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Primitive n-th roots

  1. Jan 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that primitive n-th roots of unity have the form [itex]e^{i2\pi k/n}[/itex] for [itex]k\in\mathbb{Z},n\in\mathbb{N}[/itex], [itex]k[/itex] and [itex]n[/itex] coprime.

    The attempt at a solution

    So the n-th roots of unity [itex]z[/itex] have the property [itex]z^{n}=1[/itex]. I have previously shown that [itex](e^{2\pi ik/n})^{n}=e^{2\pi ik}=(e^{2\pi i})^{k}=1^{k}=1 [/itex]. However, I'm not sure where to start in proving that primitive n-th roots of unity have that property. Any ideas on where I could get started?
     
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  3. Jan 20, 2012 #2

    Dick

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    The problem isn't to show that if z is a primitive n-th root of 1 that z^n=1. You know that. A primitive n-th root of 1 is also an n-th root of 1. What's the definition of a primitive root?
     
  4. Jan 21, 2012 #3
    A primitive n-th root has the smallest such n that z^n = 1. So if k and n aren't coprime then they would have a common factor except 1, because if they did, you could have a smaller n s.t. z^n = 1.
     
  5. Jan 21, 2012 #4

    Dick

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    Well, yes, that's the really sloppy explanation.
     
  6. Jan 21, 2012 #5
    We never covered it in class so I've had to make do with various online sources and have had some difficulty coming up with a formal definition to help me with the proof (though I basically understand what's going on). Do you have a moreformal definition that I could use as a starting point? Thanks!
     
  7. Jan 21, 2012 #6

    Dick

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    z is a primitive nth root of unity if [itex]z^n=1[/itex] and [itex]z^k \ne 1[/itex] for [itex]0<k<n[/itex].
     
  8. Jan 21, 2012 #7
    How's this for starters?

    Let [itex]z\in\mathbb{C}[/itex] be a primitive n-th root of unity. Then, for [itex]n\in\mathbb{N}[/itex] and [itex]k\in\mathbb{Z}[/itex] s.t. [itex]0<k<n[/itex] ,[itex]z^{n}=1[/itex] and [itex]z^{k}\neq1[/itex]. Suppose [itex]n=jk[/itex] for some j[itex]\in\mathbb{Z}[/itex]. Then, since z is a root of unity, [itex]z=e^{2k\pi i/n}[/itex] and [itex]z^{k}=z^{n/j}=e^{(2\pi ik/n)^{n/j}}=e^{(2\pi ik)/j}=(e^{2\pi i})^{k/j}=1^{k/j}=1[/itex]. But [itex]z^{k}\neq1[/itex].

    I know I've only proven (or tried to) that [itex]n[/itex] is not a multiple of [itex]k[/itex]. Any ideas on what I'm missing?
     
  9. Jan 21, 2012 #8

    Dick

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    If n and k are not coprime then there is an s>0 such that n=sn' and k=sk'.
     
  10. Jan 21, 2012 #9
    How about this?

    Let [itex]z\in\mathbb{C}[/itex] be a primitive n-th root of unity. Then, for [itex]n\in\mathbb{N}[/itex] and [itex]k\in\mathbb{Z}[/itex] s.t. [itex]0<k<n[/itex] ,[itex]z^{n}=1[/itex] and [itex]z^{k}\neq1[/itex]. Suppose k and n are not coprime. Then, [itex]\exists s\in\mathbb{N}[/itex] s.t. [itex]n=sn'[/itex] and [itex]k=sk'[/itex] for some [itex]k'\in\mathbb{Z},n'\in\mathbb{N}[/itex]. Then, since [itex]z[/itex] is a root of unity, [itex]z=e^{2k\pi i/n}[/itex] and [itex]z^{k}=z^{nk'/n'}=e^{(2\pi ik/n)^{nk'/n'}}=e^{(2\pi ik)k'/n'}=(e^{2\pi i})^{kk'/n'}=1^{kk'/n'}=1[/itex]. But [itex]z^{k}\neq1[/itex]. Contradiction, so k, n coprime.
     
  11. Jan 21, 2012 #10

    Dick

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    That 'proof' should be making you feel at least a little uncomfortable. Pick n=6 and k=2. You are claiming z^2=1? That's not true. Tell me what's wrong with it.
     
  12. Jan 21, 2012 #11
    Do I need to exclude 2 somehow? It seems like 2 is the only prime number where the proof would fail.
     
  13. Jan 21, 2012 #12

    Dick

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    Pick n=15 and k=3. z^3 still isn't 1. Put the numbers in and trace through your steps in the proof. Figure out where something fishy happens.
     
  14. Jan 21, 2012 #13
    Ah, the problem seems to be that I should have [itex]s\neq k [/itex].
     
  15. Jan 21, 2012 #14

    Dick

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    You are just guessing. Look at this. [itex]e^\frac{2 i \pi}{3}={(e^{2 i \pi})}^\frac{1}{3}=1^\frac{1}{3}=1[/itex]. True or false?
     
  16. Jan 21, 2012 #15

    [itex]e^\frac{2 i \pi}{3}\neq{(e^{2 i \pi})}^\frac{1}{3}[/itex]. That was some sloppy algebra on my part -- [itex]x^{2/3}=(x^{2})^{1/3}[/itex] iff [itex]x>0[/itex].
     
  17. Jan 21, 2012 #16

    Dick

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    You have to be careful. Fractional powers of complex numbers aren't well defined. It's only 'safe' to say (a^b)^c=a^(b*c) if c is an integer. You are throwing fractional powers around in your 'proofs' in a bad way.
     
  18. Jan 21, 2012 #17
    Okay, so going back to [itex]k=sk'[/itex] and [itex]n=sn'[/itex], can I use the fact that for s>1, [itex]n/s \leq n[/itex] but [itex]z^{n/s}=e^{2\pi i(k/s)}=1[/itex]? Or am I still using fractional powers improperly?
     
  19. Jan 21, 2012 #18

    Dick

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    No. That's fine. n/s=n' is an integer. So your are allowed to bring that into the power of the exponent. And k/s=k' is an integer. So, sure, you get 1.
     
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