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Math Help - Field of Quotients

  1. #1
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    Field of Quotients

    If F is a field, prove that the field of quotients, Q, is isomorphic to F
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  2. #2
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    Can you clear it up a bit?

    What do you mean by the field of quotients? I am only familiar with the field of fractions when dealing with rings. In a field it doesnt really make sense to talk about a field of fractions because your object is already a field. \frac{a}{b}=ab^{-1}\in F, so I am not entirely sure what this question is asking for you to show
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  3. #3
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    I checked this problem in wiki, "The field of fractions of a field is isomorphic to the field itself" (link).

    You only need a well-defined isomorphism \phi_s:F \rightarrow S^{-1}F given by f \mapsto fs/s for any s in S, where S is the set of all non-zero elements of a field F, and \phi_s(s) is a unit in S^{-1}F for any s \in S. It is easy to check that \phi_s is an injective homomorphism. To see \phi_s is surjective, for any f/s \in S^{-1}F, we have f/s = \phi_s(fs^{-1}).
    Last edited by aliceinwonderland; October 23rd 2009 at 01:10 PM. Reason: typo
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  4. #4
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    I'm a little confused on how you're showing the isomorphism, phi, is well-defined. What is the set S^-1 F defined to be?
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  5. #5
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    Quote Originally Posted by Gamma View Post
    What do you mean by the field of quotients? I am only familiar with the field of fractions when dealing with rings. In a field it doesnt really make sense to talk about a field of fractions because your object is already a field. \frac{a}{b}=ab^{-1}\in F, so I am not entirely sure what this question is asking for you to show
    Yes, it does make sense (albeit trivial) to speak of the field of fractions of a field.
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  6. #6
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    Quote Originally Posted by Coda202 View Post
    If F is a field, prove that the field of quotients, Q, is isomorphic to F
    If you consider the definition and construct of Field of quotients, Q as a field of all equivalence classes (equivalence class of (a,b), i.e. [a,b] are all (x,y) where ay=bx and b,y<>0. Think of [a,b] as a/b), then

    Consider phi(x)=[x,1] for all x in F.

    It is easy to show that above is an isomorphism from F onto Q.

    Note if F were an Integral Domain, we can define an isomorphism from F into Q. In field the fact that every non-zero element has an multiplicative inverse turns this isomorphism from 'into' -> 'onto'
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  7. #7
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    I'm having trouble showing that phi is an isomorphism, especially the well-defined part.
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  8. #8
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    Quote Originally Posted by Coda202 View Post
    I'm having trouble showing that phi is an isomorphism, especially the well-defined part.
    Can you be more specific plz? Where exactly are you having trouble.
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  9. #9
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    I'm trying to show that phi: D --> Q defined by phi(a) = [a,1] is an isomorphism.
    I don't have a problem showing that phi is onto. I'm having a problem showing that phi is 1-1 and preserves the operations + and *
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  10. #10
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    Quote Originally Posted by Coda202 View Post
    I'm trying to show that phi: D --> Q defined by phi(a) = [a,1] is an isomorphism.
    I don't have a problem showing that phi is onto. I'm having a problem showing that phi is 1-1 and preserves the operations + and *
    For onto
    Consider [a,b] in Q where a,b belong to F.
    [a,b]=[ab^-1,bb^-1]=[ab^-1,1]
    F being a field and b<>0 ensure there exists b^-1.
    So phi(ab^-1)=[a,b]
    Hence onto

    For 1-1
    Let phi(a) = phi(b)
    [a,1]=[b,1]
    By definition of equality in Q a.1=1.b => a=b
    Hence 1-1
    For 1-1 you can also show phi(a)=[0,1] => a=0

    For +,* well defined
    Note phi(a+b) = [a+b,1] = [a,1]+[b,1] = phi(a)+phi(b)
    Similarly for *
    These result follow directly from the definition of +,* in Q.

    Hope this clarifies
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  11. #11
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    Very much so, thanks.
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