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Math Help - About to lose it: rings, polynomials, and ideals.

  1. #1
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    About to lose it: rings, polynomials, and ideals.

    Questions are attached. This isn't homework or to be graded; it is exam preparation, and I feel so helpless with it. The only question I am positive I have a solution for is "c." Clearly if we are given two polynomials f(x) and g(x) with leading terms a*x^n and b*x^m, the leading term of f(x)g(x) is a*b*x^(m+n). a and b are elements of an integral domain, therefore a*b != 0, which also shows that D[x] is an integral domain.

    Now, "b" seems simple, but I can't quite figure it out. My professor only said "there's a theorem for that." I see that in the field F, n*1 != 0 for any n in the integers. It would seem reasonable to assume that the set n*1 forms a subring over addition and multiplication. A simple function from the subring S to Z (and Z to S) can be constructed, but he informed me that this wasn't sufficient to prove S was isomorphic to Z. Following this failed aspect of the proof, it is rather simple to show that some subfield of F is isomorphic to Q, as Q is a quotient ring of Z: this means that there is some quotient ring of S isomorphic to Q.
    Attached Thumbnails Attached Thumbnails About to lose it: rings, polynomials, and ideals.-asdfjkl.jpg  
    Last edited by pantsaregood; April 3rd 2013 at 10:54 AM.
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  2. #2
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    Re: About to lose it: rings, polynomials, and ideals.

    Hi,
    For your specific question, the set \{(m\cdot1_F)(n\cdot1_F)^{-1}\,:\,m,n\in\mathbb Z\,\text{with}\,n\ne0\} is easily shown to be a subfield of F. ( n\cdot1_F)^{-1} exists since F has characteristic 0). Next the map:
    f\,:\,(m\cdot1_F)(n\cdot1_F)^{-1}\mapsto{m\over n} is an isomorphism onto Q -- you have to show f is well defined and then is an isomorphism, grunge work but easy.
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