An element c is irreducible if it is a nonzero nonunit, and c = a*b only when a or b is a unit.
Note that 2 is prime in Z6, but 2 = 2×4, so 2 is not irreducible. Similarly, irreducible elements need not be prime. We demonstrated this by adjoining the square root of -5 to Z.
Don't be confused by the inconsistent definitions. A prime integer was defined as a number that had no proper divisors, in other words, a number that could not be reduced. We then proved that these irreducible numbers are indeed prime; that p into s×t means p divides s or p divides t. Since prime and irreducible mean the same thing in the integers, we didn't worry about it too much. But now we have to get use to the general definitions of irreducible (no proper divisors) and prime (always divides some element in the product).
In an integral domain, p prime implies p irreducible. Write p = ab, as though p could be reduced. Now p divides a or b; say pc = a. Thus p = pcb, and p*(1-cb) = 0. Remember there are no zero divisors in an integral domain. With p nonzero, 1-cb = 0, cb = 1, and b is a unit. Thus p is irreeucible after all.