Integral domain that is not a factorization domain

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?

Incidentally, integral domains in which every nonzero nonunit can be factored into irreducibles are usually called atomic (originating from the idea that irreducibles are "atoms" of a sort).

As for examples of non-atomic domains, there are many.

• Let $R=\overline{\mathbb{Z}}$ be the integral closure of $\mathbb{Z}$ in $\mathbb{C}$. In other words, $R$ consists of every complex number that is the root of a nonzero monic polynomial in $\mathbb{Z}[x]$ (And yes, $R$ is a ring, for an elegant proof of why integral closures are rings, check out Kaplansky's book Commutative Rings). Note that every element of $\mathbb{Q}\setminus \mathbb{Z}$ is not in $R$ (eg there is no monic polynomial in $\mathbb{Z}[x]$ with $\frac{1}{2}$ as a root), thus $R$ is not a field.

  So, pick any nonzero nonunit $r\in R$. Then, there exist $a_0,a_1,\cdots,a_{n-1}\in \mathbb{Z}$ (not all zero) such that

$$a_0+a_1 r + a_2 r^2 + \cdots + a_{n-1}r^{n-1} + r^n = 0$$

Note that $\sqrt{r}$ is a complex number and in fact

$$a_0 + a_1 (\sqrt{r})^2 + a_3 (\sqrt{r})^6 + \cdots + a_{n-1}(\sqrt{r})^{2n-2} + (\sqrt{r})^{2n}=0$$

Therefore $\sqrt{r}\in R$ and we have $r=\sqrt{r}\sqrt{r}$. Thus no nonzero nonunit element of $R$ is irreducible and in fact $R$ is an antimatter domain.

• As another example, let $K$ be any field, and let $$R=K[x,y/x,y/x^2,y/x^3,\cdots,y/x^n,\cdots]$$

In a sense, you can think of elements of $R$ as being "polynomials" in $x$ and $y$, except that you can divide $y$ by any power of $x$ that you'd like.

Let $$S=\{f(x,y)\in R\,\vert\,f\textrm{ has a nonzero constant term}\}$$

It is clear that $S$ is multiplicatively closed (ie for all $f,g\in S$, $fg\in S$), so let $T=R_S$ be the localization of $R$ at $S$. In other words,

$$T=\left\{\frac{f(x,y)}{g(x,y)}\,\vert\,f\in R, g\in S\right\}$$

It's not too tough to show that any nonzero element of $T$ is of the form $\frac{y^n}{x^m}u$ for $u\in U(T)$, $n\geq 0$, and $m\leq 0$ if $n=0$. And once you have that, it's easy to show that for all nonzero $\alpha,\beta\in T$ either $\alpha \vert \beta$ or $\beta \vert \alpha$--ie $T$ is a valuation domain.

In fact, the (unique, as Bill pointed out in his answer) atom of $T$ is $x$. Thus the only elements of $T$ that can be written as a product of irreducibles are of the form $ux^n$ for $n>0$ and $u\in U(T)$. In particular, $y$ cannot be written as a product of irreducibles. Therefore $R$ is not atomic (but not antimatter, either).

Sure. What you want is an integral domain which does not satisfy the ascending chain condition on principal ideals (ACCP), meaning that there is an infinite chain of principle ideals $P_0\subset P_1\subset \cdots$ where each inclusion is strict. Not all of these are not factorization domains, but it's a good place to start. Consider the ring $A$ of algebraic integers. It is not hard to see that $$(2)\subset (2^{1/2})\subset (2^{1/3})\subset \cdots$$ the details of which I leave to you. I'm not certain that this is not a factorization domain, but I would be surprised if it is.

Any valuation domain contains at most one atom (irreducible) up to associates since, by definition, $\rm\ p\ |\ q\ \ or\ \ q\ |\ p\ .\:$ So any such atom generates the unique maximal ideal. Hence a valuation domain $\rm\:D\:$ whose maximal ideal is not principal has no atoms, so it is not a factorization domain. If you seek examples that also have atoms then simply enlarge $\rm\:D\:$ to $\rm\:D[x]\:,\:$ which has the atom $\rm\:x\:.$

Domains with no atoms are known as antimatter domains. Google it for related literature.

The example below is perhaps is fairly simple, but requires a fair amount of background information. It is offered as a possibly useful technique for constructing counterexamples. The presentation is of necessity abbreviated. Details about the ultraproduct and ultrapower constructions can be found in many places.

Let $I$ be an infinite index set, for definiteness the set of positive integers. Let $D$ be a non-principal ultrafilter on $I$. Let $\mathbb{Z}$ be the integers with the usual addition and multiplication.

We give a quick definition of the ultrapower $\mathbb{Z}^I/D$. Consider the set $\mathbb{Z}^I$ of all functions $f\colon I\to \mathbb{Z}$. Call two such functions $f$ and $g$ equivalent modulo $D$ if $\{i\colon f(i)=g(i)\} \in D$. On the equivalence classes, define addition and multiplication pointwise modulo $D$. For example, if $f/D$ and $g/D$ are elements of $\mathbb{Z}^I/D$, define $h$ by $h(i)=f(i)+g(i)$, and define $f/D+g/D$ by $f/D+g/D=h/D$. Multiplication is defined analogously. It is easy to verify that the sum and product are well-defined, that is, independent of the choice of representatives.

If $D$ is a non-principal ultrafilter, then $\mathbb{Z}^I/D$ is not isomorphic to $\mathbb{Z}$. However, by Los's Theorem, $\mathbb{Z}^I/D$ is elementarily equivalent to $\mathbb{Z}$. That means that for any sentence $\varphi$ in the usual first-order language of ring theory, $\varphi$ is true in $\mathbb{Z}^I/D$ iff $\varphi$ is true in $\mathbb{Z}$.

As a simple consequence, $\mathbb{Z}^I/D$ is an integral domain, since the property of being an integral domain is easily expressible as a sentence of first-order ring theory. But much more is true. Since, for example, Bezout's Theorem can be stated as a sentence in our first-order language, we can conclude that it holds in $\mathbb{Z}^I/D$.

Now we show that some non-zero non-unit elements of $\mathbb{Z}^I/D$ cannot be expressed as a product of a finite number of irreducibles. Let $f(i)=2^i$, and consider the object $f/D \in \mathbb{Z}^I/D$. Suppose that $f/D=f_1/D\cdot f_2/D \cdots f_n/D$ for some positive integer $n$. Since the intersection of finitely many sets in $D$ is also in $D$, for some $k \le n$ the set of $i$ such that $f_k(i)$ is divisible by $4$ is in the ultrafilter $D$. It follows that $f_k/D$ is not irreducible.

Comment: How is this consistent with the fact that $\mathbb{Z}^I/D$ is elementarily equivalent to $\mathbb{Z}$? After all, $\mathbb{Z}$ is a "factorization domain." The answer is that being a factorization domain is not a first-order property in the elementary theory of integral domains. We can say in our language that $x$ is the product of at most $17$ irreducibles, but not that it is the product of a finite number of irreducibles.

An ultrafilter $D$ on $I$ can be thought of as a $\{0,1\}$-valued finitely additive "measure" on the power set of $I$. (Sets in the ultrafilter have "measure" $1$, sets not in the ultrafilter have measure $0$, and every set has measure $1$ or $0$.) Now we can think of functions that agree on a set in $D$ as being equal "almost everywhere." That point of view can make the construction of examples more natural.

Instead of constructing an example based on $f(i)=2^i$, we could make other choices. For example, $f(i)$ could be the product of the first $i$ primes.