# About the number of irreducible elements in UFD ring

by julypraise
Tags: elements, irreducible, number, ring
 P: 110 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?
P: 606
 Quote by julypraise 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 $\,\,c_1,...,c_n\,\,$, 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 ***
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 P: 110 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?
P: 606

## About the number of irreducible elements in UFD ring

 Quote by julypraise 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?
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 P: 714 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?
 P: 110 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.
P: 110
 Quote by Sankaku 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).
P: 714
 Quote by julypraise 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.

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