Does any integral domain contain an irreducible element? Let $R$ be an integral domain which is not a field. 

Does $R$ necessarily have an irreducible element? 

I suspect the answer is no, but I couldn't find an example showing that... 
 A: It is proven here that the ring of all algebraic integers has no irreducible elements.
A: Another class of examples is given by valuation rings whose value group is divisible.
A: Hint $\ $ There are rings, not fields, closed under square roots (e.g. the ring of all algebraic integers). In such a ring there are no irreducibles since every element factors: $\rm\ a = \sqrt{a} \sqrt{a}.$
A concrete example: $ $ let $\rm\, D = \Bbb Q[x,x^{1/2},x^{1/4},x^{1/8},\ldots].\,$ Every element $\ne0$ is writable in the form $\rm\ d_i x^{i}\! + d_j x^j\! +\cdots\! + d_k x^k\! = x^i\, f\ \,$ for $\rm\,\ f\in D,\,\ d_i\in\Bbb Q,\,\ d_i\ne 0,\:$ and $\rm\:i < j < \ldots < k,\:$ e.g.
$$\rm a\, x^{1/8}\! + b\, x^{1/4}\! + c\, x^{1/2} =\, x^{1/8} (a + b\, x^{1/8}\! + c\, x^{3/8})\, =\, x^{1/8}\, f,\quad a\ne 0$$
All exponents of $\rm\,x\,$  in terms of $\rm\,f\,$ remain of form $\rm\,n/2^m\!,\,$ being differences of such.  We can force all the cofactors $\rm\,f\,$ to be units by adjoining inverses of all such elements, as follows. Since $\rm\,f\,$ has "constant" term $\rm\:a\ne 0,\,\ f\,$  is not in the maximal ideal $\rm\, M = (x,x^{1/2},x^{1/4},x^{1/8},\ldots).$ Therefore, localizing at $\rm\,M,\,$ i.e. adjoining inverses of all elements $\rm\not\in M,\,$ turns all the $\rm\,f\,$ cofactors into units, yielding  a domain $\rm\,\bar D,\,$ not a field $\rm(x^{-1}\!\not\in \bar D),\,$ where all elements have form $\rm\,x^k\, u,\,$ for some unit $\rm\,u.\,$ Since $\rm\ a = x^k = (x^{k/2})^2 = \sqrt{a}\sqrt{a},\ $ this domain has no irreducibles.  
