Decomposition using roots of unity

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1. Apr 1, 2015

ichabodgrant

1. The problem statement, all variables and given/known data
Decompose x5 - 1 into the product of 3 polynomials with real coefficients, using roots of unity.

2. Relevant equations
As far as I know, for xn = 1 for all n ∈ ℤ, there exist n distinct roots.

3. The attempt at a solution

So, let ω = e2πi/5. I can therefore find all the 5th roots of unity:

ω1 = e2πi/5
ω2 = ω2 = e4πi/5
ω3 = ω3 = e6πi/5
ω4 = ω4 = e8πi/5
ω5 = ω5 = e5πi/5 = 1

As far as I can get all the roots, I still don't quite understand how to decompose it into a product of 3 polynomials... What does it mean?

2. Apr 1, 2015

ichabodgrant

I can get (x - 1)(x4 + x3 + x2 + x + 1), but then what to do?

To decompose (x4 + x3 + x2 + x + 1) into 2 more?

3. Apr 1, 2015

Dick

Your polynomial is $(x-\omega_1) (x-\omega_2) (x-\omega_3) (x-\omega_4) (x-\omega_5)$. Try looking at a pair of factors corresponding to complex conjugate roots.

4. Apr 1, 2015

ichabodgrant

You mean this pair are conjugates to each other?

5. Apr 1, 2015

Dick

I mean if $r$ is a complex number and $r^*$ is its conjugate then $(x-r) (x-r^*)$ is a real polynomial. Use that.

6. Apr 1, 2015

ichabodgrant

You mean I can make two pairs into the form (x−r)(x−r∗) that makes them real?

7. Apr 1, 2015

Staff: Mentor

The last one in your list is wrong. e5πi/5 = eπi = -1.

8. Apr 1, 2015

Dick

I mean that if you multiply that out the coefficients of each power of x will be real. Try it. Do you see why?

9. Apr 1, 2015

Ray Vickson

Just try it out for yourself!

10. Apr 1, 2015

ichabodgrant

I can solve it now. Thanks.

11. Apr 1, 2015

ichabodgrant

Sorry. My typo... It should be (e2πi/5)5, so it equals 1.

12. Apr 1, 2015

Staff: Mentor

I figured as much. To keep the same form as the other roots in your list, you could write it as e10πi/5