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Math Help - Field of Quotients

  1. #1
    Junior Member hercules's Avatar
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    Field of Quotients

    Lemma 30.4: ( D is an integral domain, FD is the field of quotients)
    Let D1 denote the subset of FD consisting of all [a,e] for each a in D. Then D1 is a subring of FD and D is isomorphic to D1.


    Question: Assume that D, D1, and FD are as in Lemma 30.4. Show that each element of FD is a solution of some equation ax=b with a,b element of D1.


    since a and b are elements of D1 they have the form a/1 and b/1. Then putting this into the equation ax=b, the solution is x= (a^-1)b which is exactly b/a or [b,a]. [b,a] is element of
    FD. This solution exists because the nonzero elements of FD form a group and ax=b has solution in a group.


    Somehow it seems too simple. Or i might be wrong and totally confused.

    Please point out all the properties of groups, rings and fields employed in the solution.
    Thank You.
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  2. #2
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    It looks okay.
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