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The number-theoretic Möbius function is multiplicative on coprime positive integers
Statement
If are coprime positive integers, then
Facts & Assumptions
Given: Coprime positive integers (Coprime integers: ).
Unique prime factorisation implies that coprime positive integers have disjoint prime supports, and every divisor of has a unique product form with and (The fundamental theorem of arithmetic: every integer is a product of primes, and the factorisation is unique up to order — if with every and prime, then and for some ).
Möbius functions multiply on product posets (The Möbius function of a product poset is the product of the Möbius functions).
for every positive integer (The number-theoretic Möbius function is the poset Möbius function of divisibility: ).
Möbius values transport through poset isomorphisms because the diagonal and recurrence conditions uniquely determine them (The Möbius recurrence: and both interval sums of vanish when ).
Proof
By [L1], is a bijection from to . It preserves and reflects divisibility coordinatewise, again by the disjoint prime supports, so it is a poset isomorphism.
Apply the product theorem at the endpoints and transfer along the isomorphism by [L4]: .
Replacing each poset value in step 2.1 by its number-theoretic value using [L3] yields .
Depends on
- The number-theoretic Möbius function is the poset Möbius function of divisibility: $\mu(n)=\mu_{\mid}(1,n)$
- The Möbius function of a product poset is the product of the Möbius functions
- Coprime integers: $\gcd(a,b) = 1$
- The fundamental theorem of arithmetic: every integer $n \ge 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i<r} p_i = \prod_{j<s} q_j$ with every $p_i$ and $q_j$ prime, then $r = s$ and $q_i = p_{\pi(i)}$ for some $\pi \in \operatorname{Sym}(r)$
- The Möbius recurrence: $\mu_P(x,x)=1$ and both interval sums of $\mu_P$ vanish when $x<y$
Used by
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Sources
- Stanford Pairing-Based Cryptography notes, Möbius inversion (standard reference, not scraped)