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Math Help - reducibility in Q implies reducibility in Z.

  1. #1
    Senior Member abhishekkgp's Avatar
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    reducibility in Q implies reducibility in Z.

    If , then factors into a product of two polynomials of lower degrees if and only if it has such a factorization with polynomials of the same degrees

    .

    How to prove it??
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by abhishekkgp View Post
    If , then factors into a product of two polynomials of lower degrees if and only if it has such a factorization with polynomials of the same degrees

    .

    How to prove it??
    Assume it factorises into two polynomials in Q, f=gh. Then f=(1/a)g'h', where a is the product of all the denominators of all the coefficients of g and h, with g' and h' in Z[x]. Then k=g'h' is in Z[x], so f=(1/a)k with k in Z[x]. Can you see where to go from here? The point is that you must be able to `spread out' the integer a between g' and h'.
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  3. #3
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Swlabr View Post
    Assume it factorises into two polynomials in Q, f=gh. Then f=(1/a)g'h', where a is the product of all the denominators of all the coefficients of g and h, with g' and h' in Z[x]. Then k=g'h' is in Z[x], so f=(1/a)k with k in Z[x]. Can you see where to go from here? The point is that you must be able to `spread out' the integer a between g' and h'.
    i will take a special case and do it.
    let .

    hence,.

    this easily forces

    doing the spreading out thingy you told.

    this doesn't do it since all coefficients are not proved to be integers; the terms attached to 'x' are integers but the others i can't comment upon. i cannot figure out how to exploit

    help!!!!
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