# Why the prime/irreducible distinction?

1. Jan 14, 2007

### StatusX

I was going through a chapter on unique factorization domains (UFDs). They use the following definitions:

irreducible: An element r in a ring R is irreducible if r is not a unit and whenever r=ab, one of a or b is a unit.

prime: an element is prime if the ideal it generates is a prime ideal.

Then they show that in any commutative ring, all primes are irreducible, and in a principle ideal domain (PID), irreducibles are also prime. Then they go through a bunch of stuff to show PIDs are UFDs, and finally that, in a UFD, it's also the case that irreducibles are prime. In other words, in a UFD, which is the setting for which irreducibles were originally defined, primes and irreducibles are the same thing. Why the distinction then? Or at least, when defining a UFD, why not do it in terms of primes instead of irreducibles? The end result is the same, isn't it?

2. Jan 14, 2007

### Hurkyl

Staff Emeritus
(1) Generality: that same definition of irreducible works in more general rings, and the basic notion of irreducibility works in very general contexts.

(2) Tradition: one learns to speak of irreducible polynomials before learning ring theory. It makes sense to be able to retain that choice of language in the ring theoretic context.

(3) Connotation: opting to say "P follows because a is prime" versus "P follows because a is irreducible" (or vice versa) is suggestive of what the suppressed details are, and thus makes a proof easier to follow.

3. Jan 14, 2007

### StatusX

But still, why not define UFDs in terms of factorization into primes rather than into irreducibles? If a ring is a UFD with respect to irreducibles, ie, every element has a unique factorization into irreducibles, then it is also one with respect to primes (since in a UFD, irrdeducibles are the same as primes). So a definition in terms of primes would be at least as general. I'm not sure if it would actually be more general.

4. Jan 14, 2007

### Hurkyl

Staff Emeritus
Mathematically, there are many equivalent ways to characterize any notion. Which is chosen as a definition and which are relegated to theorems is irrelevant.

From a learning standpoint, it seems to me that the usual definition more directly captures the idea of a unique factorization domain, so it would make sense to choose that as your starting point.

If we used your definition, you still have to prove the theorem that in any ring with unique factorization into arbitrary irreducibles, the irreducibles are prime, and thus the ring is a StatusX-UFD.

[color=#aaaaaaa](Your definition cannot be more general: if you have unique factorization into primes, that automatically tells you that you have unique factorization into irreducibles)[/color]

(edit: grayed out flawed argument)

Last edited: Jan 14, 2007
5. Jan 14, 2007

### StatusX

OK thanks. I went back and tried to reproduce the proof that PIDs are UFDs, but for prime factorization instead of irreducible factorization. I was able to show uniqueness, and in fact, that whenever an element in any integral domain has a prime decomposition, it is unique up to associates.

But for existence I ran into trouble. The way the book did it, you need to show that for any element r such that there are no a,b, both non-units, with r=ab, then r is prime. In other words, you need to show that irreducibles are prime. It's strange that after going through all this you end up showing that, when the ring is a UFD, irreducibles are prime anyway. I'm gonna try to find a direct proof that PIDs have a unique prime decomposition.

Last edited: Jan 14, 2007
6. Jan 14, 2007

### StatusX

Also, I see why a factorization into primes gives a factorization into irreducibles, but how do you know that if a factorization into primes is unique, then there isn't another factorization into irreducibles, some of which aren't also prime? (ie, so the ring isn't a UFD)

7. Jan 14, 2007

### Hurkyl

Staff Emeritus
Oh! You have a good point there! That was a rather silly mistake on my part.

I think the usual idea for showing unique factorization works here, though: if we had two factorizations, one into primes, and one into irreducibles, then if we choose any prime in the factorization, we can prove that one of those irreducibles must be divisible by that prime, and thus equal to that prime times a unit.

Last edited: Jan 14, 2007
8. Jan 14, 2007

### StatusX

Yea, that sounds right. Another way to see it is like this. Let a UPFD be what you think it is. Then in such a UPFD, irreducibles are prime, since if r is irreducible, its prime factorization can't have more than one element, and so it is a prime. The argument follows from there.

This just makes it all the more confusing why we refer to UFDs and not UPFDs if they are exactly the same thing. I mean, we end up showing that in a UFD, the factorization we end up with is not just a factorization into irreducibles, but more specifically is one into primes.

Moreover, its easy to show that if an element in a ring has any prime decomposition, this decomposition is unique up to associates. Thus showing a ring is a UPFD would just amount to showing the existence of a (prime) decomposition, and uniqueness would follow.

On the other hand, when showing a ring is a U(I)FD, uniqueness does not follow from existence. You would need to first show every element has a factorization into irreducibles, and then in addition show this factorization is unique. You'd end up seeing that what you have is in fact a factorization into primes, and if you had shown this in the first place, you wouldn't have had to do that second step.

Sorry Hurkyl, I'm not holding you personally responsible for these definitions, I'm just trying to understand these things better. Thanks for your help.

Last edited: Jan 14, 2007
9. Jan 14, 2007

### mathwonk

the defn of prime is such that it is difficult to show existence of them.

any idot can prove unqiuenes of prime factorization. but irreducibles are defined so that frequently one has existence of afctorization. so one wants both existence and uniqueness.

thus one wants a condition that says irreducible and prime are the same. besides primes are always irreducible.

so in any noetherian ring, factorization into irreducibles is unique iff irreducibles are prime.

i.e. factorization into primes is trivially unique, but may not exist. on the other hand factorization into irreducibles frequently exists. so in order to know that the irreducibles into which factorization exists, is also unique, one needs to know those irreducibles are prime.

so factorization into irreducibles, equivalently primes, exists and is unique, precisely when principal ideals have the ACC, and all irreducibles are prime.

Last edited: Jan 14, 2007
10. Jan 14, 2007

### StatusX

I realize that, my point was just that both definitions should be presented so that we have both techniques. For example, the proof in my book that a PID was a UFD first showed existence (by using the ascending chain condition along with the fact that irreducibles are prime) to show existence, which is fine. But then they proved uniqueness by mirroring the (idiot) proof for prime decompositions, where as they could have just pointed out that what we have is a prime factorization, and show these are always unique (ie, slightly adapt the proof to get a much more general result). Maybe my criticism is more specific to my book (Dummit & Foote) than I thought.

Last edited: Jan 14, 2007
11. Jan 14, 2007

### mathwonk

well DF is kind of a mediocre book in my view. try jacobson for a good treatment.

12. Jan 19, 2007

### fopc

Are you referring to the three volume set, "Lectures in Abstract Algebra"
by Nathan Jacobson? If true, can you tell me whether he treats
subjects concerning infinite dimensional vector spaces (obviously from the
algebraic point of view), or not? Example: Hamel (algebraic) basis. Thanks for any help.

13. Jan 25, 2007

### mathwonk

sorry for not noticing your post fopc. I have been away a while.

I have not read jacobsons 3 volume set. But I would guess from what I know of him that the answer is yes.

A Hamel basis is probably just a Q vector basis for the reals , as far as I know so in a sense anyone treats them.

I.e. By the zorn lemma, every vector space over any field has a maximal independent set which is trivially a basis. bingo.

these books are very formal and maybe hard to read without faslling asleep but they are by an expert and are cheap used.