Re: JSH: The "Published" paper he dosen't what you to know about.


Arturo Magidin wrote:
In article <1158896704.800509.198470@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,

Now, in some special cases he was able to show his argument work; it
"works" when v = -1 because then you get a linear polynomial and two
of the ai are formally set to 0. It works when we "set" m = 1,
f=sqrt(2), because then v=-1+2 = 1, so we have

2W^3 - 3W + 1 = (W-1)(2W^2 + 2W - 1)

and so one ai is equal to 1, and the other two are divisible by
sqrt(2). And so on.

Ah. And this is where I came in. We have a polynomial over Q(v), namely

(v^3+1)W^3 - 3vW + 1. From its derivation from the Fermat curve
x^3+y^3=z^3, it is of genus one with j invariant zero considered as an
algebraic curve. It has a couple of singularities at infinity; one of
these corresponds to v = -1, where two of the branches go to infinity
and one to W = -1/3.

It really makes more sense to use W = 1/x here, so that you are working
with the polynomial x^3 - 3vx^2 + v^3+1. Now the polynomial is monic in
x, and as a curve is of total degree three, with no singularities; it
still has genus zero and is isomorphic to the Fermat elliptic curve.
You still get the ramification at v = -1, where as a function of v x
increases to -1 and then decreases again. At v = -1, one branch passes
through x = -3, and the other two descend on either side from x = 0, v
= -1. For values of v locally close
to v = -1, such as v = 4 5-adically, we get that the x = -3 branch
leads to a 5-adically distinguished root. None of this seems to have
much to do with the question of rational points, however.

.

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