Units of Ring of Arithmetic Functions

Theorem
Let $f$ be an arithmetic function.

Then $f$ is a unit in the Ring of Arithmetic Functions $\mathcal A$ :
 * $f \left({1}\right) \ne 0$.

Sufficient Condition
Let $f$ be a unit in the Ring of Arithmetic Functions $\mathcal A$

Aiming for a contradiction, suppose $f \left({1}\right) = 0$.

Then for all arithmetic functions $g$:

Since the identity element $\iota$ of $\mathcal A$ satisfies:
 * $\iota \left({1}\right) = 1$

it follows that $f$ has no inverse in $\mathcal A$.

That is, $f$ is not a unit.

Thus by Proof by Contradiction:
 * $f \left({1}\right) \ne 0$.

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

We want to find an arithmetic function $f^{-1}$ such that:


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

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

So taking:
 * $f^{-1} \left({1}\right) = f \left({1}\right)^{-1}$

part $(1)$ is satisfied.

Now suppose we have constructed $f \left({k}\right)$ for $k = 1, \dotsc, n-1$.

We compute:

So to satisfy $(2)$, we require:


 * $\displaystyle f \left({1}\right) f^{-1} \left({n}\right) + \sum_{\substack{d \mathop \backslash n \\ d \mathop > 1}} f \left({d}\right) f^{-1}\left(\frac n d \right) = 0$

To obtain this we take:


 * $\displaystyle f^{-1} \left({n}\right) = -\frac 1 {f \left({1}\right)} \sum_{\substack {d \mathop \backslash n \\ d \mathop > 1}} f \left({d}\right) f^{-1}\left({\frac n d}\right)$

Thus we have inductively constructed the required function.