Invertibility of Arithmetic Functions

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

Then $f$ has a Dirichlet inverse $f(1) \neq0$.

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 $f \left({1}\right) \ne 0$.

Sufficient Condition
Now suppose that $f \left({1}\right) \ne 0$.

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


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

We have:
 * $\left({f * g}\right) \left({1}\right) = f \left({1}\right) g \left({1}\right)$

So we have no choice but to define:
 * $g \left({1}\right) = f \left({1}\right)^{-1}$

and condition $(1)$ is satisfied.

Condition $(2)$ can be written as:

That is:
 * $g \left({n}\right) = -\dfrac1{f(1)} \displaystyle \sum_{\substack{d \mathop \backslash n \\ d \mathop > 1}} f \left({d}\right) g\left(\frac n d \right)$

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

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

Also see

 * Units of Ring of Arithmetic Functions