3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ That's right, we do not care about the question of factorization for elements which have multiplicative inverses. They aren't very interesting -- generally, factorizations are only interesting "up to units". Intuitively, think of playing a game where you start with $1$ and are allowed to multiply with an element of your ring; your goal is to get to a given element $a$. A factorization of $a$ would be a way to win the game in finitely many steps. Now, steps in which you multiply by a unit are not interesting -- you can always undo them by multiplying by its inverse. And you can make them ... $\endgroup$ – darij grinberg Mar 23 '15 at 5:08
  • 1
    $\begingroup$ ... as often as you wish. What carries real information are the irreversible steps -- i.e., steps in which you multiply by non-units. (I am not saying that you can never reverse such steps -- i.e., multiplying $0$ with $2$ still gives you $0$ --, but there is no way to reverse such a step by another step in general.) These kind of steps really tell you "what is in $a$". $\endgroup$ – darij grinberg Mar 23 '15 at 5:10
  • 2
    $\begingroup$ Why should fields be regarded as UFDs? Well, for example you can recall that a polynomial ring over a field is a UFD. If we disallow fields, then this result would need an awkward extra condition that the polynomial ring have at least one variable. $\endgroup$ – darij grinberg Mar 23 '15 at 5:14
  • $\begingroup$ Thank you for this response. I found your game analogy very helpful and interesting. But I still do not understand why an element having a multiplicative inverse implies that it doesn't make sense to talk about factorization for that element. In particular, I'm asking on an element-by-element basis, if that makes sense. I feel there must be some connection between invertibility and factorization not making sense here $\endgroup$ – Ryan Mar 23 '15 at 6:08
2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Thank you this is very close. I'd agree - you said yourself that the trouble arises, but because because EVERY number has an inverse. However, we've used nothing about the invertibility of 3/10 in this answer. I'm interested in better understanding the connection between one given element being invertible and that one given element not being able to be factored. Think rings where not all elements are units, just some strict subset. $\endgroup$ – Ryan Mar 23 '15 at 6:01
  • $\begingroup$ In the factorization $3/10$ 7 occurs because I can cancel it (7 has an inverse). This way any prime can be made to occur in the factorization of 3/10 as all primes have inverses. In general if $x\ne0$ then $x = p\times x/p$ happens because $p$ has inverse $1/p$ in the rationals. $\endgroup$ – P Vanchinathan Mar 23 '15 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.