1. Activity 2-18: Cyclotomic polynomials
Activity 2-18: Cyclotomic polynomials

2. We know that a polynomial in X is
a sum of powers of X, where the powers are natural numbers.
There is a special set of polynomials in X
that deserve our attention.
Start by picking a natural number n,
and then draw the n roots of unity
on the circle centre the origin, radius 1.
Suppose we pick n = 8.

3. eight complex roots of unity.
So we have
a regular octagon,
made up of the
eight complex roots of unity.
Now a root is called primitive
if when we take successive powers,
it takes n repetitions to return to 1.
For example:
So w3 is primitive,
while w6 is not.

4. So for n = 8, w, w3, w5 and w7 are primitive.
What happens if we form the polynomial
(X  w)(X  w3)(X  w5 )(X  w7) = 8(X)?
If we multiply this out, we get:
X4 - X3(w7 + w5 + w3 + w) + w4X2(w4 + 1)(w4 + w2 + 1)
- w9X(w6 + w4 + w2 + 1) + w16
Now we know that the polynomial w8 – 1 can be factorised into
(w-1)(w7+w6+w5+w4+w3+w2+w+1)
Since w  1, w8 – 1 = 0  w7+w6+w5+w4+w3+w2+w+1 = 0.

5. X4 - X3(w7 + w5 + w3 + w) + w4X2(w4 + 1)(w4 + w2 + 1)
w7+w6+w5+w4+w3+w2+w+1 = (w + 1)(w6+w4+w2+1), and so also if w8 - 1 = 0, w6+w4+w2+1 = 0, since w  -1.
Multiplying by w, w7+w5+w3+w = 0, since w  0.
And w6+w4+w2+1 = (w2 + 1)(w4 + 1), and so w4 + 1 = 0, since w  ±i.
In the light of this, what happens when we simplify
X4 - X3(w7 + w5 + w3 + w) + w4X2(w4 + 1)(w4 + w2 + 1)
- w9X(w6 + w4 + w2 + 1) + w16 ?
This expression becomes
X4 + 1.

6. (X  w)(X  w3)(X  w5)(X  w7) = X4 + 1.
Is there a better way?
Clearly X8  1 = (X4  1)(X4 + 1) = (X  1)(X  w)...(X  w7)
So the roots that are eighth roots of 1,
while not being fourth roots of 1
(the primitive roots of X8 – 1 = 0)
are w, w3, w5 and w7.
So (X  1)(X  w2)(X  w4)(X  w6) = X4  1, and
(X  w)(X  w3)(X  w5)(X  w7) = X4 + 1.
And so 8(X) = X4 + 1.

7. (X  w)(X  w2)(X  w4 )(X  w5) (X  w7)(X  w8) = 9(X),
What happens if we try the same thing with n = 9?
We have the expression
(X  w)(X  w2)(X  w4 )(X  w5) (X  w7)(X  w8) = 9(X),
where w this time is the first ninth root of 1.
Simplifying very much as we did before, we arrive at
X6 + X3 + 1.
So what do we notice here?
Each time we have arrived at a polynomial expression where the coefficients are integers;
w does not appear in either case.

8. Remarkably, these observations remain true whatever n is.
It is also true that neither expression factorises over the integers. We say that X and X6 + X3 + 1 are irreducible in Z.
Remarkably, these observations remain true whatever n is.
Can we notice anything further here?

9. It seems that p(X), where p is prime, is always
Xp-1 + Xp-2 + Xp-3…+ X2 + X + 1.
Define (n) to be the number of numbers
In 1, 2, …, n  1 that are coprime with n.
It seems that
the degree of n(X) is (n),
and reflecting
on how we have
constructed n(X) ,
this makes sense,
since (n) is the
number of primitive roots.

10. It is tempting to conjecture that every coefficient in n(X) must be either 1 or -1, but this isn’t true:
Note that 105 = 3 x 5 x 7.
Theorem: Migotti If n has at most two distinct odd prime factors, the coefficients of n(X) are all in the set {1, −1, 0}.

11. Is the converse true? That if the coefficients of n(X) are all in the set {1, −1, 0},
then n has at most two distinct odd prime factors?
Untrue!
651(X) = 3x7x31(X) only has coefficients in {1, −1, 0}.
255255(X)=3x5x7x11x13x17(X)
has coefficients up to ±532.

12. So X15 – 1 = d|15 d(X) = 15(X) 5(X) 3(X) 1(X)
One last thing:
Xn – 1 = d|n d(X).
So X15 – 1 = d| d(X) = 15(X) 5(X) 3(X) 1(X)
So 15(X) = (X15-1)/(5(X) 3(X) 1(X) )
= X8 - X7 + X5 - X4 + X3 - X + 1.
So the next cyclotomic polynomial can always be calculated
from the ones that have gone before.
Task: if n is odd, 2n(X) = n(-X).
Can you prove this?

13. Carom is written by Jonny Griffiths, hello@jonny-griffiths.net
With thanks to: Wikipedia, for a helpful article, Graham Everest, and Shaun Stevens.
Carom is written by Jonny Griffiths,