Integral domain that is not a factorization domain

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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).

Answer 4

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.