Irreducible elements need not be prime

Counterexample

In the integral domain $\mathbb{Z}[\sqrt{-5}] = \{a + b\sqrt{-5} : a, b \in \mathbb{Z}\}$, the element $2$ is irreducible but not prime. Let $N(a + b\sqrt{-5}) = a^2 + 5b^2$. This norm is multiplicative, and the units are the elements of norm $1$, that is $\pm 1$. If $2 = x y$ with nonunits $x, y$, then $N(x) N(y) = 4$, so $N(x) = 2$. But $a^2 + 5b^2 = 2$ has no integer solution. So $2$ is irreducible. The element $2$ divides $(1 + \sqrt{-5})(1 - \sqrt{-5}) = 6$. It does not divide $1 \pm \sqrt{-5}$, because $(1 \pm \sqrt{-5})/2$ is not in $\mathbb{Z}[\sqrt{-5}]$. So $2$ is not prime.

Remarks

The same ring shows that factorization is not unique: $6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$. See [[$\mathbb{Z}[\sqrt{-5}]$ is not a UFD|0BSJYCG7NM]]. In a unique factorization domain, every irreducible element is prime.

Explanations

L1 human

In the integers, "cannot be factored" and "divides a factor of each product it divides" are the same property. In $\mathbb{Z}[\sqrt{-5}]$, the number $2$ cannot be factored, but it divides $6 = (1 + \sqrt{-5})(1 - \sqrt{-5})$ and neither factor. So the two properties are different in general rings.