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It seems natural to me to generalize the notion of perfect number to rings of integers more general that ${\Bbb Z}$. I'll want to think of number fields concretely, as subfields of ${\Bbb C}$. For a mutatis mutandis definition, a given algebraic integer must have a distinguished finite set of divisors. That makes two tricky parts: figuring out what to say if the group of units is richer than $\{-1,1\}$ making due with the order if the field is not a subfield of ${\Bbb R}$.

Do definitions occur in the literature? Do nontrivial examples exist?

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Check out the references in emis.ams.org/journals/JIS/VOL15/Palimar/palimar5.pdf –  Franz Lemmermeyer Jul 21 '12 at 16:46

2 Answers 2

Definition 2.0.5 in the following dissertation gives one possible definition of perfect number:

http://eprints.maths.ox.ac.uk/741/1/smallbone2.pdf

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The interesting question seems to me not what a perfect number is but what an arithmetic function is (generalizing the divisor function, the sum-of-divisors, totient, etc.) over a number field.

One answer is as follows. My opinion is that arithmetic functions should not be thought of as having domain $\mathbb{N}$ but as having domain the poset of nonzero ideals in $\mathbb{Z}$ (which can be identified with $\mathbb{N}$ under division, but my point is that the additive structure of $\mathbb{N}$ does not enter into this). Dirichlet convolution $$f \ast g = \sum_{d | n} f(d) g (n/d)$$

can then be seen as a special case of the multiplication in a subalgebra of the incidence algebra of the poset of ideals (the subalgebra which assigns the same number to an interval $[a, b]$ as it does to the interval $[1, \frac{b}{a}]$). The assignment of a Dirichlet series $$\sum \frac{f(n)}{n^s}$$

to an arithmetic function merely instantiates this algebra as an actual algebra of functions. Now, the poset of nonzero ideals in $\mathbb{Z}$ is a product of posets, one for each prime, and the arithmetic functions which are multiplicative (in the sense that $f(mn) = f(m) f(n)$ if $(m, n) = 1$) are just the ones that are built from corresponding functions on each factor. In the Dirichlet series picture this is reflected in the existence of an Euler product $$\sum \frac{f(n)}{n^s} = \prod_p \left( \sum_{k \ge 0} \frac{f(p^k)}{p^{ks}} \right).$$

The generalization to number fields is as follows. In any Dedekind domain $D$, we may consider the incidence algebra of the poset of nonzero ideals in $D$. This algebra has a subalgebra of functions which assign the same number to an interval $[I, J]$ as to the interval $[1, I^{-1} J]$, and this gives us the appropriate generalization of Dirichlet convolution $$(f \ast g)(K) = \sum_{IJ = K} f(I) g(J)$$

where $I, J, K$ are ideals. When $D$ is the ring of integers in a number field, we can define norms $N(I)$ of ideals and also assign Dirichlet series. This assignment $$\sum_I \frac{f(I)}{N(I)^s}$$

is now not faithful as there can be different ideals with the same norm, but it is still true that the multiplicative functions (the ones such that $f(IJ) = f(I) f(J)$ if $I, J$ generate the unit ideal) admit an Euler product $$\sum_I \frac{f(I)}{N(I)^s} = \prod_P \left( \sum_{k \ge 0} \frac{f(P^k)}{N(P)^{ks}} \right)$$

where the product is now over all prime ideals in $D$.

This is all a long-winded way of setting the stage for what I actually wanted to say:

For number fields, an arithmetic function now manifestly has different domain and codomain: it takes input an ideal and returns a number.

In particular, to me the correct generalization of the sum-of-divisors function is the function $\sigma(I)$ which sums the norms of ideals dividing a given ideal $I$, and it no longer makes sense to ask whether this is equal to $I$ or $2I$. Of course there are other interesting questions to ask regarding comparing different functions, e.g. one might ask whether $\sigma(I) = c N(I)$ for some constant $c$ (this reduces to the perfect number question over $\mathbb{Q}$ taking $c = 2$, but I think from this perspective it is now clear that there is nothing special about $c = 2$).

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Perfect numbers aren't about the number of divisors, but the sum of divisors. You want to sum the norms of ideals dividing the ideal $I$, probably. –  Will Sawin Jul 22 '12 at 18:59
    
@Will: right, thanks for the correction. –  Qiaochu Yuan Jul 22 '12 at 19:37
    
Nothing special about 2?! :-) –  Mariano Suárez-Alvarez Jul 23 '12 at 3:30

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