Wikipedia gives the definition of a Unique Factorisation Domain as one where every element "can be written as a product of prime elements (or irreducible elements)" which suggests that in a UFD prime and irreducible elements are the same.
However, I thought that only in a PID were prime and irreducible elements the same and in a UFD it is true that all prime elements are irreducible, but not visa versa.
What's going on here?
What the wikipedia article says is correct: a domain has the property that every (nonzero nonunit) element factors uniquely into prime elements iff it has the property that every (...) element factors uniquely into irreducible elements.
Although in a general domain prime elements are irreducible but irreducible elements need not be prime, in a UFD irreducible elements are prime. Indeed, this is much of the point of being a UFD: e.g. a Noetherian domain in which irreducible elements are prime is a UFD. See for instance $\S 4.3$ of this expository piece on factorization in domains for more details.
The definition of a UFD involves two parts, existence and uniqueness of factorizations into atoms (irreducibles). Your quote mistakenly omits uniqueness. More precisely, D is a UFD iff every nonzero nonunit in D has a factorization into atoms, unique up to order and associates (unit factors). The uniqueness condition is easily seen to be equivalent to the fact that atoms are prime. Indeed, generally one may prove that in any domain, if an element has a prime factorization, then that is the unique atomic factorization, up to order and associates. The proof is straightforward - precisely the same as the classical proof for $\mathbb Z$.
Hence one can equivalently define a UFD as a domain where every nonzero nonunit has a prime factorization since, as said, such factorizations are necessarily unique.
