Every Irreducible Element is Prime in a Unique Factorisation Domain

Proof

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

Then, there exists some $k$ such that

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

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

for some unit $u_i$ or $u_j$ and some positive integer $i\le n$ or $j\le m$.

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.

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.