Isn't this problem somewhat akin to defining mod(p,q), p,q complex numbers? The problem is that almost every definition is arbitrary: besides |mod(p,q)|<|q|, and the fact that mod(p+n*q,q)=mod(p,q) (n an integer), and perhaps mod(p+i*n*q,q)=mod(p,q), what other criteria should we use to pin down this function? We could require the image of mod(p,q) to be a parallelogram in (1,i) space, but that might be far too confining: we might want to use one of Gosper's dragons as the basic "rep-tile". BTW, I assume that there is some sort of connection between mod(p,q) and elliptic functions? I've never seen such a connection written down, but the basic 2-D periodicity would seem to scream it out. At 02:04 PM 6/8/2011, Dan Asimov wrote:
I was going to ask the same question.
It seems the canonicalest choice would be elements z of Z[i] with Re(z) > 0 and Im(z) >= 0. But there's something arbitrary about this.
But even more natural might be to define the sum of aliquot parts in something like Z[i] / z ~ iz. I would hope this object would retain some kind of useful algebraic structure, that would be related to the ring structure of Z[i] as Z+ is related to the group structure of Z.
--Dan
Gene wrote:
<< In the ring Z[i], the units (divisors of 1) are +1, -1, +i, -i. Thus if d is a divisor of 3+i, so are -d, id, -id. In taking the sum of divisors, how do you make a canonical choice among the associates of each divisor? If you take them all, the sum is zero. In the rational integers, one chooses the positive divisor.
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, including a lot of references. Victor On Wed, Jun 8, 2011 at 4:10 PM, Henry Baker <hbaker1@pipeline.com> wrote:
Isn't this problem somewhat akin to defining mod(p,q), p,q complex numbers? The problem is that almost every definition is arbitrary: besides |mod(p,q)|<|q|, and the fact that mod(p+n*q,q)=mod(p,q) (n an integer), and perhaps mod(p+i*n*q,q)=mod(p,q), what other criteria should we use to pin down this function?
We could require the image of mod(p,q) to be a parallelogram in (1,i) space, but that might be far too confining: we might want to use one of Gosper's dragons as the basic "rep-tile".
BTW, I assume that there is some sort of connection between mod(p,q) and elliptic functions? I've never seen such a connection written down, but the basic 2-D periodicity would seem to scream it out.
At 02:04 PM 6/8/2011, Dan Asimov wrote:
I was going to ask the same question.
It seems the canonicalest choice would be elements z of Z[i] with Re(z) > 0 and Im(z) >= 0. But there's something arbitrary about this.
But even more natural might be to define the sum of aliquot parts in something like Z[i] / z ~ iz. I would hope this object would retain some kind of useful algebraic structure, that would be related to the ring structure of Z[i] as Z+ is related to the group structure of Z.
--Dan
Gene wrote:
<< In the ring Z[i], the units (divisors of 1) are +1, -1, +i, -i. Thus if d is a divisor of 3+i, so are -d, id, -id. In taking the sum of divisors, how do you make a canonical choice among the associates of each divisor? If you take them all, the sum is zero. In the rational integers, one chooses the positive divisor.
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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.
This is interesting, but I'm still having trouble with "THE first quadrant". Isn't the usual thing to take as the canonical quadrant for Gaussian factors the quarter plane to the right of the origin bounded by x=y and x=-y, rather than the all-non-negative upper-right quadrant bounded by the axes? The choice impacts what the "sum of divisors" is, right?
"The first quadrant" is well-defined, except possibly concerning what part of the boundary is included. It is what we learned in analytic geometry. I'm not aware that there is a canonical choice among the four associates of a Gaussian integer. The choice suggested here does have one remaining ambiguity: 1+i or 1-i ? Perhaps a better definition for a sum of divisors function would be to use the norm of the divisor rather than the divisor itself. (The norm of an algebraic number is the product of all its conjugates. For imaginary quadratic fields, it's the absolute value squared.) Now it's independent of the choice of associate. This easily generalizes to the sum of the n-th powers of divisors. This also extends to other rings of algebraic integers. When the ring does not possess unique factorization, there seems to be a choice: use only integer divisors, or allow ideals as divisors. -- Gene
________________________________ From: Marc LeBrun <mlb@well.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Thursday, June 9, 2011 1:06 PM Subject: Re: [math-fun] Some Sum
This is interesting, but I'm still having trouble with "THE first quadrant".
Isn't the usual thing to take as the canonical quadrant for Gaussian factors the quarter plane to the right of the origin bounded by x=y and x=-y, rather than the all-non-negative upper-right quadrant bounded by the axes?
The choice impacts what the "sum of divisors" is, right?
="Eugene Salamin" <gene_salamin@yahoo.com>
"The first quadrant" is well-defined, except possibly concerning what part of the boundary is included. It is what we learned in analytic geometry.
Of course I have no quibble with the usual definition of "first quadrant". My confusion with THIS thread has come from mixing that with the definition of "positive", which I've understood, for Gaussian integers, means a region roughly symmetrical about the positive x-axis. Please see Figure 8.7(a) in Conway & Guy's "The Book of Numbers" (it's on p218 in my 1996 copy)
Perhaps a better definition for a sum of divisors function would be to use the norm of the divisor rather than the divisor itself.
Interesting idea.
="Marc LeBrun" <mlb@well.com> This is interesting, but I'm still having trouble with "THE first quadrant".
Risking belaboring the obvious in the pursuit of clarity:
= Kieran Smallbone 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.
But x=1-i has no "positive" Gaussian integer divisors at all. Are we then to take the sum of the divisors of x to be zero? Similarly, is the sum of the divisors of 2 then to be 1+(1+i)+2 = 4+i? (Heh. Then there are no even perfect naturals. Are there any at all?)
On Thu, Jun 9, 2011 at 3:01 PM, Marc LeBrun <mlb@well.com> wrote:
But x=1-i has no "positive" Gaussian integer divisors at all.
That's because it's (-i)(1+i). -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
="Mike Stay" <metaweta@gmail.com>
-Marc LeBrun <mlb@well.com> wrote: But x=1-i has no "positive" Gaussian integer divisors at all. That's because it's (-i)(1+i).
Doh, you're right! I stand corrected, so the sum of first-quadrant divisors of 1-i is 2+i. Sigh, I probably should've stuck with 5. So, are not 1, 2+i and 5 the "first quadrant" Gaussian integer divisors of 5, with sum 8+i? In contrast with the "positive" Gaussian integer divisors of 5 (as I recalled from Conway&Guy), which also include 2-i, and thus sum to 10? I can rotate 2+i by a unit vector to get 2-i, but I'm durned if I can see how to factor that vector into first-quadrant Gaussian integers. But then, I've been wrong before...
2+i is not a multiple of 2-i. They are independent prime divisors of 5. 1+i and 1-i _are_ associates; each is a multiple of the other. Rich ---- Quoting Marc LeBrun <mlb@well.com>:
="Mike Stay" <metaweta@gmail.com>
-Marc LeBrun <mlb@well.com> wrote: But x=1-i has no "positive" Gaussian integer divisors at all. That's because it's (-i)(1+i).
Doh, you're right!
I stand corrected, so the sum of first-quadrant divisors of 1-i is 2+i.
Sigh, I probably should've stuck with 5.
So, are not 1, 2+i and 5 the "first quadrant" Gaussian integer divisors of 5, with sum 8+i?
In contrast with the "positive" Gaussian integer divisors of 5 (as I recalled from Conway&Guy), which also include 2-i, and thus sum to 10?
I can rotate 2+i by a unit vector to get 2-i, but I'm durned if I can see how to factor that vector into first-quadrant Gaussian integers.
But then, I've been wrong before...
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="rcs@xmission.com" <rcs@xmission.com>
2+i is not a multiple of 2-i. They are independent prime divisors of 5. 1+i and 1-i _are_ associates; each is a multiple of the other.
OK, thanks Rich, good, I didn't goof that. Originally I wrote my message in terms of 2-i and then at the last minute screwed it up by "simplifying" to 1-i. Sorry. The point, lest it get lost in all my stupid arithmetic mistakes, was that where Havermann says that Kieran replied that
Spira's idea was that any integer can be written as a product of a
unit (±1) and positive primes
This is true if you define "positive" as Conway & Guy say that Gauss did, because then everything is indeed an associate of a unique factorization. BUT then the part with the example that leads up to the statement
where the primes 1+i and 1+2i are "positive" because both they lie
in the top right quadrant.
incorrectly (doubtless inadvertently) veers off and mis-defines "positive" as "top-right-quadrant". Specifically 1+2i *isn't* "positive" in the good sense of Gauss. Conversely you can't fully factorize 5 into "top-right-quadrant" primes exclusively. As I said, I think the overall idea is a FINE approach to defining sum of divisors, you just have to be careful to interpret Spira's "positive" as Gauss did, and not as Kieran did in the example. If I'm still unintelligible or wrong please forgive me; I will desist now.
Marc LeBrun:
you can't fully factorize 5 into "top-right-quadrant" primes exclusively
Mathematica factors Gaussian integers and I just plotted all factors of x + y i for integers x and y from -100 to +100. With the exception of -1 and -i, all other factors are in the top-right quadrant. More specifically: In[1]:= FactorInteger[5,GaussianIntegers->True] Out[1]= {{-I,1},{1+2 I,1},{2+I,1}} I'm going to guess that the "positive" Gaussian quadrant illustrated by Conway and Guy can be rotated into the *top-right* quadrant without loss of the unique-factorization principle.
="Hans Havermann" <pxp@rogers.com> I'm going to guess that the "positive" Gaussian quadrant illustrated by Conway and Guy can be rotated into the *top-right* quadrant without loss of the unique-factorization principle.
OK, you've convinced me; it's not broken, just different. So with the top-right quadrant the sum-of-divisors of 5 is 9+3i, while with Gauss's quadrant it's 10.
More generally, let S be any set of Gaussian primes such that the set of all Gaussian primes is the disjoint union of S, -S, iS, -iS. Then we may call S the "standard primes", and any Gaussian integer can be uniquely factored into standard primes and a unit. There doesn't seem to be any natural choice of S, and "top right quadrant" or "Conway-Guy" are as good as any other convention. The sum of divisors function, which evaluates to a single number, because it depends on the choice of S, seems to me to be an uninteresting object. If instead the function evaluates to the finite set of sums of all possible choices of associates, then it would not involve a choice of S. Perhaps, in this form, a sum of divisors set function would have interesting mathematical properties. -- Gene
________________________________ From: Hans Havermann <pxp@rogers.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Thursday, June 9, 2011 9:30 PM Subject: Re: [math-fun] Some Sum
Marc LeBrun:
you can't fully factorize 5 into "top-right-quadrant" primes exclusively
Mathematica factors Gaussian integers and I just plotted all factors of x + y i for integers x and y from -100 to +100. With the exception of -1 and -i, all other factors are in the top-right quadrant. More specifically:
In[1]:= FactorInteger[5,GaussianIntegers->True] Out[1]= {{-I,1},{1+2 I,1},{2+I,1}}
I'm going to guess that the "positive" Gaussian quadrant illustrated by Conway and Guy can be rotated into the *top-right* quadrant without loss of the unique-factorization principle.
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It's a bad notation to define a number to be "positive" if it lies within some sector, for then products of positive numbers need not be positive. The only situation I can think of where it's sensible to adopt a convention for the choice of associate is in a computer algebra factorization function. Then a number will always factor the same way into primes and a unit, assuming of course that we are working in a UFD (unique factorization domain). -- Gene
________________________________ From: Marc LeBrun <mlb@well.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Thursday, June 9, 2011 3:01 PM Subject: Re: [math-fun] Some Sum
="Marc LeBrun" <mlb@well.com> This is interesting, but I'm still having trouble with "THE first quadrant".
Risking belaboring the obvious in the pursuit of clarity:
= Kieran Smallbone 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.
But x=1-i has no "positive" Gaussian integer divisors at all.
Are we then to take the sum of the divisors of x to be zero?
Similarly, is the sum of the divisors of 2 then to be 1+(1+i)+2 = 4+i?
(Heh. Then there are no even perfect naturals. Are there any at all?)
participants (7)
-
Eugene Salamin -
Hans Havermann -
Henry Baker -
Marc LeBrun -
Mike Stay -
rcs＠xmission.com -
Victor Miller