Find the Dirichlet inverse of the identity function I thought I got this one but now I'm having doubts. I have that if $f$ is a function such that $(f\ast\text{id})(n)=(\text{id}\ast f)(n)=\iota$ then $f$ is multiplicative since it is the inverse of a multiplicative function, and so it suffices to examine its values at prime factors. So letting $n=\prod_{i=1}^{k}p_{i}^{\alpha_{i}}$
  we require $$\left(f\ast\text{id}\right)(n) = \sum_{ab=n}f\left(a\right)b
 = \sum_{i=1}^{k}f\left(p_{i}^{\alpha_{i}}\right)\prod_{j\ne i}p_{j}^{\alpha_{j}}
 = 0
 $$
And since $\prod_{j\ne i}p_{j}^{\alpha_{j}}\ne0$
  for all $i$, we need $f\left(p_{i}^{\alpha_{i}}\right)=0$ for all primes $p_i$. Then for $n=1$ we require $\left(f\ast\text{id}\right)\left(1\right) = f(1)1
 = 1
 $ so $f(1)=1$. This implies that $f(n)=\cases{1 \ n=1 \\ 0 \ n\ne 1}$, or, in other words, $f=\iota$.
But I am doubting this now, since $\iota$ is supposed to be the identity with respect to the Dirichlet convolution, so that $\iota\ast \text{id}$ should be $\text{id}$. But I can't see where I've made a mistake. Can anyone help?
 A: It is not true that the only divisors of a composite are prime powers - generally divisors of a composite number will themselves be products of more than one prime - so your expression after the $\sum_{ab=n}$ sum is not correct. Just because it is sufficient to look at prime powers does not mean one can suppose no other kinds of factors of a composite number exist!
You want to "localize": just look at the case where $n=$ prime power. Observe
$$\delta_{r}=\iota(p^r)=(f*{\rm id})(p^r)=\sum_{l=0}^rf(p^l)p^{r-l}$$
for all primes $p$ and integers $r\ge0$. In particular,
$$\begin{array}{lll} f(1)=1 & \implies & f(1)=1 \\ 
f(p)+pf(1)=0 & \implies & f(p)=-p \\
f(p^l)+\cdots+p^{l-1}f(p)+p^lf(1)=0 & \implies & ?~~?~~?~~?~~? \end{array}$$
the third line is the same for $l\ge2$. Prove it using induction. Can you now write down the general form for $f$ at all integer arguments? Hint: consider the $\mu(\cdot)$ function, squarefreeness and signs.
A: Let us use Dirichlet generating functions (g.f.) for this problem. Define
$\, \zeta_a(s) := \sum_{n} a_n/n^s \,$ for a sequence $\,a_n.\,$
Then Dirichlet convolution corresponds to multiplication. That is, $\,\zeta_{a*b}(x) = \zeta_a(s)\zeta_b(s).\,$ Also, the Dirichlet convolution inverse is the reciprocal of the g.f.. We have $\, \text{id}_n := n \,$ with g.f. $\, \zeta_{\text{id}}(s) = \zeta(s-1) \,$ and let the Dirichlet inverse of $\,\text{id}\,$ be denoted by $\,f.\,$ Then the g.f. of$\, f \,$ is $\, \zeta_f(s) = 1/\zeta(s-1).\,$ Now $\, \zeta_\mu(s) = 1/\zeta(s) \,$ which implies that 
$\, f_n = n\mu_n.\,$ Note that is is OEIS sequence A055615 "Dirichlet inverse of n".
This clearly generalizes. If $\, \zeta_b(s) := 
\zeta_a(s-k)\,$ so that $\, b_n = n^k a_n, \,$
then the Dirichlet convolution inverses have the same relationship.
A: I believe this is known and proved on Wikipedia now. It's also given more generally as an identity in Apostol's book. If you let $\operatorname{Id}_k(n) := n^k$, then the Dirichlet inverse of this identity function is given by the pointwise product $\operatorname{Id}_k^{-1} = \mu \cdot \operatorname{Id}_k$. 
