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Loser66

  • one year ago

A primitive nth root of unity is a complex number a such that 1,a,a^2,....., a^(n-1) are distinct roots of unity. Show that if a, b are primitive nth and mth roots of unity, respectively, then ab is a kth root of unity for some integer k. What is the smallest value of k? What cn be said if a and b are nonprimitive roots of unity? Please, help

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  1. Loser66
    • one year ago
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    I don't know why I feel it is trivial:) \(a=|a| e^{i\theta}\\b=|b|e^{i\tau}\\ab= |ab|e^{i(\theta +\tau)}\) it shows that ab is a root of unity, right?

  2. imqwerty
    • one year ago
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    ?i don't get the question :) unity is 1 nd roots of real numbers r real

  3. Loser66
    • one year ago
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    "A primitive nth root of unity is a COMPLEX number a............"

  4. beginnersmind
    • one year ago
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    If a is \(a = e^{i\phi}\) is a primitive root of unity then \(\frac{\phi}{2\pi}=\frac{p}{q}\) such that p and q are natural numbers and p<q.

  5. beginnersmind
    • one year ago
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    And gcd(p,q) = 1

  6. Loser66
    • one year ago
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    then?

  7. Loser66
    • one year ago
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    My computer is on and off!! if you don't see I respond, please forgive me.

  8. beginnersmind
    • one year ago
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    BTW, want to point out that here: \[a=|a| e^{i\theta}\\b=|b|e^{i\tau}\\ab= |ab|e^{i(\theta +\tau)}\] |a| = |b| = 1, since they are roots of unity. Anyway, we write \[a=e^{i\theta} = e^{i\frac{p}{q}2\pi} \] \[b=e^{i\tau} = e^{i\frac{r}{s}2\pi} \] \[ab= e^{i2\pi*(\frac{p}{q}+\frac{r}{s})} = e^{i2\pi*\frac{ps+rq}{sq}}\] So ab is a (sq)th root of unity.

  9. beginnersmind
    • one year ago
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    p,q,r,s are natural number s.t p<q, r<s

  10. beginnersmind
    • one year ago
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    \[a^n= 1 = e^{2 \pi \ i} \\ a= e^{\frac{2\pi}{n} i }\] Did you mean: \[ \Large a= e^{\frac{2\pi k}{n} i }\] and now the constraint that a is a primitive root is equivalent to gcd(k,n) =1

  11. beginnersmind
    • one year ago
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    Anyway, I think if m and n are distinct a*b will be a primitive root of (m*n)th order. I don't want to go through this because it's a pain to write out in latex. There's also some value in thinking it through for yourself. But really, it's mostly because I'm lazy :) I'd rather go through the main idea again: There is a bijection between primitive roots of unity and proper fractions k/n, where gcd(k,n) = 1. Multiplying two roots of unity is equivalent to adding the fractions they are paired with. So if a and b are primitive roots, associated with k/n and l/m respectively, ab is associated with (k/n + l/m). To find the order of this primitive root write this fraction (km+ln)/mn and write it in its simplest form. The order of ab is the denominator. In particular, when I said that a*b will have an order of (m*n), when m =/= n, I might as well have said that: gcd(k,n) = gcd(l,m) = 1 and m =/= n implies gcd((km+ln)/(mn)) = 1

  12. beginnersmind
    • one year ago
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    "i.e. k=8 = least common multiple of 4 and 8" You're right. "also, because we have 3 in the numerator this is not a "primitive root"" The definition for primitive root is that it's a root for some degree n but not for any degree m<n. The numerator doesn't need to be 1. In the e^(i*2pi*(k/n)) form it's equivalent to gcd(k,n) = 1.

  13. beginnersmind
    • one year ago
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    There is no contradiction. Take the equation z^4 = 1. The solution set for this is {1,i,-1,-i}. Of course i is a primitive root. But so is -i, which you can check 1 -i (-i)^2 = -1 (-i)^3 = i That's 4 distinct roots of unity, exactly as the definition requires.

  14. beginnersmind
    • one year ago
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    "I guess you are missing the point. The definition says "a" is the primitive nth root. in the list of roots: a^0,a^1, a^2, ... a^(n-1)" No, that's not what it says. Read it again carefully. It says that you can check if a number is a number is a primitive n-th root by veriftying that its first n powers are all distinct (and roots of unity). If you don't believe read the wikipedia article on roots of unity or the planetmath article I linked earlier. It's a standard definition, not a word made up for specifically this problem.

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