[ksbrown swiped my usual pattern of separating distinct posts/mail with "===="
so to be clear I am using "&&&&" -- djr ]
From: ksbrown@ksbrown.seanet.com (Kevin Brown)
Newsgroups: sci.math
Subject: Simple Complex Quadratic Fields
Date: Thu, 17 Aug 1995 18:10:47 GMT
For any odd prime p compute the sequence of squares k^2 (mod p) with
k=0,1,2,3,...,p. For example, with p=19 the squares are
0,1,4,9,16,6,17,11,7,5,5,7,11,17,6,16,9,4,1,0
Letting C and P denote "composite" and "prime" respectively, this
sequence of squares looks like this:
0,1,C,C,C,C,P,P,P,P,P,P,P,P,C,C,C,C,1,0
Of course, the sequence of squares is symmetrical because
k^2 = (p-k)^2 (mod p), so we really only need to consider the
squares with k=0,1,2,..,(p-1)/2. Over this range we notice
that the squares (mod 19) begin with a sequence of composites
(after 0 and 1) and then the descending sequence of primes
17,11,7,5.
This pattern of a pure sequence of composites followed by a pure
sequence of primes is really quite unusual. Normally the prime and
composite residues are scatterred about randomly in the sequence of
squares (mod p). The only primes (less than 70000) for which the
sequence of square residues has this special structure are listed
below, along with the corresponding central sequences of prime square
residues. (I'll include the moduli 1 and 2, since their sequences
trivially possess the property of no composites after the first
prime.)
p central sequence
--- -----------------------------------------
1: -
2: 1
3: 1
7: 2
11: 5 3
19: 17 11 7 5
43: 41 31 23 17 13 11
67: 59 47 37 29 23 19 17
163: 151 131 113 97 83 71 61 53 47 43 41
The numbers 1,2,3,7,11,19,43,67,163 are precisely the values d such
that the complex quadratic field K[sqrt(-d)] is 'simple', i.e., the
fields in which the fundamental theorem of arithmetic is true (unique
factorization).
Does anyone know of a proof that the field K[sqrt(-p)] (where p is an
odd prime) is simple if and only if the sequence of square residues
(mod p) has the structure noted above? Also, is there an analagous
structure that distinguishes the primes p such that the real field
K[sqrt(+p)] has unique factorization? Can the class number ( > 1) of
any given quadratic field be inferred from the corresponding sequence
of square residues? What is the significance of the primes in the
central sequence?
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Date: Wed, 23 Aug 1995 22:08:09 -0700
To: rusin@math.niu.edu (Dave Rusin)
From: ksbrown@ksbrown.seanet.com (Kevin Brown)
Subject: Re: Simple Complex Quadratic Fields
At 10:47 PM 8/23/95 CDT, Dave Rusin wrote:
>I have never seen this before, although surely your questions are on
>target. Have you had any responses?
I received two replies, one from Dave Carlton and one from Kurt Foster.
In case you're interested, here's a copy of their comments:
===============================================================
From: david carlton
Subject: Simple Complex Quadratic Fields
I confess, I thought the thing that you mentioned was some stupid
random occurrence, and was really floored when you connected it with
unique factorization in those rings of integers of number fields! But
when I recovered from my shock, I remembered something similar, and
after digging it up it did turn out to be quite relevant, though it
doesn't answer all of your questions, and in particular doesn't say
anything about the real quadratic fields:
Assume p > 3, p = 3 (mod 4).
1) The class number of Q(sqrt(-p)) is 1. (I.e. it's a UFD.)
2) (l/p) = -1 for all primes l such that l < p/4
2') Ditto for l < sqrt(p/3)
3) (For p > 7) p = 3 (mod 8) and R_p - N_p = 3, where R_p (resp. N_p)
is the number of quadratic residues (resp. non-residues) in the
interval [1, (p-1)/2].
4) x^2 + x + (p+1/4) is prime for 0 <= x <= (p-7)/4.
4') Ditto for 0 <= x <= 1/2(sqrt(p/3) - 1)
That's from appendix A to Serre's _Lectures on the Mordell-Weil
Theorem_, p. 190. 2 and 2' are clearly related to what you're working
on, since they explain those composite values at the beginning. And
the prime residues that you're getting are an initial sequence of the
numbers that criterion 4 claims are prime. And, indeed, (p-1-2x/2)^2
is congruent to x^2 + x + (p+1)/4 mod p, so that explains why those
are prime.
======================================================================
From: kfoster@rmii.com (Kurt Foster)
Subject: Re: Simple Complex Quadratic Fields
Offhand I'd say it has something to do with results on "reduction"
of quadratic forms. As you undoubtedly know, Q(sqrt(-163)) having only
one ideal class means there is only one (pos. def.) reduced quadratic
form of discriminant -163. This, in turn, is intimately related to the
fact that n^2 - n + 41 is prime for n = 1 to 40, a connection being
provided by standard inequalities which determine whether a form is
reduced. I think "Number Theory" by Borevich and Shafarevich has the
relevant results on quadratic forms.
Of course, as p increases without bound, the number of quadratic
residue classes which are represented by a prime q < p, becomes an
infinitesimal proportion of all quadratic residue classes, since the
number of primes q available is only pi(p), which is asymptotically
Li(p), or, more crudely, p/ln(p), whereas the number of quadratic
residue classes is (p-1)/2. The Polya-Vinogradov inequality might also
have something to say about the situation. See, for example,
"Multiplicative Number Theory" (2nd ed) by Harold Davenport.
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