About the number of irreducible elements in UFD ring

In summary: If you can't, then... you probably have the answer to this question.In summary, the definition of a unique factorization domain ring, as described in Hungerford's text, states that any nonzero nonunit element can be written as a product of finitely many irreducible elements. This does not necessarily mean that there are only finitely many irreducible elements in the ring, as it is possible for a nonzero nonunit element to be a product of infinitely many irreducible elements. However, in a UFD, the irreducible elements involved in the formation of a product are associates of a finite number of irreducibles. While some rings may allow for the concept of an infinite product, it is not generally applicable in a UFD
  • #1
julypraise
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When chracterizing the definition of unique factorization domain ring, the Hungerford's text, for example, states that

UFD1 any nonzero nonunit element x is written as x=c_1. . .c_n.

Does this mean any nonzero nonunit element is always written as a product of finitely many irreducible elements?

I think it is not the case. Because if it were then it implies that any R has finitely many irreducible elements.

So is it just for convenience's sake? So even if a nonzero nonunit element is a product of infinitely many irreducible elements, we can just put it as x=c_1. . .c_n? Even if it is uncountably infinitely many?
 
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  • #2
julypraise said:
When chracterizing the definition of unique factorization domain ring, the Hungerford's text, for example, states that

UFD1 any nonzero nonunit element x is written as x=c_1. . .c_n.

Does this mean any nonzero nonunit element is always written as a product of finitely many irreducible elements?

I think it is not the case. Because if it were then it implies that any R has finitely many irreducible elements.


*** Uh? The above does not mean there're only n irreducible elements [itex]\,\,c_1,...,c_n\,\,[/itex], but that for any element x in the

ring there exists some finite number of irreducible elements (that'll depend on x) s.t...etc. ***



So is it just for convenience's sake? So even if a nonzero nonunit element is a product of infinitely many irreducible elements, we can just put it as x=c_1. . .c_n? Even if it is uncountably infinitely many?


*** There is no such thing as infinite product or infinite series in a general ring, UFD or not, unless you can introduce a topological

structure on the ring that'd make that infinite stuff acceptable in some way. It is ALWAYS a finite number, both in product and sum.

DonAntonio ***

...
 
  • #3
My argument for that goes like this: Let R' be the set of all irreducible elements in R. Form a product of all of them. Call this product x. Then x is nonzero nonunit. Thus x is represented as c_1...c_n where c_i is irreducible. Then as R is a UFD, the irreducible element in the formation of the product x are associates of c_1...c_n, meaning there are finitely many irreducibles.

By the way what do you mean by a product can never be infinite? Can we just do 'times' infinitely many in any case?
 
  • #4
julypraise said:
My argument for that goes like this: Let R' be the set of all irreducible elements in R. Form a product of all of them. Call this product x.


*** Once again, this makes no sense unless you can define mathematically (as opposed to wishfully) an infinite

product in a general ring...and believe me: you can't.

DonAntonio ***




Then x is nonzero nonunit. Thus x is represented as c_1...c_n where c_i is irreducible. Then as R is a UFD, the irreducible element in the formation of the product x are associates of c_1...c_n, meaning there are finitely many irreducibles.

By the way what do you mean by a product can never be infinite? Can we just do 'times' infinitely many in any case?

...
 
  • #5
Just think about the Integers as a basic example of a UFD and your intuition should show you whether you are on the right track.

Pick any nonzero nonunit. It is an infinite product of irreducible elements?

Are there only finitely many irreducible elements?
 
  • #6
Okay thanks guys. Btw DonAntonio I've thought about what you said. At least among the structures that I know of, the infinite product cannot be intuitively conceived, as you said, except some trivial things like 1 = 11111111... and 0 = 0000000000000... or for a unit u = uuu^-1uu^-1uu^-1..., motivating some kind of stability notion..., which in turn motivates me to think that for some kind of rings and ring objects, this infinite product notion may be possible. But anyway this now is off the topic, so I may end up here. Anyway thanks.
 
  • #7
Sankaku said:
Just think about the Integers as a basic example of a UFD and your intuition should show you whether you are on the right track.

Pick any nonzero nonunit. It is an infinite product of irreducible elements?

Are there only finitely many irreducible elements?


What I can say right now is that for the integer ring, this infinity case is not the case. But I'm not sure if for any infinite ring, the number of irreducible elements is also infinite (possibly regardless of the cardinarlity).
 
  • #8
julypraise said:
What I can say right now is that for the integer ring, this infinity case is not the case. But I'm not sure if for any infinite ring, the number of irreducible elements is also infinite (possibly regardless of the cardinarlity).

Well, go looking and see if you can find it.
 

What is a UFD ring?

A UFD (Unique Factorization Domain) ring is a type of ring in abstract algebra where every nonzero element can be written as a unique product of irreducible elements. This means that there are no multiple ways to break down an element into smaller factors.

What are irreducible elements?

Irreducible elements in a UFD ring are elements that cannot be broken down any further. They cannot be factored into smaller elements within the same ring. In other words, they are the building blocks of the ring.

How do you determine the number of irreducible elements in a UFD ring?

This can vary depending on the specific ring and its properties. However, in general, one method is to use the fundamental theorem of arithmetic, which states that every positive integer can be expressed as a unique product of primes. This can be extended to UFD rings by using the definition of irreducible elements.

Why is the number of irreducible elements important in a UFD ring?

The number of irreducible elements is important because it tells us about the structure of the ring. For example, if a UFD ring has a large number of irreducible elements, it may be more complicated and have more diverse factorizations. On the other hand, a UFD ring with a small number of irreducible elements may have simpler factorizations and be easier to work with.

Can the number of irreducible elements in a UFD ring change?

Yes, the number of irreducible elements in a UFD ring can change. It can change if the ring itself changes, such as through a ring homomorphism. It can also change if we consider different subsets of the ring, as the number of irreducible elements may differ within different subsets.

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