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Unique Factorisation Domains/Fields

  1. Dec 5, 2011 #1
    In my notes/online I have that for a set to be a field is a stronger condition than for a set to be a unique factorisation domain (obviously), but I'm confused about the concept of irreducibility in a field. For example in R\{0} there are no irreducible elements as far as I can tell? I'm trying to show whether a different ring is a UFD or not but again I find the concept of irreducibility difficult as here again the ring seems to have no irreducible elements.

    Thanks in advance.
     
  2. jcsd
  3. Dec 6, 2011 #2

    mathwonk

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    in any ring there are three classes of non zero elements: units, zero divisors, non unit non zero divisors. irreducibles are always, by definition, a subset of the non units. in a field, all non zero elements are units, hence there is no possibility of irreducibles.
     
  4. Dec 6, 2011 #3

    Deveno

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    perhaps you should tells us which ring you are trying to prove is a UFD.
     
  5. Dec 6, 2011 #4
    Ok that makes sense, so is the statement that all fields are UFDs correct? Or just nonsense because the concept of factorisation only makes sense with regards to irreducibles?

    Let 2^(1/n) be the real positive nth root of 2. If we consider Z [ 2^(1/n) ] as the image of Z[X] under the evaluation map f(X) -> f(2^(1/n)), then let R= union from n=1 to infinity of (Z [ 2^(1/n) ]). I have shown already that R is a subring of the reals but as far as I can see it has no irreducible elements-
    If a is a member of Z[2^(1/p)] for some natural number p, say a = g(2^(1/p)) for some polynomial g in Z[X]. Then we can always take out a factor of 2^(1/2p) to give

    a = 2^(1/2p) * h(2^(1/2p))

    where h is some polynomial in Z[X] (with the same integer coefficients as g but different powers). So a is reducible.

    Am I on the completely wrong track?
     
  6. Dec 6, 2011 #5

    lavinia

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    Isn't the number 3 irreducible in this ring?
     
  7. Dec 6, 2011 #6
    Ok I completely missed that the polynomial can have a constant term (being slow) so yes I think you're right, 3 is irreducible. But I still can't see whether or not it is a unique factorisation domain as any element where the polynomial has a zero constant term is reducible. For example there definitely is not a unique factorisation of the element 2 into irreducibles- infact there isn't any factorisation of 2 into irreducibles as far as I can see, nor (do I think!) is it a unit.

    (I've just found a bit on wikipedia saying 'any field is trivially a UFD' which now makes sense to me but obviously my example is not a field).
     
  8. Dec 7, 2011 #7

    lavinia

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    None of the fractional powers of 2 have inverses in this ring since negative powers are excluded. So two factorizations will be in-equivalent. How about two factorizations of the square root of 2?
     
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