Sigma function is multiplicative

By Anonymous on Tuesday, May 1, 2001 - 11:22 pm :

How does one show that the sigma function (the sum of all the positive divisors of an integer) is multiplicative?
i.e. $σ (p \times q) = σ (p) \times σ (q)$ where $p$ and $q$ are coprime.

By Kerwin Hui (Kwkh2) on Wednesday, May 2, 2001 - 03:43 pm :

It suffices to prove that

$σ (p^{a} n) = σ (p^{a}) σ (n)$

where $p$ is a prime and $n$ is coprime to $p$ .

But if $b | n$ , then $p^{i} b | p^{a} n$ , where $0 \leq i \leq a$ , and if we have a list of positive divisors of $n$ as $b_{1}$ , $b_{2}$ , ..., $b_{m}$

then ${p^{i} b_{j} : 0 \leq i \leq a, 1 \leq j \leq m}$ are distinct and are the only divisors of $p^{a} n$ . Hence sigma function is multiplicative.

Kerwin