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Field of Fractions

  1. Sep 8, 2010 #1
    Hi guys,

    I know that for integral domains with finte elements that if we show that each element has a multiplicative inverse then it is a field.

    I need to show that the field of fractions is a field.

    As the domain is not finite how does that effect the proof of being a field?


    regards
    Brendan
     
  2. jcsd
  3. Sep 8, 2010 #2

    Office_Shredder

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    Whether the domain is finite or not will be irrelevant unless you have a pretty wacky proof.

    Your proof should probably be inspired by how you would prove that the rational numbers are a field, which is obviously infinite
     
  4. Sep 9, 2010 #3

    Landau

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    The definition of a field is an integral domain in which each non-zero element has an inverse. I don't see where you get the finiteness condition from.
    So just take an arbitrary non-zero element in the field of fractions and show that it has an inverse.
     
  5. Sep 12, 2010 #4
    Is this what you meant by the first question? "If an integral domain is finite then it is a field." This is fairly well-known. To prove it: Suppose x is a nonzero element of a finite integral domain; then find positive integers k > l such that xk = xl. Then prove that xxk-l-1 = 1.
     
  6. Sep 12, 2010 #5
    If adriank is correct, and you mean prove A. "If an integral domain is finite then it is a field." and B. "Why does this fail if the integral domain is not finite?" then consider the following:

    First, if you have an integral domain D any a (not 0) inside D then the function
    mult_a: D --> D given by
    mult_a(x) = a*x
    is injective -exactly- because D is an integral domain:
    mult_a(x) = mult_a(y) => a*x = a*y => x=y by cancellation property of integral domains.

    But if is D finite, mult_a is also surjective, so for some x, mult_a(x) = 1. So what does this mean about x?

    Second, if D is not finite then you can't conclude anything: the function f: Z --> Z (Z = integers) given by f(x) = 2*x is injective but certainly not surjective, and Z is an integral domain.

    Anyways, that an injective function f: A --> A is surjective if A is finite is a very important fact in mathematics. That this can fail if A is infinite is equally important. You can use this last fact to construct a (nontrivial) group G so that G x G is isomorphic to G, for example.


    Skolem
     
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