Why the prime/irreducible distinction?

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In summary, the authors showed that in any ring with unique factorization into arbitrary irreducibles, the irreducibles are prime. They also showed that any ring that is a UFD with respect to irreducibles is also a UFD with respect to primes. However, they made an argument that is flawed, and they introduced a new term that is not defined well.
  • #1
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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?
 
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  • #2
(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
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
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)
 
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  • #5
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 going to try to find a direct proof that PIDs have a unique prime decomposition.
 
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  • #6
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
Oh! You have a good point there! That was a rather silly mistake on my part. :frown:

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.
 
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  • #8
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.
 
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  • #9
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.
 
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  • #10
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.
 
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  • #11
well DF is kind of a mediocre book in my view. try jacobson for a good treatment.
 
  • #12
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
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.
 

1. Why is the distinction between prime and irreducible numbers important in mathematics?

The distinction between prime and irreducible numbers is important because it helps us understand the fundamental building blocks of numbers and their relationships. Prime numbers are those that can only be divided by 1 and themselves, while irreducible numbers are those that cannot be factored into smaller numbers. This distinction allows us to better understand the properties and patterns of numbers, and is essential in many areas of mathematics, such as number theory and algebra.

2. How are prime and irreducible numbers different?

Prime numbers are a subset of irreducible numbers. All prime numbers are irreducible, but not all irreducible numbers are prime. This means that while all prime numbers have no factors other than 1 and themselves, there are other numbers that have no factors, but are not considered prime. For example, the number 2 is both prime and irreducible, but the number 6 is not prime, but is irreducible.

3. What are some real-world applications of the prime/irreducible distinction?

The prime/irreducible distinction has many practical applications, especially in cryptography. Prime numbers are used in encryption algorithms to generate secure keys and protect sensitive information. Additionally, the study of prime and irreducible numbers has also led to advancements in computer science, coding theory, and other fields.

4. How can we determine if a number is prime or irreducible?

To determine if a number is prime, we can use the method of trial division, where we divide the number by all smaller numbers and see if there are any remainders. If there are no remainders, then the number is prime. To determine if a number is irreducible, we need to check if it can be factored into smaller numbers. If it cannot, then it is irreducible.

5. Are there any famous unsolved problems related to prime and irreducible numbers?

Yes, there are several unsolved problems related to prime and irreducible numbers, such as the famous Goldbach's conjecture, which states that every even integer greater than 2 can be expressed as the sum of two prime numbers. Another unsolved problem is the Twin Prime conjecture, which states that there are an infinite number of pairs of prime numbers that are only two apart from each other. These and other unsolved problems demonstrate the complexity and importance of prime and irreducible numbers in mathematics.

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