Sum of divisors

Multiplicativity. If $\gcd(m,n)=1$, then every divisor of $mn$ is uniquely $d_1d_2$ with $d_1\mid m$, $d_2\mid n$. So $\sigma(mn)=\sigma(m)\sigma(n)$.

Prime power. $\sigma(p^\alpha)=1+p+\cdots+p^\alpha=\frac{p^{\alpha+1}-1}{p-1}$.

Conclusion. $\sigma(n)=\prod_i\frac{p_i^{\alpha_i+1}-1}{p_i-1}$.

Example. $\sigma(12)=\sigma(2^2)\sigma(3)=7\cdot4=28$, consistent with $1+2+3+4+6+12=28$.