# 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.

• "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
• 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

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