Note: I am not an expert on this subject. However, I have enough of a background to be able (barely) to follow the explanations of others. I have seen explanations of what I said above given by Larry Washington, Karl Rubin, and J.P. Serre. If anyone thinks they are out to discredit Mr. Benschop then I suggest that said person needs a reality check.

I have seen Mr. Benschop's arguments. I am unconvinced by them. Hensel lifting a solution of the Fermat equation from mod p^n to Q does not and CAN NOT work. One can not prove FLT from local considerations.
**]

You say you've seen my arguments. They are rather simple, based on an exception namely:

and x, y in the p-1 cyclic subgroup A_k of units where x^p=x, y^p=y, (x+y)^p=x+y (mod p^k),

called the Core of semigroup Z(.) mod p^k. It exists for every k >1 because the units group G_k (mod p^k) has order (p-1).p^{k-1} , so G_k is a direct product, and in fact an arithmetic product G_k = A_k . B_k of residue subgroups in G_k. In the Core A_k holds: x^p = x (mod p^k) for each x in A_k, so it is the extension of FST (Fermat's Small Theorem) to k>1.

Now any solution mod p^2 of [1] can indeed be "lifted" to mod p^k for any k>2, hence also the solutions that have z=x+y -- I do NOT contest that: Hensel's extension lemma remains valid all right. So your Selmer group condition holds, but the conclusion for such special solutions in Core is wrong, because:

which implies INequality for the p-th powers of the corresponding integers < p^k, all k>1, since: (x+y)^p > x^p + y^p -- so INequality despite the Hensel lift. ( since the "mixed terms" with x and y in the expansion are > 0 ).

Do you notice the difference?
Hensel's lift STILL holds, but does NOT always block derivation of INequality for integers. Of course, to settle all of FLTcase1 it needs to be shown that ALL solutions of [1] have this EDS property, ( or a variation of it, for k=2: the Triplets: see my homepg ref[1] ). Which in fact is precisely what I do in my paper.
Notice that the cubic root solution has form: a+1 = -1/a ( mod p^k, p=1 mod 6 ) where apparently "+1" and "-n" and "1/n" are essential, and a loop of type: a+1=-1/b, b+1=-1/c, c+1=-1/d, d+1=-1/e, etc... z+1=-1/a can hold. To show that such loop can ONLY have length 1 or 3, apply function composition ( associative but not nec. commutative ) to the three elementary functions: Complement C(n)=-n, Inverse I(n)=1/n and Successor S(n)=n+1: each of the four possible compostions using all three: ICS, SIC, ISC, CSI has period 3 upon iteration, implying that no loops of length >3 can exist in arithmetic mod p^k ( the period 3 is one in excess of the number of symmetries: I and C commute IC=CI ). For looplength 1 we have special case a+1=-1/a: the cubic root solution with a^2 = 1/a (a^3=1, a!=1). For p-th power residues the first prime with a triplet is p=59 (in fact two triplets): not easily found without computer. And p=79 is the smallest prime with a cubic root solution (p=1 mod 6) *and* a triplet.

--- Fermat might have found the cubic root solution easily --- by experimenting mod p^2 for p=3,5,7, with for p=7 the first occurrance. But he kept quiet about it, because he could not prove they are the ONLY solutions of FLTcase1 for residues (which they are indeed NOT: the triplets for some primes p \geq 59, of both types p = \pm 1 mod 6). -- Clearly this is speculation, but with the means of those times (1640) not unrealistic: one needs no semigroup techniques for the cubic root solution (yes for the triplets), on which I wrote a paper ( homepg ref[5] ), presented at the NMC33 (Netherlands Mathematics Congress, apr98 - digest p39).

Your attitude (and that of about all experts I was confronted with) reminds me of the church prelates invited by Galileo to look through his little telescope, and see for themselves the motion of satellites around Jupiter: There *is* a center of motion other than the Earth, despite what the old greeks (Aristotle?) claimed. It ONLY requires another than an egocentric viewpoint, which apparently was too much for most: they even refused to have a look. The courage of program committee members of the three conferences where I was allowed to speak (with varying emphases), especially the first one in Prague ("Semigroups & Applications"), is to be admired: they most probably were "blasted" by their colleagues (of your prejudiced type, pardon the expression).

No bad feelings, though.... A word equivalent to "crank" must have been used for instance to Fourier (my favourite, next to Boole, Shannon and Tellegen) - trying to convince his contemporaries that every periodic function is the sum of a sequence of Sines and Cosines -- with an unbelievable prejudice against this, even by Lagrange + Laplace + Legendre (1807), and by many other well known scientists & mathematicians. . . . . So what's new?

Ciao, Nico Benschop. -- http://www.iae.nl/users/benschop . . . 21may98

______________if stuck@closure(mod m), use the carry__________
______________or stuck@Selmer's group, . . . use EDS __________
______________ if n has k digits, then n^p has pk digits __________

Subject:       Re: Wiles' proof of FLT
Author:        benschop_nf@my-dejanews.com
Date:          Thu, 28 Jan 1999 (sci.math)

In article <78nkqa$flv$1@gannett.math.niu.edu>,
  rusin@vesuvius.math.niu.edu (Dave Rusin) wrote:
> In article <36AF07A7.1C74@iae.nl>, Nico Benschop   wrote:
> >[ from Simon Singh's book I understand that Wiles proved something
> >    like: 'every' Elliptic Curve is has a Modular Form (EC=MF).
> >  Using a spectral method, comparing the number of solutions of
> >  each particular EC(a,b,c) mod p, yielding an "L-spectrum"
> >  of numbers for all primes p <--> comparing this to a similar
> >  "L-spectrum" for corresponding MF(...)
> >
> >My problem with this approach: being a residue method, *why* does this
> >not suffer from the known "Hensel lift" -- which has always been the
> >main objection against a direct FLT proof via residues ?
> >     (saying: a solution of x^p + y^p = z^p mod p^2 can be extended
> >     to any precision k>2, hence cannot provide integer inequality).
>
> Why would you think there is any relation? Just because both of them
> refer to "reducing mod p"?

True, of course: "for all primes p", yielding for each EC(a,b,c) [for
fixed a,b,c coefficients] an infinite but countable series of naturals
counting the # solutions for each mod p_k (k-th prime). This, I understand,
is Dirichlet's L-series of a particular EC, *and* of a particular MF to
match. This 1-1 correspondence Wiles showed (for semistable EC's ?),
so the general Tanyama conjecture (for all EC's) is not proven yet(?)

> The equation  x^p + y^p = z^p  describes a projective curve; the goal is
> to show it has no integer points with xyz <> 0. As you have observed, it
> is impossible to do this by looking modulo p^k: given integers xyz with
> x^p + y^p = z^p mod p^2  then there exist integers  XYZ  with
> X=x mod p^2, etc., and  X^p + Y^p = Z^p  mod  p^k.
>
> Wiles' result has nothing to do with this curve, but rather (as you
> observe) applies to elliptic curves

I got that, Dave. But I'm now not concentrating on FLT, but on the *method*
of proof via residues mod p (*all* primes p;-)  It somehow reminded me of
the Hensel-lift problem (mistakenly assumed to block a direct FLTcase1 proof
via residues mod p^k, see http://www.iae.nl/users/benschop/campaign.htm )

Clearly,  analysis via residues mod p^k (all k>0, any fixed prime) differs
from the L-series approach:
          analysis via residues mod p_k (all k>0) .. [p_k is the k-th prime]

I guess mod p residues are preferred because arithmetic Z(+,*) mod p is a
field, having no non-zero divisors of 0 (re: the main thm of arithmetic!)
Which does not hold for mod p^k with k>1.

But the L-series approach seems equivalent, via the Chinese Remainder Thm,
to residues mod m_k (k -->inf), where m_k = \prod first k primes, is'nt it?

In that sense it *still* is a residue method, no matter how large k gets,
_not_ yielding enough info for *integer* (cq rational) solutions, does it?

> (only for p=3 is the Fermat curve an EC).
> Given an elliptic curve y^2= x^3 + A x^2 + B x + C for some integers A,B,C
> one may look for rational points on it. As in the previous paragraph one
> may get some information about possible solutions by working mod p (though,
> as one eventually discovers, this information is not sufficient to answer
> questions about rational or integral solutions).

> But among other differences with the previous paragraph one should observe
> that here we may take restrictions mod p for _every_ prime p.

This still does not suffice: it remains a residue method mod m_k (all k>0).
see above.  By what *extra* mechanism does the Dirichlet L-seris become a
(non-residue) *integer* method??  Note: in the FLTcase1 problem the Hensel
lift is escaped due to exponent p distributing over a sum. Since for any
solution of x^p+y^p=z^p mod p^2  has  z=x+y (!) so (x+y)^p=x^p+y^p mod p^2,
which blocks msd 0-extension to integers, because (X+Y)^p > X^p+ Y^p for
the corresponding 2-digit X,Y,X+Y integers with p-th powers < p^{2p}.

> It is the collection of
> data for the many primes, but one fixed curve, which is encapsulated in
> the L-functions of which you spoke. The point of modularity is that the
> number of solutions mod p should behave in a fairly understandable way
> as p changes. This is what Wiles proved: that all elliptic curves over Q
> will behave in this way. (Well, his proof only covers _almost all_ of them.)
>
>  [...the Frey/Ribet FLT counter-example as EC...]  -- Dave Rusin

Just curious,
  Nico Benschop -- http://www.iae.nl/users/benschop/scimat98.htm