Sigma Function is Multiplicative

Theorem

The sigma function:

$\displaystyle \sigma: \Z_{>0} \to \Z_{>0}: \sigma \left({n}\right) = \sum_{d \backslash n} d$

is multiplicative.

Proof

Let $I_{\Z_{>0}}: \Z_{>0} \to \Z_{>0}$ be the identity function:

$\forall n \in \Z_{>0}: I_{\Z_{>0}} \left({n}\right) = n$.

Thus we have:

$\displaystyle \sigma \left({n}\right) = \sum_{d \backslash n} d = \sum_{d \backslash n} I_{\Z_{>0}} \left({d}\right)$.

But from Identity Function is Completely Multiplicative, $I_{\Z_{>0}}$ is multiplicative.

The result follows from Sum Over Divisors of Multiplicative Function.

$\blacksquare$

Sources

• George F. Simmons: Calculus Gems (1992), Chapter $\text {B}.2$