In a unique factorization ring with unity (I am not considering commutativity and zero divisors in definition of UFD) irreducible implies prime.
And it was proved in ring with unity without zero divisors (commutativity not necessary) prime implies irreducible in question Prime which is not irreducible in non-commutative ring with unity without zero divisors.
So question is:
Does in a ring with unity prime implies irreducible or not?
Definitions: $p$ is prime iff $p|ab$ implies that $p|a$ or $p|b$, and $x$ is irreducible iff $x = ab$ implies that either $a$ or $b$ is a unit.
Best Answer
In a ring with zero divisors, a prime element need not be irreducible. A simple example is $\mathbb{Z}/(6)$, where we see that the elements $\overline{2},\, \overline{3}$ and $\overline{4}$ are all prime - the primes $\overline{2}$ and $\overline{4}$ are associated, $\overline{4} = \overline{5}\cdot\overline{2}$ - as one verifies, but we have
$$\overline{2} = \overline{2}\cdot \overline{4},\quad \overline{4} = \overline{2}\cdot \overline{2}\quad\text{and}\quad \overline{3} = \overline{3}\cdot \overline{3},$$
so none of them is irreducible.