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Math Help - Quotient Fields

  1. #1
    Jen
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    Quotient Fields

    The question is, Let n be a positive integer and let D be the subset of all rational numbers of the form \frac{a}{n^k} with a \in \mathbb{Z} and k any positive integer. Show that D is an integral domain whose quotient field is isomorphic to the field of rational numbers.


    So I have already shown that D is an integral domain.

    I am having issues proving that its quotient field is isomorphic to the field of rational numbers. I don't really even know where to start.

    Any hints would be nice. I don't want a full proof I would like to do my own work. Just a little push in the right direction would be great.

    Thank you!
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  2. #2
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    Well, I would suggest computing the characteristic of the quotient field. If this number is zero, then your field would contain a copy of the rational numbers.
    For the other direction, maybe direct computations could work (seeing that the rationals is itself a field)
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  3. #3
    Jen
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    I appreciate your help Jose, but I haven't learned about characteristics yet. Not that you could have known that. So thank you.

    Here is what I have so far. (which isn't much)

    So I think that the quotient field of D is, F = \left\{ \left[\frac{a}{n^{k_1}}, \frac{b}{n^{k_2}} \right] : \mbox{ }\frac{a}{n^{k_1}}, \frac{b}{n^{k_2}} \in D \mbox{ }and \mbox{ }\frac{b}{n^{k_2}} \ne 0 \right\} (I hope that this is all standard notation, most of this stuff is fairly new to me so I am just mimicking the notation from my book)

    Or, in other words, \frac{a}{n^{k_1}}\over \frac{b}{n^{k_2}}

    So, I have no idea how to even define \theta: F \to \mathbb Q such that it would isomorphically map the field of quotients from D to the field of rationals.

    Am I even thinking of this right??

    I am really lost.
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  4. #4
    Jen
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    I can see that it is isomorphic. I see that each element of D's quotient field will simplify down into an element of the rationals (since the integers are a ring). Is that all I need to do? Just map each element of D onto the equivalent element of the rationals?
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  5. #5
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    If you know that your identification is onto, and that it is a homomorphism then yes, otherwise you would only be proving that your field is contained in the rationals (Notice that every field homomorphism is automatically inyective).
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  6. #6
    Jen
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    I haven't been able to walk through a proof of isomorphism yet.

    Did I define the guotient field of D correctly?

    I had a discussion with another person in my class who defined it as,

         \left[    \frac {a}  {n^{k_1}}, 1    \right]      (once again I don't know if this is standard notation, if not let me know)

    and said that this and it's inverse represents the quotient field of D.

    Is this correct? If so, how? It seems to me like this should just be a subset of the field of quotients of D.
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  7. #7
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    Hadn't noticed, but your classmate is right: D is actually a field (ie. it's field of quotients is D), and the only thing to do is check that the inverse of an element in D over Q is in D. This proves that the function defined in your post covers D, now one only has to prove that it is in fact a field isomorphism, but by what I said earlier you need only check that it is a ring homomorphism.
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