Activity 2-18: Cyclotomic polynomials www.carom-maths.co.uk Activity 2-18: Cyclotomic polynomials

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.

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.

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.

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.

(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.

(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.

Remarkably, these observations remain true whatever n is. It is also true that neither expression factorises over the integers. We say that X5 + 1 and X6 + X3 + 1 are irreducible in Z. Remarkably, these observations remain true whatever n is. Can we notice anything further here?

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.

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}.

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.

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|15 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?

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, hello@jonny-griffiths.net