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Factorise elements in Z[x]?

  1. Apr 1, 2007 #1
    1. The problem statement, all variables and given/known data
    How do you factorise polynomials in Z[x]? Z is not a field so you can't use the theorem 'a is a root of f <=> (x-a) divides f'

    3. The attempt at a solution
    Would you map the polynomials from Z[x] to Q[x] by multiplying by 1, since all elements in Z[x] are in Q[x]. Then factorise in Q[x] usiong the theorem above. If the factors have coefficients in Z than map (multiply by 1) these back to Z[x] so you have factorised these polynomials and they exist in Z[x]. If the factors in Q[x] have coefficients not in Z then you can't map back to Z[x] hence these are not factorisable in Z[x].



    Is this how you would factorise polynomials in Z[x], formally?
     
    Last edited: Apr 2, 2007
  2. jcsd
  3. Apr 1, 2007 #2

    StatusX

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    Look up Gauss' lemma.
     
  4. Apr 2, 2007 #3
    How does Gauss' lemma help?
     
  5. Apr 2, 2007 #4

    matt grime

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    What makes you think passing to Q[x] suddenly makes things easier? What is the factorization ox x^2-2 in Z[x]? And in Q[x]? The x-a a root thing is still as valid/not valid in Z[x] as in Q[x]. In fact that is what (one of) Gauss's Lemmas states: a poly is irreducible over Z iff it is irreducible over Q.

    You factorise it 'by doing it'. I.e. trying to find a factorization. In general this is *very hard* and can't be done by any decent analytic means. It is however possible to show there are no integral roots by exhaustion.
     
    Last edited: Apr 2, 2007
  6. Apr 2, 2007 #5
    I like to pass over to Q[x] so I can use the theorem I quoted in the OP because Q is a field. I like to justify every step I take instead of relying on intuitition.

    The version you quoted isn't in here
    http://en.wikipedia.org/wiki/Gauss's_lemma_(polynomial)
     
  7. Apr 2, 2007 #6

    matt grime

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    Really? Second bullet point. If a poly is irreducible over Z it is irreducible over Q. (Obviously if it is reducible over Z it is reducible over Q).
     
  8. Apr 2, 2007 #7
    Right. But the second point only has one way from the integers to the rationals. You had iff in your previous post?! Consider f(x)=2x^2+2x+2 is irreducible in Q[x] but is reducible in Z[x] i.e f(x)=2(x^2+x+1).

    Anyhow, if you follow my method, and map the polynomials to Q[x] then you can (potentially) factorise by finding zeros of it (which is allowed by the theorem). Then follow two implications,
    Let f(x) by original polynomial in Z[x]
    1. f(x) primitive in Z[x] and irreducible in Q[x] => irreducible in Z[x]
    2. f(x) is monic and reducible in Q[x] => reducible in Z[x]

    It was convienent that all f(x) I was interested in factorising were monic hence also primitive so by mapping them to Q[x] I was able to tell if it was irreducible or reducible in Z[x], rigorously. If the word is appropriate here.

    But would you say 'mapping' is a good word here? I am merely factoring specific polynomials given to me which were said to be in Z[x]. But I chose to test reducibility in Q[x]. Is a map really needed here? If not than what should I say or justify me taking the polynomials to Q[x] and doing the maths?
     
    Last edited: Apr 2, 2007
  9. Apr 2, 2007 #8

    matt grime

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    I pointed out the 'other' direction. It is trivial: irreducible over Q implies irreducible over Z. That really is trivial.
     
  10. Apr 2, 2007 #9
    But what about

    f(x)=2x^2+2x+2 is irreducible in Q[x] but is reducible in Z[x] i.e f(x)=2(x^2+x+1) with neither 2 nor x^2+x+1 is a unit in Z[x]

    So you need
    f(x) primitive in Z[x] and irreducible in Q[x] => irreducible in Z[x]
    Correct?

    The definition of irreducible I use is, P is irreducible <=>
    P is non unit and if p=ab than a is a unit or b is a unit



    What about my mapping question in my previous post? Is it appropriate?
     
    Last edited: Apr 2, 2007
  11. Apr 2, 2007 #10

    matt grime

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    Ok, modulo trivialities like dividing by a constant, jeez are you missing the point.
     
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