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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?

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$k[x_1,x_2,x_3,...]$ with $x_i=x_{i+1}^2$ – yoyo Jan 6 '12 at 0:31

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.

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It is indeed not a factorization domain: if $a$ is a nonzero nonunit, then $\sqrt{a}$ (for a fixed choice of square root) is also an algebraic integer that is a nonzero nonunit. Since $a=\sqrt{a}\sqrt{a}$, no nonzero nonunit is irreducible, so no nonzero nonunit can be written as a product of irreducibles. – Arturo Magidin Jan 6 '12 at 1:09
@Alex: And of course, there are nonzero nonunits, e.g., every integer other than $1$ and $-1$ is a nonzero nonunit, since the only rational numbers that are algebraic integers are the integers... – Arturo Magidin Jan 6 '12 at 2:17

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).

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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.

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Generally for Bezout domains a more satisfying treatment can be found in "Monoid domain construction of antimatter domains" Comm Algebra 35(10)(2007) 3236-3241. For a preprint look up: – mzafrullah Jul 28 '13 at 2:35
Oh yes and the reason why I could remember the above article is the fact that in a Bezout domain too a principal prime ideal is maximal and so a Bezout domain with no maximal ideal principal is antimatter, as in a Bezout domain too an irreducible element is a prime. – mzafrullah Jul 28 '13 at 2:43

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.

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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".

  1. 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.
  2. 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].
  3. 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.)

  4. 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.
  5. 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.


[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.

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Welcome to MSE. I've fixed up your post a little bit to take advantage of the built-in MarkDown syntax for formatting. In the future I encourage making the best use of them. Also, this site supports MathJAX, which is very similar to LaTeX. I also encourage you to consult this guide and re-format your post making use of it. Lastly, if you have any questions on formatting your post (especially on how to achieve a desired outcome), don't hesitate to ask on meta. – Willie Wong Jul 31 '13 at 8:51
Thanks Willie! I would try. But as I stagger in only occasionally, I wonder if the old dog should learn new tricks! – mzafrullah Jul 31 '13 at 12:41

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