# Exhibit an integral domain $R$ and a non-zero non-unit element of $R$ that is not a product of irreducibles.

Exhibit an integral domain $R$ and a non-zero non-unit element of $R$ that is not a product of irreducibles.

My thoughts so far: I don't really have a clue. Could anyone direct me on how to think about this? I'm struggling to get my head round irreducibles.

Thanks.

• Hint: if $R$ is noetherian, any element is a product of irreducible ones. Do you know any non-noetherian ring?
– Plop
May 6 '11 at 20:06
• wouldn't all fields work for this?? May 6 '11 at 21:27
• @quanta: In a field, there are no nonzero nonunits. If the statement was "an integral domain R such that for all nonzero nonunit elements a, a is not a product of irreducibles", then a field wouldd satisfy it by vacuity. But here they are asking to exhibit an $R$ and an $a$, so fields don't work. May 7 '11 at 19:01

HINT $\$ Consider a domain, not a field, that is closed under square-roots. It has no irreducibles since every element factors $\rm\ d\ =\ \sqrt{d}\ \sqrt{d}\:.\$ For example, the domain of all algebraic integers.

A standard example of a non-Noetherian domain is the ring $R=\{f(x) \in \mathbb{Q}[x]: f(0) \in \mathbb{Z} \}$; that is, the ring of rational polynomials with integer constant term. The units of $R$ are just $\pm 1$. The polynomial $x$ is reducible, since it factors as $2 \cdot \frac{1}{2}x$, but is not a product of irreducibles, since one of the factors would have to be $qx$ for some rational $q$, and again this factors as $2 \cdot \frac{q}{2}x$.

The following is a not so simple answer, actually two related answers. They will look more complicated than the already posted answers. But it may be interesting to see how ideas from Model Theory can be brought to bear on the problem, ideas that may not be completely familiar to everyone.

Using the Compactness Theorem: Let $\mathbb{L}$ be the first-order language that has two binary function symbols $+$ and $\times$, and constant symbols $0$, $1$, and $a$. The first four symbols are the usual ones when one formalizes the theory of rings with unit, and the last symbol $a$ is for stting up the compactness argument.

We now describe a set $\mathbb{A}$ of axioms. The set $\mathbb{A}$ contains the usual axioms for integral domain. Note that there is a formula $\text{NU}(x)$ in our language that says that $x$ is not a unit.

In addition to the axioms for integral domain, $\mathbb{A}$ contains the following axioms.

First comes an axiom that says that $a$ is not equal to $0$, and is not a unit.

Add also the axiom $$\exists x_1\exists x_2(\text{NU}(x_1) \land \text{NU}(x_2) \land a=x_1 \times x_2)$$ Add also the axiom $$\exists x_1\exists x_2\exists x_3(\text{NU}(x_1) \land \text{NU}(x_2) \land \text{NU}(x_3)\land a=x_1 \times x_2 \times x_3)$$ and continue adding axioms in this fashion forever. Note that it is quite common for theories to have an infinite number of axioms: Peano Arithmetic and ZFC are standard examples.

Observe that any finite subset $F$ of $\mathbb{A}$ has a model. Indeed, that model can be chosen to be the integers, with the constant symbol $a$ interpreted suitably. Specifically, if the "largest" of the special axioms in $F$ calls for $a$ to be a product of (at least) $k$ non-units, we can interpret $a$ as $2^k$, or as the product of the first $k$ primes.

Since every finite subset of $\mathbb{A}$ has a model, then by the Compactness Theorem the full set $\mathbb{A}$ has a model $\mathcal{M}$. This model is an integral domain. The interpretation of the constant symbol $a$ in $\mathcal{M}$ then satisfies all of our special axioms, so we have found our desired example.

For people unfamiliar with basic Model Theory, I should point out that the above argument is essentially automatic. It is no exaggeration to say that it could be replaced in its entirety by "The result is obvious by compactness."

Using Ultrapowers: We give only a brief sketch. For details look first http://en.wikipedia.org/wiki/Ultraproduct. Let $\mathbb{Z}$ be the integers, under the usual addition and multiplication. Let $I$ be a countably infinite index set, like the natural numbers, and let $U$ be a non-principal ultrafilter on $I$. Look at the ultrapower $\mathbb{Z}^I/U$. The elements of the ultrapower are equivalence classes $f/U$ of functions from $I$ to $\mathbb{Z}$ (that is, equivalence classes of sequences). Two sequences $f$ and $g$ are $U$-equivalent if the set of $i$ such that $f(i)=g(i)$ belongs to $U$. Define addition and multiplication on $\mathbb{Z}^I/U$ pointwise modulo $U$.

It turns out that $\mathbb{Z}^I/U$ is a non-standard model of the "elementary theory of the integers." And if we let, for example, $f(i)=2^i$, then $f/U$ can be expressed as a product of $n$ non-units for any positive integer $n$.

More generally, the ultraproduct can be used to produce algebraic objects with interesting properties.

The two methods are in a sense quite similar, with the ultrapower providing a sort of explicit (if one can accept a non-principal ultrafilter as explicit!) way of constructing a model of the type guaranteed to exist by the Compactness Theorem. (By the way, the Compactness Theorem turns out to be equivalent to the assertion that a certain topological space is compact, though I believe the name "Compactness Theorem" was used well before this was realized. The ultraproduct can be viewd as a way of producing certain limit objects.)

Such an element can be factored, each factor can be factored, each factor can be factored, etc.

Changing the problem into an additive one, you would want to find an element that can be written as a sum of two strictly smaller numbers, each of which can be written as a sum of two strictly smaller numbers, each of which... etc.

Perhaps thinking along the lines of: $$1 = \frac{1}{2}+\frac{1}{2} = \left(\frac{1}{4}+\frac{1}{4}\right) + \left(\frac{1}{4}+\frac{1}{4}\right) = \cdots = \left(\frac{1}{2^n}+\frac{1}{2^n}\right) + \cdots + \left(\frac{1}{2^n}+\frac{1}{2^n}\right) = \cdots$$

Hmmm... Is there any way we could turn that into some kind of multiplicative, instead of additive, set of equalities?

• @user6312: You can post your solution as an "answer" (and later even accept it); it can then also be looked over by others for errors, if you want. May 6 '11 at 20:27
• It seemed best to delete, since the error was a silly one. Surely one does not accept one's own answers! May 7 '11 at 0:50
• @user6312: One can accept an answer to one's own question. I thought, given your "thanks", that you might be the OP ('userxxxxx`; didn't bother double-checking the numbers). But of course, one cannot accept one's answer to someone else's question. May 7 '11 at 5:47
• I added an answer that may amuse or interest you. May 8 '11 at 16:51

how about something like $\mathbb{C}[x_1,x_2,x_3,...]$ where $x_i^2=x_{i-1}$ for $i>1$