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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.

roi_saumon
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    "Suppose $x^2$ is irreducible. In particular, this means it isn't a prime." You meant to say it isn't a unit, right? – saulspatz Nov 17 '18 at 18:18
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    Beware $ $ There is no standard of "irreducible" in rings with zero-divisors. Various incompatible definitions are in use. e.g. see the paper linked here, where (Corollary 2.7) idempotents are irreducible $\iff$ prime. – Bill Dubuque Nov 17 '18 at 22:22
  • @BillDubuque okay, I see your point. So is it true that for all $a$, $b$ such that $x^2=ab$ we have either $x^2 \vert a$ and $a \vert x^2$ or $x^2 \vert b$ and $b \vert x^2$? Also, for the definition of prime there is no problem no? $ab \in (p) \iff p \vert ab$ – roi_saumon Nov 17 '18 at 23:37

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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

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