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Finding the minimal polynomial of primitive 15th root of 1

  1. Oct 28, 2016 #1
    1. The problem statement, all variables and given/known data
    So I need the find the minimal polynomial of the primitive 15th root of unity. Let's call this minimal polynomial m(x)

    2. Relevant equations


    3. The attempt at a solution
    I know that m(x) is an irreducible factor of x^15 - 1 and also that the degree of m(x) is equal to the Euler function (e), and e(15) = 8, so m(x) will have degree 8. I also know that given one of the primitive roots of unity, call it c, that c^7+c^6+c^5+c^4+c^3+c^2+c+1 = 0. Where can I go from here to find m(x) over Q? perhaps multiplying together (x-c^7)(x-c^6)(x-c^5)......(x-1) will give me it? Is there a faster way than this?

    Thanks PF!
     
  2. jcsd
  3. Oct 28, 2016 #2

    lurflurf

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    Homework Helper

    So x^15-1 has the minimal polynomial in it and other stuff besides
    since the factors 15 are 1,3,5,15 we need to divide by the stuff in x-1,x^3-1,x^5-1
    if we divide by x^3-1 and x^5-1 we have divided by x-1 twice we should put an extra x-1 in so we have two
    then divide by x^3-1 and x^5-1

    in other words if m(n)=minimum polynomial of x^n-1
    x^15-1=m(1)m(3)m(5)m(15)
    x^5-1=m(1)m(5)
    x^3=m(1)m(3)
    x-1=m(1)
    thus
    m(1)=x-1
    m(3)=(x^3)/(x-1)
    m(5)=(x^5)/(x-1)
    so express m(15) using division and multiplication of
    x-1,x^3-1,x^5-1,x^15-1
     
  4. Oct 30, 2016 #3
    Do you mean that m(3) = x^3-1/(x-1) and m(5)= x^5-1/(x-1) ?? and what exactly do you mean by m(n)? The minimal polynomial of some integer n? Thanks btw!
     
  5. Oct 30, 2016 #4

    lurflurf

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    I see what I wrote did not make sense I meant to write m(n) is the minimum polynomial of the nth root of unity ie
    (-1)^(2/n)

    m(3) = (x^3-1)/(x-1)=1+x+x^2
    and
    m(5)= (x^5-1/)(x-1)=1+x+x^2+x^3+x^4

    do you see how to find m(15)?
     
  6. Oct 30, 2016 #5
    Yes, thank you!
     
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