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Residue field of p-adic integers

  1. Jan 13, 2012 #1
    In the field of rationals [itex]\mathbb{Z}_{(p)}[/itex] (rationals in the ring of the p-adic integers), how is it possible to prove the residue field [itex]\mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)}[/itex] is equal to [itex]\mathbb{Z}/p\mathbb{Z}[/itex] ?

    I've narrowed it down to [itex]\mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)} = \left\{ a/b\in\mathbb{Q} : p\nmid a, p \nmid b \right\} [/itex], but can't seem to make the last step...

    Or maybe I'm barking up the wrong tree. Hmm...
     
    Last edited: Jan 13, 2012
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  3. Jan 13, 2012 #2

    micromass

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    How did you define [itex]\mathbb{Z}_{(p)}[/itex]???
     
  4. Jan 13, 2012 #3
    I defined [itex]\mathbb{Z}_{(p)}[/itex] to be

    [itex]\mathbb{Z}_{(p)} = \left\{\frac{a}{b}\in\mathbb{Q}:p\nmid b\right\}[/itex].
     
  5. Jan 13, 2012 #4

    micromass

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    That aren't the p-adic integers...

    Anyway, you could prove that the quotient is a field that contains p elements. That shows it.
     
  6. Jan 13, 2012 #5

    morphism

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    There's a really obvious map from Z_(p) onto Z/pZ whose kernel is pZ_(p). Hint: if p doesn't divide b, then b is a unit in Z/pZ.
     
  7. Jan 14, 2012 #6
    Thanks morphism. I'm not particularly up on group/ring theory etc. However, I am learning!

    From what I understand from your response, we seek a map from [itex]\mathbb{Z}_{(p)}[/itex] to [itex]\mathbb{Z}/p\mathbb{Z} = \left\{0,1,\ldots,p-1\right\}[/itex]. By kernel I think you mean the subset of [itex]\mathbb{Z}_{(p)}[/itex] that maps to the zero element in [itex]\mathbb{Z}/p\mathbb{Z}[/itex]. So the kernel is [itex]p\mathbb{Z}_{(p)}[/itex], i.e. the set of all rationals in [itex]\mathbb{Z}_{(p)}[/itex] such that [itex]p[/itex] divides the numerator. I think one of the things I'm finding difficult is to understand how [itex]\mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)}[/itex] can equal [itex]\mathbb{Z}/p\mathbb{Z}[/itex] (the set with [itex]p[/itex] elements) since [itex]\mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)}[/itex] actually contains not p elements but a whole load of rationals such that p does not divide numerator or denominator. So how can we say they are equal when one contains fractions and the other p integers?

    I'm beginning to think that [itex]\mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)} = \mathbb{Z}/p\mathbb{Z}[/itex] because if we take all elements in [itex]\mathbb{Z}_{(p)}[/itex] and modulo (congruence?) them to [itex]p\mathbb{Z}_{(p)}[/itex], the very basic set of resulting elements that results is [itex]\mathbb{Z}/p\mathbb{Z}[/itex]. Is this along the right/wrong lines?
     
  8. Jan 15, 2012 #7

    morphism

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    Z_(p)/pZ_(p) contains only p elements. Think about Z/pZ: Z is infinite, but Z/pZ only has p elements in it.

    I think your problem is stemming from the fact that you're trying to show that Z_(p)/pZ_(p) and Z/pZ are equal, when they're not (well, depending on your definition of Z/pZ). They're "isomorphic".
     
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