# Integral domain whose irreducible elements are not prime

Is there some integral domain such that none of its irreducible elements is prime?

Recall that a nonzero, non invertible element $$a$$ of an integral domain $$D$$ is said to be

• Irreducible, if for all $$b,c\in D$$ such that $$a=bc$$, then either $$b$$ or $$c$$ is a unit in $$D$$.
• Prime, if for all $$b,c\in D$$ such that $$a$$ divides $$bc$$, then either $$a$$ divides $$b$$ or $$a$$ divides $$c$$.

Clearly prime implies irreducible. The converse is not true in general, but is valid when $$D$$ is a UFD. Obviously these notions are vacuously equivalent if $$D$$ has not irreducible elements (see here for examples).

Summarizing, I want to know if there is some integral domain with at least one irreducible element, but without prime elements.

• An integral domain without prime elements is a field. I think that you don't call units irreducible. So it's impossible to have irreducible elements without primes. – T.J. Gaffney Jan 6 '14 at 9:51
• @Gaffney: can you give a proof? – Marek Jan 6 '14 at 9:52
• It's true that an integral domain with no prime ideals is a field (actually, I don't even have proof of this much weaker statement but it seems plausible). But can't it happen that a ring has only non-principal prime ideals therefore having no primes? – Marek Jan 6 '14 at 9:56
• @Marek: I'm mistaken; this is probably not true. – T.J. Gaffney Jan 6 '14 at 10:50
• Ah, a domain without non-trivial prime ideals must contain no non-trivial ideals at all (since every ideal is contained in a maximal ideal which is prime) and thus be a field. So we're indeed looking for rings having only non-principal prime ideals. – Marek Jan 6 '14 at 11:28

I found this paper which in Example 2.2 (d) (page 4) claims that $K[[X^2,X^3]]$ is an integral domain with no prime elements but many irreducible elements, for any field $K$.
Looking at $R=\mathbb{R}[X^2,X^3]$. It is atomic in the sense that every non-zero non-unit has a factorization into irreducibles, but the lengths of these factorizations can vary. This would be impossible for a factorization into primes. In a domain, if an element has two prime factorizations, they must be unique up to rearrangement and associate.
Consider the factorizations of the element $X^6=X^2\cdot X^2 \cdot X^2 = X^3 \cdot X^3$ are two factorizations into irreducibles. The first of length 3 and the second of length 2. ($X^3$ and $X^2$ are irreducible since $X \notin R$). This is an example of a bounded factorization domain which fails to be a half factorization domain.
Hint  The irreducibles of $\,\Bbb Q+x\Bbb R[[x]]\,$ are $\,rx\ne 0,\,r\in \Bbb R.\,$ But $\,rx\mid (\pi rx)^2,\,\ rx\nmid \pi r x\,$ by $\,\pi\not\in\Bbb Q$