Is Dirichlet convolution of $2$ multiplicative functions multiplicative?

Question

The task is to show that $\lambda*|\mu|=\epsilon$, can you please look at my proof and show, why the Dirichlet convolution of $2$ multiplicative functions is multiplicative ?

Let $n=p_1^{a_1}\cdot p_2^{a_2}\dots\cdot p_r^{a_r}$, and $\lambda(n)=(-1)^{a_1+a_2+\dots a_r}$ and

$\mu(n)=\begin{cases}1&\text{if}\ n=1\\(-1)^{t}&\text{if}n=p_1\cdot p_2\dots p_t, \ \text{where $p_i$ are distinct primes}\\0&\text{if $\exists p$ such that $p^2\mid n$}\end{cases}$

I have to show that the Dirichlet convolution $\lambda*|\mu|=\epsilon$ where $\epsilon$ is the Dirichlet convolution identity, i.e. $\epsilon(1)=\begin{cases}1&\text{if $n=1$}\\0&\text{else}\end{cases}$

Now if I define $F(n):=\left(\lambda*|\mu|\right)(n)$, and assume that $F$ is multiplicative then I think I can prove it;

If $\underline{n=1}$:

$\displaystyle F(1)=\left(\lambda*|\mu|\right)(1)=\sum\limits_{d\mid 1\atop n\in\mathbb N}\lambda(1)|\mu(1)|=(-1)^0|1|=1$

If $\underline{n=p^k}$:

$\displaystyle F(p^k)=\left(\lambda*|\mu|\right)(p^k)=\sum\limits_{d\mid p^k\atop n\in\mathbb N}\lambda(d)|\mu(\frac{p^k}{d})|$

$=\lambda(1)\underbrace{|\mu(p^k)|}_{=0}+\lambda(p)\underbrace{|\mu(p^{k-1})|}_{=0}+\dots+\underbrace{\lambda(p^{k-1})}_{=(-1)^{k-1}} \underbrace{|\mu(p)|}_{|-1|=1}+\underbrace{\lambda(p^k)}_{(-1)^k}\underbrace{|\mu(1)|}_{=1}$

$=(-1)^{k-1}+(-1)^k=0$

So If $F$ is multiplicative then;

$F(n)=F(p_1^{a_1})F(p_2^{a_2})\dots F(p_r^{a_r})=0\cdot0\dots\cdot0=0$

do you agree ?

Answer 1

F is multiplicative, as the Dirichlet convolution of two multiplicative arithmetic functions, take $\mu$ and $\nu$ , multiplicative arithmetic functions, then their convolution is the function defined by $\sum_{d|mn} \mu(d) \nu(mn/d)$, but taking m and n for which $(m,n) = 1$, we could devide the set of all deviders of $mn$ two into two sets, having just intersection 1, the set of dividors of m and the set of all dividors of n, so the sum becomes $\sum_{d|m}\mu(d)\nu(m/d)\sum_{d_1|n}\mu(d_1)\nu(n/d_1)$, as everything is finite we have that we can use termwise multiplication by no restrictions. Proof is straightforward.