Irreducible and not prime

Question

You know a counterexample for:

Let $R$ be a commutative ring ($1\neq{0}$). If $a\in{R}$ is irreducible then $(a)$ is prime.

Thanks.

Answer 1

The canonical example is $\mathbb{Z}[\sqrt{-5}]$. Consider: $$2 \cdot 3 = 6 = (1 + \sqrt{-5})(1 - \sqrt{-5})$$ One can verify that $2$ is irreducible, but $2$ does not divide $(1 + \sqrt{-5})$ or $(1 - \sqrt{-5})$. Therefore, $(2)$ is not prime.

Answer 2

In a UFD, irreducibles are prime (indeed, this condition plus acc on principal ideals characterizes UFDs among domains), so one should look for a ring that is not a UFD. One example would be $R = k[x,y,z,w]/(xy-zw)$. The element $x \in R$ is irreducible, but $(x)$ is not prime, as $zw = xy \in (x)$, but neither $z$ nor $w$ is in $(x)$.

Answer 3

Here is an example that is perhaps simpler than the common quadratic integer examples. $\:$ Let $\rm R = \mathbb R + x\:\mathbb C[x],\:$ i.e. the ring of all polynomials with complex coefficients and real constant coefficient. Here $\rm\:x^2\:$ has infinitely many distinct factorizations into irreducibles

$$\rm x^2\ =\ (c\: x)\: (c^{-1}\: x),\quad c = r + {\it i},\quad \forall\: r\in \mathbb R$$

The factors are nonassociate irreducibles in $\rm R$ since, for $\rm\:r,s\in \mathbb R$

$$\rm (r+{\it i})x\ |\ (s+{\it i})x\ \ in\ \ R\iff \frac{(s+{\it i})\:x}{(r+{\it i})\:x}\in R\iff \frac{s+{\it i}}{r+{\it i}}\in \mathbb R\iff r = s$$

In particular, the irreducible $\rm\,x\,$ is not prime.

Note $\:$ Such constructions are often used by ring theorists since they yield a very rich source of (counter-) examples, e.g. see $\ $ M. Zafrullah, Various facets of rings between $\rm\:D[X]\:$ and $\rm\:K[X].$