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Linking points on algebraic curves with class numbers
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Dave Rusin
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Joined: 25 Mar 2005
Posts: 487

PostPosted: Wed Jun 28, 2006 8:30 pm    Post subject: Linking points on algebraic curves with class numbers Reply with quote

A colleague asked me a question (about extensions of algebraic number fields)
which led us to do some experiments involving a parameter a . Only for
a few values of a does the experiment "succeed". Here is the list so far:
a = 1, 19, 37/17, 53/17, 89/19, 199/109, 323/197, 433/323, 629/251,
683/397, 703/487, 757/127, 901/719, 971/701, 1007/127, ...
I haven't looked as far for negative a but up to height 200 the
"successes" occur only for a = -1, -73/17, and -181/71.
My colleague is really only interested in integer solutions, I think;
I looked for them but found no more in a search up to a = 300 000.

Now the game is to decide where these numbers "come from".

I could not fail to notice that almost all these numerators and
denominators were prime.

Less obvious (but checked because of the appearance of "163") is the
fact that for every prime in this list, the class number of the
field Q(sqrt(+p)) is 1. It's also true for all the prime divisors
of the composites on this list. That's an awful lot of agreement.
That is the fact to be explained. (Note: for each composite
numerator or denominator, h(Q(sqrt(+n))) is greater than 1.)


Here is the "experiment": we have a polynomial

P := 1 + (3+9*a^3)*X^3 + (27*a^6+6+18*a^3)*X^6 + (7-27*a^6)*X^9
+ (27*a^6+6+18*a^3)*X^12 + (3+9*a^3)*X^15 + X^18;

which is irreducible in Q(a)[X]. (It's the miniminal polynomial of
some unit in some number field.) This P remains irreducible when the
parameter a is specialized to a rational value, except for some
exceptional values of a . Those are the "successes" of the experiment.


Well, the origins of the polynomial P happen to force any factorization
to be as a product of three sextics f_i, so it occurs to me that we could
take the coefficients of the sextics as 18 variables; then comparing
the coefficients of X in the equation P = f1 f2 f3 gives 18
equations in these variables together with a . So the factorizations
correspond to points on an algebraic variety. Since P is
generically irreducible, I guess this variety is a curve. The
experiments I ran amount to asking whether there is a rational point
in Q^19 lying over each tested value of a, that is, I am just
computing the projection of this curve to one of the coordinate axes.

These experiments suggest to me that the curve has an infinite number
of rational points. By Faltings' theorem, the curve must have genus
0 or 1. I guess since the values of a are not growing very rapidly
in height, I'm inclined to think the curve is of genus 0, i.e. it's
a rational curve. In particular it would be parameterizable, i.e. we
should be able to get these special values of a as values of some
rational function(s) a = a(t). (I'm just guessing that the curve is
irreducible; if not, my remarks would have to apply separately to
each component.)


OK, now compare the two ways of looking at the results of the experiment.
On the one hand, we think we're looking at the values of some rational
function, (or perhaps at the coordinates of rational points on an
elliptic curve). On the other hand, we have data that suggest all the
prime divisors of the numerators and denominators will correspond to
fields with class number 1. I can't recall such a theorem linking these
two themes. In fact, I don't even recall hearing that there are known
to be infinitely many primes for which h(sqrt(+p)) = 1 , which I
suspect could be a consequence of this observation.


So that's my question: is there any theorem known which connects
coordinates of points on an algebraic curve
with
class numbers of number fields
in this kind of fashion?


dave


PS -- it would be nice to have a curve in some low-dimensional space
which is birationally equivalent to the curve I am describing. One can
easily eliminate 7 or 8 of the variables but that gives a very messy
set of equations describing a variety which still involves nearly
a dozen variables. It's not very likely that I could compute the
genus that way, let alone a parameterization. I'm open to suggestion.
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OwlHoot
science forum beginner


Joined: 25 Jun 2006
Posts: 6

PostPosted: Wed Jun 28, 2006 10:35 pm    Post subject: Re: Linking points on algebraic curves with class numbers Reply with quote

Dave Rusin wrote:
Quote:

[..]

Here is the "experiment": we have a polynomial

P := 1 + (3+9*a^3)*X^3 + (27*a^6+6+18*a^3)*X^6 + (7-27*a^6)*X^9
+ (27*a^6+6+18*a^3)*X^12 + (3+9*a^3)*X^15 + X^18;

which is irreducible in Q(a)[X]. (It's the miniminal polynomial of
some unit in some number field.)

As this is a reciprocal polynomial, P / X^9 can be expressed as
a cubic in X^3 + 1/X^3 with coefficients which are polynomials,
with rational coefficients, in 'a'.

Maybe the discriminant of this cubic factors, or has some
property that makes it more likely (than one might expect)
to be a square.

If so, that would narrow down the possibilities. But I don't have
a computer algebra package to check. (Tried buying the student
edition of Mathematica on ebay a year or two back, but couldn't
get the wretched thing to work, with any of the bootleg codes
I could find, and I'm eternally damned if I'll spend two grand on
the "commercial" version!)


Cheers

John R Ramsden
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Gene Ward Smith
science forum Guru


Joined: 08 Jul 2005
Posts: 409

PostPosted: Wed Jun 28, 2006 11:23 pm    Post subject: Re: Linking points on algebraic curves with class numbers Reply with quote

OwlHoot wrote:
Quote:
Dave Rusin wrote:

[..]

Here is the "experiment": we have a polynomial

P := 1 + (3+9*a^3)*X^3 + (27*a^6+6+18*a^3)*X^6 + (7-27*a^6)*X^9
+ (27*a^6+6+18*a^3)*X^12 + (3+9*a^3)*X^15 + X^18;

which is irreducible in Q(a)[X]. (It's the miniminal polynomial of
some unit in some number field.)

As this is a reciprocal polynomial, P / X^9 can be expressed as
a cubic in X^3 + 1/X^3 with coefficients which are polynomials,
with rational coefficients, in 'a'.

If we set y = X^3 + 1/X^3, then y satisfies

y^3 + 3(3a^3+1)y^2 + 3(3a^3+1)y - (27a^6 + 18 a^3 -1) = 0

The discriminant of this is

-3^9 a^6 (a^4+1)^4 (a^2-a+1)^2

Quote:
Maybe the discriminant of this cubic factors, or has some
property that makes it more likely (than one might expect)
to be a square.

Evidently, the discriminant of the cubic for rational a not equal to
zero is never a square, which is also true of the original polynomial.
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Gene Ward Smith
science forum Guru


Joined: 08 Jul 2005
Posts: 409

PostPosted: Wed Jun 28, 2006 11:30 pm    Post subject: Re: Linking points on algebraic curves with class numbers Reply with quote

Gene Ward Smith wrote:

Quote:
y^3 + 3(3a^3+1)y^2 + 3(3a^3+1)y - (27a^6 + 18 a^3 -1) = 0

y^3 + 3(3a^3+1)y^2 + 3(3a^3+1)^2y - (27a^6 + 18 a^3 -1) = 0

Quadradic subfield Q(sqrt(-3)).
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Dave Rusin
science forum Guru


Joined: 25 Mar 2005
Posts: 487

PostPosted: Thu Jun 29, 2006 4:44 am    Post subject: Re: Linking points on algebraic curves with class numbers Reply with quote

In article <e7uou2$rqd$1@news.math.niu.edu>, I wrote:

Quote:
Here is the list so far:
a = 1, 19, 37/17, 53/17, 89/19, 199/109, 323/197, 433/323, 629/251,
683/397, 703/487, 757/127, 901/719, 971/701, 1007/127, ...

Now the game is to decide where these numbers "come from".

Here's what I figured out.

1. All the numbers on this (finite) list have the form
a = (u^3+18*u*v^2+9*v^3+9*u^2*v)/(-u^3+9*u*v^2+9*v^3)
for some coprime pair (u,v) of integers.

2. Any number of this form will be on the list (i.e. for all these a,
the polynomial of the previous post factors into three sextics).

3. For any specific coprime u,v the gcd of the numerator and
denominator divides 3^6. In particular

3a. The only integers on the list are -1, +1, and +19 .
3b. The primes dividing these numerators and denominators are precisely
the primes >3 for which u^3 - 9*u - 9 has a linear factor mod p.
(These are 17, 19, 37, 53, 71, 73, 89, 109, 127, 163, 179, ...)

It is still that these numerators and denominators seem to be prime,
or "nearly" so, fairly often; I'm not sure why. But it's not true
that all these primes determine real quadratic fields with class number 1.
I guess I was just distracted by the "Law of Small Numbers".

Quote:
These experiments suggest to me that the curve has an infinite number
of rational points. By Faltings' theorem, the curve must have genus
0 or 1.

The genus is indeed 0.

dave
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David Bernier
science forum Guru Wannabe


Joined: 01 May 2005
Posts: 101

PostPosted: Fri Jun 30, 2006 3:29 am    Post subject: Re: Linking points on algebraic curves with class numbers Reply with quote

Dave Rusin wrote:
Quote:
In article <e7uou2$rqd$1@news.math.niu.edu>, I wrote:

Here is the list so far:
a = 1, 19, 37/17, 53/17, 89/19, 199/109, 323/197, 433/323, 629/251,
683/397, 703/487, 757/127, 901/719, 971/701, 1007/127, ...

Now the game is to decide where these numbers "come from".

Here's what I figured out.

1. All the numbers on this (finite) list have the form
a = (u^3+18*u*v^2+9*v^3+9*u^2*v)/(-u^3+9*u*v^2+9*v^3)
for some coprime pair (u,v) of integers.

2. Any number of this form will be on the list (i.e. for all these a,
the polynomial of the previous post factors into three sextics).

3. For any specific coprime u,v the gcd of the numerator and
denominator divides 3^6. In particular

3a. The only integers on the list are -1, +1, and +19 .
3b. The primes dividing these numerators and denominators are precisely
the primes >3 for which u^3 - 9*u - 9 has a linear factor mod p.
(These are 17, 19, 37, 53, 71, 73, 89, 109, 127, 163, 179, ...)

By 3b, and the fact that the gcd of the numerator and denominator
divides 3^6, a (in lowest terms) cannot have the prime factors
3, 5, 7, 11, 13, 23, 29, 31, 41, 43, 47, 59, 61, 67, 79 ... appear
in the numerator or denominator of a. So if the numerator
of a (in lowest terms) is not prime, it must be one of:
17^2 = 289
17*19 = 323
17*37 = 629,
17*53 = 901,
....
19*19 = 361
19*37 = 703,
19*53 = 1007,
....
37*53 = 1961,
....
Apart from 1961, the numbers just above that don't appear as numerators
in the first few terms of the list at the very top are 17^2=289
and 19^2 = 361. The remainder, 323, 629, 703, 901 and 1007,
all appear as numerators. A new question is whether the
numerator and denominator of a (in lowest terms) is always
square-free?

I also tried u=1642, v=1777, and the resulting a, in lowest terms,
has a numerator which is a product of two distinct prime factors,
and a denominator also of this form:

u=1642, v=1777,
a=f/g with

f = 191378250361 = 307 * 623381923 and
g = 92739377771 = 72287 * 1282933.

David Bernier

Quote:
It is still that these numerators and denominators seem to be prime,
or "nearly" so, fairly often; I'm not sure why. But it's not true
that all these primes determine real quadratic fields with class number 1.
I guess I was just distracted by the "Law of Small Numbers".

These experiments suggest to me that the curve has an infinite number
of rational points. By Faltings' theorem, the curve must have genus
0 or 1.

The genus is indeed 0.

dave
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