Invertibility of Arithmetic Functions

Theorem
Let $f: \N \to \C$ be an arithmetic function.

Then $f$ has a Dirichlet inverse :
 * $\map f 1 \ne 0$

Proof
Let $*$ denote Dirichlet convolution.

Let $\varepsilon$ denote the identity arithmetic function.

Sufficient Condition
Let $f$ have a Dirichlet inverse $g$:
 * $f * g = \varepsilon$

Then:

Thus $\map f 1 \ne 0$.

Necessary Condition
Suppose that $\map f 1 \ne 0$.

We want to find an arithmetic function $g$ such that:


 * $(1): \quad \map {\paren {f * g} } 1 = 1$
 * $(2): \quad \map {\paren {f * g} } n = 0$ for all $n > 1$

We have:
 * $\map {\paren {f * g} } 1 = \map f 1 \map g 1$

So we have no choice but to define:
 * $\map g 1 = \paren {\map f 1}^{-1}$

and condition $(1)$ is satisfied.

Condition $(2)$ can be written as:

That is:
 * $\map g n = -\dfrac 1 {\map f 1} \ds \sum_{\substack {d \mathop \divides n \\ d \mathop > 1} } \map f d \map g {\frac n d}$

We can recursively define $g$ by this formula, starting with $\map g 1$ as above.

By definition, $g$ then satisfies:
 * $g * f = \varepsilon$

Also see

 * Units of Ring of Arithmetic Functions