Fwd: heights over function fields

[William Stein]

---------- Forwarded message ----------
From: Noam Elkies <elk...@math.harvard.edu>
Date: Thu, Dec 3, 2009 at 4:48 AM
Subject: Re: heights over function fields
To: wst...@gmail.com


> Robert Bradshaw and I computed maybe the regulator of our rank 2 curve
> over that function field using Tate's suggestion.   See

>      http://sagenb.org/home/pub/1198/

Where does this elliptic surface

 y^2 = x^3 + 4*t*(t+1)^2*x^2 - t^3*(t+1)^2*x

come from?  It looks like this is an elliptic K3 surface with
singular fibers of type I2* (= D6) at t=0 and t=infinity, of type
I0* (= D4) at t=-1, and of type I2* (= A1) at t=1.  In particular
the group of components of multiplicity 1 always has exponent 2,
so we can compute the height of any section P using just the
naive height of 2P -- explaining why the approximations using the
naive height of 2^n*P stabilize at n=1.

The root lattice has rank 6+4+6+1=17, so if you have two independent
sections it must be a supersingular elliptic surface, with another
section defined only over the quadratic extension of Z/17Z.

Does

 P = [
   E.point((t, t^3 + 3*t^2, 1), check=False),
   E.point((t^2, t^3, 1), check=False)
   ]

mean that you have a pair of sections with x=t and x=t^2?  Neither of
those seems to make y^2 the square of a polynomial.

NDE



--
William Stein
Associate Professor of Mathematics
University of Washington
http://wstein.org

[William Stein]

---------- Forwarded message ----------
From: salman <salm...@gmail.com>
Date: Fri, Dec 4, 2009 at 3:46 PM
Subject: Re: heights over function fields

To: William Stein <wst...@gmail.com>, Noam Elkies <elk...@math.harvard.edu>


Hi Noam,

Thanks for the helpful comments.

> Where does this elliptic surface
>
>  y^2 = x^3 + 4*t*(t+1)^2*x^2 - t^3*(t+1)^2*x
>
> come from?  It looks like this is an elliptic K3 surface with
> singular fibers of type I2* (= D6) at t=0 and t=infinity, of type
> I0* (= D4) at t=-1, and of type I2* (= A1) at t=1.  In particular
> the group of components of multiplicity 1 always has exponent 2,
> so we can compute the height of any section P using just the
> naive height of 2P -- explaining why the approximations using the
> naive height of 2^n*P stabilize at n=1.

It's the first quadratic twist (by t+1) of the Legendre curve

y^2 = x*(x+t)*(x+t^2)

over F_5(t) of analytic rank 2. I think the sign for the x coefficient
is supposed to be positive, for what it's worth.

> The root lattice has rank 6+4+6+1=17, so if you have two independent
> sections it must be a supersingular elliptic surface, with another
> section defined only over the quadratic extension of Z/17Z.
>
> Does
>
>  P = [
>    E.point((t, t^3 + 3*t^2, 1), check=False),
>    E.point((t^2, t^3, 1), check=False)
>    ]
>
> mean that you have a pair of sections with x=t and x=t^2?  Neither of
> those seems to make y^2 the square of a polynomial.

This may be the result of the sign discrepancy for the a4 term.

-Sal

[Noam Elkies]

salman <salm...@gmail.com> writes:

> Thanks for the helpful comments.

You're welcome. I had asked:

>> Where does this elliptic surface

>> y^2 = x^3 + 4*t*(t+1)^2*x^2 - t^3*(t+1)^2*x

>> come from? [...]

and you reply:

> It's the first quadratic twist (by t+1) of the Legendre curve

> y^2 = x*(x+t)*(x+t^2)

> over F_5(t) of analytic rank 0. [later corrected: of sign 2.]

> I think the sign for the x coefficient is supposed to be positive,
> for what it's worth.

Thanks, I think I can n ow more-or-less reconstruct what you're
trying to do. Som reconstruction is encessary because it seems that
this equation was transmitted by "broken telephone" -- the link
<http://sagenb.org/home/pub/1198> that William sent gives

Elliptic Curve defined by y^2 = x^3 + (4*t^3+8*t^2+4*t)*x^2 +
(16*t^5+15*t^4+16*t^3)*x over Fraction Field of Univariate
Polynomial Ring in t over Finite Field of size 17

which is the same as what I wrote (note characteristic 17, not 5)
except that I factored the coefficients (this displays most of the
reducible fibers) and replaced the coefficient of 16 for the
last (a4) term by the equivalent -1. Equivalent mod 17, that is,
not mod 5...

Anyway, the specifics of the intended surface may be different from
what I computed for the characteristic-17 formula, but the basic
idea is the same: use Tate's algorithm to get the types of the
reducible fibers and then use that information to get exact
rather than approximate heights and height pairings. Note that
in general "naive_height(2^n*P)/4^n" always approaches the
canonical height (that's Tate's definition) but need not actually
stabilize. Stabilization happens when and only when some 2^n*P
is in the "narrow Mordell-Weil group" of points that meet
each reducible fiber in the identity component. That's automatic
when all the component groups are 2-groups, which will indeed
happen for quadratic twists of the "Legendre curve", but that's
quite a special case.

I'm naturally reminded of Gross's nice lecture at PCMI'09 on
a different family of twists of a Legendre curve, which have
extra rank for primes congruent to 3 mod 4 (namely 1 instead of 0
over the prime field, and 2 over its quadratic cover and over its
algebraic closure); that's evidently not the same family you're
looking at.

NDE