TITLE: Integral domain that is not a factorization domain
QUESTION [13 upvotes]: I am looking for rings that are integral domains but not factorization domains, that is, rings in which it is not possible to express a nonzero nonunit element as a product of irreducible elements. 

Do you know any example?

REPLY [3 votes]: First off the question is a bit garbled. I think you wanted to say that: I am looking for rings that are integral domains but not factorization domains, with the property that it is not possible, in these domains, to express a nonzero nonunit element as a product of irreducible elements. Do you know any example?
Of course there are plenty of examples. But let me be clear I am taking "a nonzero nonunit" to mean "any nonzero nonunit element".

Take a non-discrete rank one valuation domain V. The maximal ideal M of V is not principal and it is not too hard to see that assuming any element a product of irreducible elements would make M principal as in a valuation domain an irreducible element is a prime. So no nonzero nonunit is a product of irreducible elements. A domain in which no nonzero nonunit is expressible as a product of irreducible elements is called an antimatter domain.

Let R be the semigroup ring C[Q^+] where C is the field of complex numbers and Q^+ the set of non-negative rational numbers. The explanation might take time, so let me refer you to Theorem 1 of [1].

Any Bezout domain (every finitely generated ideal is principal) whose maximal ideals are all non-principal. The argument is the same as in the case of a non-discrete rank one valuation domain or generally as in the case of a valuation domain whose maximal ideal is non-principal. The main argument is: In a Bezout domain an irreducible element is a prime and a prime element generates a maximal ideal. (That in a Bezout domain an irreducible element is a prime can be shown as follows: Let a be irreducible in a Bezout domain R and let a|bc where a divides neither of b and c then as a is irreducible a has no nonunit common factors with b or with c. Thus (a, b)= R and  (a, c) = R. But then (a, b)(a, c)= R or (a^2 , ac, ab,  bc) =R  but this give rise to a contradiction because we assumed at the outset that a|bc and that a is irreducible and hence a nonunit. (The proof is exactly the same as in showing that in a PID an irreducible element is prime.) Next let p be a prime in a Bezout domain To show that pR is maximal all you have to show is that for every x in R\pR, xR +pR =R But this is obvious as x in R\pR means that p does not divide x and so must not have a nonunit common factor with x, but xR + pR must be principal and this forces xR + pR =R.)
Now let R be a Bezout domain whose maximal ideals are all non-principal and suppose by way of contradiction that there is a nonzero nonunit x in R such that x is a product of irreducible elements then any of these irreducible factors generates a maximal ideal a contradiction to the assumption that all maximal ideals of R are non-principal. This example is a part of a Theorem in [1] (see the comment after Theorem 3.)

Any integral domain D over which every non constant polynomial splits as a product of linear factors is an example. For such an integral domain let a be irreducible and consider X^2 – a. Then by the condition X^2 –a = (X-r)(X-s), which forces s =-r  and so s^2 = a which contradicts the assumption that a is irreducible.  Note that the ring of all algebraic integers is indeed an example.
Most of the above examples are Bezout domains and hence Schreier. Here’s an example of a domain that is not Schreier and hence not Bezout yet an antimatter domain.

Let V be a valuation domain whose maximal ideal is non-principal, let L be a nontrivial field extension of K the quotient field of V and let R = V + XL[[X]] then R is antimatter and not schreier. (For illustration see Example 2.11 and Proposition 3.10 of [2].) Of course if you look up [1], [2} and references there you can get many more examples of antimatter domains.
In case you insist that you are looking for integral domains in which some elements have the property that they are not expressible as products of irreducible elements, even then the above three examples apply. However the field of examples increases considerably. I will give only one type R= Z+XQ[X], here X cannot be expressed as a product of irreducible elements as using the degree argument,
X= a_1 a_2 …a_n would force one of the a_i say a_1 to be of degree one in X and the rest of degree 0 in X and this will force X = X/(a2 a_3 …a_n) (a_2a_3…a_n). But of these X/(a2 a_3 …a_n) can be shown to be not irreducible as X/(a2 a_3 …a_n)= (X/p(a2 a_3 …a_n))p for any prime p.



References
[1] D.D. Anderson, Jim Coykendall, Linda Hill and M. Zafrullah, “Monoid domain construction of antimatter domains, Comm. Algebra
[2] D.D. Anderson and M. Zafrullah, The Schreier property and Gauss’ Lemma, Bolletino
U. M. I. (8)10-B (2007)43–62.
Note: I do owe an explanation for calling the question garbled. I do apologize but when you say the domain is not a UFD it could mean many things more than each or some element is not expressible as a (finite) product of irreducible elements and the use of "that is" seem to indicate that you think that not being a UFD is equivalent to what follows. By the way The ring of entire function is an example of a Bezout domain in which every non-zero nonunit is the associate of a countable product of finite powers of non-associated primes. Yet the ring of entire functions is not a UFD because some elements of it cannot be expressed as (finite) products of irreducible elements.