Units of Ring of Arithmetic Functions

Theorem
Let $f$ be an arithmetic function.

Then $f$ is a unit in the Ring of Arithmetic Functions $\mathcal A$ if and only if $f(1) \ne 0$.

Proof
First suppose that $f(1) = 0$.

Then for all arithmetic functions $g$,

Since the identity element $\iota$ of $\mathcal A$ satisfies $\iota(1) = 1$, $f$ has no inverse in $\mathcal A$.

Now suppose that $f(1) \neq 0$.

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

1. $(f * f^{-1})(1) = 1$ 2. $(f * f^{-1})(n) = 0$ for all $n > 1$

We have


 * $(f * f^{-1})(1) = f(1)f^{-1}(1)$

So taking $f^{-1}(1) = f(1)^{-1}$, part 1. is satisfied.

Now suppose we have constructed $f(k)$ for $k = 1,\ldots, n-1$. We compute

So to satisfy part 2. we require:


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

To obtain this we take:


 * $\displaystyle f^{-1}(n) = -\frac1{f(1)}\sum_{\substack{d \mathop \vert n \\ d \mathop > 1}} f(d) f^{-1}\left(\frac n d \right)$

Thus we have inductively constructed the required function.