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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.

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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. –  Gaffney Jan 6 at 9:51
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@Gaffney: can you give a proof? –  Marek Jan 6 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 at 9:56
    
@Marek: I'm mistaken; this is probably not true. –  Gaffney Jan 6 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 at 11:28
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3 Answers

up vote 5 down vote accepted

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 $K$.

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Unfortunately I do not have the rep to comment on your answer above, but I just wanted to point out that this example is very similar to another interesting example of factorization properties.

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.

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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 $\,\pi\not\in\Bbb Q$

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