Dirichlet convolution of multiplicative functions is multiplicative

Statement

Let $f$ and $g$ be multiplicative functions. Then their Dirichlet convolution $f * g$ is a multiplicative function.

Remarks

Let $m$ and $n$ be coprime. The proof uses that each divisor of $mn$ is a unique product of a divisor of $m$ and a divisor of $n$. With $f = \mathrm{id}_k \colon n \mapsto n^k$ and $g = \mathbf{1}$, the theorem gives the multiplicativity of the divisor function, since $\sigma_k = \mathrm{id}_k * \mathbf{1}$. The Dirichlet inverse of a multiplicative function is also multiplicative.

Explanations

L1 human

Multiplicative functions stay multiplicative when you convolve them. So a new function built from known ones by divisor sums keeps the property, and you can compute it from prime powers.