Coefficient of $n$th cyclotomic polynomial equals $-\mu(n)$

Question

For a polynomial $f=X^n+a_1X^{n-1}+\ldots+a_n \in \mathbb{Q}[X]$ we define $\varphi(f):=a_1 \in \mathbb{Q}$. Now I want to show that for the $n$th cyclotomic polynomial $\Phi_n$ it holds that $$\varphi(\Phi_n)=-\mu(n)$$ where $\mu(n)$ is the Möbius function. What I know is that $\displaystyle\Phi_n=\prod_{d|n} (X^{\frac{n}{d}}-1)^{\mu(d)}$.

Answer 1

Note that the Möbius function $\mu(n)$ can be defined as the unique arithmetic function $f$ that fulfills $$ \forall n\in\mathbb{N}: \sum_{d\mid n} f(d) = \begin{cases}1 & \text{for $n=1$}\\ 0 &\text{for $n>1$}\end{cases}\tag{1}$$

Now define $$g(d) = \left[\Phi_d(X)\right]_{-1}$$ where $[\cdot]_{-1}$ is the next-to-leading coefficient. It follows that $$\sum_{d\mid n} g(d) = \sum_{d\mid n}\left[\Phi_d(X)\right]_{-1} = \left[\prod_{d\mid n} \Phi_d(X)\right]_{-1} = \left[X^n-1\right]_{-1} = \begin{cases} -1 & \text{for $n=1$}\\ 0 &\text{for $n>1$} \end{cases}$$ And by the above definition of $\mu(n)$ you can conclude $g(n)=-\mu(n)$, which implies $\left[\Phi_n(X)\right]_{-1}=-\mu(n)$.

Update: For monic univariate polynomials $f_1,f_2$ of degree at least $1$ we have $$\left[f_1(X)\,f_2(X)\right]_{-1} = \left[f_1(X)\right]_{-1}+\left[f_2(X)\right]_{-1}$$ because the coefficient in question is the negative sum of the roots of $f_1$ and $f_2$ (with multiplicity). Even without considering roots, this follows from looking at how the product expands.

Update: Proof that $\mu$ is the unique arithmetic function with property $(1)$:

For $n=1$ we obtain $f(1)=1=\mu(1)$. For $n>1$, let $n=p_1^{e_1}\cdots p_r^{e_r}$ where $p_1,\ldots,p_r$ are pairwise distinct primes and $e_1,\ldots,e_r$ are positive integers. Then $$\begin{align} \sum_{d\mid n} \mu(d) &= \sum_{j_1=0}^{e_1}\cdots\sum_{j_r=0}^{e_r} \mu(p_1^{j_1}\cdots p_r^{j_r}) = \sum_{j_1=0}^{e_1}\cdots\sum_{j_r=0}^{e_r} \begin{cases} 0 & \text{if any $j_i>1$}\\ (-1)^{j_1+\cdots+j_r} & \text{otherwise}\end{cases}\\ &= \left(\sum_{j_1=0}^{1}(-1)^{j_1}\right)\cdots \left(\sum_{j_r=0}^{1}(-1)^{j_r}\right) = 0 \end{align}$$ Thus $\mu$ has property $(1)$.

Now for uniqueness: Let $\mu_1,\mu_2$ be arithmetic functions with property $(1)$. Then necessarily $\sum_{d\mid n}\mu_1(d)=\sum_{d\mid n}\mu_2(d)$ for all positive integers $n$. Suppose $\mu_1\neq \mu_2$, then there exists a minimal positive integer $m$ such that $\mu_1(m)\neq \mu_2(m)$. But then $\sum_{d\mid m}\mu_1(d)\neq\sum_{d\mid m}\mu_2(d)$ which contradicts the hypothesis. (All $d<m$ do not make any difference since $m$ is minimal in that respect, but $d=m$ does make a difference.) The only remaining possibility is $\mu_1=\mu_2$.

Answer 2

Working out the following definition of the Cyclotomic Polynomial $$ {\displaystyle \Phi _{n}(x)=\prod _{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}\left(x-e^{2i\pi {\frac {k}{n}}}\right),} $$ you'll get $$ {\displaystyle x^{\varphi(n)}+x^{\varphi(n)-1} \left({ -\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}}} \right)} + \dots + 1, $$ because every coefficient of the expanded polynomial may be represented as an elementary symmetric polynomial, which is $e_1(\cdot)$ in your case.

Using the defintion here, you'll spot right away that this gives the Möbius function: $$ {\displaystyle -\mu (n)=-\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}}} $$

Answer 3

Hopefully more understandable for beginners:

The Möbius function $\mu$ is characterized by: $$\forall n\in\Bbb N\quad\sum_{d\mid n}\mu(d)=\begin{cases}1&\mbox{ if } n=1\\ 0&\mbox{ if } n>1.\end{cases}$$ Let $S(n)$ denote the sum of complex primitive $n$-th roots of $1.$ In order to prove that $S=\mu,$ all we have to do is to prove that $S$ satisfies the same equation.

Now, $\sum_{d\mid n}S(d)$ is the sum of all $n$-th roots of $1:$ $$\sum_{d\mid n}S(d)=\sum_{k=0}^{n-1}(e^{2\pi i/n})^k=\begin{cases}1&\mbox{ if } n=1\\ \frac{1-(e^{2\pi i/n})^n}{1-e^{2\pi i/n}}=0&\mbox{ if } n>1,\end{cases}$$ and we are done.

Edit. Elementary explanation (if needed) for why $\sum_{d\mid n}S(d)$ is the sum of $n$-th roots of $1$: if $z^d=1$ and $d\mid n$ then obviously $z^n=1$. Conversely, if $z^n=1$, let $d$ be the integer s.t. $z$ is a primitive $d$-th root of $1$, then $d\mid n$ (write $n=dq+r$ the Euclidean division of $n$ by $d$ and notice that $r$ is necessarily $0$).

Answer 4

This is equivalent to proving the sum of primitive $n$th roots equals $\mu(n),$ which we can prove with principle of exclusion-inclusion. Let $\text{ord}(\omega)$ denote the order of $\omega.$ First, recall that $\sum\limits_{\text{ord}(\omega)|n} \omega = \sum\limits_{i=0}^{n-1} z^i = \frac{z^n - 1}{z-1} = 0$ for $n > 1$ where $z$ is a primitive $n$th root of unity.

Let $p_1, p_2, \dots, p_r$ be the prime factors of $n.$ Notice that $$\sum\limits_{\text{ord} = n} = \sum\limits_{\text{ord}|n} - (\sum\limits_{\text{ord}|\frac{n}{p_1}} + \dots + \sum\limits_{\text{ord}|\frac{n}{p_r}}) + \sum\limits_{i, j} \sum\limits_{\text{ord}|\frac{n}{p_i p_j}} - \sum\limits_{i, j, k} \sum\limits_{\text{ord}|\frac{n}{p_i p_j p_k}} + \dots$$

By our initial observation, every sum is $0$ except when $m = 1$ in $\text{ord}|m.$ This term appears iff $n$ is squarefree, and the sign is the number of prime factors. Thus, the sum we seek is $\begin{cases} 0, n \text{ squarefree} \\\ (-1)^k, n = p_1 \cdots p_k \end{cases},$ which is the definition of the Mobius function.

Answer 5

This is exercise 4.3.28 of Niven, Zuckerman, Montgomery's An Introduction to the Theory of Numbers 5th Ed.

Define $$S(n)=\sum_{\substack{1\leq k\leq n \\ (k,n)=1}} e^{2\pi i\frac{k}{n}}.$$

It suffices to prove that $S$ is multiplicative and agrees with $\mu$ at prime powers.

Let $n,m$ be coprime. We now show $S(mn)=S(m)S(n)$.

Note that each summand in the expansion of $S(m)S(n)$ is of the form $e^{2\pi i\frac{a}{m}+2\pi i\frac{b}{n}}=e^{2\pi i\frac{an+bm}{mn}}$, with $(a,m)=(b,n)=1$. Clearly $(an+bm,mn)=1$. By CRT, we see that each term in the expansion must be distinct. Let $k$ be such that $(k,mn)=1$. Let $a',b'$ be such that $a'n\equiv1\pmod m$ and $b'm\equiv1\pmod n$. Let $k_1,k_2$ be such that $k\equiv k_1\pmod m$ and $k\equiv k_2\pmod n$. Let $1\leq c_1\leq m,1\leq c_2\leq n$ be such that $c_1\equiv a'k_1\pmod m$ and $c_2\equiv b'k_2\pmod n$. Clearly $(c_1,m)=1$ and $(c_2,n)=1$. Hence $e^{2\pi i\frac{c_1}{m}+2\pi i\frac{c_2}{n}}=e^{2\pi i\frac{nc_1+mc_2}{mn}}$ is in the expansion. By CRT and reducing the exponent, $e^{2\pi i\frac{nc_1+mc_2}{mn}}=e^{2\pi i\frac{k}{mn}}$. Thus $S(mn)=S(m)S(n)$.

Let $p$ be a prime. We now show $\mu(p^n)=S(p^n)$.

Clearly $S(p)=-1=\mu(p)$. Let $n>1$. Note that \begin{align*} &\sum_{1\leq k\leq p^n}e^{2\pi i k}\\&=\sum_{\substack{1\leq k\leq p\\(k,p)=1}}e^{2\pi i\frac{kp^{n-1}}{p^n}}+\sum_{\substack{1\leq k\leq p^2\\(k,p)=1}}e^{2\pi i\frac{kp^{n-2}}{p^n}}+\cdots+\sum_{\substack{1\leq k\leq p^{n-1}\\(k,p)=1}}e^{2\pi i\frac{kp}{p^n}}+\sum_{\substack{1\leq k\leq p^n\\(k,p)=1}}e^{2\pi i\frac{k}{p^n}}\\ &=S(p)+S(p^2)+\cdots+S(p^{n-1})+S(p^n)+1 \end{align*} Since $\sum_{1\leq k\leq p^n}e^{2\pi i k}=0$, we have $\sum_{k=2}^{n}S(p^k)=0$. By induction on $n$, we have $S(p^n)=0=\mu(p^n)$ for $n>1$.

Therefore $$\mu(n)=\sum_{\substack{1\leq k\leq n \\ (k,n)=1}} e^{2\pi i\frac{k}{n}}.$$

Answer 6

The proof below is outlined in exercise 2.14 of Apostol's Introduction to Analytic Number Theory.

Let $f(x)$ be defined for all rational $x$ in $0\leq x\leq 1$ and let $$ F(n)=\sum_{k=1}^n f\left(\frac{k}{n}\right),\,\,\,F^*(n)=\sum_{\substack{k=1\\(k,n)=1}}^nf\left(\frac{k}{n}\right).$$ (a) Prove that $F^*=\mu*F$, the Dirichlet product of $\mu$ and $F$.

(b) Use (a) or some other means to prove that $\mu(n)$ is the sum of the primitive $n$th roots of unity: $$\mu(n)=\sum_{\substack{k=1\\(k,n)=1}}^ne^{2\pi ik/n}.$$

(a) Note that \begin{align*} \sum_{d\mid n}F^*(d)&=\sum_{d\mid n}\sum_{\substack{k=1\\(k,d)=1}}^df\left(\frac{k}{d}\right)\\ &=\sum_{d\mid n}\sum_{\substack{k=1\\(k,d)=1}}^df\left(\frac{kn/d}{n}\right)\\ &=\sum_{k=1}^{n}f\left(\frac{k}{n}\right)\\ &=F(n), \end{align*} where the second to last equality follows from the fact that $\{kn/d| 1\leq k \leq d, (k,d)=1, d|n\}=\{1,2\cdots, n\}$.

By Möbius inversion, we have $F^*=\mu*F$.

(b) Let $f=\exp$. By (a), we have \begin{align*} F^*(n)&=\sum_{d\mid n}\mu(d)\sum_{k=1}^{n/d} e^{2\pi ik/(n/d)}. \end{align*} Since $\sum_{k=1}^me^{2\pi ik/m}=e^{2\pi i/m}\frac{\left(e^{2\pi ik/m}\right)^m-1}{e^{2\pi ik/m}-1}=0$ for $m>1$, $\sum_{k=1}^{n/d} e^{2\pi ik/(n/d)}=0$ for all divisors $d$ of $n$ such that $d<n$. So $F^*(n)=\mu(n)e^{2\pi i\cdot1/1}=\mu(n)$, or $$\mu(n)=\sum_{\substack{k=1\\(k,n)=1}}^ne^{2\pi ik/n}.$$