Invertibility of Arithmetic Functions

Theorem

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

Then $f$ has a Dirichlet inverse if and only if:

$\map f 1 \ne 0$

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:

\(\ds 1\)

\(=\)

\(\ds \map \varepsilon 1\)

Definition of Identity Arithmetic Function

\(\ds \)

\(=\)

\(\ds \map {\paren {f * g} } 1\)

\(\ds \)

\(=\)

\(\ds \sum_{d \mathop \divides 1} \map f d \map g {d^{-1} }\)

Definition of Dirichlet Convolution

\(\ds \)

\(=\)

\(\ds \map f 1 \map g 1\)

Divisors of One

Thus $\map f 1 \ne 0$.

$\Box$

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:

\(\ds 0\)

\(=\)

\(\ds \sum_{d \mathop \divides n} \map f d \map g {\frac n d}\)

\(\ds \)

\(=\)

\(\ds \map f 1 \map g n + \sum_{\substack {d \mathop \divides n \\ d \mathop > 1} } \map f d \map g {\frac n d}\)

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:

$g * f = \varepsilon$

$\blacksquare$

Also see

• Units of Ring of Arithmetic Functions