Re: Langlands and FLT

galathaea <galathaea@xxxxxxxxx> writes:

On Aug 13, 8:29 am, Allan Adler <a...@xxxxxxxxxxxxxxxxxxxx> wrote:
No offense, but I think that you're not addressing my questions. I think it
is important to answer this question: let rho0 be a representation of
Gal(Qpbar/Qp) into GL(2,k), where k is a finite field; does rho0 always
arise from a finite flat k-vector scheme G over Qp (resp. over Zp)?

This is a yes or no question.

The candidate for G would be the algebra of all Gal(Qpbar/Qp)-equivariant
functions from k^2 to Qpbar (resp. Zpbar, the ring of integers of Qpbar),
where Gal(Qpbar/Qp) acts on k^2 via rho0 and on Qpbar via the Galois action.

I don't think I can proceed until this question is answered. Please try to
suppress the big picture until that has been done. If you don't know the
answer, just say so.

I thought I had answered this question.

I think the answer is no.

I had written:

I believe all one needs is an elliptic curve at a place
with bad reduction over Q_p. There is a theorem which
states that if E[p^r] is the generic fiber of a flat
group scheme over Z_p for all r > 1, then E_/Q_p has
good reduction.

In other words, I believe that if you take a curve at a
place with bad reduction, you can find an r for which the
representation of E[p^r] is not the generic fiber of a
flat group scheme over Z_p.

OK, thanks for the clarification. Now, I think that E[p^r] is not a
k-vector scheme unless r < 2. So, even if the theorem you cite is true
(I don't know one way or the other), it doesn't address the particular
case of representations into GL(2,k), where k is a finite field. E[p^r]
would afford a representation into GL(2,Z/p^rZ). On the other hand, the
theorem you cite does seem to imply that if H is a finite abelian group
and rho0 is a representation of Gal(Qpbar/Qp) into Aut(H), then the
construction I outlined doesn't necessarily produce a group scheme flat
over Zp.

So, let me first think a little bit more about that construction.

Where do you think the theorem you mentioned is proved?

I do not know how to explicitly construct an example, but
I think this result at least provides a starting point.

OK, thanks.

I am earnestly trying to answer your questions, which is one
of the reasons I took my time in response. When I rush to
respond I make many more errors of irrelevance because I get
too focussed on my own pursuits. I am trying to understand
your own questions because I suspect they will force me to
tie together the concrete and abstract elements I have such
a tenuous grasp on. Please forgive the excess exposition that
accompanies.

Thanks, I appreciate your efforts on this.

Actually, let's look at my construction again. The Galois group Gal acts
on H and we can write H as a union of orbits. Choose one point from each
orbit. If h is an orbit representative, let Gal_h denote the subgroup of Gal
that fixes h. Let K_h denote the fixed field of Gal_h. Then as a vector space
over Qp, the algebra of Gal-equivariant functions from H to Qpbar looks
like the direct sum of the fields K_h as h runs over the chosen set of
orbit representatives. If we had looked at the ring of Gal-equivariant
functions from H to the ring of integers of Qpbar, then instead we would
get a ring whose underlying Zp module would look like the direct sum of the
rings R_h, where R_h is the ring of integers of K_h; as a Zp module, it
is therefore finitely generated and torsion free. Since Zp is a principal
ideal domain, every finitely generated torsion free module over Zp is free.
So my construction seems to lead to a Zp algebra whose underlying Zp module
is free and therefore flat. Therefore, the construction does seem to give
rise to a finite flat group scheme over Zp.

Do you accept this argument?
--
Ignorantly,
Allan Adler <ara@xxxxxxxxxxxxxxxxxxxx>
* Disclaimer: I am a guest and *not* a member of the MIT CSAIL. My actions and
* comments do not reflect in any way on MIT. Also, I am nowhere near Boston.
.

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