Definition
$$ \begin{align*} \mu(n) = \begin{cases} 1 & \text{if } n = 1, \\ (-1)^k & \text{if } n \text{ is the product of } k \text{ distinct primes.} \\ 0 & \text{Otherwise} \end{cases} \end{align*} $$

Example

Take any prime \(p\), then \(\mu(p) = -1\). Take any \(n\) such that it has two prime factors that are equal, then \(\mu(n) = 0\).

\(\mu(p)\) is Multiplicative

Theorem
Given integers \(m\) and \(n\). If \(\gcd(m,n) = 1\), then $$ \begin{align*} \mu(mn) = \mu(m)\mu(n) \end{align*} $$

Proof

We have three cases:
Case 1: Either \(m = 1\) or \(n = 1\). Suppose without the loss of generality that \(n = 1\). Then, the left hand side is

$$ \begin{align*} \mu(mn) = \mu(m \cdot 1) = \mu(m) \end{align*} $$

while the right hand side is

$$ \begin{align*} \mu(m)\mu(n) = \mu(m) \cdot 1 = \mu(1) \end{align*} $$

Thus, \(\mu(mn) = \mu(m)\mu(n)\).

Case 2: Suppose that either \(n\) or \(m\) has at least one prime factor with multiplicity greater than 1. Suppose without the loss of generality that \(m\) has one prime factor \(p^{\alpha}\) where \(\alpha \geq 2\). This implies that \(\mu(m) = 0\) by definition. Thus,

$$ \begin{align*} \mu(m)\mu(n) = 0 \cdot \mu(n) = 0 \end{align*} $$

Furthermore, note that \(mn\) has the same prime factor \(p\) where \(\alpha > 0\). This implies that \(\mu(mn) = 0\). Thus, \(\mu(mn) = \mu(m)\mu(n)\).

Case 3: Suppose that both \(n\) and \(m\) consists of distinct prime factors. Let the prime factorization of \(m\) be \(m = p_1p_2 \ldots p_k\) and let \(n\)’s prime factorization be \(n = q_1 \ldots q_l\). Then

$$ \begin{align*} \mu(m) = (-1)^k \quad \text{ and } \quad \mu(n) = (-1)^l \end{align*} $$

Hence \(\mu(m)\mu(n) = (-1)^k(-1)^l = (-1)^{k+1}\). Moreover, we know that \(\gcd(n,m) = 1\). This implies that

$$ \begin{align*} \{p_1,\ldots, p_k\} \cap \{q_1,\ldots, q_k\} = \emptyset \end{align*} $$

Thus, \(mn\) has distinct prime factors and \(\mu(mn) = (-1)^{k+1}\). Therefore, \(\mu(mn) = \mu(m)\mu(n)\) as desired. \(\blacksquare\)

References

  • Math310 Class Notes by Matthias Beck