# Examples of UFD2 but not UFD1

I overcame a theorem which says:
Let R be an Integral Domain. If every Irreducible element is Prime then it satisfies UFD 2. Converse (i.e, UFD 2 implies every Irreducible element is Prime) is true if it is UFD 1.

where,
UFD 1 is basically the Existence of a Factorization of every element.
UFD 2 is the Uniqueness of the Factorization for every factorization of every element.
We know that being UFD satisfies both UFD 1 and UFD 2.

Now I was having trouble finding examples for the situation when UFD 2 satisfies but NOT UFD 1.

Can anyone help me with this? Thanks in advance. ( Please write in a bit details if possible)

My basic idea was to figure out examples for the case that UFD 2 does not imply that Every Irreducible element is Prime by avoiding UFD 1 ( if factorization exists then it has to be unique but it is not mandatory that every element will have factorization).

• UFD 2 is, in fact, equivalent to every irreducible element being prime. – Geoffrey Trang Aug 22 '20 at 16:32
• But that is for UFD.....UFD2 does not guarantee the existence. – tom_choudhurry Aug 22 '20 at 16:36

Note that a ring has to be non-Noetherian to fail UFD 2, since if $$x$$ fails to have a factorization into irreducibles, then it must be that $$x$$ does factor somehow as $$x=ab$$ where $$a$$ and $$b$$ are not units and either $$a$$ or $$b$$ lacks a factorization into irreducibles - and we can then look at a factorization of that an so on to yield an infinite chain $$x_1,x_2,x_3,\ldots$$ where each term strictly divides the last - and then $$(x_1)\subsetneq (x_2) \subsetneq (x_3) \subsetneq\ldots$$ would be an infinite ascending chain of ideals.

So, we can just start by looking at our favorite non-Noetherian ring and see what happens - the example that comes to my mind would be letting $$R$$ be the set of polynomial expressions with rational coefficients in terms of the form $$x^{n/2^k}$$ for $$n,k\in\mathbb N$$ - or, equivalently, the direct limit of the rings $$\mathbb Q[x]\rightarrow\mathbb Q[x^{1/2}] \rightarrow \mathbb Q[x^{1/4}]\rightarrow\ldots .$$ The element $$x$$ lacks a factorization into irreducibles in this ring. This follows from noting that the factors of $$x^{\alpha}$$ are just the terms of the form $$x^{\beta}$$ where $$\beta \leq \alpha$$ - but none of these are irreducible.

However, it is true that every irreducible element $$p$$ is prime. In particular, suppose that we had some $$a,b\in R$$ such that $$p|ab$$. Then, it must be that all of $$p$$ and $$a$$ and $$b$$ are in some ring $$\mathbb Q[x^{1/2^k}]$$ and that $$p$$ is irreducible in this ring - but this ring is isomorphic to $$\mathbb Q[x]$$ which is a PID meaning $$p$$ is prime in it, thus $$p$$ must divide either $$a$$ or $$b$$ in this ring and therefore also in $$R$$. Thus, this ring has uniqueness of prime factorizations, but not existence.

One idea would be to find a ring where no element has a factorisation. An example is the ring of algebraic numbers $$\overline{\mathbb Z}$$, which has no irreducible elements at all: if $$a$$ is algebraic non-unit, then so is $$\sqrt a$$; since $$a = \sqrt a\cdot\sqrt a$$, $$a$$ is not irreducible.

For a less vacuous example, consider the ring $$\overline{\mathbb Z}[X]$$ of polynomials with algebraic coefficients. In this ring, elements of the form $$aX + b$$ are irreducible, where $$a, b\in \overline{\mathbb Z}[X]$$ are coprime. If an element $$f\in \overline{\mathbb Z}[X]$$ has a factorisation, then it must be primitive: its coefficients must have no common factor. Such an element can be factorised uniquely as a product of primitive linear factors.

• If I have understood correctly ...your first example is basically UFD 1 but not UFD 2 – tom_choudhurry Aug 22 '20 at 18:31
• The opposite. There are no irreducibles, so no element can be factorised as a product of irreducibles. UFD2 is vacuously true. – Mathmo123 Aug 22 '20 at 18:44
• ohh okkk ...I see your point ...Thanks...The examples were nice – tom_choudhurry Aug 22 '20 at 19:52