Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to prove that for each positive integer $n$ the sum of the primitive $n$th roots of unity in $\mathbb{C}$ is $\mu(n)$, where $\mu$ is the Möbius function.

share|improve this question
    
So what's the question? –  Patrick Da Silva Oct 20 '12 at 0:02
    
Yes...or not: what's your question? ¿Cuál es tu pregunta? –  DonAntonio Oct 20 '12 at 0:03
    
I need one proof to the given proposition –  Elmo goya Oct 20 '12 at 0:11
    
Well explained... not wikipedia... please!!! –  Elmo goya Oct 20 '12 at 0:33
    
You can try to prove it for primes, then $p^n$, then use multiplicativity. –  Berci Oct 20 '12 at 0:36
show 2 more comments

2 Answers

up vote 3 down vote accepted

Do you know $$\sum_{d\mid m}\mu(d)=1{\rm\ if\ }m=1,\,\,=0{\rm\ else}$$ The sum of the primitive $n$th roots of unity is $$\sum_{\gcd(k,n)=1}e^{2\pi ik/n}=\sum_1^n\sum_{d\mid\gcd(k,n)}\mu(d)e^{2\pi ik/n}=\sum_{d\mid n}\mu(d)\sum_0^{(n/d)-1}e^{2\pi idk/n}$$ The inner sum os the sum of all the $m$th roots of unity where $m=n/d$, so it's zero except for $d=n$ when it's $1$. So, the original sum evaluates to $\mu(n)$.

share|improve this answer
add comment

Let $\theta$ denote the first $n$th primitive root: $\theta:=e^{2\pi i/n}$.

  1. If $n=p$ is prime, $\mu(p)=-1$ and each $0\ne a<p$ is relatively prime to $p$, so this $\theta^{a}$ is primitive $p$th root. The sum of all $n$th roots is always $0$ (because if we multiply it by $\theta$, it doesn't change). So we miss only the $\theta^0=1$, hence the sum is $-1$.
  2. If $n=p^k$ ($k\ge 2$), then $\mu(n)=0$ and exactly the $p\cdot a$ elements have common divisor with $n$, so $$\sum_{\theta^u\text{ prim.root}}\theta^u=\sum_{u\ne a\cdot p}\theta^u = \sum_{u=0}^{n-1}\theta^u-\sum_{v=0}^{\frac np-1} \theta^{pv} $$ Can you continue?
  3. You also need to show that both functions in question are multiplicative, i.e., whenever $\gcd(a,b)=1$, we have $$\mu(ab)=\mu(a)\cdot\mu(b) $$ and same for the other function.

From these the proposition follows.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.