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Classical Möbius inversion over positive divisors

Statement

Let RR be a commutative ring and let f,g:Z>0Rf,g:\mathbb Z_{>0}\to R. Then

g(n)=dnf(d)for every n1g(n)=\sum_{d\mid n}f(d)\quad\text{for every }n\ge1

if and only if

f(n)=dnμ(n/d)g(d)=dnμ(d)g(n/d)for every n1.f(n)=\sum_{d\mid n}\mu(n/d)g(d)=\sum_{d\mid n}\mu(d)g(n/d)\quad\text{for every }n\ge1.

All divisors in the sums are positive.

Facts & Assumptions

Given: A commutative ring RR and functions f,gf,g on the positive integers.

[L1]

Lower-finite poset inversion says g(n)=dnf(d)g(n)=\sum_{d\mid n}f(d) exactly when f(n)=dnμ(d,n)g(d)f(n)=\sum_{d\mid n}\mu_{\mid}(d,n)g(d) (Möbius inversion on a lower-finite poset, with the dual upper-finite form).

Proof

technique · direct
1.1

Apply [L1] to the lower-finite divisibility poset from [L3] and substitute [L2]. This gives f(n)=dnμ(n/d)g(d)f(n)=\sum_{d\mid n}\mu(n/d)g(d).

L1L2L3
2.1

The map dn/dd\mapsto n/d is a bijection of the positive divisors of nn with itself and is its own inverse. Reindexing the sum in step 1.1 by [L4] gives f(n)=dnμ(d)g(n/d)f(n)=\sum_{d\mid n}\mu(d)g(n/d).

step 1.1L4
3.1

Since [L1] is an equivalence, steps 1.1 and 2.1 prove both directions and both standard indexings.

step 1.1step 2.1L1

Depends on

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