For a polynomial $f=X^n+a_1X^{n1}+\ldots+a_n \in \mathbb{Q}[X]$ 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 $\displaystyle\Phi_n=\prod_{dn} (X^{\frac{n}{d}}1)^{\mu(d)}$.
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2Do you know the relation between $a_1$ and the roots of $f$? – Gerry Myerson Apr 03 '14 at 08:46

1To elaborate on @GerryMyerson's hint, recall that for a monic polynomial of degree $n$, the coefficient of the degree $n1$ term is equal to the negative of the sum of the roots of the polynomial. – heropup Apr 03 '14 at 08:57
2 Answers
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 nexttoleading coefficient. It follows that $$\sum_{d\mid n} g(d) = \sum_{d\mid n}\left[\Phi_d(X)\right]_{1} = \left[\prod_{d\mid n} \Phi_d(X)\right]_{1} = \left[X^n1\right]_{1} = \begin{cases} 1 & \text{for $n=1$}\\ 0 &\text{for $n>1$} \end{cases}$$ 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 $\mu_1=\mu_2$.
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Working out the following definition of the Cyclotomic Polynomial $$ {\displaystyle \Phi _{n}(x)=\prod _{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}\left(xe^{2i\pi {\frac {k}{n}}}\right),} $$ 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: $$ {\displaystyle \mu (n)=\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}}} $$
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@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 9780198531715 – draks ... Nov 07 '18 at 06:19