Decomposition using roots of unity

[ichabodgrant]

Homework Statement

Decompose x⁵ - 1 into the product of 3 polynomials with real coefficients, using roots of unity.

Homework Equations

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

The Attempt at a Solution

[/B]
So, let ω = e^2πi/5. I can therefore find all the 5th roots of unity:ω₁ = e^2πi/5
ω₂ = ω² = e^4πi/5
ω₃ = ω³ = e^6πi/5
ω₄ = ω⁴ = e^8πi/5
ω₅ = ω⁵ = e^5πi/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?

 

[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?

 

[Dick]

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?

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.

 

[ichabodgrant]

You mean this pair are conjugates to each other?

 

[Dick]

You mean this pair are conjugates to each other?

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

 

[ichabodgrant]

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

 

[Mark44]

So, let ω = e^2πi/5. I can therefore find all the 5th roots of unity:ω₁ = e^2πi/5
ω₂ = ω² = e^4πi/5
ω₃ = ω³ = e^6πi/5
ω₄ = ω⁴ = e^8πi/5
ω₅ = ω⁵ = e^5πi/5 = 1

The last one in your list is wrong. e^5πi/5 = e^πi = -1.

 

[Dick]

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

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

 

[Ray Vickson]

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

Just try it out for yourself!

 

[ichabodgrant]

I can solve it now. Thanks.

 

[ichabodgrant]

The last one in your list is wrong. e^5πi/5 = e^πi = -1.

Sorry. My typo... It should be (e^2πi/5)⁵, so it equals 1.

 

[Mark44]

Sorry. My typo... It should be (e^2πi/5)⁵, so it equals 1.

I figured as much. To keep the same form as the other roots in your list, you could write it as e^10πi/5