible in an integral domain.