Decomposition using roots of unity

  • #1

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


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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?
 

Answers and Replies

  • #2
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
Dick
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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.
 
  • #4
You mean this pair are conjugates to each other?
 
  • #5
Dick
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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.
 
  • #6
You mean I can make two pairs into the form (x−r)(x−r∗) that makes them real?
 
  • #7
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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
The last one in your list is wrong. e5πi/5 = eπi = -1.
 
  • #8
Dick
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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?
 
  • #9
Ray Vickson
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You mean I can make two pairs into the form (x−r)(x−r∗) that makes them real?

Just try it out for yourself!
 
  • #11
The last one in your list is wrong. e5πi/5 = eπi = -1.


Sorry. My typo... It should be (e2πi/5)5, so it equals 1.
 
  • #12
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Sorry. My typo... It should be (e2πi/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 e10πi/5
 

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