# Cyclotomic polynomial

I want to prove basic results of cyclotomic polynomials over $$\mathbb{Q}$$.

$$(1)\ \Phi_n(0)=1$$ for all $$n>1$$.

$$(2)\ \Phi_{2n}(x)=\Phi_n(-x)$$ for all odd numbers $$n>1$$.

I want use this result as definition: For every positive integer $$n$$,

$$x^{n}-1=\prod_{d|n} \Phi_{d}(x)$$

where $$d$$ runs over all positive divisors of $$n$$.

1. Remark that $n=1$ is an exception, since $\Phi_1(0) = -1$. Since $x^n-1 = \prod_{d \mid n} \Phi_d(x)$, putting $x=0$ on both sides yields $$-1 = \prod_{d\mid n} \Phi_d(0).$$ Now use induction on $n$, and the fact that $\Phi_1(0)$ accounts for a factor of $-1$ on the right-hand side.

2. Remark that $n=1$ is again an exception, because $\Phi_2(x) = x+1 = -\Phi_1(-x)$. Now, for a general odd $n$,

$$\prod_{d \mid 2n} \Phi_d(x) =x^{2n}-1 = (x^n-1)(x^n+1) = -(x^n-1)((-x)^n-1) = -\prod_{d \mid n} \Phi_d(x)\Phi_d(-x).$$

Now we use induction on $n$. The divisors of $2n$ come in pairs $(d, 2d)$, where $d$ is a divisor of $n$. By the induction hypothesis, $\Phi_{2d}(x) = \Phi_d(-x)$ for each proper divisor of $n$, except when $n=1$ when there is an extra minus sign. By comparing both sides it follows that $\Phi_{2n}(x) = \Phi_n(-x)$.

• in the right hand side(last term) how can i get $\Phi_{n}(-x)$?i had done induction but i get two times product on divisors. Nov 2, 2013 at 5:22
• @SiddhantTrivedi The right-hand side can be written as $-\Phi_n(x)\Phi_n(-x) \prod_d \Phi_n(d)\Phi_n(-d)$ where the product is over the proper divisors of $n$. The left-hand side can be written as $\Phi_n(x)\Phi_{2n}(x) \prod_d \Phi_d(x)\Phi_{2d}(x)$, where the product is also over the proper divisors of $n$. Now apply induction... Nov 2, 2013 at 5:31
• very nice, thanks Bruno! Feb 2, 2014 at 19:41
• My pleasure, @1LiterTears ! Feb 3, 2014 at 1:44
• @BrunoJoyal For second part i have done in exactly same way but i ended up with $Φ_{2n}(x)=−Φ_n(x)$.. I guess you also got the same thing....i do not know how could you get rid off that negative sign.. :( :(
– user87543
Nov 1, 2014 at 7:45

Here is another answer that uses the basic definition $$\Phi_n(x) = \prod_{1\le k\le n \atop \gcd(k,n)=1} (x-e^{2\pi i k/n}).$$ This gives $$\Phi_n(0) = (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1} e^{2\pi i k/n}.$$ Now recall that for $n>1$ we have $$\sum_{1\le k\le n \atop \gcd(k,n)=1} k = \frac{1}{2} \varphi(n) n.$$ This gives $$\Phi_n(0) = (-1)^{\varphi(n)} e^{(2\pi i/n) \times 1/2\times \varphi(n) \times n} = (-1)^{\varphi(n)} e^{\pi i \varphi(n)} = (-1)^{2\varphi(n)} =1.$$

For the second part we get that $$\Phi_n(-x) = \prod_{1\le k\le n \atop \gcd(k,n)=1} (-x-e^{2\pi i k/n}) = (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1} (x+e^{2\pi i k/n}) \\ = (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1} (x-e^{\pi i + 2\pi i k/n})= (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1} (x-e^{2\pi i (k/n+1/2)})\\ = (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1} (x-e^{2\pi i (2k+n)/(2n)}).$$ This almost concludes the proof. We can ignore the sign because $\varphi(n)$ is even.

We now claim there is a bijection between the primitive roots modulo $2n$ call them of class $\mathcal{P}_1$ given by $e^{2\pi i k/(2n)}$ where $\gcd(k,2n)=1$ and $1\le k\le 2n$ and the values produced by $e^{2\pi i (2k+n)/(2n)}$ call them of class $\mathcal{P}_2$ where $\gcd(k,n)=1$ and $1\le k\le n.$ (These two conditions will be our domain of $k.$)

Let $\mathcal{Q}$ be the class of primitive roots modulo $n$. The $\mathcal{P}_1$ roots can be obtained from these by taking $q$ as is if it is odd and taking $q+n$ if it is even (recall that $n$ is odd). Now to establish a bijection to $\mathcal{P}_2$ we must show that $q\equiv 2k+n\bmod (2n)$ and $q+n\equiv 2k+n\bmod (2n)$ always have a unique solution $k$ in the domain of $k$. In the first case we take $k=(q+n)/2$ which is easily seen to be in this domain. In the second case take $k=q/2,$ also clearly in this domain. Moreover we clearly hit each value of $k$ exactly once. This finishes the proof.

• for $n>1$ how can i prove this? $$\sum_{1\le k\le n \atop \gcd(k,n)=1} k = \frac{1}{2} \varphi(n) n.$$ Nov 2, 2013 at 5:29
• I got it.Thanks. Nov 2, 2013 at 6:19

The OP does not say what additional information may be used, so here are two answers that use Möbius inversion as in $$\Phi_n(x) = \prod_{d|n} (x^d-1)^{\mu(n/d)}$$ since this is a instant consequence of the formula you were given to work with.

Using this formula we have immediately that $$\Phi_n(0) = \prod_{d|n} (-1)^{\mu(n/d)} = (-1)^{\sum_{d|n} \mu(n/d)}.$$ Now for $n>1$ we have $\sum_{d|n} \mu(n/d) = 0$, and hence $\Phi_n(0) = (-1)^0 = 1.$ For $n=1$ we get $\Phi_1(0) = (-1)^1 = -1.$ For the second part start with $$\Phi_n(-x) = \prod_{d|n} ((-x)^d-1)^{\mu(n/d)}.$$ Now since $d$ is odd we have $(-x)^d = (-1)^d x^d = -x^d$, giving $$\prod_{d|n} (-x^d-1)^{\mu(n/d)} = \prod_{d|n} \left(\frac{x^{2d}-1}{-x^d+1}\right)^{\mu(n/d)} = \prod_{d|n} (-1)^{\mu(n/d)} \prod_{d|n} \left(\frac{x^{2d}-1}{x^d-1}\right)^{\mu(n/d)}\\ = \prod_{d|n} (-1)^{\mu(n/d)} \prod_{d|n} \left(\frac{x^d-1}{x^{2d}-1}\right)^{\mu(2n/d)} \\= \prod_{d|n} (-1)^{\mu(n/d)} \prod_{d|n} \left(\frac{x^d-1}{x^{2d}-1}\right)^{\mu(2n/d)} \prod_{d|n} \left(\frac{x^{2d}-1}{x^{2d}-1}\right)^{\mu(2n/d/2)}\\ = \Phi_{2n}(x) \prod_{d|n} (-1)^{\mu(n/d)} \prod_{d|n} \frac{1}{(x^{2d}-1)^{\mu(2n/d)}(x^{2d}-1)^{\mu(n/d)}}\\ = \Phi_{2n}(x) \prod_{d|n} (-1)^{\mu(n/d)} \prod_{d|n} \frac{1}{(x^{2d}-1)^{\mu(2n/d)+\mu(n/d)}}.$$ Now since $n>1$ we have $\prod_{d|n} (-1)^{\mu(n/d)}=1$ and $\mu(2n/d)+\mu(n/d)=0$ so that the final answer is $$\Phi_{2n}(x) \times 1 \times \prod_{d|n} \frac{1}{(x^{2d}-1)^0} = \Phi_{2n}(x).$$

• In the 2nd proof how to get 3rd setp?(in the power you changed $\mu(2n/d)$) Nov 2, 2013 at 6:32
• What a proof!really great. Nov 2, 2013 at 7:33
• Can u give me any proof of first formula? or any link for this. Nov 2, 2013 at 7:36