Galois representations attached to elliptic curves

[David Kohel]

Hi,

Below I forward a question of Peter Stevenhagen regarding
the Galois representation attached to an elliptic curve over Q.
I think it would be relatively easy to solve, and even better
to write a class of elliptic curve Galois representations which
could be interactively investigated modulo p^n (without losing
information already computed modulo p^(n-1)), although for
the \hat{Z} representation, there are subtleties of the diagonal
image of the mod p^n and mod r^m Galois representations.

I indicated that this would be an ideal problem to discuss at
a Sage Days.

I'm CC'ing Peter, but maybe he should invite himself to join
the sage-nt mailing list (normally low traffic when P^1(Z/NZ)
is not being discussed).

Cheers,

David

Forwarded message:

Is there an easy way using Magma/GP/sage or whatever to
figure out what the Galois representation over Q is of the
elliptic curve E: y^2= (x+1)(x^2+4)?

Ideally, I would like to know what the open subgroup of GL_2(Zhat)
is that arises as the Galois group of Q(E^tors) over Q.
Next best, I would like to know the p-power torsion representation
for the `critical' primes p= 2 and 3 (rational torsion points),
and maybe 5 (bad reduction).

As you see, E has a rational 2-torsion point, and the 2-torsion
field already contains a 4th root of unity.

Is there a button-click giving me G(4)=Gal(Q(E[4])/Q), or even G(8)?

[William Stein]

On Wed, May 27, 2009 at 5:53 AM, David Kohel <drk...@gmail.com> wrote:
>
> Hi,
>
> Below I forward a question of Peter Stevenhagen regarding
> the Galois representation attached to an elliptic curve over Q.
> I think it would be relatively easy to solve, and even better
> to write a class of elliptic curve Galois representations which
> could be interactively investigated modulo p^n (without losing
> information already computed modulo p^(n-1)), although for
> the \hat{Z} representation, there are subtleties of the diagonal
> image of the mod p^n and mod r^m Galois representations.
>
> I indicated that this would be an ideal problem to discuss at
> a Sage Days.
>
> I'm CC'ing Peter, but maybe he should invite himself to join
> the sage-nt mailing list (normally low traffic when P^1(Z/NZ)
> is not being discussed).
>
> Cheers,
>
> David
>
> Forwarded message:
>
> Is there an easy way using Magma/GP/sage or whatever to
> figure out what the Galois representation over Q is of the
> elliptic curve E: y^2= (x+1)(x^2+4)?

I played around for a few minutes and created this Sage worksheet that
does *not* answer your question, as far as I can tell:

http://nt.sagenb.org/home/pub/12/

>
> Ideally, I would like to know what the open subgroup of GL_2(Zhat)
> is that arises as the Galois group of Q(E^tors) over Q.
> Next best, I would like to know the p-power torsion representation
> for the `critical' primes p= 2 and 3 (rational torsion points),
> and maybe 5 (bad reduction).
>
> As you see, E has a rational 2-torsion point, and the 2-torsion
> field already contains a 4th root of unity.
>
> Is there a button-click giving me G(4)=Gal(Q(E[4])/Q), or even G(8)?
>
> >
>

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

[N. Bruin]

> Is there an easy way using Magma/GP/sage or whatever to
> figure out what the Galois representation over Q is of the
> elliptic curve E: y^2= (x+1)(x^2+4)?

[...]

> Is there a button-click giving me G(4)=Gal(Q(E[4])/Q), or even G(8)?

Funnily enough, this is something I've been working with myself
recently. In magma, the steamroller approach of constructing the 4-
torsion as a subscheme, projecting down to a line and asking for the
Galois group of the defining polynomial there actually works.
(yes, Claus has worked hard on getting magma to do the bookkeeping in
computing galois groups of polynomials with non-transitive galois
action)

_<x>:=PolynomialRing(Rationals());
E:=EllipticCurve((x+1)*(x^2+4));
E4:=ReducedSubscheme((E!0)@@map<E->E|DefiningPolynomials
(MultiplicationByMMap(E,4))>);
assert Degree(E4) eq 16;
L:=Projection(E4,Ambient(E4)![0,1,1]);//some generic projection
direction
assert IsReduced(L);
g:=Evaluate(DefiningPolynomial(L),[x,1]);
assert Degree(g) eq 16 and IsSquarefree(g);
G4:=GaloisGroup(g);

This only gives you the Galois group as a permutation group on 16
elements. You'd have to work a bit to get it as a subgroup of GL(2,Z/
4).

[Peter Stevenhagen]

More pedestrian: the division polynomials on William's worksheet
show
that the field generated by the x-coordinates of the 4-torsion
points
is dehidral over Q of degree 8. If you pick the smallest
prime p splitting completely in it (61 I believe) you find that
(E mod p) does not have complete 4-torsion, so Q(E[4]) is indeed
of
degree 16. It is the subgroup of GL(2,Z/4Z) consisting of
those
matrices M that reduce mod 2 to (1 * over 0
1),
AND the * is 0 mod 2 if and and only if det(M) is 1 mod
4.
(That's because Q(E[2])=Q
(i).)

If you look at the primes modulo which the 8-division
polynomial
om William's worksheet splits completely, you find that for all
these
primes (E mod p) has complete 8-torsion. So Q(E[8]) is the
splitting
field of the 8-division polynomial, of degree 128 over Q,
which
only has degree 8 over Q(E[4]), not
16.
And the Galois group G(8) of Q(E[8]) over Q has index 3*2*2=12
inside
GL(2,Z/8Z).

[Chris Wuthrich]

Unfortunately I do not have time to do these computations right now
(I am in the middle of marking), but there might be another way of
getting the Galois group with working on the division polynomials.

For each prime v, the characteristic polynomial of the matrix
rho(Frob_v) in GL_2(Z_2) or GL_2(Z/8Z) can be computed. This gives
plenty of conjugacy classes in GL_2(Z/8) that intersect your Galois
group. (At least when there are no roots or distinct roots, we can
identify the conjugacy class in this way.) On the other hand you know
an upper bound on the image from the isogeny over Q. Hopefully you can
prove that it is all of this. Sometimes one can also use Tate-curves
at multiplicative primes v to get the image of the decomposition group
D_v to generate more. But here c_5 is divisible by 2 so it is not
straight forward. Otherwise one gets elements like [1 1\\ 0 1] mod p
for free.

But as I said, I have not done the computations here at all and maybe
this is all irrelevant.

> (That's because Q(E[2])=Q

are we talking about the same curve ? For 20a1 it is Q(i), isn't it ?

Chris.

[Peter Stevenhagen]

Yes, the method you sketch is the one employed by Lang & Trotter
in their Springer Lecture Notes. It works fine in case the Galois
group of Q(E[p^{k+1}]) over Q(E[p^k]) is of full order p^4, but it
is a bit harder if your Galois group is smaller, like in this case
when going from 2 to 4 and from 4 to 8, where we `lose' a factor 2
in each case. I now know G(8), and from 8 to 16 (and further) everything
is maximal size again, so the 2-power torsion Galois group
Gal(Q(E[2^\infty]/Q) is the inverse image in GL_2(ZZ_2) of G(8).

I should look in the Tate-curve-use - Serre also uses it in the way
you mention.

Sorry for the lousy typography in my post, I never do this via a
browser-interface that happily cuts lines, after the line

> > (That's because Q(E[2])=Q

the next line started with (i)....
So yes we are talking about the same curve 20a1!

Happy marking!
Peter