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Math Help - Quotient Field

  1. #1
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    Quotient Field

    Determine Q(D) for D={m+ni|m,n in Z} contained in C

    Note: D is an integral domain and Q(D) is its quotient field



    I know for Q(D), we have [a,b]+[c,d]=[ad+bc,bd] and [a,b]*[c,d]=[ac,bd]

    I just don't know how to start this.
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  2. #2
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    Quote Originally Posted by mathematic View Post
    Determine Q(D) for D={m+ni|m,n in Z} contained in C

    Note: D is an integral domain and Q(D) is its quotient field



    I know for Q(D), we have [a,b]+[c,d]=[ad+bc,bd] and [a,b]*[c,d]=[ac,bd]

    I just don't know how to start this.
    Hint: What is \text{Frac}(\mathbb{Z}\right) and use this to try to answer your question.
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  3. #3
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    I've never seen the notation frac(Z).

    I was thinking of looking at something like:
    (m+ni)/(a+bi) and looking at at both addition and multiplication
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  4. #4
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    Because for just regular numbers, we can do:
    m/n+p/q=(mq+pn)/nq
    m/n*p/q=mp/nq

    Would I look at m+ni/a+bi or just something like m/a+bi?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathematic View Post
    Because for just regular numbers, we can do:
    m/n+p/q=(mq+pn)/nq
    m/n*p/q=mp/nq

    Would I look at m+ni/a+bi or just something like m/a+bi?
    What they want you to do is figure out what subfield of \mathbb{C} is \text{Frac}\left(\mathbb{Z}\right) (the quotient field) is isomorphic to. Try proving that it's isomorphic to \mathbb{Q}(i).
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  6. #6
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    I was given to this hint to try.
    Q(D) ∼= Q(i) = {r + si | r, s ∈ Q} by showing that the
    map f : Q(i) → Q(D) given by
    f(a/b+c/di) = [ad + bci, bd]

    We want to show 1-1, onto, f(a+b)=f(a)+f(b) an f(ab)=f(a)f(b)
    1-1:Assume a/b+c/di=e/f+g/hi
    f(a/b+c/di)=f(e/f+g/hi)
    [ad+bci,bd]=[eg+fgi,fh]
    onto:I always get hung up onto.
    f(a+b)=f((a/b+c/di)+(e/f+g/hi))=f(af+be/bf+chi+gdi/-dh)=[(af+be)(-dh)+bf(chi+gdi)(-dh)]
    f(a*b):
    Consider f((a/b+c/di)(e/f+g/hi))=f(ae/bf+ga/hi+ce/di+cg/-hd)
    This is where I've gotten so far
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  7. #7
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    f(a+b)=f((a/b+c/di)+(e/f+g/hi))=f(af+be/bf+chi+gdi/-dh)=[(af+be)(-dh)+bf(chi+gdi)(-dh)]=[-dhaf-dhbe+bfchi+bfgdi,-bfdh]
    f(a*b):
    Consider f((a/b+c/di)(e/f+g/hi))=f(ae/bf+ga/hi+ce/di+cg/-hd)
    This is where I've gotten so far
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  8. #8
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    Is that the right idea or no?
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  9. #9
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    Quote Originally Posted by mathematic View Post
    Is that the right idea or no?
    Honestly, what you have written is really hard to follow.

    We are trying to determine the quotient field of \mathbb{Z}[i] (what you have written as D).

    First, notice that \mathbb{Z}\subseteq \mathbb{Z}[i], so that Q( \mathbb{Z})\subseteq Q(\mathbb{Z}[i]). Since \mathbb{Q} is the quotient field of \mathbb{Z}, this means \mathbb{Q}\subseteq Q(\mathbb{Z}[i]).

    Obviously i\in Q(\mathbb{Z}[i]). This and the previous containment force \mathbb{Q}(i)\subseteq Q(\mathbb{Z}[i]) (if \mathbb{Q},i are both inside of the field Q(\mathbb{Z}[i]), then so is the smallest field containing \mathbb{Q} and i).

    But then \mathbb{Q}(i) is a field satisfying \mathbb{Z}[i]\subseteq \mathbb{Q}(i) \subseteq Q(\mathbb{Z}[i]). Since the quotient field of an integral domain is the smallest field containing that integral domain, this forces equality: \mathbb{Q}(i)=Q(\mathbb{Z}[i]).
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