Finding the minimal polynomial of primitive 15th root of 1

[PsychonautQQ]

Homework Statement

So I need the find the minimal polynomial of the primitive 15th root of unity. Let's call this minimal polynomial m(x)

Homework Equations

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!

 

[lurflurf]

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

 

[PsychonautQQ]

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

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!

 

[lurflurf]

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)?

 

[PsychonautQQ]

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)?

Yes, thank you!