Is there an integral domain such that none of its irreducible elements is 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.
 A: 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 $\,\pi\not\in\Bbb Q$
A: 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.
A: 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 $K$.
