# Is there a “pseudo-UFD” without unity?

It is known that every euclidean domain has a product identity element. Does the same happens to the UFDs?

I tried to find a counter example playing with the powers of $$2$$ with strange operations, but I get stuck. Any idea?

Important details

By domain I mean, a commutative ring (maybe without unity) with no zero divisors.

In a domain without unity, we can define the divisibility relation ($$a$$ divides $$b$$ if and only if there exists $$c$$ so that $$ac=b$$), also the irreducibility concept ($$a\not=0$$ is reducible if and only if there exists $$b,c\not=0$$ such that $$a=bc$$, so, an irreducible element is an element which is not reducible), and the "associates" relation ($$a$$ and $$b$$ are associates if and only if $$a|b$$ and $$b|a$$).

So we can define something like an UFD, which we call a "pseudo-UFD", like this

Let $$A$$ be a commutative ring without unity and without zero divisor, we will call it "pseudo-UFD" if

• For each $$a\in A$$, there exists irreducible elements $$p_1,\dots,p_r$$ (not necessarily different) so that $$a = p_1\cdots p_r$$.
• If $$a=q_1\cdots q_s$$ (being $$q_1,\dots,q_s$$ irreducible elements not necessarily different), then $$r=s$$ and, maybe after reordering, $$p_i|q_i$$ and $$q_i|p_i$$ for every $$i\in\{1,\dots,r\}$$.

As @rschwieb pointed out, in an unitless ring without zero divisors, the associates relation is empty, so the second item of the definition will change to

• If $$a=q_1\cdots q_s$$ (being $$q_1,\dots,q_s$$ irreducible elements not necessarily different), then $$r=s$$ and, maybe after reordering, $$p_i=q_i$$ for every $$i\in\{1,\dots,r\}$$.

So, is there a "pseudo-UFD" without unity?

• Comments are not for extended discussion; this conversation has been moved to chat. – Aloizio Macedo Nov 21 '18 at 17:42

Personally, I would hope that any proposed definition would permit something like $$2\mathbb Z$$ to be a pseudo-UFD.
The main problem is to define what "uniqueness of factorization" means, and that is what you've attempted with the criterion mentioned. Requiring two factorizations into irreducibles to be the same length is a must. The main barrier is to deal with associate elements without mentioning units. The way you have things now, we have to allow $$\pm x$$ to be non-associate to one another, but this is unattractive.
One can patch this artificially by saying that $$\pm x$$ are associates to one another, and then perhaps we do get $$2\mathbb Z$$ to be a pseudo-UFD. But unfortunately, this convention does not extend the normal definition of UFDs, since lots of UFDs have many more associates!
So, one can see here the challenges posed by talking about factorization in domains without identity. Perhaps it would be most fruitful to explore which domains without identity can be embedded in domains such that the only units are $$\pm 1$$, so that the ad-hoc definition of associates mentioned above works.
Definition of a UFD is: $$R$$ is an integral domain ($$R$$ is a commutative ring having unity and no zero-divisors) such that each $$a\in R \setminus (R^* \cup {0})$$ factors into a product of irreducible elements.Definition clears your doubt