Irreducible but not prime：Element which is prime but not irreducible.

Using $$ \mathbb{Z}[x]/(x2-x) $$

Notice that the elements of this ring are of the form $a + bx$.To show that $x$ is prime, suppose $x$ divides a non-zero number $m$ :

then $m = bx$ for some $b$. if $m = (c + dx)(c+dx)$, then without loss of generality (wlog) $c = 0$ and $d \neq 0$. Hence $x$ divides $c + dx = dx$. That is, $x$ is prime.

But $x$ is not ``irreducible since $x = x2$ and $x$ is not a unit (to show it is not a unit, show that $x(a+bx) = 1$ cannot be solved.

However, using the reference in the comments above, this example would not apply depending on the used definition of irreducible. The correct definition is:

In a ring $R$ (with out without zero divisors) an element $a$ is irreducible if whenever $a = bc$ then either $(a) = (b)$ or $(a) = (c)$. (See abstract of ref given in the comments above [1]).

Using the above definition, then every prime element is irreducible:

Suppose $p$ is a prime element. If $p = p...