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Math Help - quick question on rings of polynomials.

  1. #1
    Senior Member abhishekkgp's Avatar
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    quick question on rings of polynomials.

    If is a field then the units in are precisely the units in . True or false?

    The answer is not given in the book.
    My answer: True.

    Proof:

    let
    where that is degree of and degree of
    now, can be written as
    .


    This gives (doesnt it?)
    since is a field one of is 0 contradicting the fact that the degrees of the polynomials were
    is this correct?
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  2. #2
    Super Member TheChaz's Avatar
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    F field => F integral domain, so the coefficient of the largest term x^(m + n) will never be zero.
    The units are constants.
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  3. #3
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    what you have shown is that if deg f > 0 or deg g > 0, fg ≠ 1.

    therefore, if fg = 1, it must be that deg f = deg g = 0, hence f(x) = a0, and g(x) = b0, and neither a0 nor b0 can be 0.
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  4. #4
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Deveno View Post
    what you have shown is that if deg f > 0 or deg g > 0, fg ≠ 1.

    therefore, if fg = 1, it must be that deg f = deg g = 0, hence f(x) = a0, and g(x) = b0, and neither a0 nor b0 can be 0.
    so doesn't it prove that if F is a field then F[x] has precisely the units in F??
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  5. #5
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    by the very definition of a field, every non-zero element is a unit, because for any non-zero a, 1/a exists, and we have a(1/a) = 1.

    so, yes, the non-zero constant polynomials are the units of F[x], which can be considered "the units of F" through the isomorphism F-->F[x]:

    a--->f(x) = a.

    (there is a slight distinction between a field element, a, and f(x) = a, which is a "polynomial expression". the constant polynomials of F[x]

    "act just like" field elements, and the isomorphism is often understood implicitly).
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