Victor Miller:
This paper: http://eprints.maths.ox.ac.uk/741/1/smallbone2.pdf has a lot of interesting things to say about the problem of extending the sum of divisors function to number fields...
Thank you for this. I was able to track down Kieran Smallbone (the author of that paper), who appears to be doing mathematical modelling in biological systems (in Manchester UK) these days. I emailed him:
As a recreational (very amateur) mathematician, I am trying to square the concept of the sum of aliquot (proper) divisors of a number (the sum of all divisors of n, with the exception of n itself) with Spira's 1961 "complex sum of divisors" extension to Gaussian integers. Spira's sum-of-divisors function would include (I imagine) the number itself. So, to get only the sum of the "proper" divisors out of Spira's function I have seen some folk subtracting n from it (for example, here: < http://oeis.org/A102924 >), which seems reasonable at first glance.
But because Spira's definition is a product (involving the number's factorization, if I understand it correctly), I'm unsure if simply subtracting n is a valid approach to reducing Spira's sum-of- divisors to a sum-of-proper-divisors. Any insight you might have on this would be greatly appreciated.
Kieran replied:
Overall: no, I don't think σ(z) - z is the way to go. It's easiest to think of your question first in terms of the normal integers. Spira's idea was that any integer can be written as a product of a unit (±1) and positive primes
e.g. -28 = (-1).(2^2).(7^1)
Then to calculate the sum-of-divisors, forget the unit and add the sums of powers of positive primes:
σ(-28) = (1+2+2^2).(1+7) = 1 + 2 + 4 + 7 + 14 + 28 = 56.
Notice that -28 doesn't appear in the sum on the right. I think the natural choice is to take σ(-28) - 28 = 28, instead of σ(-28) - (-28) = 84. In other words, σ(n) - pa(n) where (I've invented this notation) pa(n) is the positive associate of n.
In the normal integers, pa(n) is either ±n. Now going to your Gaussian integers, there are units ±1 and ±i, so the positive associate pa(z) is either ±z or ±i.z.
For example
2+4i = (-i).((1+i)^2).(1+2i)
where the primes 1+i and 1+2i are "positive" because both they lie in the top right quadrant.
σ(2+4i) = (1 + (1+i) + (1+i)^2).(1 + 1+2i) = -2+10i
Its positive associate is the product of the positive primes without the unit
pa(2+4i) = ((1+i)^2).(1+2i) = -4+2i
and so
σ(2+4i) - pa(2+4i) = 2+8i.