16
$\begingroup$

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 $\displaystyle\Phi_n=\prod_{d|n} (X^{\frac{n}{d}}-1)^{\mu(d)}$.

$\endgroup$
  • 2
    $\begingroup$ Do you know the relation between $a_1$ and the roots of $f$? $\endgroup$ – Gerry Myerson Apr 3 '14 at 8:46
  • 1
    $\begingroup$ To elaborate on @GerryMyerson's hint, recall that for a monic polynomial of degree $n$, the coefficient of the degree $n-1$ term is equal to the negative of the sum of the roots of the polynomial. $\endgroup$ – heropup Apr 3 '14 at 8:57
13
$\begingroup$

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 $\mu_1=\mu_2$.

$\endgroup$
1
$\begingroup$

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: $$ {\displaystyle -\mu (n)=-\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}}} $$

$\endgroup$
  • $\begingroup$ The last equality is far from obvious at this point. $\endgroup$ – darij grinberg Nov 6 '18 at 22:06
  • $\begingroup$ @darijgrinberg that's true. I refer to: Hardy, G. H.; Wright, E. M. (1980), An Introduction to the Theory of Numbers (5th ed.), Oxford: Oxford University Press, ISBN 978-0-19-853171-5 $\endgroup$ – draks ... Nov 7 '18 at 6:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy