Why Can't we Factor Invertible Elements?

Question

I'm currently studying Herstein's Algebra; specifically, UFDs and the abstract notion of factorization. This is perhaps more of an intuitive question than one with a hard answer.

We define irreducible elements, first and foremost, as elements which are not units. That is, elements which do not have multiplicative inverses. Implicitly, then, wouldn't this imply that, in some way, we do not care about the question of factorization for elements which have multiplicative inverses? Why is this question not interesting?

Separate but related question: Moreover, it is clear that any field is a UFD. Yet, every element in a field is unit so how can we, intuitively, talk about fields being UFDs? I understand the definition vacuously holds for non-unit elements yet it seems wrong to me to slap this Unique Factorization label on a structure where "factorization is not interesting" due to all elements being units. Would highly appreciate any thoughts on the matter.

Answer 1

Why can't we factor invertible elements?

If by factoring we just mean writing something as a product of ring elements, then of course we can factor units. In fact, you can do that in many ways. So many ways, in fact, that it doesn't really tell us anything.

implicitly [...] we do not care about the question of factorization for elements which have multiplicative inverses? Why is this question not interesting?

This is close to a related question: Why are irreducible elements non-units?

The mindset to get into is that units don't tell us anything interesting about factorization. Part of the spirit of factoring $xy=z$ is that $x$ and $y$ are somehow measurably "smaller" than $z$. The way to do this is to require that $(x)$ and $(y)$ properly contain $(z)$.

So, to take examples from $\Bbb Z$, $(3)$ and $(5)$ both contain $(15)$ properly, but if I were to factorize as $(-1)(-15)$, then $(-15)=(15)$ and $-15$ isn't significantly smaller than $(15)$. We haven't gained anything, so to speak, by this rewriting.

If you were trying to factorize a unit, you would never get anything interesting, because if $u,v,w$ are all units in a domain $D$ such that $uv=w$, then $(u), (v), (w)$ are all $D$.

Even if we chose not to adopt this philosophy of sizes, allowing units in factorizations would be a total end to unique factorization in any domain. Every "different" factorization of $1$ would provide a "different" factorization of any other element. For this reason, factorizations that differ by units are considered identical in the theory of domains.

Yet, every element in a field is unit so how can we, intuitively, talk about fields being UFDs?

(Every element is a unit except zero, of course.) The vacuous explanation is often anticlimactic, but it makes the logical picture consistent. For instance, it lets you say that $D[x]$ is a UFD if and only if $D$ is a UFD. Why would we want to make exceptions when $D$ is a field when the statement is already so good?

Answer 2

We should go to the spirit of the definition: irreducubility is a concept that arises out of the concept of prime numbers. Let us take the field of rational numbers and take one factorization of, e.g., 3/10 as below: $$\frac3{10} = 7^2\times \frac37\times \frac1{70}$$
This shows 7 is a factor of $3/10$ well the same trick will lead us to show that any prime is a factor of $3/10$. This trouble arises because every number has inverse. So we avoid that and define irreducubility assuming that element does not have inverse.