Prime always implies irreducible; in a UFD, irreducible also implies prime.Suggest a slogan
Let \(R\) be an integral domain.
Every prime element of \(R\) is irreducible.
If \(R\) is a unique factorization domain, a nonzero nonunit \(r \in R\) is prime if and only if \(r\) is irreducible.
In the integral domain \(\ZZ[\sqrt{-5}]\), the element \(2\) is irreducible but not prime: \(2\) divides \(6 = (1+\sqrt{-5})(1-\sqrt{-5})\) but divides neither factor.