Modular Elliptic Curves and Fermat's Last Theorem

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p = 3. The only one of these with a quotient group of order p is PSL2(F3)
when p = 3. It follows that p [G : G ] except in this one case which we treat
separately. So assuming now that p [G : G ] we see that G contains a non-
trivial unipotent element u. Since G has no fixed line there must be another
noncommuting unipotent element v in G . Pick a basis for �0|G consisting
of their fixed vectors. Then let � be an element of Gal(Q�/Q) for which the
image of �0(�) in G/G is prescribed and let �0(�) =(a b). Then
c d
a b 1 s� 1
� =
c d 1 r� 1
has det (�) = det �0(�) and trace � = s�(ra� + c) +br� + a + d. Since p e" 3
we can choose this trace to avoid any two given values (by varying s) unless
ra� + c = 0 for all r. But ra� + c cannot be zero for all r as otherwise
a = c = 0. So we can find a � for which the ratio of the eigenvalues is not
�(�), det(�) being, of course, fixed.
MODULAR ELLIPTIC CURVES AND FERMAT S LAST THEOREM 477
Now suppose that im �0 does not have order divisible by p but that the
associated projective representation has image isomorphic to S4 or A5, so
�0
necessarily p =3. Pick an element � such that the image of �0(�) in G/G is
any prescribed class. Since this fixes both det �0(�) and �(�) we have to show
that we can avoid at most two particular values of the trace for �. To achieve
this we can adapt our first choice of � by multiplying by any element og G . So
pick � " G as in (i) which we can assume in these two cases has order 3. Pick
�
a basis for �0, by expending scalars if necessary, so that � �! ( ). Then one
�-1
checks easily that if �0(�) =(a b) we cannot have the traces of all of �, �� and
c d
�2� lying in a set of the form {"t} unless a = d = 0. However we can ensure
that �0(�) does not satisfy this by first multiplying � by a suitable element of
G since G is not contained in the diagonal matrices (it is not abelian).
In the A4 case, and in the PSL2(F3) A4 case when p = 3, we use a
different argument. In both cases we find that the 2-Sylow subgroup of G/G
is generated by an element z in the centre of G. Either a power of z is a suitable
candidate for �0(�) or else we must multiply the power of z by an element of
G , the ratio of whose eigenvalues is not equal to 1. Such an element exists
because in G the only possible elements without this property are {"I} (such
elements necessary have determinant 1 and order prime to p) and we know
that #G > 2 as was noted in the proof of part (i).
Remark. By a well-known result on the finite subgroups of PGL2(Fp) this
lemma covers all �0 whose images are absolutely irreducible and for which
�0
is not dihedral.
"
Let K1 be the splitting field of �0. Then we can view W� and W� as
Gal(K1(�p)/Q)-modules. We need to analyze their cohomology. Recall that
we are assuming that �0 is absolutely irreducible. Let be the associated
�0
projective representation to PGL2(k).
The following proposition is based on the computations in [CPS].
Proposition 1.11. Suppose that �0 is absolutely irreducible. Then
"
H1(K1(�p)/Q, W� ) =0.
Proof. If the image of �0 has order prime to p the lemma is trivial. The
subgroups of GL2(k) containing an element of order p which are not contained
in a Borel subgroup have been classified by Dickson [Di, �260] or [Hu, II.8.27].
Their images inside PGL2(k ) where k is the quadratic extension of k are
conjugate to PGL2(F ) or PSL2(F ) for some subfield F of k , or they are
isomorphic to one of the exceptional groups A4, S4, A5.
"
Assume then that the cohomology group H1(K1(�p)/Q, W� ) =0. Then
by considering the inflation-restriction sequence with respect to the normal
478 ANDREW JOHN WILES
subgroup Gal(K1(�p)/K1) we see that �p " K1. Next, since the representation
is (absolutely) irreducible, the center Z of Gal(K1/Q) is contained in the
diagonal matrices and so acts trivially on W�. So by considering the inflation-
restriction sequence with respect to Z we see that Z acts trivially on �p (and
"
on W� ). So Gal(Q(�p)/Q) is a quotient of Gal(K1/Q)/Z. This rules out all
cases when p =3, and when p = 3 we only have to consider the case where the
image of the projective representation is isomporphic as a group to PGL2(F )
for some finite field of characteristic 3. (Note that S4 PGL2(F3).)
Extending scalars commutes with formation of duals and H1, so we may
assume without loss of generality F �" k. If p = 3 and #F > 3 then
H1(PSL2(F ), W�) = 0 by results of [CPS]. Then if is the projective
�0
representation associated to �0 suppose that g-1im g = PGL2(F ) and let
�0
"
H = gPSL2(F )g-1. Then W� W� over H and
� �
(1.18) H1(H, W�) �" F H1(g-1Hg, g-1(W� �" F )) =0.
F F
"
We deduce also that H1(im �0, W� ) =0.
Finally we consider the case where F = F3. I am grateful to Taylor for the
following argument. First we consider the action of PSL2(F3) on W� explicitly
by considering the conjugation action on matrices {A " M2(F3) : trace A =0}.
One sees that no such matrix is fixed by all the elements of order 2, whence
2
H1(PSL2(F3), W�) H1(Z/3, (W�)C �C2) =0
where C2 �C2 denotes the normal subgroup of order 4 in PSL2(F3) A4. Next
�
we verify that there is a unique copy of A4 in PGL2(F3) up to conjugation.
�
For suppose that A, B " GL2(F3) are such that A2 = B2 = I with the images
�
of A, B representing distinct nontrivial commuting elements of PGL2(F3). We
can choose A =(1 0) by a suitable choice of basis, i.e., by a suitable conju-
0 -1
gation. Then B is diagonal or antidiagonal as it commutes with A up to a
scalar, and as B, A are distinct in PGL2(F3) we have B =(a -a-1) for some [ Pobierz całość w formacie PDF ]

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