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Quotient Field of the Positive Rationals

  1. Oct 27, 2011 #1
    So earlier this year I came here to discuss about having fun with groups, rings and isomorphisms and such. I fell upon the idea of finding an isomorphism of the positive rationals to the sequence of the exponents found in their prime factorization. I didn't know what much to do with it since I hadn't taken any modern algebra courses at the time and had no comfort in actually defining anything or proving anything to be true.

    When I was playing around with it, I defined a binary operation that basically, now that I've refined it, takes any two positive rationals, takes the polynomials associated with them in Z[X], multiply those polynomials, then take the inverse of my original map back to the positive rationals.

    That is, I defined a map [itex]\phi :\mathbb{Q}^{+}\to\mathbb{Z}[X][/itex] such that if [itex]a\in\mathbb{Q}^{+}[/itex], then there exists [itex]k\in\mathbb{N}[/itex] and [itex]e_{0},e_{1},\dots ,e_{k}\in\mathbb{Z}[/itex] such that [itex]a=\prod^{k}_{i=0} p_{i}^{e_{i}}[/itex] and thus [itex]\phi (a)=(e_{0},e_{1},\dots,e_{k},0,0,\dots)[/itex]. Its easy to show that this an isomorphism and that [itex]\phi(ab)=\phi(a) +\phi(b)[/itex].

    I then defined a binary operation [itex]\oplus :\mathbb{Q}^{+}\times\mathbb{Q}^{+}\to \mathbb{Q}^{+}[/itex] such that [itex]a\oplus b=\phi^{-1} [\phi(a) \cdot \phi(b)][/itex].

    It is then easy to show that [itex](\mathbb{Q}^{+},\cdot,\oplus,1,2)[/itex] is an integral domain and that [itex]\phi[/itex] is an isomorphism of integral domains.

    I was wondering now if I could find the field of fractions for the integral domain I just defined. What field would I get out of it?

    Anyone have an answer?

    Its a little late to work it out but I'll see where I get and post it up here :)
  2. jcsd
  3. Oct 27, 2011 #2
    Like I can easily show that the relation [itex]~[/itex] over [itex]\mathbb{Q}^{+}\times\mathbb{Q}^{+}[/itex] defined such that for all [itex]a,b,c,d\in\mathbb{Q}^{+}[/itex], we have [itex](a,b)~(c,d)[/itex] iff [itex]a\oplus d=b\oplus c[/itex] is an equivalence relation.

    Now comes the hard part of defining some sort of binary relation [itex]*:(\mathbb{Q}^{+}\times\mathbb{Q}^{+})\times (\mathbb{Q}^{+}\times\mathbb{Q}^{+})\to\mathbb{Q}^{+}\times\mathbb{Q}^{+}[/itex].

    I feel like I'm on to something.

    Could I be making some sort of [itex]\mathbb{Q}_{p_{1},p_{2},\dots}[/itex] extension? As if the [itex]p_{1},p_{2},\dots[/itex]-adic field? Could it even be that I'm defining some sort of field that has to do with logarithms?

    I'm so happy, excited and confused all at once.
  4. Oct 28, 2011 #3
    Anyone have anything?
  5. Oct 29, 2011 #4


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    just use the standard operations defined for the field of fractions:

    [(a,b)] * [(c,d)] = [((ad)(bc), b⊕d)] for the additive operation, and

    [(a,b)] ● [(c,d)] = [(a⊕c, b⊕d)], for the multiplicative operation.

    you'll want to exclude {1} from being the second element in your positive rational pairs, and you'll want to be able to show that the mapping Q+ ---> (Q+ x Q+\{1}) given by q---> [(q,2)] is an isomorphism.

    some random thoughts that occured to me: your field of fractions should be Z(x), the rational functions with integer coefficients, which i think is the same as Q(x). which leads to an interesting question: Q(x) contains Q[x] as a sub-ring, what extension of Q+ does this correspond to?

    it appears that your correspondence leads to a total order on Z[x]. is this order preserved when we go to Z(x)? what do we get when we consider the cauchy completion of Q+ (is there a corresponding convergent sequence of polynomials in Z[x]? does this lead to Z[[x]] (integral power series))?
  6. Oct 30, 2011 #5
    I was wondering about order and completion as well in Z[X]. Another thing I was asking is if this new extension of Q+ is complete itself. Which would be interesting considering the fact that this new set is countable.

    This also has me testing a few things with respect to treating Z[X] as a metric space that I'll type up when i have more time.
  7. Oct 30, 2011 #6
    and also, you can generalize this extension of Q+ to any permutation of primes. Like if you choose the first coefficient of your polynomial to correspond to the power of some other prime than 2, then you get a whole different ordering, whole different everything. Thats why I feel it has something to do with p-adics.
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