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Math Help - Chinese Remainder Theorem.

  1. #1
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    Chinese Remainder Theorem.

    I need to prove problem 3.87 (The Chinese Remainder Theorem) on Rotmans book.

    Prove that if k is a field and f(x), and f'(x) are in k[x] and are relatively prime, then given b(x) and b'(x) in k[x], there exists c(x) in k[x]. with

    c-b in (f) and c-b' in (f'); more over if d(x) is another common solution then c-d is in (ff').

    I have an good outline of the proof but I need help filling in major gaps.

    Here is what I have: Every solution c-b in (f) has the form c=b+gf where g(x) is in k[x]. Hence we must find g such that gf - (b'-b) is in (f')

    The first question I have is what is (f). then why a solution there has to have the form b+ gf? and lastly why is it enough to find such a g.

    Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Chinese Remainder Theorem.

    Quote Originally Posted by jcir2826 View Post
    I need to prove problem 3.87 (The Chinese Remainder Theorem) on Rotmans book.

    Prove that if k is a field and f(x), and f'(x) are in k[x] and are relatively prime, then given b(x) and b'(x) in k[x], there exists c(x) in k[x]. with

    c-b in (f) and c-b' in (f'); more over if d(x) is another common solution then c-d is in (ff').

    I have an good outline of the proof but I need help filling in major gaps.

    Here is what I have: Every solution c-b in (f) has the form c=b+gf where g(x) is in k[x]. Hence we must find g such that gf - (b'-b) is in (f')

    The first question I have is what is (f). then why a solution there has to have the form b+ gf? and lastly why is it enough to find such a g.

    Thanks.
    I think something got a little muddled when you transcribed this problem, this isn't 3.87 in Rotman, what edition are you using?
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  3. #3
    Junior Member
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    Re: Chinese Remainder Theorem.

    The first edition is 3.87. The second edition is 2.95. Thanks.
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