Element which is prime but not irreducible. I wanted to find an element which is irreducible but not prime. I found on wiki the example that says that $x^2$ is prime but not irreducible in  $\mathbb{Q}[x]/(x^2+x)$.
My reasoning why this is not a irreducible element is as following.
Suppose $x^2$ is irreducible. In particular, this means it isn't a unit.
But $x^2=xx=(-x^2) (-x^2)=x^2x^2$ And so this would be a product of two non-units. Does this makes sense?
To show that it is not a prime I have to show that $x^2 \vert ab \implies x^2 \vert a$ or $x^2 \vert b$ but I don't know how to show this.
 A: Using 
$$
  \mathbb{Z}[x]/(x^2-x)
$$
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 $m = (c + dx)(c'+d'x)$, 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 $x = x^2$ 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 $p = p_1 p_2$ then wlog $p$ divides $p_1$ since $p$ is prime.  We have that $(p_1) = (p)$ since $p_1 \in (p)$ ($p$ divides $p_1$) and $p \in (p_1)$ ($p_1$ divides $p$).    
[1] Anderson and Chun, Irreducible Elements in Commutative rings with non-zero divisors, http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.442.1455&rep=rep1&type=pdf
