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About the number of irreducible elements in UFD ring

  1. May 6, 2012 #1
    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?
  2. jcsd
  3. May 6, 2012 #2
  4. May 7, 2012 #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?
  5. May 7, 2012 #4
  6. May 7, 2012 #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?
  7. May 8, 2012 #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.
  8. May 8, 2012 #7

    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).
  9. May 9, 2012 #8
    Well, go looking and see if you can find it.
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