Science:Math Exam Resources/Courses/MATH312/December 2009/Question 05 (a)

MATH312 December 2009

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Question 05 (a)
Let $\displaystyle \mu$ denote the Mobius function. If $\displaystyle n>1$ is an integer, show that $\displaystyle \sum _{d\mid n}\mu (d)=0$

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Recall that the Mobius function is the function that evaluates to 1 if n is square-free and has an even number of prime factors, -1 if n is square-free and has an odd number of factors, and 0 if n has a square factor. First show/recall that the Mobius function is multiplicative.

Hint 2
Let $\displaystyle S(n)=\sum _{d\mid n}\mu (d)$. Show that this function is multiplicative then evaluate it on prime powers. Note: One could solve this problem by a more direct approach however learning about the theory of multiplicative functions is quite important and worth doing.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First we show that the Mobius function is multiplicative. Let m and n be coprime integers. Then let $\displaystyle m=\prod _{i=1}^{M}p_{i}^{\alpha _{i}}$ and $\displaystyle n=\prod _{j=1}^{N}q_{j}^{\beta _{j}}$. We wish to show that $\displaystyle \mu (mn)=\mu (m)\mu (n)$. Clearly, if m or n contains a square prime factor, then both sides evaluate to 0 so without loss of generality, suppose that each $\displaystyle \alpha _{i},\beta _{j}=1$. We break into cases based on the number of prime factors. Case 1: M and N are both odd or both even.. In these cases, $\displaystyle \mu (mn)=1$. Now, since the number of prime factors are both odd or both even, we have that $\displaystyle \mu (m)=\mu (n)$ and since these are now either plus or minus 1 (we've eliminated the case when the Mobius function is 0 on either side), we have $\displaystyle \mu (mn)=1=\mu (m)^{2}=\mu (m)\mu (n)$ Case 2: M and N are of opposite parity.. Suppose that M is even (the argument is symmetric). In this case $\displaystyle \mu (mn)=-1$ since we have an odd number of factors. Now, we have that $\displaystyle \mu (m)=-1$ and $\displaystyle \mu (n)=1$. Thus, $\displaystyle \mu (mn)=-1=\mu (m)\mu (n)$ Hence the Mobius function is multiplicative. Now, let $\displaystyle S(n)=\sum _{d\mid n}\mu (d)$. I claim that this function is also multiplicative. For this, again let m and n be coprime integers (notice that $\displaystyle S(1)=1$). We see that $\displaystyle S(m)S(n)=\sum _{d\mid m}\mu (d)\sum _{e\mid n}\mu (e)=\sum _{d\mid m}\sum _{e\mid n}\mu (d)\mu (e)=\sum _{d\mid m}\sum _{e\mid n}\mu (de)$ Now as m and n are coprime, we see that for any divisor $\displaystyle h\mid mn$, we can write $\displaystyle h$ uniquely of the form $\displaystyle h=de$ where $\displaystyle d\mid m$ and $\displaystyle e\mid n$ and vice versa. Thus, we have $\displaystyle S(m)S(n)=\sum _{d\mid m}\sum _{e\mid n}\mu (de)=\sum _{h\mid mn}\mu (h)=S(mn)$ showing that this function is also multiplicative. So, by multiplicativity, it suffices to show that $\displaystyle S(p^{a})=0$ where p is a prime and a is an integer. Here, we have $\displaystyle S(p^{a})=\sum _{d\mid p^{a}}\mu (d)=\sum _{i=0}^{a}\mu (p^{i})=\mu (1)+\mu (p)+\sum _{i=2}^{a}\mu (p^{i})$. Notice that the last sum might be non-existent if a is 1. In any case, the last sum evaluates to 0 since all the terms contain a $\displaystyle p^{2}$ factor. Notice that $\displaystyle \mu (1)+\mu (p)=1-1=0$ and so $\displaystyle S(p^{a})=0$ so long as a is not 0 (which is allowed since n was an integer greater than 1). Thus we have a multiplicative function which evaluates to 0 on all prime powers, hence it must be zero for all values greater than 1 since $\displaystyle S(n)=S\left(\prod _{j=1}^{N}q_{j}\right)=\prod _{j=1}^{N}S\left(q_{j}\right)=0$. This completes the proof.

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