Let be $$ \phi: \mathbb{N} \rightarrow \mathbb{C} $$ an arithmetic function.
How can I show that $ \phi $ is only invertible concerning convolution if $ \phi (1) \neq 0 $ holds?
So ..:
$\phi $ invertible concerning convolution $ \leftrightarrow $ $ \phi(1) \neq 0 $
for the $ \leftarrow $ part I thought about constructing an inverse $ \mu = \phi^1 $. But then how do I choose $ \mu (1)$ ? And how do I choose $\mu (n) $ ?
very thankful for any help ! :-)
This is done by induction.
First, we recall the definition of the convolution of f with g:
$$ (f*g)(n) = \sum_{d \mid n} f(d) g(n/d)$$
Now suppose $\phi$ is invertible. This means that there exist $\psi$ such that $(\phi * \psi)(n) = 1$ if $n=1$, and zero otherwise. Then $1= (\phi*\psi)(1) = \phi(1) \psi(1) $
This implies $\phi(1)\neq 0$.
On the other hand, if the latter condition holds, you can find such a $\psi$ by induction. In fact by the last equation we have $\psi(1) =1/\phi(1)$. For bigger n we have
$$ 0 = \phi * \psi (n)= \sum_{d \mid n}\phi(d) \psi(n/d) $$ Which implies, by isolating the term $\psi(n)$: $$ \psi(n) \phi(1) = - \sum_{d \mid n, d \neq 1} \phi(d)\psi(n/d)$$ But we know all the terms on the right hand side by inductive hypothesis, and $\phi(1)$ ia different by zero, so we can isolate $\psi(n)$!