This second version includes some corrections in the English language, an in depth study of the case p = 3 (especially Theorem 8), further details on some conjectures, and some minor mathematical improvements.

, the pth power of an ideal), congruent to 1 modulo p; so, from [Gr1, Theorem 2.2, Remark 2.3, (ii)], α is locally a pth power in K giving easily α ≡ 1 (mod p p+1 ), then c (ζ-ζ -1 ) a+c ζ -1 ≡ 0 (mod p p+1 ), hence c ≡ 0 (mod p 2 ).

Some works of Furtwängler and Vandiver revisited and Fermat ′ s last theorem by Georges Gras and Roland Quême

Abstract. From some works of P. Furtwängler and H.S. Vandiver, we put the basis of a new cyclotomic approach to Fermat ′ s last theorem for p > 3 and to a stronger version called SFLT, by introducing governing fields of the form Q(µ q-1 ) for prime numbers q. We prove for instance that if there exist infinitely many primes q, q ≡ 1 (mod p), q p-1 ≡ 1 (mod p 2 ), such that for q | q in Q(µ q-1 ), we have q 1-c = a p (α) with α ≡ 1 (mod p 2 ) (where c is the complex conjugation), then Fermat ′ s last theorem holds for p. More generally, the main purpose of the paper is to show that the existence of nontrivial solutions for SFLT implies some strong constraints on the arithmetic of the fields Q(µ q-1 ). From there, we give sufficient conditions of nonexistence that would require further investigations to lead to a proof of SFLT, and we formulate various conjectures. This text must be considered as a basic tool for future researches (probably of analytic or geometric nature).

Résumé. Reprenant des travaux de P. Furtwängler et H.S. Vandiver, nous posons les bases d'une nouvelle approche cyclotomique du dernier théorème de Fermat pour p > 3 et d'une version plus forte appelée SFLT, en introduisant des corps gouvernants de la forme Q(µ q-1 ) pour q premier. Nous prouvons par exemple que s'il existe une infinité de nombres premiers q, q ≡ 1 (mod p), q p-1 ≡ 1 (mod p 2 ), tels que pour q | q dans Q(µ q-1 ), on ait q 1-c = a p (α) avec α ≡ 1 (mod p 2 ) (où c est la conjugaison complexe), alors le théorème de Fermat est vrai pour p. Plus généralement, le but principal de l'article est de montrer que l'existence de solutions non triviales pour SFLT implique de fortes contraintes sur l'arithmétique des corps Q(µ q-1 ). A partir de là, nous donnons des conditions suffisantes de non existence qui nécessiteraient des investigations supplémentaires pour conduire à une preuve de SFLT, et nous formulons diverses conjectures. Ce texte doit être considéré comme un outil de base pour de futures recherches (probablement analytiques ou géométriques).

Introduction

This paper is devoted to the study of the following phenomenon. Consider the maximal abelian extension Q nr of Q, unramified (= nonramified) at a given prime p > 2; from class field theory, we get Q nr = n, p∤n Q(µ n ).

Then denote by H Q nr the maximal p-ramified (i.e., unramified outside p)

1991 Mathematics Subject Classification. 11D41, 11R18. Mots clefs. Fermat ′ s last theorem, Furtwängler ′ s theorems, cyclotomic fields, cyclotomic units, class field theory, Čebotarev density theorem.

abelian p-extension of Q nr ; this extension is given by n, p∤n H Q(µn) where H Q(µn) is the maximal p-ramified abelian p-extension of Q(µ n ).

Then consider H Q nr [p] := n, p∤n H Q(µn) [p], the maximal p-elementary pramified extension of Q nr , union of the corresponding maximal p-elementary p-ramified extensions of Q(µ n ).

We have found that any nontrivial solution (u, v) of a classical diophantine equation, associated to Fermat ′ s equation, and called the SFLT equation 1 , implies some constraints on the law of decomposition of every prime q = p in H Q nr [p]/Q nr .

These constraints may be characterized at some finite steps via the law of decomposition of q in a canonical family F n of p-cyclic subextensions of H Q(µn) [p]/Q(µ n ), where n | q -1 depends on q, u, v (see Theorem 4).

Some aspects needed to prove this relation can be found in some former technics of Furtwängler and Vandiver, in a different viewpoint from ours, to try to give a classical cyclotomic proof of Fermat ′ s last theorem (FLT).

Of course the problem is now empty for Fermat ′ s equation, except if we wish to prove FLT by this way; but we will see that for the SFLT equation the result is unknown for p > 3 (but conjecturally similar) and, moreover, leads to infinitely many solutions for p = 3. But as we will show, the case p = 3 is exceptional and we will explain in Section 9 for what reasons.

Unfortunately, we have no deep results to propose, but only some material which may be helpful for those interested in going further.

2. Generalities on the method -The ω-SFLT equation 2.1. Prerequisites on Fermat ′ s last theorem. Let p be a prime number, p > 2. Let a, b, c be pairwise relatively prime nonzero integers, such that a p + b p + c p = 0.

We can find for instance in [START_REF] Gras | Analysis of the classical cyclotomic approach to Fermat ′ s Last Theorem[END_REF][START_REF] Ribenboim | 13 Lectures on Fermat ′ s Last Theorem[END_REF][START_REF] Washington | Introduction to cyclotomic fields[END_REF] the following obvious properties concerning such a speculative counterexample to FLT, where ζ is a primitive pth root of unity, K := Q(ζ), p := (ζ -1)Z [ζ], and N K/Q is the norm map in K/Q (for a detailed proof, a more complete bibliography, and an analysis of the classical cyclotomic approach to FLT, we refer to [START_REF] Gras | Analysis of the classical cyclotomic approach to Fermat ′ s Last Theorem[END_REF]):

(i) We have (where ν ≥ 0 is the p-adic valuation of c) 2 :

a + b = c p 0 or p νp-1 c p 0 with ν ≥ 2, and

N K/Q (a + b ζ) = c p 1 or p c p 1 ,
with -c = c 0 c 1 or p ν c 0 c 1 , and p ∤ c 0 c 1 . By permutation, since p ∤ ab, we have the following analogous relations:

b + c = a p 0 , N K/Q (b + c ζ) = a p 1 , with -a = a 0 a 1 , c + a = b p 0 , N K/Q (c + a ζ) = b p 1 , with -b = b 0 b 1 . (ii) We have: (a + b ζ) Z[ζ] = c p
1 or p c p 1 , with N K/Q (c 1 ) = c 1 Z, where c 1 is an integer ideal of K prime to p, and the analogous relations:

(b + c ζ) Z[ζ] = a p 1 , with N K/Q (a 1 ) = a 1 Z, (c + a ζ) Z[ζ] = b p 1 , with N K/Q (b 1 ) = b 1 Z. (iii)
The positive numbers a 1 , b 1 , c 1 have prime divisors all congruent to 1 modulo p.

Lemma 1. We can choose x, y, z ∈ {a, b, c} in the following manner:

(i) First case of FLT, p > 3:

yx ≡ 0, y + x ≡ 0 (mod p), yz ≡ 0, y + z ≡ 0 (mod p), x + z ≡ 0 (mod p). (ii) First case of FLT, p = 3: yx ≡ 0, y + x ≡ 0 (mod 3), yz ≡ 0, y + z ≡ 0 (mod 3), xz ≡ 0, x + z ≡ 0 (mod 3). (iii) Second case of FLT, p ≥ 3: y ≡ 0 (mod p), yx ≡ 0, y + x ≡ 0 (mod p), yz ≡ 0, y + z ≡ 0 (mod p), xz ≡ 0, x + z ≡ 0 (mod p).

Proof. Consider the differences ab, bc, ca in the first case of FLT. If two of them are divisible by p, we obtain a ≡ b ≡ c ≡ 0 (mod p), then since a + b + c ≡ 0 (mod p), we get 3 a ≡ 0 (mod p) which implies p = 3. So, if p > 3, there exist two differences having the first required property, and called yx, yz.

The second condition is satisfied for any sum and any p ≥ 3. The case p = 3 in the first case of FLT is clear since a ≡ b ≡ c ≡ ±1 (mod 3).

In the second case of FLT, we take y = c ≡ 0 (mod p) so that all the conditions in (iii) are satisfied (we put y = c instead of z = c, to get, for x + y ζ, a p-primary pseudo-unit instead of a number x + y ζ ∈ p).

Note that for p > 3 in the first case, xz may be divisible by p under some circumstances (e.g. under the necessary condition 2 p-1 ≡ 1 (mod p 2 ) since, from x p + y p + z p = 0, we get 2 z p + y p ≡ 0 (mod p 2 )).

2.2. Statement of a stronger conjecture than FLT. We have given in [START_REF] Gras | Analysis of the classical cyclotomic approach to Fermat ′ s Last Theorem[END_REF] a conjecture which implies FLT and which is not covered by Wiles proof; we recall here the statement, which will be called the strong Fermat last theorem (SFLT).

Conjecture 1. Let p be a prime number, p > 2. Then for u, v ∈ Z, with g.c.d. (u, v) = 1, the equation:

(u + v ζ) Z[ζ] = w p
1 or p w p 1 (depending on whether u + v ≡ 0 (mod p) or not), equivalent to:

N K/Q (u + v ζ) = w p 1 or p w p 1 , w 1 = N K/Q (w 1 ) ∈ 1 + p Z,
where w 1 is an ideal of K (necessarily prime to p), has no solution for p > 3 except the trivial ones: u + v ζ = ±1, ±ζ, ±(1 + ζ), and ±(1ζ).

The difference between FLT and SFLT is the following. A solution of Fermat ′ s equation u p + v p + w p = 0 comes from a solution of (u+ v ζ) Z[ζ] = w p 1 or p w p 1 (with the same u, v as above), if and only if there exists w 0 ∈ Z such that u + v = w p 0 or p νp-1 w p 0 since N K/Q (u + v ζ) = w p 1 or p w p 1 , giving w := -w 0 w 1 or -p ν w 0 w 1 for a solution of Fermat ′ s equation.

As for FLT we can speak of the first case of the conjecture (or of the equation) when: u v (u + v) ≡ 0 (mod p) and of the second case when: u v ≡ 0 (mod p) (which implies u or v ≡ 0 (mod p 2 ) as for Fermat ′ s equation); then the case:

u + v ≡ 0 (mod p) will be called the special case for SFLT.

For the first case of SFLT, we have not necessarily uv ≡ 0 (mod p), 3 except for p = 3 since u v (u+v) ≡ 0 (mod 3) implies u ≡ v ≡ ±1 (mod 3), hence uv ≡ 0 (mod 3). See the forthcoming Remark 1 for p = 3.

In the sequel, we will assume that (x, y, z) is a solution of Fermat ′ s equation such that the conditions of Lemma 1 are satisfied (i.e., yx and yz are prime to p when p > 3, and if p | xyz, we suppose that p | y).

3 If u -v ≡ 0 (mod p), then α := uζ+v u+vζ is a pseudo-unit congruent to 1 modulo p; so, from [START_REF] Gras | Analysis of the classical cyclotomic approach to Fermat ′ s Last Theorem[END_REF]Theorem 2.2,Remark 2.3,(ii)], α is locally a pth power giving α ≡ 1 (mod p p+1 ), then (u-v) (ζ-1) u+vζ ≡ 0 (mod p p+1 ), hence u -v ≡ 0 (mod p 2 ). This is valid in the Fermat case if

x -z ≡ 0 (mod p), and gives x -z ≡ 0 (mod p 2 ).

In that case we will have two similar counterexamples to the above conjecture: (x + y ζ) Z[ζ] = z p 1 , (z + y ζ) Z[ζ] = x p 1 (this concerns the first or second case of SFLT). Then it will exist the third counterexample (x + z ζ) Z[ζ] = y p 1 (if p ∤ y) or p y p 1 (if p | y). More precisely, the first case of SFLT implies the first case of FLT, the second or the special case of SFLT implies the second case of FLT, and FLT holds as soon as first and second cases, or first and special cases of SFLT, hold.

Remark 1. Conjecture 1 is false for p = 3 since for ζ = j of order 3 we have the six kind of parametric formulas giving all solutions: u + v j = j h (s + t j) 3 , or j h (1j) (s + t j) 3 , s, t ∈ Z, s + t ≡ 0 (mod 3), g.c.d. (s, t) = 1, and 0 ≤ h < 3. These solutions concern all the cases: -first case (for which uv ≡ 0 (mod 9)):

• (u, v) = (-s 3t 3 + 3s 2 t, -s 3t 3 + 3st 2 ), from u + v j = j 2 (s + t j) 3 ; -second case (for which u or v ≡ 0 (mod 9)):

• (u, v) = (3st 2 -3s 2 t, s 3 + t 3 -3s 2 t), from u + v j = j (s + t j) 3 ; • (u, v) = (s 3 + t 3 -3st 2 , 3s 2 t -3st 2 ), from u + v j = (s + t j) 3 ; -special cases (for which u + v ≡ 0 (mod 3)):

• (u, v) = (s 3 + t 3 + 3s 2 t -6st 2 , -s 3t 3 + 6s 2 t -3st 2 ), from u + v j = (1j) (s + t j) 3 ;

• (u, v) = (s 3 + t 3 -6s 2 t + 3st 2 , 2s 3 + 2t 3 -3s 2 t -3st 2 ), from u + v j = j (1j) (s + t j) 3 ;

• (u, v) = (-2s 3 -2t 3 + 3s 2 t + 3st 2 , -s 3t 3 -3s 2 t + 6st 2 ), from u + v j = j 2 (1j) (s + t j) 3 . The special cases are not similar since for the first solution u + v ≡ 0 (mod 9) and for the others, u + v ≡ ±3(s 3 + t 3 ) ≡ ±3(s + t) ≡ ±3 (mod 9).

Contrary to the case of Fermat ′ s equation, we will not take into account the symmetries of the writing of the solutions (u, v), especially for the second case (this will be important in Section 9) but we will not distinguish (u, v) from (-u, -v).

Thus a proof of SFLT must eliminate, in a natural way, the case p = 3 which is an obstruction for the relevance of the method developed here. We will explain later (Section 9) for what reasons this case is exceptional and finally does not matter, a priori, for the general theory; we are obliged to differ this justification because we need many general material. Meanwhile, for a more comprehensive information, we do not always suppose p > 3 in the development of the first parts of the study.

2.3. The cyclotomic field Q(ζ) and the character ω. We first recall the algebraic context concerning the cyclotomic field K = Q(ζ).

Definition 1. (i) Let g := Gal (K/Q) and let ω be the character of Teichmüller of g (i.e., the character with values in µ p-1 (Q p ) such that for the

s k ∈ g defined by s k (ζ) = ζ k , k ≡ 0 (mod p), ω(s k ) (also denoted ω(k)) is the unique (p -1)th root of unity in Q p , congruent to k modulo p).
(ii) The idempotent corresponding to ω is:

e ω := 1 p-1 s∈g ω -1 (s) s = 1 p-1 p-1 k=1 ω -1 (k) s k ∈ Z p [g].
(iii) We represent e ω in Z[g] modulo p and still denote it e ω (this means that e ω s k ≡ ω(k) e ω ≡ k e ω (mod p Z[g]) and that e ω (1

-e ω ) ∈ p Z[g]). Put e ω := p-1 k=1 u k s k , u k ∈ Z, u k ≡ 1 p-1 ω -1 (k) ≡ k -1 p-1 (mod p). We have ω -1 (s p-k ) = -ω -1 (s k ) since ω(s -1 ) = -1; thus we can suppose that u p-k = -u k for 1 ≤ k ≤ p-1 2 . Then we have e ω = (1 -s -1 ) e ′ ω with e ′ ω = p-1 2 k=1 u k s k . In that case, if an element A of a multiplicative Z[g]-module M is fixed by the complex conjugation s -1 of K, we then have A eω = 1 M (the unit element of M).
(iv) We have ζ eω = ζ for any representative e ω .

Example 1. For p = 3 we have e ω = 1 2 (1s), with s := s -1 . Thus a representative with integer coefficients may be e ω = s -1.

For p = 5, we have for instance

e ω = -1 + 2 s 2 -2 s 3 + s 4 = -1 + 2 s + s 2 -2 s 3 = (1 -s 2 ) (2 s -1), with s := s 2 .
Recall that the unit group E of K is equal to ζ ⊕ E + , where E + is the group of units of the maximal real subfield K + of K (see [START_REF] Washington | Introduction to cyclotomic fields[END_REF]Prop. 1.5]).

Thus if

ε = ζ h ε + , ε + ∈ E + , we get ε eω = ζ h .
2.4. The principles of the method -The fundamental relation. The purpose of this text is to examine some properties of the arithmetic of the fields Q(µ q-1 ), in relation with a nontrivial solution of the SFLT equation:

(u + v ζ) Z[ζ] = w p 1 or p w p 1 , with g.c.d. (u, v) = 1,
for prime numbers q such that q ∤ u v and the order n of v u modulo q is prime to p. The cases where n ≤ 2 (i.e., q | u 2v 2 ) are particular, especially in the case where (u, v) is a part of a solution (x, y, z) of Fermat ′ s equation, and give Furtwängler ′ s theorems [Fur] (see Corollaries 2 and 3 to Lemma 3 for a generalization of Furtwängler ′ s theorems for the SFLT equation, and Remark 3 for the classical case of the FLT equation; see also [Mih] in the context of a Nagell-Ljunggren equation, which is the particular case of the SFLT equation with v = 1).

The cases where n is divisible by any nontrivial power of p give technical complications and are of a different nature. Some complements in this direction are developed in [Que] where similar studies are proposed.

Lemma 2. Let u, v be relatively prime integers, let n ≥ 1, and let q be a prime number. Then the two following properties are equivalent:

(i) q ∤ n and q | Φ n (u, v) := ξ ′ of order n (u ξ ′ -v); (ii) q ∤ u v and v u is of order n modulo q. Proof. Suppose that q | Φ n (u, v) and q ∤ n. Then q ∤ u v since Φ n (u, v) is an homogeneous form u φ(n) ± • • • ± v φ(n) in coprime integers u, v, where φ(n) is the Euler indicator. For ξ of order n fixed, the ideal (q, u ξ -v) of the field Q(µ n ) is a prime ideal dividing q because of the relation q Φ n (u, v) = ξ ′ of order n (u ξ ′ -v); moreover (q, u ξ -v) is of degree 1, unramified in Q(µ n )/Q (since q ∤ n), thus we get q ≡ 1 (mod n) and the fact that v u is of order n modulo q. If q ∤ u v and v u is of order n modulo q, then u n -v n ≡ 0 (mod q). From the relation u n -v n = d | n Φ d (u, v) we deduce that there exists m | n such that q | Φ m (u, v), which implies q | u m -v m ,
hence m = n by definition of the order; since we have ( v u ) q ≡ v u (mod q), it is clear that the order n cannot be divisible by q, proving the lemma.

Corollary 1. Consider the set of numbers of the form Φ n (u, v) when n varies in N\{0}.

Then a prime number q divides one of the numbers Φ n (u, v), n ≡ 0 (mod q), if and only if q ∤ u v. When these conditions (q ∤ n, q | Φ n (u, v)) are satisfied, then n is unique.

If q is an arbitrary given prime number, to have q | Φ n (u, v) with n > 2 and q ∤ n, we must first verify that q ∤ u v(u 2v 2 ) and then compute the order n of v u modulo q which is then a divisor of q -1. 4 Definition 2. Let q = p be a prime number.

(i) Fermat quotients. Let f be the residue degree of q in K/Q and let κ :=

q f -1 p . Since f | p -1, we have κ ≡ 0 (mod p) if and only if q p-1 ≡ 1 (mod p 2 ).
4 It is clear that the trivial solutions u + v ζ = ±1, ±ζ, ±(1 ± ζ) of the SFLT equation are precisely such that u v (u 2 -v 2 ) = 0, in which case such primes q do not exist, which has perhaps a significant meaning.

The integer κ := q p-1 -1 p is called the Fermat quotient of q. We have the

relation κ ≡ p -1 f κ ≡ -1 p log(q) (mod p)
, where log is the p-adic logarithm. (ii) Power residue symbols. Let us recall the definition and properties of the pth power residue symbols

• • in K and M := Q(µ n )K, n | q -1, with values in µ p . Let q be a prime ideal dividing q in Q(µ n ). If α ∈ M is prime to Q | q in M , then let α be the image of α in the residue field Z M /Q ≃ Z K /q K ≃ F q f for q K = Z K ∩ Q (indeed, q totally splits in M/K); since ζ ∈ Z M , the image ζ of ζ is of order p (since ζ ≡ 1 (mod Q)) and we can put α κ = ζ µ , µ ∈ Z/pZ,
∈ Gal(M/Q) one obtains, from α κ ≡ ζ µ (mod Q), τ α κ ≡ τ ζ µ (mod τ Q), thus: τ α τ Q M = τ α Q M = α Q ω(τ ) M = ζ µ ω(τ ) . If α ∈ K, for any q K | q in K we get α q K K = α Q M for any Q | q K in M .
In particular this implies We know, from Stickelberger, that the ω-component of the p-class group of K is trivial (also an application of the reflection theorem, see [START_REF] Gras | Class Field Theory: from theory to practice[END_REF]II.5.4.6.3]); so the ideal class cℓ(w 1 ) eω is trivial.5

ζ q K K = ζ κ (the symbol of ζ does not depend on the choice of q K | q).
Write:

w eω 1 = δ ω Z[ζ], δ ω ∈ K × .
Then we have:

γ ω := (u + v ζ) eω = ε ω δ p ω or (ζ -1) eω ε ω δ p ω ,
where ε ω ∈ E. To simplify, we put π := ζ -1.

Lemma 3. (The fundamental relation). Let (u, v) be a solution of the equation g.c.d. (u, v) = 1 (since the cases where u v (u + v) = 0 are obvious directly, we exclude them).

(u + v ζ)Z[ζ] = w p 1 or p w p 1 ,
(i) In the nonspecial cases (u + v ≡ 0 (mod p)) for p ≥ 3, we have

γ ω = ( u v + ζ) eω = (1 + v u ζ) eω = 1 + v u+v π eω ∈ ζ v u+v • K ×p . (ii) In the special case (u + v ≡ 0 (mod p)) for p > 3, we have γ ω = (1 + v u ζ) eω ∈ ζ 1 2 • K ×p . 6 (iii) In the special case (u + v ≡ 0 (mod 3)) for p = 3, then γ ω = (1 + v u ζ) eω ∈ ζ 1 2 -u+v 3 v • K ×3 . Proof. (i) We have γ ω = ε ω δ p ω with ε ω = ζ h ε + , ε + ∈ E + ,
for some h; then applying again e ω we get

γ eω ω = ε eω ω δ eωp ω ∈ ζ h • K ×p . Since e 2 ω ≡ e ω (mod p Z[g]), any factor A e 2
ω may be written A eω up to a pth power; thus

γ ω ∈ ζ h • K ×p . Since u + v ζ = (u + v) 1 + v u+v π , (u + v ζ) eω ∈ ζ h • K ×p is equivalent to (1 + v u+v π) eω ∈ ζ h • K ×p ; then using [Gr1, Remark 3.4]: 1 + v u+v π eω ≡ 1 + v u+v π (mod π 2 ), we get immediately h ≡ v u+v (mod p). Similarly we have u + v ζ = v ( u v + ζ) = u (1 + v u ζ) for which (u + v ζ) eω = ( u v + ζ) eω = (1 + v u ζ) eω , proving the point (i). (ii) Suppose that u + v ≡ 0 (mod p); put u v = -1 + λ p, then u v + ζ = π + λ p = π α, where α := 1 + λ p π ≡ 1 (mod π p-2 ). Then we get γ ω := (u + v ζ) eω = (1 + v u ζ) eω = ( u v + ζ) eω = π eω α eω . But from the relation (u+v ζ) Z[ζ] = (π) w p 1 , we obtain (u+vζ) eω ∈ π eω ζ h K ×p , for some h, giving α eω ∈ ζ h K ×p hence h ≡ 0 (mod p) in that case since p > 3. Then 1 + v u ζ eω ∈ π eω K ×p . Put α ∼ β in K × if α β -1 ∈ K ×p . From (ζ -1) (ζ + 1) = ζ 2 -1, we get: (ζ -1) eω (ζ + 1) eω = (ζ 2 -1) eω = (ζ -1) s 2 eω ∼ (ζ -1) 2eω , giving (ζ + 1) eω ∼ (ζ -1) eω . But ζ + 1 = ζ 1 2 (ζ 1 2 + ζ -1 2 ) yields (ζ + 1) eω ∼ ζ 1 2 since ζ 1 2 +ζ -1 2 ∈ K + . Then we have the relation (ζ -1) eω ∼ (ζ +1) eω ∼ ζ 1 2
, hence the point (ii) of the lemma.

(iii) If p = 3 in the special case, we get from the computations in the proof of (ii),

γ ω = π eω α eω ∈ π eω ζ h K ×3 , for some h, with α = 1 + 3 λ π and λ = u+v 3 v . Thus α = 1 + (ζ 2 -1) u+v 3 v ≡ 1 -π u+v 3 v (mod π 2 ), giving the congruence h ≡ -u+v 3 v (mod 3) and γ ω ∈ ζ 1 2 -u+v 3 v . K ×3 . 6 Where ζ 1 2 is the unique pth root of unity such that (ζ 1
2 ) 2 = ζ; this convention will be used in a systematic way in the paper.

In the second case of SFLT we have γ ω ∈ K ×p (resp. ζ . K ×p ) if p | v (resp. p | u) since in this case v u+v ≡ 0 (resp. v u+v ≡ 1) (mod p). Note that the condition u + v ≡ 0 (mod 9), when p = 3 in the special case, is not necessarily satisfied for the SFLT equation (use Remark 1) but is true when (u, v) is a part of a solution (u, y, v) or (v, y, u) of Fermat ′ s equation when 3 | y and more generally when p > 3, p | y (see Subsection 2.1, (i)).

Corollary 2. (Generalization of the first theorem of Furtwängler). Let q = p be a prime number such that q | u v for a nontrivial solution of the equation

(u + v ζ)Z[ζ] = w p 1 or p w p 1 , g.c.d. (u, v) = 1. Then, in the nonspecial cases, u κ ≡ 0 (resp. v κ ≡ 0) (mod p) if q | u (resp. q | v
), for p ≥ 3; in the first case we get κ ≡ 0 (mod p). For p > 3 in the special case, then κ ≡ 0 (mod p). For p = 3 in the special case, we get u-2v

3v κ ≡ 0 (resp. v-2u 3v κ ≡ 0) (mod 3) if q | u (resp. q | v); thus if u + v ≡ 0 (mod 9), then κ ≡ 0 (mod 3). If u + v = 3 e, e ≡ 0 (mod 3), then κ ≡ 0 (mod 3) if q | u (resp. q | v) when u ≡ e (resp. v ≡ e) (mod 3). Proof. We have (u+v ζ) eω ∈ ζ h . K ×p with h ≡ v u+v (mod p) in the nonspe- cial cases, p ≥ 3, h ≡ 1 2 (mod p) in the special case, p > 3, and h ≡ 1 2 -u+v 3 v
(mod 3) in the special case, p = 3. Let q K be any prime ideal of K dividing q. We use the pth power residue symbol in K (see Definition 2, (ii)).

Since

u+v ζ ≡ v ζ (resp. u+v ζ ≡ u) (mod q) if q | u (resp. q | v), we get (u+v ζ) eω q K K = ζ κ (resp. 1) if q | u (resp. q | v); but we have ζ h q K K = ζ v u+v κ (resp. ζ 1 2 κ , ζ ( 1 2 -v+u 3 v ) κ
) in the nonspecial cases (resp. in the special case, p > 3, p = 3).

This gives in the nonspecial cases for q | u, v u+v κ ≡ κ (mod p) equivalent to u u+v κ ≡ 0 (mod p), hence u κ ≡ 0 (mod p). If q | v, we get v u+v κ ≡ 0 (mod p) giving v κ ≡ 0 (mod p).

The special case for p > 3 yields to 1 2 κ ≡ κ (resp. 1 2 κ ≡ 0) (mod p) if q | u (resp. q | v), giving κ ≡ 0 (mod p) in any case.

For p = 3 in the special case we get

( 1 2 -u+v 3 v ) κ ≡ κ (resp. ( 1 2 -u+v 3 v ) κ ≡ 0) (mod 3) if q | u (resp. q | v), giving u-2v
3 κ ≡ 0 (resp. v-2u 3 κ ≡ 0) (mod 3). The case u + v ≡ 0 (mod 9) is clear as well as the case u + v ≡ ±3 (mod 9). Corollary 3. (Generalization of the second theorem of Furtwängler). Let q = p be a prime number such that q | u 2v 2 for a nontrivial solution of the equation

(u + v ζ)Z[ζ] = w p 1 or p w p 1 , g.c.d. (u, v) = 1.
Then, in the nonspecial cases, (vu) κ ≡ 0 (mod p) for p ≥ 3; in particular, in the second case, κ ≡ 0 (mod p). In the first case for p = 3, the information is empty since u ≡ v ≡ ±1 (mod 3).

For p > 3 in the special case, the information is empty. For p = 3 in the special case we get u+v 3 v κ ≡ 0 (mod 3). Thus κ ≡ 0 (mod 3) as soon as v + u ≡ 0 (mod 9).

Proof. We have (u+v ζ) eω ∈ ζ h . K ×p with h ≡ v u+v (mod p) in the nonspecial cases, p ≥ 3, h ≡ 1 2 (mod p) in the special case, p > 3, and h ≡ 1 2 -u+v 3 v in the special case, p = 3. Then we have (1

+ v u ζ) eω ζ -1 2 ∈ ζ h . K ×p with h ≡ 1 2 v-u
u+v (mod p) in the nonspecial cases, h ≡ 0 (mod p) in the special case, p > 3, and h ≡ -u+v 3 v (mod 3) in the special case, p = 3.

Let q K be any prime ideal of K dividing q. If q | u 2v 2 , then v u ≡ ±1 (mod q) and we get (1

+ v u ζ) eω ζ -1 2 ≡ (1 ± ζ) eω ζ -1 2 ≡ 1 (mod q K ) since (1 ± ζ) eω ∼ ζ 1 2
(see proof of Lemma 3). Thus we obtain h κ ≡ 0 (mod p) in every case.

The nonspecial cases yield to v-u u+v κ ≡ 0 (mod p), hence κ ≡ 0 (mod p) if uv ≡ 0 (mod p). Thus the case p = 3 is empty since u ≡ v ≡ ±1 (mod 3).

The special case for p > 3 is empty. The special case for p = 3 gives v+u 3 v κ ≡ 0 (mod 3). 2.5. Consequences of Lemma 3. We make the following comments on the fundamental Lemma 3 and its corollaries to introduce the ω-SFLT equation and suitable cyclotomic units.

For arbitrary relatively prime integers u, v, we still have (excluding the obvious cases where u v (u + v) = 0):

γ ω := (u + v ζ) eω = u v + ζ eω = 1 + v u ζ eω = 1 + v v + u π eω
and the various congruences of Lemma 3, γ ω ≡ ζ h (mod π 2 ), with h = v u+v (nonspecial cases, p ≥ 3), h = 1 2 (special case, p > 3), and h = 1 2 -u+v 3 v (special case, p = 3). Then we obtain γ ω ζ -h ≡ 1 (mod π 2 ), which implies easily that γ ω ζ -h is a p-primary number (use [START_REF] Gras | Analysis of the classical cyclotomic approach to Fermat ′ s Last Theorem[END_REF]Lemma 3.15]); but since this number is not in general the pth power of an ideal it may not be a global pth power. 7 So, from class field theory, there exist infinitely many prime ideals q K of K, prime to u v, such that (1 + v u ζ) eω ζ -h is not a local pth power at q K , except if we have a counterexample (u, v) to SFLT in which case such primes do not exist.

The pth power residue symbol of this number is invariant by conjugation of q K since:

(1

+ v u ζ) ζ -h eω κ ≡ ζ ′ (mod q K ) 7 In the case where (u + v ζ) Z[ζ] = w p 1 , Lemma 3 shows that (1 + v u ζ) eω ζ -h ∈ K ×p
; so in this particular case, where (1 + v u ζ) eω ζ -h is a pseudo-unit, local pth power at p, we get a necessary and sufficient condition to get a global pth power.

implies, by conjugation by s k ∈ g:

(1 + v u ζ k ) ζ -k h eω κ ∼ (1 + v u ζ) ζ -h k eω κ ≡ ζ ′k (mod s k (q K ))
equivalent (up to pth powers) to:

(1 + v u ζ) ζ -h eω κ ≡ ζ ′ (mod s k (q K ));
so the symbol only depends on q, the prime number under q K which does not divide u v.

We suppose v u of order n modulo q (which is equivalent to q | Φ n (u, v) and q ∤ n), and we suppose n prime to p.

Let q be a prime ideal above q in Q(µ n ). We have equality of the pth power residue symbols of (1 + v u ζ) eω ζ -h at any q K in K, and of the cyclotomic unit (1

+ ξ ζ) eω ζ -h at Q in Q(µ n )K,
where ξ is a suitable nth root of unity and Q is any prime ideal above q in Q(µ n )K (ξ is characterized by the congruence ξ ≡ v u (mod q) in Q(µ n ), and

q K must be Q ∩ Z[ζ]
). Of course h is a priori unknown (but constant with respect to q) and the local study of (1 + ξ ζ) eω ζ -h is uneffective in general, but we may use some partial informations, as the following ones in the context of FLT.

Let (x, y, z) be a solution of Fermat ′ s equation (first or second cases).

a) Take for instance u := x, v := y (so we are in the nonspecial cases of SFLT) which gives h = y y+x . • If ζ is not a local pth power at q K (which is equivalent to κ ≡ 0 (mod p)), we will consider the pth power residue symbol at Q of the real cyclotomic unit η 1 := (1 + ξ ζ) eω ζ -1 2 (see Definition 4 in Subsection 3.1) which must be that of

ζ h-1 2 = ζ 1 2 y-x y+x .
For FLT we have some informations on the differences like yx, yz, which are prime to p for p > 3 or p = 3 in the second case; in these cases a contradiction to the existence of such a solution of Fermat ′ s equation is that the unit (1 + ξ ζ) eω ζ -1 2 be a local pth power at Q or does not give the "good" symbol.

For p = 3 in the first case, we know that x ≡ y ≡ z ≡ ±1 (mod 3); so a contradiction is that this unit be not a local 3th power at Q.

• If ζ is a local pth power at q K (which is equivalent to κ ≡ 0 (mod p)), a contradiction is that the unit η 1 be not a local pth power at Q. b) In the second case of FLT (p | y) with u = x, v = z (special case of SFLT) we have different but similar reasonings using the value of h -1 2 given by Lemma 3 for p ≥ 3 since x + z ≡ 0 (mod 9) when p = 3. The hope in this attempt is that, the arithmetical properties of the fields Q(µ n ) ⊆ Q(µ q-1 ) being a priori independent of the SFLT problem, they may give valuable indications on the local properties of η 1 = (1+ξ ζ) eω ζ -1 2 , especially in an analytic point of view. In some sense the fields Q(µ q-1 ) will play the role of governing fields for this problem. Indeed, under a solution of the SFLT equation, the residue symbol above q of this unit is, independently of the choice of q, equal to the symbol of a constant power of ζ, which may be absurd.

These cyclotomic fields have been introduced by Vandiver in some papers as [START_REF] Vandiver | A property of cyclotomic integers and its relation to Fermat ′ s Last Theorem[END_REF][START_REF] Vandiver | Summary of results and proofs concerning Fermat ′ s Last Theorem, proceedings of National Academy of Sciences[END_REF][START_REF] Vandiver | Application of the Theory of Relative Cyclic Fields to both Cases of Fermat ′ s Last Theorem[END_REF] to generalize classical results of Kummer and some congruences giving Furtwängler ′ s theorems and Wieferich ′ s criteria; these papers essentially depend on the Stickelberger element S := 1 p p-1 k=1 k s -1 k , related to the generalized Bernoulli numbers and the annihilation of the p-class group of K.

Meanwhile some of the relations of Lemma 3 are considered by Vandiver for other purposes than our ′ s. In Vandiver papers, analytic results (like Čebotarev ′ s theorem) or class field theory are not used, and it seems that no method of contradiction can be deduced from these computations which are essentially local at p, and it has been explained in [START_REF] Gras | Analysis of the classical cyclotomic approach to Fermat ′ s Last Theorem[END_REF] the probable inefficiency of such local studies. Our present work is mainly global and does not concern the arithmetic of K as in the historical researches.

Lemma 4. The equation

(u + v ζ) Z[ζ] = w p 1 or p w p 1 , in integers u, v, with g.c.d. (u, v) = 1, is equivalent to the equation (u + v ζ) eω ∈ ζ h K ×p with h ≡ v
u+v (mod p) in the nonspecial cases, p ≥ 3, h ≡ 1 2 (mod p) in the special case, p > 3, and h ≡ 1 2 -u+v 3 v (mod 3) in the special case, p = 3. Proof. A direction being proved (Lemma 3), let l = p be any prime ideal dividing the ideal (u

+ v ζ) Z[ζ]; the use of the congruences u + v ζ ≡ 0 (mod l) implies that (u + v ζ) Z[ζ] = p δ
ℓ l α ℓ , δ = 0 or 1, α ℓ ≥ 1, for distinct prime numbers ℓ with a single l | ℓ (otherwise, using appropriate conjugations, we get u ≡ v ≡ 0 (mod l)). Moreover, it shows that l is of degree 1 and that g operates transitively on the set of conjugates of l; hence, since p eω = Z[ζ] and since (u + v ζ) eω Z[ζ] = ℓ l α ℓ eω is a pth power by assumption, we get α ℓ ≡ 0 (mod p) for all ℓ.

From any relation (u

′ + v ′ ζ) eω ∈ ζ . K ×p , u ′ , v ′ ∈ Z, we deduce the solution (u, v) := 1 g.c.d. (u ′ ,v ′ ) (u ′ , v ′ )
of the SFLT equation with a unique h. We call the second equation the ω-SFLT equation; the corresponding form of the SFLT conjecture for p > 3 seems reasonable as soon as p is sufficiently large since it enunciates (for u v (u 2v 2 ) = 0) that there exists

a sum p-1 k=1 λ k ζ k , λ k ∈ Q, whose pth power is of the form: (u ζ -v u+v + v ζ u u+v ) eω (resp. of the form (u ζ -1 2 + v ζ 1
2 ) eω ), depending on two coefficients u, v instead of p -1 in general. It will be interesting to have the response at least for p = 5.

So for p > 3 the truth of SFLT would imply FLT; in this paper we concentrate our attention mainly on SFLT, using the simpler ω-SFLT context which does not concern mainly the arithmetic of K, the nerve center of the unsuccessful classical theory.

For a recent critical history on FLT see [Co]. For some complements on these cyclotomic technics, see [START_REF] Helou | Norm residue symbol and cyclotomic units[END_REF][START_REF] Helou | Proof of a conjecture of Terjanian for regular primes[END_REF][START_REF] Terjanian | Sur la loi de réciprocité des puissances ℓ-èmes[END_REF][START_REF] Ribenboim | 13 Lectures on Fermat ′ s Last Theorem[END_REF].

3. Introduction of the governing fields Q(µ q-1 ) 3.1. Furtwängler and Vandiver revisited. Consider the SFLT equation:

(u + v ζ) Z[ζ] = w p 1 or p w p 1 , in integers u, v, with g.c.d. (u, v) = 1, independently of FLT.
Let q be a prime number such that q ∤ u v and such that v u is of order n modulo q (which is equivalent from Lemma 2 to q ∤ n and q | Φ n (u, v)), with n prime to p.

In another point of view, for a given n prime to p, the primes q | Φ n (u, v) are solutions (i.e., v u is of order n modulo q), if and only if q ∤ n. In practice the condition q ∤ n is always satisfied because we are only concerned with large primes q, so that q ≡ 1 (mod n).

Consider the following diagram, where

L := Q(µ n ), M := LK, and G = Gal(M/L) ≃ g (we have L ∩ K = Q): M L = Q(µ n ) Q K = Q(ζ) G g Definition 3.
The prime q ≡ 1 (mod n) being totally split in L/Q, if q is a prime ideal of L over q, there exists a unique primitive nth root of unity ξ such that ξ ≡ v u (mod q). Reciprocally, if ξ is a primitive nth root of unity, there exists a unique prime ideal q of L over q such that ξ ≡ v u (mod q). This ideal is (q, u ξv) and will also be denoted q ξ (it depends on u, v).

We associate with q (for u, v fixed) a pair (ξ, q) where the prime ideal q := q ξ above q and the primitive nth root of unity ξ are characterized by the congruence ξ ≡ v u (mod q) in L. This pair is defined up to Q-conjugation since ξ ≡ v u (mod q ξ ) is equivalent to ξ t ≡ v u (mod q t ξ = q ξ t ), for all t ∈ Gal(L/Q). We obtain an equivalence relation. The class only depends on q for u, v given.

Taking a representative pair, we will fix ξ (for instance ξ = exp(2iπ/n)) which defines q ξ . Since v u modulo q is unknown but well-defined, we must note that, in what follows, the class is uneffective among φ(n) possible classes, and for each n prime to p dividing q -1. This explains that, in some circumstances, we will have to take q ≡ 1 (mod p) since, if not, it is not possible to assert that v u is of order modulo q prime to p.

Definition 4. For the given nth root of unity ξ, n ≡ 0 (mod p), we consider the cyclotomic number of M associated to ξ:

8 η := η(ξ) := (1 + ξ ζ) ζ -1 2 ∈ M,
where ζ 1 2 is the unique pth root of unity such that (ζ

1 2 ) 2 = ζ (so, the exponent 1
2 is seen as a p-adic integer or as an element of (Z/pZ) × ). This is coherent with the context of p-Kummer theory above M .

Then we put:

η 1 := η eω = (1 + ξ ζ) eω ζ -1 2 ∈ M ; we have η 1 ∈ M +
, where M + is the maximal real subfield of M : indeed, if c is the complex conjugation, then:

η c 1 = (1 + ξ -1 ζ -1 ) eω ζ 1 2 = (1 + ξ ζ)ξ -1 ζ -1 ζ 1 2 eω = (1 + ξ ζ) eω ζ -1 2 = η 1 , since ξ eω = 1 and ζ ′eω = ζ ′ for any ζ ′ ∈ µ p .
We note that η 1 is a cyclotomic unit and that η 1 ≡ 1 (mod π Z M ). Starting from ξ ≡ v u (mod q) and extending q to M we obtain:

9 η 1 ≡ 1 + v u ζ eω ζ -1 2 (mod Q | q Q).
We note that these prime ideals Q of M may be written Q ξ since they are above q ξ ; for ξ fixed, they are conjugated by the elements of G.

From Lemma 3, we get 1

+ v u ζ eω = ζ v v+u • δ p ω (nonspecial cases, p ≥ 3) or ζ 1 2 • δ p ω (special case, p > 3) or ζ 1 2 -v+u 3 v • δ 3 ω (special case, p = 3), with δ ω ∈ K × , giving η 1 ≡ ζ 1 2 v-u v+u • δ p ω or δ p ω or ζ 1 2 v+u 3 v • δ 3 ω (mod Q | q Q).
From the congruences on η 1 , Definition 2, Corollaries 2 and 3, we then have obtained in the only context of SFLT the following result which includes the case p = 3: Theorem 1. Let p be a prime number, p ≥ 3. Suppose given the relation

(u + v ζ) Z[ζ] = w p 1 or p w p 1 in coprime integers u, v, where w 1 is an ideal of K := Q(ζ), ζ p = 1, ζ = 1, and p := (ζ -1) Z[ζ]. Let q = p, q ∤ u v, be a prime number such that v u is of order n modulo q, with n prime to p; put η := (1 + ξ ζ) ζ -1 2 , η 1 := η eω , where ξ is a primitive nth root of unity (see Definition 4). Put q := (q, u ξ -v) in L := Q(µ n ).
We get in M := LK:

8 We know that 1 + ξ ζ is a (cyclotomic) unit except if -ξζ is of prime power order, which is the case if and only if ξ = -1 (i.e., n = 2) in which case 1 + ξ ζ is a p-unit.
9 Warning: if for instance v u is of order dp modulo q, with p ∤ d, q is totally split in M/Q and we have the congruence

v u ≡ ξ =: ψ ζ 1 (mod Q), for some Q | q in M , where ψ is of order d and ζ 1 of order p; but in the relation 1+ v u ζ ≡ 1+ξ ζ (mod Q), the root ξ = ψ ζ 1 is not invariant by G so that the congruence 1 + v u ζ eω ≡ (1 + ξ ζ) eω (mod Q) does not exist. From γ := 1 + v u ζ we get s k (γ) := 1+ v u ζ k and if eω = k u k s k we obtain instead γ eω = 1+ v u ζ eω ≡ k (1+ψ ζ 1 ζ k ) u k (mod Q)
, in which the term 1 + ψ is not always a cyclotomic unit (see [Que]).

η 1 Q M = ζ 1 2 v-u v+u κ , ∀Q | q, in the nonspecial cases (p ∤ u + v), p ≥ 3, 10 η 1 Q M = 1, ∀Q | q, in the special case (p | u + v), p > 3, η 1 Q M = ζ 1 2 v+u 3 v κ , ∀Q | q, in the special case, p = 3.
These relations show that

η 1 Q M
only depends on the Fermat quotient of q once u, v are given. Note that the class of the pairs (η t 1 , Q t ), t ∈ Gal(M/K), for any choice of Q | q in M , corresponds canonically to the class of the (ξ t , q t ), since we have the relation Definitions 2, 3, and4). The symbol

η 1 Q t M = η 1 Q M = η t 1 Q t M , where Q t | q t , and η t 1 = (1 + ξ t ζ) eω ζ -1 2 (see
η 1 Q M may be different from η 1 Q t M , t = 1, since there is no local information on 1 + ξ t ζ 1 + ξ ζ
. But as we have seen,

η 1 Q M = η 1 s Q M for any s ∈ G. Remark 2. Since for g.c.d. (u, v) = 1 the equation (u + v ζ) Z[ζ] = w p 1 or p w p 1 is equivalent to the equation N K/Q (u + v ζ) = w p 1 or p w p 1 , we deduce from u + v ζ ≡ u (1 + ξ ζ) (mod Q) for all Q | q, that (the case n = 2 giving u q K K = v q K K = p q K K in the nonspecial cases, = 1 otherwise): N M/L (u+v ζ) ≡ N M/L (u (1+ξ ζ)) ≡ u p-1 1 + ξ p 1 + ξ = u p-1 (1+ξ) tp-1 (mod Q),
for all Q | q, where t p is the Frobenius automorphism of p in L. This gives:

(1 + ξ) tp-1 Q M = u Q M = v Q M resp. = pu Q M = pv Q M ,
in the nonspecial cases (resp. the special case), for all Q | q, with q = q ξ .

From a solution (x, y, z) of Fermat ′ s equation, we get the three relations (see Subsection 2.1):

(x + y ζ) Z[ζ] = z p 1 , (z + y ζ) Z[ζ] = x p 1 , (x + z ζ) Z[ζ] = y p 1 or p y p 1 . For p > 3, the conditions: p ∤ x 2 -y 2 , p ∤ z 2 -y 2 , p ∤ x + z (resp. p ∤ x -z),
in the first (resp. second) case, and the conditions:

p ∤ x 2 -y 2 , p ∤ z 2 -y 2 ,
in the second case, are satisfied by choice of the notations (see Lemma 1). 10 In the first case of SFLT for p > 3 we may have u -v ≡ 0 (mod p) (then u -v ≡ 0 (mod p 2 )) contrary to FLT with v := y and u := x or z. For p = 3, u -v ≡ 0 (mod 9) in the first case.

If n ≤ 2 about the relation (x + y ζ) Z[ζ] = z p 1 or (z + y ζ) Z[ζ] = x p 1 (i.e., q | x 2 -y 2 or q | z 2 -y 2 ), then M = K, Q = q K | q in K. Since η 1 = (1 ± ζ) eω ζ -1 2 ∈ K ×p , we get: ζ 1 2 y-x y+x κ = 1 or ζ 1 2 y-z y+z κ = 1.
Then these two values of n give again the second theorem of Furtwängler [Fur] in the context of FLT for p > 3, that is the fact that when q | x 2y 2 or q | z 2y 2 , then ζ κ = 1, which means κ ≡ 0 (mod p) (see Corollaries 2 and 3 generalizing the FLT situation to SFLT).

We have the same conclusion in the first case of FLT, with the supplementary condition p ∤ xz, if q | x 2z 2 (in the second case of FLT, this does not work for (x, z) since p | x + z).

Remark 3. (Furtwängler ′ s theorems and FLT; see e.g. [START_REF] Gras | Analysis of the classical cyclotomic approach to Fermat ′ s Last Theorem[END_REF]Appendix] or [START_REF] Ribenboim | 13 Lectures on Fermat ′ s Last Theorem[END_REF]IX,3]). Let (x, y, z) be a solution of Fermat ′ s equation for p > 3, under the conditions of Lemma 1.

(i) Recall that the first theorem of Furtwängler giving Wieferich criteria is that for any prime number q = p, if q | x or z (or y in the first case), then κ ≡ 0 (mod p).

Of course, if q | x + y or z + y (or x + z in the first case), then from Subsection 2.1, (i) with obvious notations, q | z 0 or x 0 (or y 0 in the first case), giving κ ≡ 0 (mod p) (from the first theorem of Furtwängler) what we can call the first part of the second theorem of Furtwängler, the second part being that if q | xy or zy (or xz in the second case), then κ ≡ 0 (mod p).

(ii) If q | x or z (or y in the first case) when q ≡ 1 (mod p), then from Subsection 2.1, (iii), q | x 0 or z 0 (or y 0 in the first case). Then we deduce that q p | y + z = x p 0 or x + y = z p 0 (or x + z = y p 0 in the first case). This means, since q ∤ y z or x y (or x z in the first case), that y z or y x (or x z in the first case) is of order 2 modulo q, giving again the first part of the second theorem of Furtwängler and κ ≡ 0 (mod p).

The two results are not independent in the case q ≡ 1 (mod p). For some other remarks on Furtwängler ′ s theorems, see [Que].

(iii) So, if we choose q ≡ 1 (mod p) such that κ ≡ 0 (mod p), this implies that q ∤ xyz in the first case of FLT, and q ∤ xz in the second case of FLT. Thus, under these assumptions on q, the hypothesis q ∤ xyz (in the first case) or q ∤ xz (in the second case) are useless for the development of our method and give effective tests in practice.

It remains the case q | y in the second case (p | y). When q ≡ 1 (mod p), q | y 0 , then q | x + z; we obtain that q ∤ x z and q | x + z but we cannot conclude, except that the root ξ ′′ associated to x z is -1. To eliminate the case q | y in the second case we must suppose q large enough, which is not effective.

(iv) In any case of FLT we have the following result (see [START_REF] Ribenboim | 13 Lectures on Fermat ′ s Last Theorem[END_REF]IV.3]). If q = p divides y and does not divide x + z then q ≡ 1 (mod p 2 ). Indeed, since q ∤ x + z = y p 0 or p ν p-1 y p 0 , we have q | y 1 and q = 1 + d p. Suppose that p ∤ d; since y + x = z p 0 and y + z = x p 0 , we see that x and z are pth powers modulo q and that x d ≡ z d ≡ 1 (mod q) giving x dz d ≡ 0 with x p + z p ≡ 0 (mod q). Since d is even this may be written x d ≡ (-z) d and x p ≡ (-z) p (mod q) with g.c.d. (d, p) = 1 which yields to x ≡ -z (mod q) (absurd). So q ≡ 1 (mod p 2 ).

This result is valid by cyclic permutation of x, y, z, only in the first case of FLT since p (in p ν p-1 y p 0 ) may not be a pth power modulo q. If we suppose that (x, y, z) (with the choices of Lemma 1) is a solution of Fermat ′ s equation, we obtain, from Theorem 1 and the fact that in the second case for p = 3, x + z ≡ 0 (mod 9) (special case (u, v) = (x, z) with u + v ≡ 0 (mod 9)):

Corollary 4. Suppose that the prime q = p is given such that q ∤ xyz and such that y

x , y z , x z are of orders n, n ′ , n ′′ (modulo q) prime to p. Let ξ, ξ ′ , ξ ′′ in Q(µ q-1 ), of orders n, n ′ , n ′′ , and let q, q ′ , q ′′ dividing q in

L = Q(µ n ), L ′ = Q(µ n ′ ), L ′′ = Q(µ n ′′ ), built from y
x , y z , x z ; then consider the corresponding cyclotomic units η 1 , η ′ 1 , η ′′ 1 . We then have: (i) First case of FLT for p > 3:

η 1 Q M = ζ 1 2 y-x y+x κ , η ′ 1 Q ′ M ′ = ζ 1 2 y-z y+z κ , η ′′ 1 Q ′′ M ′′ = ζ 1 2 x-z x+z κ ,
with yx ≡ 0 and yz ≡ 0 (mod p). 11 (ii) First case of FLT for p = 3:

η 1 Q M = η ′ 1 Q ′ M ′ = η ′′ 1 Q ′′ M ′′ = 1. (iii) Second case of FLT for p ≥ 3 (y ≡ x + z ≡ 0 (mod p)): η 1 Q M = ζ -1 2 κ , η ′ 1 Q ′ M ′ = ζ -1 2 κ , η ′′ 1 Q ′′ M ′′ = 1.
Remark 4. (a) Suppose that we are in the first case of FLT for p > 3; let q = p be a prime number such that κ ≡ 0 (mod p), and let n and n ′ be the orders of y x and y z modulo q; we suppose that p ∤ n n ′ (we have n, n ′ > 2 from the second theorem of Furwängler, and from Remark 3, (i), on the first theorem of Furtwängler, we know that q ∤ xyz).

If we find, with independent reasons, that at least one of the symbols

η 1 Q M or η ′ 1 Q ′ M ′ is trivial,
this is absurd since by definition xy ≡ 0 and zy ≡ 0 (mod p) under a solution of Fermat ′ s equation (cf. (i)).

11 Recall that for p > 3 we have no information on x -z modulo p in the first case, so that we cannot consider the third symbol in some reasonings using the above property.

The reasoning on the third symbol does not work since xz can be divisible by p.

(b) For p = 3 in the first case, all the right members are trivial under a solution of the first case of FLT and the above reasoning is different but a contradiction arises as soon as an independent fact implies the nontriviality of one of these symbols (cf. (ii)).

(c) In the second case for p ≥ 3, when κ ≡ 0 (mod p), we know that q ∤ xz. Since p ∤ n n ′ , we deduce that p ∤ n ′′ . The symbol

η ′′ 1 Q ′′ M ′′ is trivial under a solution of Fermat ′ s equation (cf. (iii)
) and a contradiction arises if not.

To have a similar reasoning as for the first case with the two other nontrivial symbols associated to ξ and ξ ′ , we need the condition q ∤ y, so that we must suppose q large enough (in practice, to get a contradiction, we need the existence of infinitely many q such that at least one of the symbols

η 1 Q M , η ′ 1 Q ′ M ′ is trivial). (d) If κ ≡ 0 (mod p)
, in any case all the symbols are trivial under a solution of Fermat ′ s equation. So a contradiction supposes that for infinitely many such q we get, independently, nontrivial symbols.

(e) We can use the above remarks to give the following reciprocal statements, for p > 3 to simplify; we suppose that any solution (x, y, z) of Fermat ′ s equation satisfies the conventions of Lemma 1.

Let ξ be a primitive nth root of unity with n ≡ 0 (mod p), let η 1 := (1+ξ ζ) eω ζ -1 2 (Definition 4), and let q ≡ 1 (mod n). Consider an arbitrary ideal q | q in L := Q(µ n ) and any prime ideal Q | q in M := LK.

We suppose given integers u, v with g.c.d. (u, v) = 1, such that q ∤ u v and v u ≡ ξ (mod q).

• If κ ≡ 0 (mod p) and

η 1 Q M
= 1, we then have:

If u + v ≡ 0 (mod p), (u, v) cannot be a part of a solution (x, y, z) = (u, v, z), (v, u, z), (x, v, u), or (x, u, v) of Fermat ′ s equation.
• If κ ≡ 0 (mod p) and

η 1 Q M
= 1, we then have:

If u+v ≡ 0 (mod p), (u, v) cannot be a part of a solution (x, y, z) = (u, y, v)
or (v, y, u) of the second case of Fermat ′ s equation.

• If κ ≡ 0 (mod p) and

η 1 Q M
= 1, we then have:

The pair (u, v) cannot be a part of a solution (x, y, z) of any case of Fermat ′ s equation.

Proposition 1. Let (x, y, z) be a solution of Fermat ′ s equation; then let q ∤ xyz be such that y x , y z , x z are of orders n, n ′ , n ′′ (modulo q) prime to p, and let L := Q(µ d ) where q =: 1 + d p r , r ≥ 0, p ∤ d.

Let ξ, ξ ′ , ξ ′′ , of orders n, n ′ , n ′′ , be such that ξ ≡ y x (mod q ξ ), ξ ′ ≡ y z (mod q ξ ′ ), ξ ′′ ≡ x z (mod q ξ ′′ ) in L, L ′ , L ′′ , respectively. Then there exist a prime ideal q | q of L and t ′ , t ′′ ∈ Gal( L/Q) such that the following congruences hold:

(i) ξ ′t ′ ≡ -ξ ξ+1 (mod q), (ii) ξ ′′t ′′ ≡ -1 ξ+1 (mod q). Proof. Since L, L ′ , L ′′ are subfields of L, there exist prime ideals q 0 , q ′ 0 , q ′′ 0 of L dividing q ξ , q ξ ′ , q ξ ′′ , respectively, such that:

ξ ≡ y x (mod q 0 ), ξ ′ ≡ y z (mod q ′ 0 ), ξ ′′ ≡ x z
(mod q ′′ 0 ). The ideals q ′ 0 and q ′′ 0 are some conjugates of q 0 and there exist t ′ , t ′′ ∈ Gal( L/Q) such that ξ ≡ y x , ξ ′t ′ ≡ y z , ξ ′′t ′′ ≡ x z (mod q 0 ). From x p + y p + z p = 0 we get x y p + z y p = -1 giving:

ξ -p + (ξ ′t ′ ) -p ≡ -1 (mod q 0 ).
Since p ∤ d, we can use the inverse of the Frobenius automorphism t p of p in L/Q, which gives easily the relation (i) (for q := t -1 p ( q 0 )). From the obvious relation ξ ′′t ′′ ξ ′-t ′ ξ ≡ 1 (mod q 0 ), which implies the equality ξ ′′t ′′ ξ ′-t ′ ξ = 1, we obtain the point (ii) since ξ = -1.12 Corollary 5. Let m := l.c.m. (n ′ , n ′′ ); then we have φ(m) > log(q) log(3) . Proof. We have ξ ′′t ′′ + ξ ′t ′ + 1 ≡ 0 (mod q); then m) , giving the result. Same results for m ′ := l.c.m. (n, n ′′ ) and m ′′ := l.c.m. (n, n ′ ).

ξ ′′t ′′ + ξ ′t ′ + 1 ∈ Q(µ m ) by definition of m, and N Q(µm)/Q (ξ ′′t ′′ + ξ ′t ′ + 1) = q N , N ≥ 1. Since N Q(µm)/Q (ξ ′′t ′′ + ξ ′t ′ + 1) < 3 φ(m) , we get N < 1 q 3 φ(
Corollary 6. We can choose the representative pairs (ξ, q), (ξ ′ , q ′ ), (ξ ′′ , q ′′ ) such that ξ ′ ≡ -ξ ξ+1 and ξ ′′ ≡ -1 ξ+1 (mod q) for a suitable q | q in L. In such a way, we have ξ ′′ = ξ -1 ξ ′ .

3.2. Case of an odd character χ = ω. We suppose that χ is an odd character of g distinct from ω; then χ = ω k , k odd, k ≡ 1 (mod p -1), which excludes the case p = 3.

As for the case k = 1, we can represent modulo p the corresponding idempotent by an element in Z

[g] of the form e χ = (1 -s -1 ) e ′ χ , e ′ χ ∈ Z[g] (see Subsection 2.3).
We suppose that the χ-component of the p-class group of K is trivial; for this, a necessary and sufficient condition is that the Bernoulli number B p-k be prime to p (see e.g. [Gr1, Section 2] for more details).

So, for any relation of the form (u + v ζ) Z[ζ] = w p

1 or p w p 1 where g.c.d. (u, v) = 1, we get immediately:

(u + v ζ) eχ = δ p χ , δ χ ∈ K × , since any χ-unit of K is a pth power for χ odd distinct from ω (moreover in the special case (1 -ζ) eχ is a χ-unit).
It is clear that Lemma 4 is valid for the character χ and that the two equations are equivalent.

The relation (u + v ζ) eχ = δ p χ may be considered as the χ-SFLT equation associated to SFLT under the triviality of the χ-class group.

As in the previous subsection, let q = p be a prime number such that q ∤ u v and v u is of order n modulo q, with n prime to p (see Lemma 2). Then let ξ of order n and q :

= q ξ | q in L = Q(µ n ), characterized by the relation ξ ≡ v u (mod q). From η = (1 + ξ ζ) ζ -1 2 , put η k := η eχ ∈ M , where M := LK; then η k = (1 + ξ ζ) eχ , since ζ eχ = 1. Thus η k ∈ M + and η k ∈ K ×p if n ≤ 2.
We deduce the congruence in M :

η k ≡ 1 + v u ζ eχ = δ p χ (mod Q | q Q).
We then have the relation

η k Q M = 1, for all Q | q, so
that, in this situation, a contradiction to the existence of a nontrivial solution of the SFLT equation is that this symbol be nontrivial for some q.

Here the value of κ does not enter.

From Kummer duality, the extension M ( p √ η k )/M is splitted, by means of a p-cyclic extension, over the extension LK χ * , where χ * = ω 1-k and K χ * is the subfield of K fixed by the kernel of χ * ; this field

K χ * is real. Of course, LK χ * = L if and only if K χ * = Q, i.e., χ = ω.
This criterion may be used for any odd character χ = ω such that the χ-component of the p-class group of K is trivial, which may have some interest. In some sense, it is similar to the case κ ≡ 0 (mod p) of the preceding case χ = ω, the symbols being trivial independently of q.

But unfortunately, the corresponding extensions M ( p √ η k )/L are metabelian (nonabelian) extensions and do not define intrinsic arithmetic properties of the field L as with the use of the single character ω to which we will return to study its properties.

3.3. Computation of the F p -dimension of a group of units. Since η 1 is considered in (E M /E p M ) eω , it is necessary to precise the F p -dimension of this group. The computation is the same for any odd character χ (this may be useful for Subsection 3.2).

Proposition 2. Let M = LK, where L = Q(µ n ), n > 2, p ∤ n. Let E M be the group of units of M and let χ = ω k be an odd character of g.

Then the F p -dimension of (E M /E p M . µ p ) eχ is equal to 1 2 [L : Q] = 1 2 φ(n). Proof. Put Γ := Gal(M/Q) = G ⊕ H where G := Gal(M/L)
and where H := Gal(M/K). Let Γ = G ⊕ H be the group of irreducible characters of Γ; for any ψ ∈ Γ, let ε ψ be the idempotent:

ε ψ := 1 | Γ | σ∈Γ ψ -1 (σ) σ ∈ C p [Γ]. If ψ = ω i . θ, ω i ∈ G, 1 ≤ i ≤ p -1, θ ∈ H, then ε ψ = ε ω i • ε θ .
From the Dirichlet-Herbrand theorem on units (see e.g. [START_REF] Gras | Class Field Theory: from theory to practice[END_REF]I.3.7]) we know that the representation C p ⊕(C p ⊗ Z E M ) is given by the representation of permutation:

C p [Γ] 1 2 (1 + c) = ψ even C p [Γ] ε ψ .
Then, since the character χ is odd,

(C p ⊕ (C p ⊗ Z E M )) εχ = C p ⊗ Z E M εχ is the representation ψ even C p [Γ] ε ψ • ε χ . Put ψ = ω i . θ; then ε ψ = ε ω i . ε θ and ε ψ . ε χ = 0 except if i = k.
The sum is over ψ = χ θ with θ odd since ψ must be even. Then:

C p ⊗ Z E M εχ ≃ θ∈ H, odd C p [Γ] ε χ. θ .
We deduce that the

C p -dimension of C p ⊗ Z E M εχ is 1 2 [L : Q].
Hence the proposition follows since ε χ ≡ e χ (mod p Z p [g]).

In particular, the

F p -dimension of (E M /E p M • µ p ) eω is equal to 1 2 [L : Q].
Thus the subgroup of (E M /E p M • µ p ) eω generated by the images of the units

t η 1 , t ∈ Gal(M/K)/ t -1 , is of F p -dimension less or equal to 1 2 [L : Q] = 1 2 φ(n).
4. Study of the cyclotomic units η 1 and the extensions F ξ

In this Section we use some classical elements of Kummer theory and of decomposition of a Kummer extension over a subfield.

η := (1 + ξ ζ) ζ -1
2 , where ξ is a primitive nth root of unity with p ∤ n, and the real cyclotomic unit η 1 := η eω defined in Definition 4. We exclude the cases n = 1 and n = 2 seen above for which η 1 ∈ K ×p .

For n > 2, L := Q(ξ) is an imaginary cyclotomic field, hence we can consider the biquadratic extension M/L + K + , where M := L K; then M + is the subfield of M of relative degree 2, distinct from LK + and from KL + .

Let f be the residue degree of q in K/Q. We note that the residue degree of q in M + /Q is equal to f . Since η 1 is a unit, the extension M ( p √ η 1 )/M is p-ramified (i.e., unramified outside p). Put π = ζ -1; π is still an uniformizing parameter at p in M (indeed, p is not ramified in M/K). We have:

η ≡ 1 + ξ + 1 2 (ξ -1) π (mod π 2 ),
giving by the usual computation modulo π 2 :

η 1 := η eω ≡ 1 + 1 2 ξ -1 ξ + 1 π (mod π 2 ); since n > 2, ξ -1 ξ + 1
is a local unit at p, showing that η 1 is not p-primary; thus in particular, the extension M ( p √ η 1 )/M is cyclic of degree p.

Kummer theory shows that the conductor of M ( p √ η 1 )/M is p p extended to M (see [START_REF] Gras | Class Field Theory: from theory to practice[END_REF]II.1.6.3]). In some sense, M ( p √ η 1 )/M is maximally wildly p-ramified and has the same conductor as

M ( p √ ζ)/M .
Moreover, this extension does not depend on the choice of ζ since we have:

(

1 + ξ ζ k ) (ζ k ) -1 2 eω = (1 + ξ ζ) ζ -1 2 s k eω ,
with s k e ω ≡ k e ω (mod p Z[g]) for any k prime to p, giving the same radical. This extension, as extension of L, only depends on ξ of order n. The family (F ξ ′ ) ξ ′ of order n is canonical.

Since η 1 is real, η 1 = (1 + ξ -1 ζ -1 ) eω ζ 1 2 which defines the same extension as (1 + ξ -1 ζ) eω ζ -1
2 as we have seen at the end of Subsection 4.1. Then we get F ξ = F ξ -1 . In the cases n ≤ 2, we have L = Q, η 1 ∈ K ×p , and

F ±1 = Q.
It is easy to see that for any t ∈ Gal(L/Q) we have the relation F ξ t = tF ξ , where by abuse of notation tF ξ means t ′ F ξ for any Q-automorphism t ′ of F ξ such that t ′ L = t. Indeed, we have in the same way, t ′ ( p √ η 1 ) = p t (η 1 ) (up to a pth root of unity) where t (η

1 ) = (1 + ξ t ζ) eω ζ -1 2 .
13 Suppose now that we have chosen a prime number q such that q ≡ 1 (mod n), p ∤ n, and let q be a fixed prime ideal above q in L; later, we will have q = q ξ when ξ is associated to the usual integers u, v, but in this subsection q is arbitrary.

13 We use the same notations for the elements of the Galois groups Gal(M/K) and Gal(L/Q), then G = Gal(M/L) and g = Gal(K/Q) and similarly for Gal(M ( p √ η 1 )/M ) and Gal(F ξ /L).

Consider the symbol

η 1 Q M (which is independent of the choice of Q | q
in M ); this symbol is equal to 1 if and only if the image of η 1 in the residue field Z M /Q is a pth power, thus if and only if Q splits in M ( p √ η 1 )/M (Hensel ′ s Lemma) which is equivalent to the splitting of q in F ξ /L.

Let H L be the maximal abelian p-ramified p-extension of L; it contains all the extensions F ξ ′ , ξ ′ of order n, the cyclotomic Z p -extension L ∞ = LQ ∞ of L which is abelian over Q, and 1 2 [L : Q] other independent Z p -extensions of L. This extension H L will be studied in more details in Section 5.

Since q totally splits in L/Q, the decomposition field of q in L ∞ /Q is L e = L Q e , where Q e ⊂ Q ∞ is the stage of degree p e over Q where q f = 1 + p e+1 d, e ≥ 0, p ∤ d; note that e = 0 is equivalent to κ ≡ 0 (mod p).

Study of the extensions

H L /L and F n /L
In this section we recall some class field theory results concerning the abelian p-ramification over L.

5.1. Class field theory and p-ramification. Let H L be the maximal abelian p-ramified p-extension of L := Q(µ n ) in the case n > 2, p ∤ n (so that L is an imaginary cyclotomic field of even degree) and let H L [p] ⊆ H L be the maximal p-elementary p-ramified extension of L.

We consider its Galois group as a vector space over F p .

Its dimension is given by the following Šafarevič formula (see e.g. [START_REF] Gras | Class Field Theory: from theory to practice[END_REF]II.5.4.1, (ii)]):

dim Fp (Gal(H L [p]/L)) = dim Fp (V L /L ×p ) + 1 2 [L : Q] + 1,
where V L is the group of pseudo-units of L which are local pth powers at each place dividing p in L.

Lemma 5. The conductor of H L [p]/L divides (p 2 ) as ideal of L.

Proof. From Hensel ′ s Lemma, since p > 2 is not ramified in L/Q (p ∤ n), the modulus (p 2 ) is sufficient for any α ∈ L × , α ≡ 1 (mod p 2 ), to be locally a pth power at each place dividing p in L.

Thus H L [p]
is contained in the ray class field L(p 2 ) and this yields:

Gal(H L [p]/L) ≃ I/I p R,
where I is the group of fractional ideals of L prime to p and R is the ray group modulo p 2 , i.e., (α) ∈ I, α ≡ 1 (mod p 2 ) .

5.2. The subextension F n . Let t -1 be the element of order 2 of the group Gal(M/L + K) and s -1 ∈ G be the element of order 2 of Gal(M/K + L) (the complex conjugation is c = s -1 t -1 as generator of Gal(M/M + )).

Since we have the relations

η c 1 = η 1 , η s -1 1 = η eω • s -1 = η -1 1 , giving the relation η t -1 1 = η -1 1 , we deduce that: Gal(M ( p √ η 1 )/L + K) ≃ Gal(F ξ /L + ) ≃ D 2p ,
the diedral group of order 2p. 14

In other words, Gal(L/L + ) acts on Gal(

F ξ /L) by σ t -1 := t ′ -1 • σ • t ′ -1 = σ -1 for all σ ∈ Gal(F ξ /L) and any extension t ′ -1 of t -1 in Gal(F ξ /L +
). It will be necessary to consider the compositum of all the extensions M ( p √ η 1 ) when ξ (of order n) varies. Indeed, in the situation of a nontrivial solution (u, v) of the SFLT equation, for any n > 2, p ∤ n, the root ξ such that ξ ≡ v u (mod q), for q = q ξ , is uneffective and the properties of the symbols

η 1 Q M , Q | q in M
, for the pairs (η 1 , Q), can be studied in this extension.

Let F n be the compositum of the corresponding extensions F ξ ′ , ξ ′ of order n, so that F n is also the compositum of the

F ξ t , t ∈ Gal(M/K), ξ fixed; since t -1 η 1 = η -1
1 , we can consider the tη 1 with t modulo t -1 (this is coherent with the relation F ξ = F ξ -1 ).

We have the equality F n M = M p t η 1 t mod <t -1 > .

Then as above Gal(L/L + ) acts on Gal(F n /L) by σ t -1 = σ -1 for all σ ∈ Gal(F n /L).

Lemma 6. The Galois closure of F ξ over Q is F n which is linearly disjoint from L ∞ /L.
Proof. Over the field K, the Galois closure of M ( p √ η 1 ) is given by the

radical t η 1 t mod <t -1 > with t η 1 = (1 + ξ t ζ) eω ζ -1 2 , t ∈ Gal(M/K), giving the first part of the lemma. The relation L 1 ⊆ F n is equivalent to: M ( p ζ ) ⊆ M p t η 1 t mod <t -1 > ,
then to the existence of a relation of the form

t mod <t -1 > (t η 1 ) λt = ζ δ p , λ t ∈ Z, δ ∈ M × ; but
since the left member is real, the use of complex conjugation implies ζ 2 ∈ M ×p , which is absurd.

14 Let A := Gal(M/L + ) = G ⊕ t -1 . Let χ 1 be the character of A defined by χ 1 (s) = 1
for all s ∈ G and χ 1 (t -1 ) = -1. Put χ = ω χ 1 ; is is easy to see that χ is the character of the radical η 1 M ×p /M ×p as A-module, since η 1 = η eω and η

t -1 1 = η -1 1 . From Kummer duality, the character of Gal(M ( p √ η 1 )/M ) is χ * := ω χ -1 = χ 1 proving that Gal(M ( p √ η 1 )/L + ) ≃ G × Gal(F ξ /L + ), with Gal(F ξ /L + ) ≃ D 2p . We also have Gal(M ( p √ η 1 )/M + ) ≃ D 2p .
Remark 5. The F p -dimension of the above radical depends on the study of the relation

t mod <t -1 >
(t η 1 ) λt ∈ M ×p ; this yields to (see Subsection 4.1):

t mod <t -1 > 1 + 1 2 ξ t -1 ξ t + 1 π λteω ≡ 1 + t mod <t -1 > λ t 1 2 ξ t -1 ξ t + 1 π (mod π 2 ).
Thus if the numbers

ξ t -1 ξ t + 1
, t mod t -1 , are linearly independent modulo p,

we get the dimension 1 2 [L : Q] and dim Fp (Gal(F n /L)) = 1 2 [L : Q] = 1 2 φ(n).
Since η 1 is a cyclotomic unit of M , the classical study of the whole group of cyclotomic units of M (of finite index in E M ) may give the exact F pdimension of the radical (see Washington book,Chap. 8); but this study depends, in a complicate manner, on the Galois group of M/Q and the law of decomposition of the prime divisors of n in this extension.

Canonical decomposition of

C L = C + L ⊕ C - L , with C + L := C 1 2 (1+t -1 ) L , C - L := C 1 2 (1-t -1 ) L .
We denote by

H - L [p] the subfield of H L [p] fixed by C + L and by H + L [p] the subfield of H L [p] fixed by C - L . We then have F n ⊆ H - L [p]
and the diagram:

H L [p] H + L [p] L H - L [p] C - L C + L Lemma 7. Put V L := V L /L ×p (see Subsection 5.1) and V L = V + L ⊕ V - L as above. Then V + L ≃ V L + /(L + ) ×p giving: dim Fp (C + L ) = dim Fp (V + L ) + 1; dim Fp (C - L ) = dim Fp (V - L ) + 1 2 [L : Q]. Proof. Since p = 2, we have C + L ≃ Gal(H L + [p]/L + ) for which the Šafarevič formula is dim Fp (C + L ) = dim Fp (V + L ) + 1, proving the lemma.
By this way, the case where dim Fp (Gal(

F n /L)) = 1 2 [L : Q] is compatible with the F p -dimension of C - L since when the invariant C - L is minimal (which is equivalent to dim Fp (V - L ) = 0) then F n = H - L [p] as soon as the tη 1 , t mod < t -1 >, are independent in M × /M ×p .
Note that the group of pseudo-units Y L := α ∈ L × , (α) = a p is elucidated by the following obvious exact sequence: For L + we get the analogous exact sequence:

1 -→ E L /E p L ---→ Y L ---→ p Cℓ L -→ 1,
1 -→ E L + /E p L + ---→ Y L + ---→ p Cℓ L + -→ 1. We have, with usual notations ±, the relations (E L /E p L ) + ≃ E L + /E p L + and (E L /E p L ) -= 1, so that Y - L ≃ p Cℓ - L and V - L ⊆ Y - L
only depends on the minus part of the p-class group of L and is often trivial.

The group V + L ≃ V L + ⊆ Y L + depends on the p-class group of L + (in general trivial) and more essentially on the units locally pth power at p in the group of units

E L + of L + which is of Z-rank 1 2 [L : Q]-1; but ε ∈ E L + is a local pth power at each place dividing p if and only if ε p δ -1 ≡ 1 (mod p 2 ), where δ | 1 2 φ(n)
is the residue degree of p in L + , which is also very rare, giving often a trivial V + L . Remark 6. Suppose that the group V L is trivial. 15

Then dim Fp (C + L ) = 1 and dim Fp (C - L ) = 1 2 [L : Q]. In this case H L is the compositum of the Z p -extensions of L which is of the form H + L H - L where H + L = L ∞ is the cyclotomic Z p -extension of L and H - L the compositum of 1 2 [L : Q] indepen- dent relative Z p -extensions of L (i.e., which are pro-diedral over L + ). Then H L [p]
is the compositum of the first stages of these Z p -extensions, the extension

H + L [p] is L 1 , and H - L [p]
M may be the Kummer extension defined by the radical generated by the tη 1 as soon as its F p -dimension is 1 2 [L : Q]. See Subsection 3.3 about these questions of dimensions. 5.4. Conclusion. We have established, from Corollary 4 and Remark 4 (Subsection 3.1), that, under a solution of Fermat ′ s equation (p > 3), for infinitely many particular prime numbers q in the case κ ≡ 0 (mod p), there exist privilegiate pairs (F ξ , q ξ ), (F ξ ′ , q ξ ′ ) for the first case (resp. (F ξ , q ξ ), (F ξ ′ , q ξ ′ ), (F ξ ′′ , q ξ ′′ ) for the second case), defined up to conjugation, with p-cyclic p-ramified extensions F ξ /L, F ξ ′ /L ′ , F ξ ′′ /L ′′ and prime ideals q ξ , q ξ ′ , q ξ ′′ , for which q ξ , q ξ ′ are inert for the first case (resp. q ξ , q ξ ′ are inert and q ξ ′′ splits, for the second case) in the corresponding extensions

F ξ /L, F ξ ′ /L ′ , F ξ ′′ /L ′′ .
In the case κ ≡ 0 (mod p), for all the above pairs, the ideals split in the corresponding extensions. This intricacy may be in contradiction, for mostly primes q, since the arithmetical properties of the governing fields Q(µ q-1 ) are independent of the Fermat problem; more precisely, a general philosophy is that the decomposition groups of prime ideals in Galois extensions do not fulfill any other laws than standard ones, and may be analyzed in a statistical point of view (see Section 7 for a direct study of these aspects).

About this, we will explain in Section 9 that the case p = 3 is precisely an exceptional counterexample to the above claim, since some constraints do exist; but we will show that these constraints are not in contradiction with statistical considerations because of the structure of the infinite set of solutions.

One may object that F ξ comes from the radical:

(1 + ξ ζ) eω ζ -1 2 M ×p
over M , which is directly associated to a problem of SFLT type, and in a standard algebraic point of view the above circumstances on the laws of decomposition may be equivalent to a contradiction to SFLT. Thus it will be necessary to obtain some analytic or geometrical informations on the splitting of q in H L [p]/L (especially in the canonical family (F ξ ′ /L) ξ ′ of order n ) so as to prove that the above particularities do not exist.

A sufficient condition proving Fermat ′ s last theorem

In this section we study a sufficient condition for FLT, which only involves congruential properties of prime ideals over q in Q(µ q-1 ).

6.1. Main result. We suppose that p > 3 and that the primes q considered are such that f > 1 and κ := q f -1 p ≡ 0 (mod p); we will then use Remark 3 using Furtwängler ′ s theorems. Thus any divisor n of q -1 is prime to p.

From a nontrivial solution (u, v) of the SFLT equation, for which v u is of order n > 2 modulo q ∤ u v, we consider the pair (ξ, q ξ ), defined up to Q-conjugation in L := Q(µ n ) (see Definition 3).

The integer n and the pair are uneffective since if we fix an ideal q | q in L, the root ξ such that q = q ξ is unknown, or if we fix a primitive nth root ξ, then the ideal q | q such that q = q ξ is unknown.

Let Q ξ be any prime ideal of M := LK above q ξ . Then the pair (η 1 , Q ξ ), where

η 1 = (1 + ξ ζ) eω ζ -1 2 ∈ M +
, is also unknown in the same manner. So if we ensure that, for instance for q 0 fixed arbitrarily in L, for

Q 0 | q 0 in M , η 1 tQ 0 M
= 1 for all t ∈ Gal(M/K)/ t -1 , then in particular for the "good" value of the pair (η 1 , Q) (i.e., such that Q | q ξ ), we get:

η 1 Q M = ζ 1 2 v-u v+u κ = 1,
giving vu ≡ 0 (mod p) which is absurd in the case of a solution (x, y, z) of Fermat ′ s equation by choice of the difference v -u = ±(y -x) or ±(y -z) (see Corollary 4, (i)).

The problem is to know if there exist such prime numbers q with F n in the splitting field of q in H L [p]/L, for all n | q -1, n > 2. If so, this will prove FLT (in the first case we know that q ∤ xyz and a single q is sufficient; for the second case where we know that q ∤ xz, it is necessary to have infinitely many such primes q to be certain that q ∤ y).

For the proof of SFLT, we must suppose uv ≡ 0 (mod p) in the first case.

If F n is in the splitting field of q, then this does not depend on the choice of q | q in L, which is a convenient simplification. In other words, q totally splits in F n /Q.

Since

F n ⊆ H - L [p],
a sufficient condition to have the total splitting of q in F n is that the Frobenius ϕ of q in H L [p]/L be an element of C + L , which is equivalent to ϕ t -1 = ϕ, hence to ϕ t -1 -1 = 1. Note that ϕ is of order p since its restriction to L 1 is of order p by assumption.

The image of ϕ ∈ C L by the isomorphism Gal(H L [p]/L) ≃ I/I p R of class field theory, is given by the class of q in I/I p R; thus the condition ϕ t -1 -1 = 1 is equivalent to q t -1 -1 ∈ I p R, i.e.,

q t -1 -1 = a p (α), α ≡ 1 (mod p 2 ),
for an ideal a of L. We must realize this for any divisor n > 2 of q -1.

For n := q -1, L := Q(µ q-1 ), we suppose that the above condition q t -1 -1 = a p ( α), α ≡ 1 (mod p 2 ), is satisfied (for q | q in L/Q).

Then let n | q -1, n > 2; since L = Q(µ n ) is imaginary, L + is fixed by the restriction t -1 of t -1 to L, and taking the norm N L/L we get:

N L/L ( q t -1 -1 ) = N L/L ( a) p N L/L ( α).
Since q is totally split in L, we have by definition N L/L ( q) = q for some q | q in L, and the above relation is of the form q t -1 -1 = a p (α), with α ≡ 1 (mod p 2 ), as expected (this coherent choice of the ideals q is possible since the required condition of splitting at each stage is independent of the choice of the ideal). So the whole condition for our purpose is given by the single condition for n = q -1, L = Q(µ q-1 ).

We have obtained the following criterion, where c is the complex conjugation:

Theorem 2. Let p be a prime number, p > 3. If there exists at least a prime number q, q ≡ 1 (mod p), q p-1 ≡ 1 (mod p 2 ), such that for a prime ideal q | q in Q(µ q-1 ), we have q 1-c = a p (α) for an ideal a and an element α of Q(µ q-1 ) with α ≡ 1 (mod p 2 ), 16 then the first case of FLT (or the 16 Since the multiplicative groups of the residue fields of L at p are of order prime to p, in any writing a p (α) we can suppose α = 1 + p β, β p-integer of L. The condition q 1-c = a p (α), α ≡ 1 (mod p 2 ), is equivalent to q 1-c = a p (1 + p β), where β ≡ β + (mod p), for a p-integer β + of L + ; indeed, this last condition implies q 2(1-c) = a (1-c)p (1 + p β) 1-c where (1 + p β) 1-c ≡ 1 + p (1 -c) β ≡ 1 (mod p 2 ), which gives the result thanks to a Bézout relation between 2 and p.

The condition q 1-c = a p (α), α ≡ 1 (mod p 2 ), is also equivalent to q = b 1+c a ′p (α ′ ), α ′ ≡ 1 (mod p 2 ); indeed, a direction being trivial, from q 1-c = a p (α) we get q 2 = q 1+c a p (α).

first case of SFLT under the supplementary condition uv ≡ 0 (mod p)) holds for p. The second case of FLT (or of SFLT) holds as soon as there exist infinitely many such primes q.

From the Čebotarev ′ s theorem, there exist infinitely many prime ideals l of Q(µ q-1 ) such that their Frobenii ϕ l lie in C + Q(µ q-1 ) (which is at least of dimension 1); the problem is to be sure that there is no obstruction to the fact that it is sometimes possible for l = q | q.

It is clear that such a set of prime numbers q would be of Dirichlet density 0, as for the set of prime numbers q, such that the ring Z[µ q-1 ] contains a principal ideal of norm q, a result proved by Lenstra in [START_REF] Lenstra | Rational functions invariant under a finite abelian group[END_REF]Cor. 7.6].

Theorem 2 may be of empty use due to an excessive condition on the primes q. So we intend, in the forthcoming subsection, to try to give a weaker form of this result (see Conjecture 2).

Proposition 2 shows that the extension

F q-1 ⊆ H - L [p], for L = Q(µ q-1 ), is of degree less or equal to 1 2 [L : Q] = 1 2 φ(q -1). So, if the torsion group V L is trivial, the equality F q-1 = H - L [p]
is possible and the sufficient condition of Theorem 2 is also necessary; thus if there is any hope of success of the method, this condition cannot be improved in practice.

6.2. Some related viewpoints. We will examine if some effective (or numerical) aspects allow us to justify the method of proof of FLT based on Theorem 2 for p > 3.

a) In this first approach, we fix q and q | q in L = Q(µ q-1 ), and we try to find some suitable values of p for which ϕ q ∈ C + Q(µ q-1 ) . Suppose that q k = (α) in L = Q(µ q-1 ) for some k > 0 and suppose that we find d > 0 such that: α d ≡ α + (mod p 2 ), for some prime p such that p ∤ k d, and some α + ∈ L + ; then α d(1-c) ≡ 1 (mod p 2 ) giving a solution of the problem for the prime p (then a posteriori k may be chosen as the order of the ideal class of q and d as a suitable divisor of the order of the multiplicative group of the residue field of L at p).

Of course this relation looks like the general problem of the Fermat quotients of algebraic numbers as studied by Hatada in [Hat]. Considering the work of Hatada and others, a serious conjecture would be that there exist infinitely many solutions p for q fixed.

Since the numerical values of p are out of range of any computer, this conjectural property is not of a practical use, but connect FLT to deep properties of algebraic numbers.

The condition q 1-c = a p (α), α = 1 + p β, is satisfied as soon as the class of q 1-c is of order prime to p, which is a weak condition; it remains to get the stronger condition β ≡ β + (mod p).

Meanwhile, we have found the following example which gives a very partial illustration but shows that there is, a priori, no systematic obstruction for this question.

Example 2. Let q = 5 and p = 463. We then have L = Q(µ 4 ) = Q(i), where i := √ -1, and q = (2 + i). We see that q is totally inert in K (i.e., f = 462) and that p is also inert in L.

We obtain the following numerical informations:

• (5 463-1 -1)/463 ≡ 0 (mod 463) (i.e., κ ≡ 0 (mod p)),

• (2 + i) 463+1 ≡ 43990 (mod 463 2 ).

This implies immediately:

q 1-c = 2 + i 2 -i and q (p+1)(1-c) = 2 + i 2 -i p+1 ≡ 1 (mod p 2 ),
giving the relation q 1-c = a p (α) with a = q c-1 and α ≡ 1 (mod p 2 ).

b) In a slightly different point of view, we must consider that in general, for a solution (u, v) of the SFLT equation, the order n of v u modulo q may be a strict divisor of q -1, even if it is clear directly that n tends to infinity with q (Corollary 5).

Thus we have the following comments.

Let m be a fixed integer, m > 2, p ∤ m.

Put K ′ := Q(µ p 2 ) ⊃ K, L := Q(µ m ), H := H - L [p]
(see Section 5), and H ′ := HK ′ . Then H/Q and K ′ /Q are linearly disjoint.

Let ϕ ∈ Gal(H ′ /H) of order pf , f | p-1. From the Čebotarev ′ s theorem, there exist infinitely many prime numbers q such that, for a suitable Q ′ | q in H ′ , the Frobenius automorphism satisfies the equality

H ′ /Q Q ′ = ϕ.
This implies the following properties: • q ≡ 1 (mod m) (since q splits in L/Q),

• q f ≡ 1 (mod p 2 ) (since q is inert in K ′ /K),
• q is totally split in H/L (since ϕ fixes H).

The condition q 1-c = a p (α), α ≡ 1 (mod p 2 ), is satisfied for any prime ideal q | q in L = Q(µ m ) but not necessarily for q in L = Q(µ q-1 ) (i.e., the Frobenius of q in H L [p] fixes H - L [p], but this is not necessarily true for the Frobenius of q in H L [p] giving possible inertia in H -

L [p]/ L H - L [p]).
The order of v u modulo q is n | q -1 and not necessarily m, and the obvious analogue of Theorem 2 applies only if n | m. In other words, we try to replace the order q -1 probably too big under the condition that the Frobenius of q lies in C + Q(µ q-1 ) by a strict divisor m (depending on q), for infinitely many q for which we hope that the Frobenius of q lies in C + Q(µm) .

Then, under a nontrivial solution (u, v) of the SFLT equation (uv ≡ 0 (mod p)), there is an obstruction to the fact that there exists at least a pair (m, q) (m and q defined as above with a Frobenius in C + Q(µm) ) such that a divisor n of m is the order of v u modulo q. This remark may constitute a way of access to a proof of FLT by means of analytic investigations and we can propose the following independent conjecture.

Conjecture 2. Let p be a prime number, p > 3, and let ρ = v u , with g.c.d. (u, v) = 1, be a rational distinct from 0 and ±1.

There exists a divisor function m : N\{0} -→ N\{0} (i.e., such that m(e) | e for all e ∈ N\{0}) such that there exist infinitely many prime numbers q, with κ ≡ 0 (mod p), totally split in F m(q-1) (see Subsection 5.2), for which the order n of ρ modulo q divides m(q-1).

Since n tends to infinity with q, this means that m(q-1) is unbounded with q. The existence of infinitely many primes q satisfying the conditions of Theorem 2 is equivalent to the conjecture with m(e) = e for all e.

The existence of such a function depends on two phenomena:

(i) The order of magnitude of the primes q discussed above from the Čebotarev ′ s theorem.

(ii) The minimal value of the order modulo q of a given rational ρ.

Example 3. For p = 5, m = 4, we have L = Q(i), and an obvious family of ideals q of L such that q 1-c = (α), α ≡ 1 (mod 25), is given by the following expression:

q = (e + 5a + 25b i) Z[i], e ∈ {1, 2, 3, 4}, a, b ∈ Z,
e, a, b being such that (e + 5a) 2 + (25b) 2 is a prime number q.

The prime numbers q < 10000, q ≡ 1 (mod 5) and q 4 ≡ 1 (mod 25), of the above form, are the following: 769, 1109, 1409, 2069, 2389, 2789, 3229, 3329, 3989, 5309, 5689, 6469, 6709, 7069, 7829, 8329, 8369, 8429. It is clear that such a construction does exist for any p and any m > 2, and the question is the following: p, u, and v being given, is it possible to find in such infinite lists of prime numbers (corresponding to arbitrary values of m), a prime q for which the order of v u modulo q is a divisor of m (which is equivalent to q | u mv m )? Note that for each m, only a finite number of q in the list can be solution.

The existence of one solution (m, q) gives the proof of the first case of FLT for p and the existence of infinitely many solutions (m, q) gives a complete proof of FLT for p.

Explicit formula for the pth power residue symbol

η 1 Q M
. We suppose that q = p is a given prime, and that n | q -1 is such that p ∤ n. Let ξ of order n and let q be a prime ideal of L = Q(µ n ) dividing q.

We consider the real cyclotomic unit η 1 := (1 + ξ ζ) eω ζ -1 2 (see Definition 4). Recall that for n ≤ 2, η 1 ∈ K ×p , so we suppose n > 2.

Let c be the complex conjugation. We suppose in this subsection that the ideal class of q 1-c is in the pth power of the class group of L, which is equivalent to q 1-c = a p (α) for an ideal a of L and an α ∈ L × prime to p. This condition is also equivalent to q = b 1+c a ′ p (α ′ ) for ideals a ′ , b of L and an α ′ ∈ L × .

We can suppose α ≡ 1 (mod p) (see the footnote in Theorem 2), so that we get q 1-c = a p (1 + p β), β p-integer in L. Taking the absolute norm gives

N L/Q (1 + p β) = N L/Q (a) -p which is a rational congruent to 1 modulo p 2 . Thus since N L/Q (1 + p β) ≡ 1 + p Tr L/Q (β) (mod p 2 )
, where Tr L/Q is the absolute trace, we obtain Tr L/Q (β) ≡ 0 (mod p). This remark will be used later.

We note that, as for the context of Theorem 2, if q-1 =: d p r , p ∤ d, and if the condition q 1-c = a p (1+p β) is satisfied for n = d and q | q in L = Q(µ n ), then it is satisfied for any divisor n > 2 of n and the corresponding ideal q = N L/L ( q); we then have β ≡ Tr L/L ( β) (mod p).

In M = LK we have:

η 1 (q) 1-c M = η 1 Q | q Q 1-c M = Q | q η 1 Q 1-c M = η 1 Q 2 p-1 f M ,
where f is the residue degree of q in K/Q; indeed, we have:

η 1 Q 1-c M = η 1 Q M . η 1 Q c -1 M = η 1 Q M . c η 1 Q -1 M = η 1 Q 2 M ,
since η 1 is real, hence the result since the symbol of η 1 does not depend on the choice of Q above q. But

η 1 (q) 1-c M = η 1 (a p ) (α) M = η 1 (α) M
. Then using the general reciprocity law (see e.g. [START_REF] Gras | Class Field Theory: from theory to practice[END_REF]II.7.4.4]) we obtain, since η 1 is a unit:

η 1 α M = η 1 α M α η 1 -1 M = P | p η 1 , α -1 P ,
product over the prime ideals P of M above p; since M/L is totally ramified at p, we will write by abuse η 1 , α p for these Hilbert symbols, where p | p in L, knowing that they are defined on M × × M × with values in µ p . 17 17 Warning: in the literature, two definitions are possible, which give the Hilbert symbol or its inverse; this is the case with the reference [Ko] used below, by comparison with our ′ s (see e.g. [START_REF] Gras | Class Field Theory: from theory to practice[END_REF]II.7.3.1]).

Thus we have obtained:

η 1 Q M = p | p η 1 , α f 2 p .
We refer now to the Brückner-Vostokov explicit formula proved in [START_REF]Number theory II, Algebraic number theory[END_REF]6.2,Th. 2.99] by giving some details for the convenience of the reader, using similar notations.

Consider the uniformizing parameter π := ζ -1 of the completions M P of M at P | p | p. The inertia field is L p . We need the formal series t(x) := 1 -(1 + x) p since 1ζ = -π here, for which t(x) -1 is the Laurent series:

- 1 x p 1 -p c 1 x + • • • + c p-1 x p-1 + p 2 c 1 x + • • • + c p-1 x p-1 2 -• • • ,
where the c i are integers.

We associates with η 1 ≡ 1 + θ π (mod π 2 ), where θ := 1 2 ξ-1 ξ+1 (see Subsection 4.1), and with α = 1 + p β, the series:

F (x) ≡ 1 + θ x (mod (x 2 )), G(x) := 1 + p β (a constant series), such that F (π) ≡ η 1 (mod π 2 ) and G(π) = α.
Recall that log is the p-adic logarithm and dlog the logarithmic derivative; so dlog(G) = 0 giving:

(F, G) = - 1 p 2 • log G p σp(G) • dlog(σ p (F )),
where σ p is the Frobenius automorphism on L p extended to series by putting σ p (x) := x p .

Thus σ p (G) = 1 + p σ p (β), σ p (F ) ≡ 1 + σ p (θ) x p (mod (x 2p )), giving: log

G p σp(G) ≡ -p σ p (β) (mod p 2 ) dlog(σ p (F )) ≡ p σ p (θ) x p-1 (mod (x 2p , p x 2p-1 )),
and finally:

(F, G) ≡ σ p (θ β) x p-1 (mod p x p-1 , x 2p-1 , x 2p p ).
Then the residue of t(x) -1 (F, G) is that of:

- 1 x p σ p (θ β) x p-1 = - 1 x σ p (θ β) (mod p x , x p-1 , x p p ),
hence it is -σ p (θ β) (mod p) since the generator ) in which case c is a multiple of p 2 (see the expression of t(x) -1 ).

To conclude we have to take the absolute local trace (which eliminates the action of the Frobenius):

Tr M P /Qp (-θ β) = (p -1) Tr Lp/Qp (-θ β) ≡ Tr Lp/Qp (θ β) (mod p). Then η 1 , α p = ζ -Tr L p /Qp 1 2 ξ-1
ξ+1 β because of our definition of the Hilbert symbol, and

p η 1 , α p = ζ -p Tr L p /Qp 1 2 ξ-1 ξ+1 β = ζ -Tr L/Q 1 2 ξ-1 ξ+1 β , the
global trace being the sum of the local ones.

We have

1 2 ξ -1 ξ + 1 β = 1 2 - 1 ξ + 1
β, so the final expression of the trace is

-Tr L/Q β ξ + 1
since that of β is zero modulo p.

This yields to

η 1 Q M = p η 1 , α f 2 p = ζ 1 2 f Tr L/Q β ξ+1 .
We have obtained the following explicit formula.

Theorem 3. Let q = p be a prime number, let n | q -1 be such that p ∤ n and n > 2. Let ξ of order n and let q be any prime ideal of L = Q(µ n ) dividing q. We suppose that the ideal class of q 1-c is the pth power of a class of L, which is equivalent to q 1-c = a p (1 + p β) for an ideal a of L and β p-integer in L. 18 Put

η 1 := (1 + ξ ζ) eω ζ -1 2 (see Definition 4).
Then for any Q | q in M := LK,

η 1 Q M = ζ 1 2 f Tr L/Q β ξ+1
, where f is the residue degree of q in K/Q and Tr L/Q the absolute trace in L/Q. This gives again the situation of Theorem 2 when β ≡ β + (mod p), β + ∈ L + , since we then have:

Tr L/Q β ξ + 1 ≡ Tr L + /Q β + ξ + 1 + β + ξ c + 1 ≡ Tr L + /Q (β + ) ≡ 0 (mod p), since Tr L/Q (β) ≡ 0 (mod p).
This theorem confirms the independence, with the SFLT problem, of the class field theory properties of the fields Q(µ n ). Meanwhile, under a nontrivial solution of the SFLT equation, for suitable values of q, n | q -1 and ξ of order n, the quantity Tr L/Q ( β ξ ξ+1 ), where β ξ corresponds to q ξ , is imposed, which yields to infinitely many conditions. But as usual we need to explain how the case p = 3 interferes appropriately with the arithmetic of the fields Q(µ n ) (see Section 9).

Remark 7. Suppose, as in Theorem 3, that q 1-c = a p (1 + p β) for an ideal a of L and β p-integer in L = Q(µ n ), with n | q -1 such that p ∤ n and n > 2. To obtain that q is totally split in F n /L, we study the equivalent condition

η t 1 Q M = 1 for all t ∈ Gal(M/K)/ t -1 ; from the theorem this is equivalent to Tr L/Q β ξ t + 1 ≡ 0 (mod p) for all t ∈ Gal(L/Q)/ t -1 .
This can be written in the following two forms:

τ ∈Gal(L/Q) β τ ξ tτ + 1 ≡ 0 (mod p), for all t ∈ Gal(L/Q)/ t -1 .
18 As we know, this condition is also equivalent to q = b 1+c a ′p (1 + p β ′ ) for ideals a ′ , b of L and β ′ p-integer in L. It is satisfied as soon as the class of q 1-c is of order prime to p. τ ∈Gal(L/Q)

β tτ ξ τ + 1 ≡ 0 (mod p), for all t ∈ Gal(L/Q)/ t -1 .
So we obtain linear systems (with "variables" β τ and 1 ξ τ +1 , respectively), whose matrices have φ(n) columns and 1 2 φ(n) lines, and the rank over F p of the first matrix (less than or equal to 1 2 φ(n)) gives a more precise approach of the required conditions on β; the condition β ≡ β + (mod p) is sufficient (use the second system) but not necessary as soon as the rank of the matrix is less than 1 2 φ(n). Let Z ′ L be the ring of p-integers of L. Then the knowledge of the image of β in Z ′ L /p Z ′ L summarizes all the needed local properties of η 1 at the prime q. Since Z ′ L /p Z ′ L is the product of the residue fields of L at the primes p | p in L, any analytic approach is available.

Example 4. Take p = 5, n = 4, and q = 5 prime congruent to 1 modulo 4. Put q = a 2 + b 2 as usual; then q = (a + i b) and q 4 = (A + i B), with A = a 4 + b 4 -6 a 2 b 2 , B = 4 ab(a 2b 2 ). We then have:

q 1-c = q 5(1-c) A -i B A + i B =: q 5(1-c) 1 + 5 β .
Since A + i B ≡ 1 (mod 5), we get A ≡ 1 and B ≡ 0 (mod 5), and a straightforward computation gives:

β ≡ - 8 i ab(a 2 -b 2 ) 5
and

β i + 1 ≡ - 4 (i + 1) ab(a 2 -b 2 ) 5 (mod 5), which yields to 1 2 Tr L/Q β i + 1 ≡ -1 2 8 ab(a 2 -b 2 ) 5
(mod 5), hence:

η 1 Q M = ζ f ab(a 2 -b 2 ) 5 .
So the symbol is trivial if and only if ab(a 2b 2 ) ≡ 0 (mod 25). We find the values q = 313 (a = 13, b = 12), q = 317 (a = 14, b = 11), . . . For q = 457 (a = 21, b = 4), we have κ ≡ 0 (mod 5). A case with 25 | ab is given by q = 641 (a = 25, b = 4).

The symbol is nontrivial for the values q = 13 (a = 3, b = 2) where

η 1 Q M = ζ 4 , q = 17 (a = 4, b = 1) where η 1 Q M = ζ 3 , . . . 7. Decomposition law of q in H Q(µ q-1 ) /Q(µ q-1 )

and conjectures

In this section we study in full generality the situation that we have encountered in the previous sections.

7.1. Law of ρ-decomposition relative to the family F n . Let p > 2 be a fixed prime number and let ρ = v u , with g.c.d. (u, v) = 1, be a fixed rational distinct from 0 and ±1. We do not suppose any relation of SFLT type between u and v.

For any prime number q = p let f be the residue degree of q in K and put κ := q f -1 p . Note that we have the relation (see Definition 2, (i)):

κ := q p-1 -1 p ≡ p -1 f κ ≡ - 1 p log(q) (mod p).
We consider the infinite set of prime numbers:

Q ρ := q, q ∤ u v (u 2 -v 2
) and the order of ρ modulo q is prime to p .

For q ∈ Q ρ , let n be the order of ρ modulo q (by definition we have

p ∤ n, n > 2); from Lemma 2, q ∈ Q ρ is equivalent to q ∤ n, q | Φ n (u, v), for n > 2, p ∤ n).
We consider the fields K := Q(µ p ), L := Q(µ n ), and M := LK which only depend on q (for ρ fixed).

We associate with q a pair (ξ, q) where the primitive nth root of unity ξ ∈ L and the prime ideal q | q of L are characterized by the congruence ξ ≡ ρ (mod q); thus, q = (q, u ξv) is also denoted q ξ as in the previous sections (see Definition 3). As we know, this pair is defined up to Q-conjugation and we obtain an equivalence relation. The class associated to q is welldefined. Of course, the classes of (ξ 1 , q 1 ) and (ξ 2 , q 2 ), corresponding to different primes q 1 and q 2 , are relative to the fields

L 1 = Q(µ n 1 ), n 1 | q 1 -1, and L 2 = Q(µ n 2 ), n 2 | q 2 -
1, and one of the main problem would be to try to connect the two situations.

From the construction of the extensions F ξ and

F n ⊆ H - L [p]
given in Subsections 4.2 and 5.2 via the real cyclotomic unit:

η 1 := (1 + ξ ζ) eω ζ -1
2 , the pair (F ξ , q ξ ) is defined up to Q-conjugation since (tF ξ , q t ξ ) = (F ξ t , q ξ t ) corresponds to (ξ t , q ξ t ); thus the class of the pair (F ξ , q ξ ) (or similarly of the pair (η 1 , Q ξ | q ξ )) characterizes the class of (ξ, q ξ ) and reciprocally. Recall that

F ξ = F ξ -1 is diedral over L + .
The following lemma is elementary but gives details on the action of Gal(L/Q) on the family of Frobenii:

Lemma 8. Let ϕ ξ := F ξ /L q ξ
be the Frobenius automorphism of the prime

ideal q ξ = (q, u ξ -v) in F ξ /L.
Then ϕ ξ t :=

F ξ t /L q ξ t = ϕ t ξ := t ϕ ξ t -1 for all t ∈ Gal(L/Q). If t = t -1 , then ϕ ξ -1 = ϕ t -1 ξ = ϕ -1 ξ in F ξ ±1 /L. Proof.
From the defining congruence ϕ ξ (α) ≡ α q (mod q ξ ) for all integers α of F ξ , we get easily t ′ ϕ ξ (α) ≡ t ′ (α) q (mod q ξ t ), for any

Q-isomorphism t ′ of F ξ such that t ′ | L = t. Put t ′ (α) =: β ∈ F ξ t ; this yields t ′ ϕ ξ t ′-1 (β) ≡ β q
(mod q ξ t ) for all integers β of F ξ t , proving the lemma by uniqueness of the Frobenius.

The Frobenius of q ξ in F ξ /L is characteristic of the class of (ξ, F ξ ) since we still have (ξ t , ϕ t ξ ) = (ξ t , ϕ ξ t ) by conjugation. This leads to give the following definition.

Definition 5. Let ρ := v u , with g.c.d. (u, v) = 1, be a fixed rational, distinct from 0 and ±1. For n > 2 prime to p, let

K := Q(µ p ), L = Q(µ n ), M = LK,
and for ξ of order n, let F ξ be such that

F ξ M = M p (1 + ξ ζ) eω ζ -1 2 . (i)
For any prime q, q ∤ n, q | Φ n (u, v) (i.e., q ∤ u v and ρ is of order n modulo q), and for q ξ = (q, u ξv) | q, we consider the class of Frobenii:

F ξ t /L q ξ t = F ξ /L q ξ t , t ∈ Gal (L/Q),
that we normalize in the following way:

-if κ ≡ 0 (mod p), we put

F * /L q * ρ,n := F ξ t /L q ξ t p log(q) t∈Gal (L/Q) ; -if κ ≡ 0 (mod p), we put F * /L q * ρ,n := 1. (ii) Call F n the canonical family (F ξ t ) t = (F ξ ′ ) ξ ′ of order n defining F n /L, where F n ⊆ H - L [p] is the compositum of the F ξ t , t ∈ Gal(L/Q)/ < t -1 >. 19 (iii) The symbol F * /L q * ρ,n
is called, by abuse of language, the law of ρ-decomposition of q for the family F n .

This object depending on ρ and n is, for each q, relative to a universal family F n which is independent of any hypothetic nontrivial solution of the SFLT equation.

Let σ be a generator of Gal(F ξ /L); since the Frobenius ϕ ξ in F ξ /L is well defined, it is of the form σ r , r ∈ Z/pZ, so that (when κ ≡ 0 (mod p)) the symbol

F * /L q * ρ,n
represents the family (or class):

σ t r p log(q) t∈Gal(L/Q) = t . σ . t -1 r p log(q) t∈Gal(L/Q) .
Thus the symbol

F * /L q * ρ,n
can take p -1 nontrivial "values" (called the cases of ρ-inertia of q for F n , when r ≡ 0 (mod p)) and a trivial one (the ρ-splitting of q for F n ). The case κ ≡ 0 (mod p) gives the ρ-splitting of q for F n .

Note that the Frobenii ϕ ξ t := Fn/L q ξ t , t ∈ Gal(L/Q), are a priori unknown and must not be confused with

F * /L q * ρ,n
; they are conjugated, of order 1 or p, and the case of order 1 is very rare since it means that q totally splits in F n /Q, i.e.,

Fn/L q ξ t = 1 for all t ∈ Gal(L/Q) (situation of Theorem 2).
19 Remark that the only knowledge of n determines the field L = Q(µn) then the family Fn.

The restriction of ϕ ξ to F ξ gives by definition ϕ ξ . Its restrictions to the other F ξ t are the

F ξ t /L q ξ = F ξ /L q ξ t -1 t .
In the previous sections, in the case κ ≡ 0 (mod p) for p > 3, we have used, as a contradiction for the existence of a solution of Fermat ′ s equation, the splitting of q ξ in F ξ for infinitely many values of q (taking for instance (u, v) = (x, y), (y, x), (z, y), or (y, z)). Same remark for a solution (u, v) of the SFLT equation under the condition uv ≡ 0 (mod p).

This property " q ξ = (q, u ξv) splits in F ξ ", independent of the choice of the representative pair as Lemma 8 shows, will be called by analogy the " ρ-splitting of q ∈ Q ρ for F n ", ρ := v u . It is equivalent to

F * /L q * ρ,n = 1.
Remark 8. In a probabilistic point of view, the ρ-splitting of q ∈ Q ρ for F n has a probability around 1 p , and we can hope a strong incompatibility for analytic reasons since Q ρ is infinite. If we ask that q be totally split in F n , this means that each q | q splits in F ξ = F -ξ (for any fixed ξ) and the probability is around 1 p 1 2 φ(n) which tends to 0 rapidly with q → ∞.

Put:

Q spl ρ := q ∈ Q ρ , q has a ρ-splitting for F n .
With a counterexample (u, v) to SFLT, we have, from a pair (ξ, q ξ ), the following results proved in Theorem 1. Put ρ := v u ; we may have u ≡ 0 (mod p) in which case ρ is not defined modulo p, but is always defined as a rational, so we preserve u and v in the congruences modulo p.

In the nonspecial cases (i.e., v + u ≡ 0 (mod p)):

η 1 Q M = ζ 1 2 v-u v+u κ , for all Q | q ξ , p ≥ 3;
In the special case (i.e., v + u ≡ 0 (mod p)):

η 1 Q M = 1, for all Q | q ξ , p > 3, η 1 Q M = ζ 1 2 v+u 3 v κ , for all Q | q ξ , p = 3.
Recall that for SFLT we cannot exclude the case uv ≡ 0 (mod p) contrary to FLT for (u, v) = (x, y), (y, x), (z, y), or (y, z). This explain that for SFLT (first case and κ ≡ 0 (mod p)) we cannot use, as a general contradiction, the ρ-splitting of q for F n . This does not matter since the existence of a nontrivial solution to SFLT is equivalent to a precise law of ρ-decomposition of q for F n , i.e., a precise value of the symbol

F * /L q * ρ,n
(which can be trivial even if κ ≡ 0 (mod p) when uv ≡ 0 (mod p)).

More precisely, we have the following lemma giving the action of the Frobenius, which determines explicitly the law of ρ-decomposition (the case p = 3 being immediate from Theorem 1, we assume for simplicity p > 3): Lemma 9. We suppose given, for the prime p > 3, a relation of the form

(u + v ζ) Z[ζ] = w p 1 or p w p 1 , with g.c.d. (u, v) = 1.
Let q be a prime number such that q ∤ u v, and such that the order n of ρ := v u modulo q is prime to p. Let Q | q ξ in M , where (ξ, q ξ ) represents the class corresponding to q.

Let M ( p √ η 1 )/M Q be the Frobenius automorphism of Q in M ( p √ η 1 )/M ,
where

η 1 := (1 + ξ ζ) eω ζ -1 2 .
We have:

(i) Nonspecial cases. If v + u ≡ 0 (mod p), then M ( p √ η 1 )/M Q . p √ η 1 = ζ 1 2 v-u v+u κ • p √ η 1 .
(ii) Special case. If v + u ≡ 0 (mod p), then

M ( p √ η 1 )/M Q • p √ η 1 = p √ η 1 . Proof. From the defining congruence p √ η 1 σ ≡ p √ η 1 q f (mod Q), for the Frobenius automorphism σ := M ( p √ η 1 )/M Q , we get: p √ η 1 σ-1 ≡ p √ η 1 q f -1 ≡ η κ 1 ≡ η 1 Q M (mod Q).
Hence the result since

η 1 Q M = ζ 1 2 v-u
v+u κ (resp. 1) in the nonspecial cases (resp. in the special case).

We intend now, in the following theorem, to translate this property into a property of the symbol

F * /L q * ρ,n
(see Definition 5), which will give the main phenomenon about the existence of a nontrivial solution to the SFLT equation (see also Remark 9).

Theorem 4. Let p be a prime number, p > 3. We suppose given a solution of the SFLT equation (u + v ζ) Z[ζ] = w p 1 or p w p 1 , with g.c.d. (u, v) = 1. Let q be a prime number such that q ∤ u v, and such that the order n of ρ := v u modulo q is prime to p and > 2. Then the symbol

F * /Q(µn) q * ρ,n
only depends on ρ and n when q varies in

Q ρ := q, q ∤ u v (u 2 -v 2
) and the order of ρ modulo q is prime to p . In other words, the law of ρ-decomposition of q ∈ Q ρ for F n only depends on ρ and n.

Proof. Let Q | q ξ in M , where (ξ, q ξ ) represents the class corresponding to q. The Frobenius automorphism of q ξ in F ξ /L is given, by restriction, by the relation

F ξ /L q ξ f = M ( p √ η 1 )/M Q | F ξ .
Indeed, in the projection Gal(M ( p √ η 1 )/M ) -→ Gal(F ξ /L), the Frobenius of the prime ideal Q gives the Artin symbol of the norm in M/L of Q, which is q f ξ ; hence the result. If κ ≡ 0 (mod p), using the relation f κ -1 ≡ -κ -1 (mod p) (see Definition 2, (i)) we get from Lemma 9 that

F ξ /L q ξ -κ -1 = M ( p √ η 1 )/M Q κ -1
| F ξ only depends on ρ and n when q varies. This proves the theorem in this case since -κ ≡ 1 p log(q) ≡ 0 (mod p) (see Definition 5). If κ ≡ 0 (mod p), we get

F ξ /L q ξ
= 1 in any case.

Remark 9. We can justify the expression "only depends on ρ and n when q varies in Q ρ " in the following way.

Let

F n := L 1 F n , where L 1 K = M ( p √ ζ )
, and let ϕ ξ :=

F n/L q ξ
; we know that ϕ ξ projects on ϕ ξ in F ξ /L and on ϕ 1 :=

L 1 /L q ξ
in L 1 /L. We treat the case κ ≡ 0 (mod p), i.e., ϕ 1 = 1.

In the same manner as in the proof of the theorem, in the projection Gal(M ( p √ ζ )/M ) -→ Gal(L 1 /L), we obtain that:

L 1 /L q ξ p log(q) = M ( p √ ζ )/M Q κ -1 | L 1
is independent of q because of the equality

M ( p √ ζ )/M Q κ -1 . p √ ζ = ζ . p √ ζ.
Moreover, this is independent of the choice of ξ (of order n) since for all t ∈ Gal(L/Q),

ϕ ξ t = t ϕ ξ t -1 projects, in L 1 /L, on ϕ ξ t | L 1 = t ϕ ξ | L 1 t -1 = t ϕ 1 t -1 = ϕ 1 since Gal(L 1 /Q) is abelian.
Which justifies the normalization and the fact that, in some sense, under the existence of a nontrivial solution to the SFLT equation, the symbol

F * /L q * ρ,n
does not depend on q but only on ρ and n (of course n depends on q, but not in a deep arithmetical manner).

From Theorem 3, when the condition q 1-c ξ = a p (1 + p β ξ ) is satisfied, for an ideal a of L and β ξ p-integer of L, then

η 1 Q M = ζ 1 2 f Tr L/Q β ξ ξ+1
, where Tr L/Q is the absolute trace in L/Q. So with a counterexample to SFLT we must have:

Tr L/Q β ξ ξ + 1 ≡ f -1 v -u v + u κ ≡ v -u v + u log(q) p (mod p) (nonspecial cases, p ≥ 3) or Tr L/Q β ξ ξ + 1 ≡ 0 (mod p) (special case, p > 3).
in the nonspecial cases (resp. in the special case).

(i) In the nonspecial cases we have:

η 1 = (1 + ξ ζ) eω ζ -1 2 -1 2 v-u v+u = (1 + ξ ζ) eω ζ -v v+u ,
which is by construction such that

η 1 Q ξ M
= 1, but the unit η 1 is not anymore real; its definition from η 1 is independent of q under a given solution of the SFLT equation.

(ii) In the special case we obtain

η 1 := η 1 = (1 + ξ ζ) eω ζ -1 2
, which is real and by construction such that

η 1 Q ξ M = 1.
The extension M ( p η 1 )/M is splitted over L by a p-cyclic p-ramified extension F ξ similar to F ξ except that it is not diedral over L + in the nonspecial cases.

We note that the relation

η 1 = η 1 ζ -1 2 v-u
v+u in the nonspecial cases shows that F ξ is a subfield of the compositum F ξ L 1 obtained in an obvious systematic way ( F ξ /L is still of degree p and p-ramified since n > 2). But F ξ is effective only if ρ is known, which is not in general the case in the nonspecial cases of the SFLT problem.

It is clear that

F ξ = F ξ if and only if u 2 -v 2 ≡ 0 (mod p).
We still have t F ξ = F ξ t . We call F n the compositum of the F ξ t , t ∈ Gal(L/Q). Hence

F n L 1 = F n L 1 .
We denote, as in Definition 5, by F n the family ( F ξ ′ ) ξ ′ of order n .

Then under a nontrivial solution of the equation attached to SFLT, we must have the splitting of q ξ in F ξ (i.e., a ρ-splitting for F n ). In other words if we define, as in Definition 5, for κ ≡ 0 (mod p), the symbol:

F * /L q * ρ,n := F ξ t /L q ξ t p log(q) t∈Gal(L/Q) , the analog of Theorem 4 is F * /L q * ρ,n
= 1 for all q ∈ Q ρ , where:

Q ρ := q, q ∤ u v (u 2 -v 2
) and the order of ρ modulo q is prime to p .

A contradiction would be that there exist prime numbers q such that F * /L q * ρ,n = 1 i.e., q ξ is inert in F ξ , which is independent of the representative pair ( F ξ t , q ξ t ) and has a probability very near from p-1 p since p -1 values of the symbol are possible. About the class of pairs ( F ξ t , q ξ t ), when

F * /L q * ρ,n
= 1, we can speak of " ρ-inertia of q for F n ".

In a similar way, in the context of Fermat ′ s equation, we deduce from the units η 1 , η ′ 1 , and η ′′ 1 (see Remark 10), the units, where r := y x ≡ ±1 (mod p):

η 1 := (1 + ξ ζ) eω ζ -r r+1 , η ′ 1 := (1 + ξ ′ ζ) eω ζ r , η ′′ 1 := (1 + ξ ′′ ζ) eω ζ 1 r , if r ≡ 0 (mod p), η ′′ 1 := (1 + ξ ′′ ζ) eω ζ -1 2
, if r ≡ 0 (mod p), giving a trivial pth power residue symbol at Q, Q ′ , and Q ′′ , respectively.

We have the same conclusion as above for the extensions

F ξ /L, F ξ ′ /L ′ , F ξ ′′ /L ′′ defined from M p η 1 M , M ′ p η ′ 1 M ′ , M ′′ p η ′′ 1 M ′′ .
Returning to SFLT with a nontrivial solution (u, v), we put as above:

Q in ρ := q ∈ Q ρ , q has a ρ-inertia for F n . Lemma 10. Suppose p > 3 and κ ≡ 0 (mod p). If u 2 -v 2 ≡ 0 (mod p) then we have Q spl ρ ⊆ Q in ρ . If u 2 -v 2 ≡ 0 (mod p) then Q spl ρ ∩ Q in ρ = ∅.
Proof. We know that F ξ is contained in the compositum L 1 F ξ and is distinct from L 1 since ξ = ±1. Suppose that F ξ is distinct from F ξ ; if q ∈ Q spl ρ , q ξ splits in F ξ /L and the Frobenius of q ξ in L 1 F ξ /L fixes F ξ and since this Frobenius must be nontrivial in L 1 /L (κ ≡ 0 (mod p)) then projects to a nontrivial Frobenius in F ξ /L. When F ξ = F ξ , the result is clear. The lemma comes from the characterization of the equality F ξ = F ξ (i.e., u 2v 2 ≡ 0 (mod p)).

It will be interesting to examine the problem, for any ρ, independently of any equation giving exceptional values of ρ.

The natural conjecture in this direction would be the following, which implies SFLT (we still put K = Q(µ p ), L = Q(µ n ), M = LK, to simplify the notations): Conjecture 3. Let p be a prime number, p > 3, and let ρ = v u , with g.c.d. (u, v) = 1, be a rational distinct from 0 and ±1. Put:

Q ρ := q, q ∤ u v (u 2 -v 2
) and the order of ρ modulo q is prime to p .

For q ∈ Q ρ , let n be the order of ρ modulo q and let F n be the family of the cyclic extensions F ξ of L, for ξ of order n, defined by the relations

F ξ K = M p (1 + ξ ζ) eω ζ -v v+u resp. F ξ K = M p (1 + ξ ζ) eω ζ -1 2 if v + u ≡ 0 (resp. v + u ≡ 0 (mod p). Say that q has a ρ-inertia for F n if F * /L q * ρ,n
= 1, i.e., q ξ := (q, u ξ-v) is inert in F ξ /L (condition independent of the choice of ξ of order n).

Then the set of primes q ∈ Q ρ having a ρ-inertia for F n , is infinite.

The extension F n (depending on ρ contrary to F n ) is a subfield of the maximal p-ramified p-elementary extension H L [p] of L, and the arithmetical properties of H L [p] and of its subextensions are, a priori, independent of any diophantine problem as Fermat ′ s equation.

Recall that to prove the first case of FLT for p, the existence of a single q ∈ Q ρ (ρ = y

x for a solution (x, y, z)) having a ρ-inertia for F n is sufficient, contrary to the second case which needs infinitely many such primes.

In the first case, p ∤ xy (x 2 -y 2 ) (from Lemma 1) and so, if κ ≡ 0 (mod p) then q ∤ xy (x 2y 2 ) from the two theorems of Furtwängler (see Corollaries 2, 3, and Remark 3).

Hence q ∈ Q ρ as soon as κ ≡ 0 (mod p) and q ≡ 1 (mod p). These two conditions on q are effective and the first case of FLT is easier than the second one because it is generally possible to check the conjecture for small values of q. The second case supposes to find q large enough; this shows that the first case is likely a weaker conjecture.

If we examine, for logical reasons, the case p = 3 for SFLT, we know that for any of the six families of solutions (u, v) of the SFLT equation (see Remark 1), we have (supposing κ ≡ 0 (mod 3) and defining η 1 in an analogous way):

(i) η 1 Q ξ M = ζ 1 2 v-u v+u κ = 1, in the first case (i.e., u v (u + v) ≡ 0 (mod 3)), which implies u -v ≡ 0 (mod 3), hence η 1 = η 1 and F ξ = F ξ ; (ii) η 1 Q ξ M = ζ ± 1
2 κ in the second case (i.e., u v ≡ 0 (mod 3)), thus If v + u ≡ 0 (mod 3) and v + u ≡ 0 (mod 9) then, for ρ := v u , we get

η 1 = η 1 ζ ∓ 1 2 and F ξ = F ξ ; (iii) η 1 Q ξ M = ζ
Q spl ρ ⊆ Q in ρ ; if v + u ≡ 0 (mod 9) or u -v ≡ 0 (mod 3) then Q spl ρ ∩ Q in ρ = ∅.
We see that uv ≡ 0 (mod 3) in case (i), u v ≡ 0 (mod 3) in case (ii); for (iii), we verify from Remark 1 that v u ∈ {-1, 2, 5} modulo 9, which gives 1 2 v+u 3 v ∈ {0, 1, 2} modulo 3. So, for q fixed we can find solutions (u i , v i ) giving the same order n of v i u i modulo q and any of the above value of v u modulo 9.

See Section 9 to go thoroughly into the exceptional case p = 3.

7.3. Construction of universal defining polynomials. The group g operates canonically on the field K(Y ) of rational fractions, where Y is an indeterminate. Consider:

F (Y ) := (1 + Y ζ) eω ζ -1 2 ∈ K(Y ).
Then if s = s r is a generator of g we have:

s.F (Y ) := (1+Y ζ s ) ζ -1 2 s eω = (1+Y ζ) ζ -1 2 s eω = (1+Y ζ) ζ -1 2 reω+pΛ ,
since s e ω = r e ω + p Λ for some Λ ∈ Z[g] (see Definition 1, (iii)). Then we obtain:

s.F (Y ) = F (Y ) r • (1 + Y ζ) ζ -1 2 pΛ . Consider the Kummer extension K(Y )( p F (Y ) )/K(Y ); since this ex- tension is abelian over Q(Y ), the K(Y )-automorphism of K(Y )( p F (Y ) ), still denoted s, defined by s • p F (Y ) := ( p F (Y ) ) r • (1 + Y ζ) ζ -1 2 Λ is of order p -1 and it is a classical result that the trace Ψ := p-1 k=1 s k • p F (Y ) defines a primitive element of the subextension cyclic of degree p contained in K(Y )( p F (Y ) )/Q(Y ).
An easy way to find Irr(Ψ, Q(Y )) is to use the Newton formulas from the computations of the traces:

Tr ( p F (Y ) ) i := p-1 k=1 s k • p F (Y ) ) i , i = 1, . . . , p -1. For instance, for p = 3, e ω = s -1, s = s 2 , s e ω = 1 -s = -e ω (thus r = 2, Λ = -e ω ), F (Y ) = (1 + Y j) eω j = ((1 + Y j) j) s-1 ; we have Ψ = (1 + Y j 2 ) j 1 + Y j 1 3 + (1 + Y j) j 2 1 + Y j 2 1 3 , for which we get Ψ 3 = (1 + Y j 2 ) j 1 + Y j + (1 + Y j) j 2
1 + Y j 2 + 3 Ψ, giving the irreducible polynomial:

Irr(Ψ, Q(Y )) = X 3 -3 X + Y 2 -4Y + 1 Y 2 -Y + 1 , or the unitary polynomial X 3 -3 (Y 2 -Y +1) 2 X +(Y 2 -4Y +1)(Y 2 -Y +1)
taking the representative idempotent e ω = s + 2.

Definition 6. The general case of degree p can be written:

Irr(Ψ, Q(Y )) = A p (Y )X p + • • • + A 1 (Y )X + A 0 (Y ), A i (Y ) ∈ Z[Y ],
and will be called a universal polynomial of degree p for the SFLT problem.

For any given nth root of unity ξ, n > 2, the polynomial:

A p (ξ)X p + • • • + A 1 (ξ)X + A 0 (ξ) ∈ L[X], L := Q(µ n ),
is the irreducible polynomial of the primitive element:

ψ := Tr M ( p √ η 1 )/F ξ ( p √ η 1 )
defining the extension F ξ , with the usual notations.

We have the following result where we recall that, for g.c.

d. (u, v) = 1, we have put Φ n (u, v) := ξ ′ of order n (u ξ ′ -v).
Theorem 5. Let p be a prime number, p > 3, and let ρ = v u , with g.c.d. (u, v) = 1, be a fixed rational distinct from 0 and ±1; suppose u-v ≡ 0 (mod p).

(i) Nonspecial cases (u + v ≡ 0 (mod p)). Let n > 2 be prime to p, and let q ∤ n be a prime number such that κ ≡ 0 (mod p) and q | Φ n (u, v).

If the polynomial A p (ρ)X p +• • •+A 1 (ρ)X+A 0 (ρ) is not irreducible modulo q, then (u, v) cannot be a solution of the equation (u + v ζ) Z[ζ] = w p
1 attached to the nonspecial cases of SFLT.

(ii) Special case (v + u ≡ 0 (mod p)). Let n > 2 be prime to p, and let q ∤ n be a prime number such that κ ≡ 0 (mod p) and q

| Φ n (u, v). If the polynomial A p (ρ)X p + • • • + A 1 (ρ)X + A 0 (ρ) is irreducible modulo q, then (u, v) cannot be a solution of the equation (u+v ζ) Z[ζ] = p w p
1 attached to the special case of SFLT.

(iii) Let n > 2 be prime to p, and let q ∤ n be a prime number such that κ ≡ 0 (mod p) and q | Φ n (u, v).

If the polynomial

A p (ρ)X p + • • • + A 1 (ρ)X + A 0 (ρ) is irreducible modulo q, then (u, v) cannot be a solution of the SFLT equation.
Proof. From Lemma 2, q ∤ n and q | Φ n (u, v) is equivalent to q ∤ u v and ρ is of order n modulo q; then ρ ≡ ξ (mod q ξ ), for any choice of the nth root of unity ξ, and in case (i) there exists a root λ ∈ Z modulo q, of the polynomial, such that:

A p (ρ)λ p + • • • + A 1 (ρ)λ + A 0 (ρ) ≡ A p (ξ)λ p + • • • + A 1 (ξ)λ + A 0 (ξ)
≡ 0 (mod q ξ ), since q divides the left member. This means that Irr(ψ, L) has the root λ modulo q ξ and that q ξ splits in F ξ /L (i.e., F * /L q * ρ,n = 1).

If we suppose that (u, v) is a counterexample to SFLT, Theorem 1 in the nonspecial cases gives

η 1 Q M = ζ 1 2 v-u v+u κ = 1 by assumption, equivalent to the inertia of q ξ in F ξ /L (contradiction).
The proofs of cases (ii) and (iii) are similar but inverted (the hypothesis implies

F * /L q * ρ,n = 1 while η 1 Q M
= 1 for a solution in these cases).

In other words, the corresponding conjecture giving a proof of SFLT under the assumption uv ≡ 0 (mod p), which implies the two cases of FLT, is the following. Conjecture 4. Let p be a prime number, p > 3, and let ρ = v u , with g.c.d. (u, v) = 1, be a rational distinct from 0 and ±1; suppose uv ≡ 0 (mod p).

(i) Case u+v ≡ 0 (mod p). There exist infinitely many prime numbers q with κ ≡ 0 (mod p) such that A p (ρ)X p + • • • + A 1 (ρ)X + A 0 (ρ) is not irreducible modulo q and q | Φ n (u, v) for suitable values of n > 2 prime to p.

(ii) Case v+u ≡ 0 (mod p). There exist infinitely many prime numbers q with κ ≡ 0 (mod p) such that A p (ρ)X p +• • •+A 1 (ρ)X +A 0 (ρ) is irreducible modulo q and q | Φ n (u, v) for suitable values of n > 2 prime to p.

(iii) There exist infinitely many prime numbers q with κ ≡ 0 (mod p)

such that A p (ρ)X p + • • • + A 1 (ρ)X + A 0 (ρ) is irreducible modulo q and q | Φ n (u, v) for suitable values of n > 2 prime to p.
Of course, without an independent approach (analytic or geometric), the problem has no longer solution since the polynomial:

A p (ρ)X p + • • • + A 1 (ρ)X + A 0 (ρ)
can be in case (i) that of a primitive element of L 1 , in which case all the primes which split in L 1 /Q are such that κ ≡ 0 (mod p), and in cases (ii) and (iii), the polynomial may be splitted over Q

. Meanwhile the universal polynomial A p (Y )X p + • • • + A 1 (Y )X + A 0 (Y ) has the nontrivial property that for any primitive nth root of unity ξ, n > 2, A p (ξ)X p + • • • + A 1 (ξ)X + A 0 (ξ) is irreducible in Q(µ n )[X] and defines a p-ramified cyclic extension of Q(µ n ).

Normic relations for cyclotomic units

In this section we give a relation between the units η 1 and η ′ 1 associated to the classes of two prime numbers q and q ′ for which the pairs (ξ, q ξ ), (ξ ′ , q ξ ′ ) are such that the order n ′ of ξ ′ divides the order n of ξ, p ∤ n.

Put n = n ′ d. We introduce the following notations:

L = Q(µ n ), L ′ = Q(µ n ′ ), M = L K, M ′ = L ′ K, η 1 = (1 + ξ ζ) eω ζ -1 2 , η ′ 1 = (1 + ξ ′ ζ) eω ζ -1 2 ;
to fix the notations, we suppose that ξ ′ = ξ d .

Since η 1 is a cyclotomic unit, the action of relative norms on this unit is well-known and we now recall the result in our particular context.

Then we have

N(η 1 ) = (η ′ 1 ) Λ ′ , where Λ ′ ≡ d . ℓ∈S (1 -ℓ -1 t ′ ℓ -1 ) (mod p), t ′ ℓ ∈ Gal(M ′ /K) being the Artin automorphism defined by t ′ ℓ (ξ ′ ) := (ξ ′ ) ℓ .
Proof. By induction we can suppose that d is a prime number ℓ.

Let ψ := ξ n ′ which is a primitive ℓth root of unity.

(i) Case ℓ | n ′ . In this case S = ∅, [M : M ′ ] = ℓ, and:

N(1 + ξ ζ) = ℓ-1 λ=0 (1 + ξ 1+λn ′ ζ) = ℓ-1 λ=0 (1 + ξ ψ λ ζ) = 1 + ξ ℓ ζ ℓ = 1 + ξ ′ ζ ℓ = (1 + ξ ′ ζ) s ℓ . Then N(η 1 ) = (1 + ξ ′ ζ) s ℓ eω N(ζ) -1 2 ∼ (1 + ξ ′ ζ) ℓ eω ζ -1 2 ℓ = (η ′ 1 ) ℓ since s ℓ e ω ≡ ℓe ω (mod p).
(ii) Case ℓ ∤ n ′ . In this case S = {ℓ} and:

N(1 + ξ ζ) = ℓ-1 λ=0, λ =λ 0 (1 + ξ 1+λn ′ ζ),
where λ 0 is the unique value modulo ℓ such that 1 + λ 0 n ′ ≡ 0 (mod ℓ), giving from the computation in (i):

N(1 + ξ ζ) = 1 + ξ ℓ ζ ℓ 1 + ξ 1+λ 0 n ′ ζ = (1 + ξ ′ ζ) s ℓ 1 + (ξ ′ ) µ ζ , where 1 + λ 0 n ′ = µ ℓ, so that µ ≡ ℓ -1 (mod n ′ ). Thus: N(1 + ξ ζ) = (1 + ξ ′ ζ) s ℓ 1 + (ξ ′ ) ℓ -1 ζ = (1 + ξ ′ ζ) s ℓ 1 + (ξ ′ ) t ′ ℓ -1 ζ = 1 + ξ ′ ζ 1 + (ξ ′ ) t ′ ℓ -1 (ζ) s -1 ℓ s ℓ = 1 + ξ ′ ζ 1 + (ξ ′ ζ) σ ′ ℓ -1 s ℓ ,
where σ ′ ℓ ∈ Gal(M ′ /Q) is the Artin automorphism defined by σ ′ ℓ (θ) = θ ℓ for any pn ′ th root of unity θ; thus, since σ ′ ℓ = s ℓ t ′ ℓ , this yields:

N(1 + ξ ζ) eω ∼ 1 + ξ ′ ζ 1 + (ξ ′ ζ) σ ′ ℓ -1 ℓ eω = 1 + ξ ′ ζ ℓ (1-σ ′ ℓ -1 ) eω ; from σ ′ ℓ -1 e ω = s -1 ℓ t ′ ℓ -1 e ω ≡ ℓ -1 t ′ ℓ -1 e ω (mod p), we get N(1 + ξ ζ) eω ∼ 1 + ξ ′ ζ ℓ (1-ℓ -1 t ′ ℓ -1 ) eω . Finally, since in this case [M : M ′ ] = ℓ -1 and N(ζ) = ζ ℓ-1 = ζ ℓ (1-ℓ -1 t ′ ℓ -1 ) , we get: N(η 1 ) ∼ (η ′ 1 ) ℓ (1-ℓ -1 t ′ ℓ -1 )
and the proposition follows.

If for instance Λ ′ is invertible modulo p, with inverse Ω ′ , then η ′ 1 ∼ N(η 1 ) Ω ′ and, over L, we can see the abelian extension F ′ n ′ (compositum of the conjugates of the F ′ ξ ′ over L ′ ) as a subfield of F n , in which case, for suitable primes q and q ′ , the properties of the Frobenii studied in this paper can be compared to give strengthened conditions. 9. Analysis of the case p = 3 versus p = 3

In this section we suppose p = 3 and consider the solutions of the equation associated to SFLT (see Remark 1) which are, for p > 3, a logical obstruction to the relevance of general statements similar to Theorem 2 and to the property of ρ-law of decomposition of Theorem 4. We intend to explain why this obstruction actually exists for p = 3 but a priori not for p > 3 when we suppose that the set of nontrivial solutions is nonempty.

The main differences between the cases p = 3 and p > 3, are that there are infinitely many solutions for the case p = 3, contrary to the case p > 3, even if we have not proved this fact (which was known for Fermat ′ s equation before Wiles proof), and that we will exhibit a group of automorphisms, acting on the set of solutions for p = 3, which creates some exceptional relations of compatibility with density theorems. So we conjecture that this fact does not exist for p > 3. 9.1. Analysis of the case p = 3 for the principle of Theorem 2. Recall Theorem 1 in that case, for the choice of a prime q = p. We have η 1 = (1+ξ ζ) eω ζ -1 2 , with ζ = j and e ω = s-1, where ξ ≡ v u (mod q ξ ) is supposed of order n ≡ 0 (mod 3) for a nontrivial solution (u, v), g.c.d. (u, v) = 1, of the SFLT equation.

Recall that if we put ρ := v u (distinct from 0 and ±1) we may have u ≡ 0 (mod 3) in which case ρ is not defined modulo 3, but is always defined as a rational. Put L = Q(µ n ) and M = LK:

(i) First case. Since u v (u+v) ≡ 0 (mod 3), we get u ≡ v ≡ ±1 (mod 3), so that

η 1 Q M = j 1 2 v-u v+u κ = 1 for any Q | q ξ in M .
(ii) Second case. We get

η 1 Q M = j ± 1 2 κ for any Q | q ξ since 3 | u v.
(iii) Special case. Then

η 1 Q M = j 1 2 v+u 3 v κ for any Q | q ξ with 3 | v + u;
we have seen, at the end of Subsection 7.2, that v+u 3 v can take any value modulo 3.

From this, we see that the existence of q totally split in H - L [3]/Q for L = Q(µ q-1 ), or at least L = Q(µ m ) for a large m | q -1, may be in contradiction with the existence of the solutions of the second and special cases when κ ≡ 0 (mod 3), i.e., 3 inert in Q 1 /Q where Q 1 = Q(µ 9 ) + . Definition 7. Consider the field k(Y ), where k is any field of characteristic distinct from 2 and 3, and the automorphism:

T : k(Y ) ---→ k(Y ) . Y -→ 2 Y -1 Y +1
Let F (Y ) := (1 + Y ζ) eω ζ -1 2 ∈ K(Y ) be the formal cyclotomic unit, yet defined in Subsection 7.3.

We intend to prove below various properties of compatibility, of this automorphism, with the method of cyclotomic units developed here. Theorem 6. (i) The automorphism T is of order 6 and we have the following orbit of Y :

T (Y ) = 2 Y -1 Y + 1 ; T 2 (Y ) = Y -1 Y ; T 3 (Y ) = Y -2 2 Y -1 ; T 4 (Y ) = -1 Y -1 ; T 5 (Y ) = -Y -1 Y -2 ; T 6 (Y ) = Y.
(ii) We have for ζ = j of order 3 and for F (Y ) = (1 + Y j) eω j, the following formulas (equalities up to 3th powers in K(Y )):

T (F (Y )) = (1 + Y j) eω ; T 2 (F (Y )) = (1 + Y j) eω j 2 ; T 3 (F (Y )) = F (Y ), summarized by the identity T i (F (Y )) ∼ F (Y ) j 1 2 i , 0 ≤ i < 3.

Proof. We have:

T (F (Y )) = 1+ 2 Y -1 Y + 1 j eω j = (Y +1+
(2 Y -1)j) eω j = (1-j +(2j +1)Y ) eω j;

since 2j + 1 = j (1j), we get finally T (F (Y )) = (1j) eω (1 + Y j) eω j; but (1j) eω = -j 2 , hence the result in this case. The other computations are obtained by induction.

We apply now the automorphism T to the solutions (u, v) in the following way. We put T ( v u ) =: V U where (U, V ) is defined up to the sign. We start for instance from the solution:

(u, v) = (-s 3t 3 + 3s 2 t, -s 3t 3 + 3st 2 ) (see Remark 1) to determine its orbit.

Theorem 7. We obtain the following identities:

T 0 v u = v u = -s 3 -t 3 + 3st 2 -s 3 -t 3 + 3s 2 t , T 1 v u = 2v -u v + u = -s 3 -t 3 -3s 2 t + 6st 2 -2s 3 -2t 3 + 3s 2 t + 3st 2 , T 2 v u = v -u v = 3s 2 t -3st 2 s 3 + t 3 -3st 2 , T 3 v u = v -2u 2v -u = -s 3 -t 3 + 6s 2 t -3st 2 s 3 + t 3 + 3s 2 t -6st 2 , T 4 v u = -u v -u = s 3 + t 3 -3s 2 t 3st 2 -3s 2 t , T 5 v u = -v -u v -2u = 2s 3 + 2t 3 -3s 2 t -3st 2
s 3 + t 3 -6s 2 t + 3st 2 , which leads to the six fundamental families of solutions of the SFLT equation for p = 3.

Remark 11. For q ≡ 1 (mod 3), q = 2, all the orbits in F q ∪ {∞} have six elements (indeed, all the equations of the form Since p > 3 we get ψ ≡ -1 or 1 modulo p, giving the solutions (a, b, c, d) = (1, 0, 0, 1) (i.e., the identity), (a, b, c, d) = (0, 1, 1, 0) (i.e., the inversion). 9.2. Analysis of the case p = 3 for the principle of Theorem 4.

We have now to explain why the phenomenon of ρ-law of decomposition (Theorem 4, i.e., F * /L q * ρ,n independent of q in the sense of Remark 9) is indeed compatible for p = 3 but (conjecturally) not for p > 3.

The following analysis suggests a suitable property of repartition (in the meaning of Čebotarev density theorem) of the values of the Frobenii, due to the infiniteness of the set of solutions of the SFLT equation for p = 3 and the fact that this set is the union of six families (see Remark 1) having complementary properties for these values.

Let q be given such that κ ≡ 0 (mod 3). As usual, for the solutions (u(s, t), v(s, t)) of the SFLT equation, put ρ := v u and call ξ any primitive nth root of unity, where n is the order of ρ modulo q, n supposed prime to 3. Put η 1 := (1 + ξ j) eω j -1 2 = (1 + ξ j) s-1 j -1 2 , then q := (q, u ξv), and denote by Q any prime ideal of M = LK above q.

Of course, in this study n is not constant when the solution (u, v) varies, so that the statistical analysis cannot be done for a fixed field L = Q(µ n ) ⊆ Q(µ q-1 ).

But up to this problem (probably not too tricky since the number of divisors n of q -1 is finite and since it is probably better to work instead in Q(µ q-1 ) for this statistical analysis), we have the following distribution of the possible cases, in a remarkable accordance with the definition of the solutions of the SFLT equation, that we summarize with the diagram of the compositum L 1 F ξ which is very simple for p = 3 (note that in the general case, L 1 F ξ /L contains p 2 -1 p-1 = p + 1 cyclic subextensions of degree p). Indeed, for p = 3 the compositum L 1 F ξ contains L 1 , F ξ , and two other cubic fields, F ′ ξ and its conjugate c F ′ ξ by the complex conjugation c (recall that F ξ /L + is diedral, L 1 /L + abelian, so that L 1 F ξ /L + is Galois). Moreover we will get F ξ among the three extensions distinct from L 1 . We denote by σ a fixed generator of Gal(F ξ /L) and call ϕ ξ the Frobenius of q ξ in F ξ /L. We refer to Theorem 1 giving the symbol η 1 Q M for p = 3, where Q | q = q ξ .

(i) First case (u v (u + v) ≡ 0 (mod 3)) corresponding to the relation u + v j = j 2 (s + t j) 3 . We have

η 1 Q M = j 1 2
v-u v+u κ = 1 since uv ≡ 0 (mod 3), F ξ = F ξ , and the diagram:

L 1 F ξ F ξ = F ξ F ′ ξ cF ′ ξ L 1 L ϕ ξ = 1
in which q is inert in F ′ ξ /L, cF ′ ξ /L, and L 1 /L. (ii) Second case (u v ≡ 0 (mod 3)) corresponding to the two relations u + v j = (s + t j) 3 and u + v j = j (s + t j) 3 . We have

η 1 Q M = j 1 2 v-u v+u κ = j ± 1
2 κ = j or j 2 ; we get F ξ = F ξ , and the two equidistributed diagrams:

L 1 F ξ F ξ F ξ = F ′ ξ cF ′ ξ L 1 L ϕ ξ = σ L 1 F ξ F ξ F ξ = F ′ ξ cF ′ ξ L 1 L ϕ ξ = σ 2
in which q is inert in F ξ /L, cF ′ ξ /L, and L 1 /L. (iii) Special case (u+v ≡ 0 (mod 3)) corresponding to the three relations u + v j = j h (1j) (s + t j) 3 , 0 ≤ h < 3. We have

η 1 Q M = j
1 2 v+u 3 v κ = 1, j, or j 2 , and the three equidistributed diagrams:

L 1 F ξ F ξ = F ξ F ′ ξ cF ′ ξ L 1 L ϕ ξ = 1 L 1 F ξ F ξ F ξ = F ′ ξ cF ′ ξ L 1 L ϕ ξ = σ L 1 F ξ F ξ F ξ = F ′ ξ cF ′ ξ L 1 L ϕ ξ = σ 2
in which the decomposition of q assembles all the above cases.

This suggests that the infiniteness of the solutions of the SFLT equation and their particular repartition into six families, is a necessary fact for the compatibility with Čebotarev ′ s density theorem. 9.3. Numerical data for the case p = 3. We give some numerical experimentations, using [PARI], in the case p = 3, to highlight the above properties of this case.

We refer to Remark 1 for the six expressions of the solutions of the SFLT equation for p = 3; when we speak of " a solution (u, v) ", we consider one of the six families (u(s, t), v(s, t)) with parameters s and t. (i) The set of primes q ≡ -1 (mod 3), q ∤ n, with κ ≡ 0 (mod 3), dividing at least one of the integers Φ n (u, v), for a solution (u(s, t), v(s, t)) of the SFLT equation, is infinite when s, t vary in Z with g.c.d. (s, t) = 1, s + t ≡ 0 (mod 3).

Then there exist numbers of the form Φ n (u, v) divisible by primes q as large as we need.

More precisely, the prime number q is solution if and only if for an e ∈ Z, of order n modulo q, the polynomial X 3 -3 e -1 X 2 -3 (1e -1 ) X + 1 splits in F q [X]; the parameters (s, t) giving the solutions (u, v) such that v u ≡ e (mod q), are given via the three roots θ k ∈ F q of the polynomial, by the relation st θ k ≡ 0 (mod q), s, t ∈ Z satisfying the above conditions, k = 1, 2, 3.

(ii) The condition q | Φ n (u, v) (q ∤ n), for a solution (u, v) of the SFLT equation, is equivalent to the ρ-splitting of q for F n i.e., it is equivalent to F * /L q * ρ,n = 1 for ρ := v u .

Proof. Let ξ of order n and let L = Q(µ n ). Since g.c.d. (s, t) = 1, this yields immediately g.c.d. (u, v) = 1 for any solution, thus u and v are not divisible by any prime q dividing Φ n (u, v) homogeneous of the form u φ(n) ±• • •±v φ(n) in coprime integers u and v.

From Lemma 2, q ∤ n and q | Φ n (u, v) is equivalent to the fact that v u is of order n modulo q, hence it is equivalent to the congruence u ξv ≡ 0 (mod q), for a suitable and unique prime ideal q | q in L; then q = (q, u ξ-v), which depends on (u, v) for ξ fixed, is one of the φ(n) prime ideals above q in L; in the previous sections it was denoted q ξ for given u, v.

We will prove that the condition q | Φ n (u, v) (q ∤ n), for a solution of the SFLT equation, can be tested independently of the choice of the solution among the six possibilities, in the following sense.

To summarize it is not excluded that the two main principles of approach of the SFLT problem that we have developped in this paper may be successful for p > 3.

However, it should be noted that results like Theorem 2 are sufficient diophantine conditions, probably too strong, and that it would be better to return to the principle of laws of ρ-decomposition of infinitely many primes q for the canonical families F n (see Subsection 7.1, Theorem 4, and Conjecture 3); this aspect can be approached from an analytic point of view with the aim to show that such a constraint is impossible for p > 3.

In this direction, an interesting fact would be that the case p = 3 would have, in some sense, a reciprocal statement, namely that the infiniteness of the set of solutions of the SFLT equation and their particular repartition into six families, is in fact necessary for the Čebotarev ′ s theorem. Thus for p > 3, in the same spirit as the case p = 3, the set of nontrivial solutions (if nonempty) would be necessarily infinite with some structural properties in order to be compatible with the above principle, which seems clearly impossible for geometrical reasons (Theorem 8 for instance).

  We return to the context of the SFLT equation (u+v ζ)Z[ζ] = w p 1 or p w p 1 , g.c.d. (u, v) = 1 (the second case corresponds to p | u v and the special case to p | u + v). Put γ ω := (u + v ζ) eω for a solution (u, v) of the above SFLT equation. In the context of a solution (x, y, z) of Fermat ′ s equation we will have analogous computations with γ ω := (x + y ζ) eω and the relation (x + y ζ)Z[ζ] = z p 1 , and also with γ ′ ω := (z+y ζ) eω and the relation (z+y ζ)Z[ζ] = x p 1 . Then in the first case, γ ′′ ω := (x + z ζ) eω with the relation (x + z ζ)Z[ζ] = y p 1 can be used knowing that zx may be divisible by p. In the second case, γ ′′ ω is of p-valuation 1 since (x + z ζ)Z[ζ] = p y p 1 and this gives a special case of the equation associated to SFLT.

4. 1 .

 1 The cyclotomic unit η 1 . We consider, independently of any relation of the form (u+ v ζ) Z[ζ] = w p1 or p w p 1 , the cyclotomic number:

4. 2 .

 2 The abelian extension F ξ /L. By definition of the character ω, whose reflect is ω * = χ 0 (the unit character), the extension M ( p √ η 1 )/M is splitted over L by means of a cyclic p-ramified extension F ξ , of degree p over L = Q(µ n ) (i.e., F ξ M = M ( p √ η 1 )).

  Gal(H L [p]/L). Consider the Galois group C L := Gal(H L [p]/L) as a module over F p [Gal(L/L + )]. Write:

  where Cℓ L is the p-class group of L, p Cℓ L the subgroup of Cℓ L of classes killed by p, E L the group of units of L, and Y L := Y L /L ×p .

  residue only with a term of the form c x p+1 of t(x) -1 (to give c p x

  in the special case (i.e., u + v ≡ 0 (mod 3)) forwhich η 1 = η 1 ζ -1 2 v+u 3 v and F ξ = F ξ if and only if v + u ≡ 0 (mod 9).

Proposition 3 .

 3 Denote by N the relative norm N M/M ′ and by S the set of distinct prime numbers dividing d and not dividing n ′ .

  ay + b cy + d = y, deduced from the Since (cY + d + (aY + b) ζ) eω and (1 + Y ζ λ ) eω each have p -1 distinct 22 roots of orders of multiplicity u k , with 1 ≤ u k ≤ p -1, it is clear that (cY + d + (aY + b) ζ) eω and (1 + Y ζ λ ) eω each are prime to A and B, then have the same roots; hence there exists ℓ ≡ 0 (mod p) such that d+b ζ c+a ζ = ζ -λℓ =: ζ ψ , ψ ≡ 0 (mod p). Then ζ ψ+1 a + ζ ψ cζ bd = 0.

22

  Indeed, for the rootsy k := -d+b ζ k c+a ζ k , 1 ≤ k ≤ p -1, y k = y k ′ is equivalent to the relation (ad -bc) (ζ k -ζ k ′ ) = 0,hence the result. The other case is trivial.

Proposition 4 .

 4 Let n > 2 be an integer not divisible by 3 and for any integers u, v with g.c.d.(u, v) = 1, let Φ n (u, v) := ξ ′ of order n (u ξ ′v).

Since here the class of w 1 is of order 1 or p, the choice of any representative eω does not affect this property.

The case ξ = -1 means y + x = z p 0 ≡ 0 (mod q), i.e., q | z, which is excluded; in the same way, ξ ′ = -1, ξ ′′ = -1.

This situation is by definition equivalent to the p-rationality of the field L (see[START_REF] Gras | Class Field Theory: from theory to practice[END_REF] IV.3.5] for some equivalent conditions).

This means that, under a nontrivial counterexample to SFLT:

and p log(q)

, if κ ≡ 0 (mod p),

and Tr L/Q

, if κ ≡ 0 (mod p), both equivalent to the knowledge of

, only depend on ρ and n for prime numbers q ∈ Q ρ . So we can hope that this fact, summarized in Theorem 4, is incompatible with the arithmetic of the cyclotomic fields Q(µ n ) for p > 3.

Remark 10. In the context of Fermat ′ s equation with r = y

x , r ′ = y z , or r ′′ = x z (supposed of orders n, n ′ , n ′′ modulo q, prime to p), we have the same conclusion as in Lemma 9 by using the units η 1 , η ′ 1 , and η ′′ 1 ; from the relation x + y + z ≡ 0 (mod p), the values r ′ , r ′′ can be computed (mod p) from r, 20 and we get the following relations valid for p ≥ 3 since in Fermat ′ s equation, the special case corresponds to x + z ≡ 0 (mod 9).

(i) If κ ≡ 0 (mod p), then:

(ii) If κ ≡ 0 (mod p), the three symbols

• are trivial.

7.2. Law of ρ-decomposition relative to the family F n , for n > 2. We still suppose p > 3. We have, under a counterexample (u, v) to SFLT, the following interpretation of the equality:

in the nonspecial cases v + u ≡ 0 (resp. the special case v + u ≡ 0) (mod p), which is also valid in the cases κ ≡ 0 or uv ≡ 0 (mod p). This will give also another formulation of Theorem 4. Consider the unit:

20 The notations r, r ′ , and

1 , for (u, v) = (x, y), (y, x), (z, y), or (y, z), and (u, v) = (x, z) or (z, x) (nonspecial cases and special case, respectively); this explains the changes of notations in the Fermat context. We obtain easily r ′ ≡ -r r+1 , r ′′ ≡ -1 r+1 modulo p.

rational fractions of Theorem 6, (i), reduce to y 2y + 1 which is irreducible over F q ). We remark the orbit of 0 which is:

this is consistent with | F q ∪ {∞} | = q + 1 ≡ 0 (mod 6).

Let q be a prime number; to simplify we suppose q ≡ 1 (mod 3). Call n i | q -1 the orders modulo q of T i v u , 0 ≤ i < 6, for any solution (u, v).

As usual we put v u ≡ ξ (mod q ξ = (q, u ξv)) and more generally:

where we recall that the pair (ξ i , q ξ i ) is defined up to conjugation, so that we can replace (ξ i , q ξ i ) by any conjugate (ξ ′ i , q ξ ′ i ). Thus we have put (u 0 , v 0 ) := (u, v) and ξ 0 := ξ.

Consider for instance

To compare the two congruences we can take a fixed prime ideal q | q in L := Q(µ q-1 ) such that q | q ξ and q | q ξ 1 by suitable conjugation of (ξ 1 , q ξ 1 ), which gives the congruences

≡ T (ξ) (mod q). More generally we can write for suitable choices of the ξ i :

which yields, for the units η i 1 associated to the ξ i (with η 0 1 = η 1 ):

(from Theorem 6, (ii)), for all Q above q in M := LK. Thus we have:

proving that the three symbols never coincide when κ ≡ 0 (mod 3).

These symbols are identical to the symbols

This proves that if for instance q ξ 0 splits in F ξ 0 /L 0 then q ξ 1 and q ξ 2 are inert in F ξ 1 /L 1 and F ξ 2 /L 2 , respectively; in other words, the three law of 21 The coherent choice of these ideals supposes that if q = (q, ξ -e) ( ξ of order q -1, e ∈ Z of order q -1 modulo q), v i u i ≡ e d i (mod q) (of order n i modulo q), we must have chosen

Some works of Furtwängler and Vandiver revisited and Fermat ′ s last theorem 53 ρ i -decomposition, or the three symbols

to the three possibilities when κ ≡ 0 (mod p).

So, since this phenomenon happens in L = Q(µ q-1 ) (even if the fields L i are distincts in general), statements like that of Theorem 6.1 are impossible for p = 3 since q cannot be totally split in F q-1 . This distribution of the three possible Frobenii, in the context of ρ idecompositions, must be compatible with the Čebotarev ′ s theorem (see Subsection 9.2 for this aspect and Subsection 9.3 for some numerical evidence and especially Example 5).

Returning to the general case, it is necessary to see whether such a nontrivial automorphism T can exist for p > 3. If not, this will be a favorable argument for our purpose.

Then, for p > 3, there does not exist any automorphism

up to a pth power in M, for some λ, µ ∈ Z, λ ≡ 0 (mod p).

Proof. Suppose that such a nontrivial automorphism does exist and put

, where D is the subgroup of diagonal matrices e I 2 , e ∈ Q × , where I 2 is the unit matrix. In particular, T is of finite order if and only if there exists n > 0 such that

For simplicity we use a representative

Then from the above identity we get the relation:

Starting from a parametric solution (u, v) such that u ξv ≡ 0 (mod q) for some q | q ξ in L = Q(µ q-1 ), consider the solution (u ′ , v ′ ) defined by:

We have the congruence u ′ ξ ′v ′ ≡ 0 (mod q) where ξ ′ is the unique (q -1)th root of unity congruent to T (ξ) = 2ξ-1 ξ+1 modulo q (the order n ′ of ξ ′ divides q -1). Then we have:

proving the equivalence of the two congruences. Hence the result by induction on the powers of T . From Theorem 7, the six families of solutions give the congruences u i ξ iv i ≡ 0 (mod q) for which v i u i := T i v u , ξ i ≡ T i (ξ) (mod q); each congruence reduces to a congruence modulo q ξ i in L i := Q(µ n i ), where q ξ i = q ∩ Z L i and n i is the order of ξ i (prime to 3 since q ≡ -1 (mod 3)).

Warning: the orders n i are divisors of q -1, not necessarily equal to n (see Example 5). But the conditions q ∤ n i and q | Φ n i (u, v), 0 ≤ i < 6, are equivalent to each other.

For instance, take the general solution of the second case 3 | v; then we have to study the congruence (s 3 + t 3 -3 st 2 ) ξ -3 st(st) ≡ 0 (mod q).

Put θ := s t

, which yields to the congruence:

Recall that for n fixed, the φ(n) ideals of L above q are the (q, ξe), where e ∈ Z, defined modulo q, is of order n in F × q ; so the congruence:

is equivalent to:

for the choice of e ≡ ξ (mod q). Since the pair (ξ, q) is defined up to conjugation, we can select e of order n, which implies suitable ξ and q.

When q is solution, there exist infinitely many (u, v) such that q | Φ n (u, v): for a root θ ∈ F q , θ ∈ Z, of the above congruence, the parameters (s, t) are obtained from the congruence s ≡ θ t (mod q) (see Example 8). At this step we have proved (i) under the existence of e such that the polynomial

The polynomial X 3 -3ξ -1 X 2 -3(1ξ -1 ) X + 1 defines the cyclic extension F ξ : indeed, with X = ξ -1 (Y + 1) one obtains the polynomial:

from the universal polynomial, irreducible of degree 3 over L (see Subsection 7.3), obtained from the cubic root of (1 + ξ j) s+2 = η 1 up to a 3th power.

Thus, the condition q | Φ n (u, v) (q ∤ n) is equivalent to the ρ-splitting of q for F n , where ρ := v u (see Subsection 7.2) or to the ρ i -splitting of q for F n i where ρ

, and n i is the order modulo q of ρ i , 0 ≤ i < 6. This proves (ii).

From the Dirichlet-Čebotarev theorem, we get a precise result taking a nontrivial Frobenius in L 1 F ξ / F ξ , and we obtain the prime ideal q ξ = (q, u ξv) where the (u, v) are obtained from the three roots θ 1 , θ 2 , θ 3 of the polynomial as explained above. We obtain infinitely many values of q with clearly a nonzero density. In other words, for κ ≡ 0 (mod 3) these primes q give again the splitting of q ξ in F ξ /L, hence its inertia in L 1 /L, F ξ /L, and in the fourth cubic subfield

). This makes clear the point (i) of the proposition.

In the case where (u, v) is for instance the general solution for the second case of the SFLT equation we get, with η 1 = (1 + ξ j) eω j, η 1 := η 1 j 1 2 = (1 + ξ j) eω , and c η 1 := η 1 j, the following diagram:

There are six analogous diagrams over each field L i .

Remark 12. Let q be a prime number such that κ ≡ 0 (mod 3). Then for a divisor m > 2 of q -1, there is not necessarily a solution (u, v) = (s 3 + t 3 -3 st 2 , 3 st(st)), s, t ∈ Z, g.c.d. (s, t) = 1, s + t ≡ 0 (mod 3), such that the order n of v u modulo q is equal to m (see Example 6). The cases m ≤ 2 correspond, for the above solution, to the congruences:

equivalent to the splitting, modulo q, of X 3 + 1 -3 X ± 3 X(X -1)). One verifies that these polynomials of Q[X] define the number field Q 1 ; so, as by assumption κ ≡ 0 (mod 3), we obtain that the orders 1 and 2 are never possible. The case m > 2 is less trivial.

Example 5. We illustrate Proposition 5 with the prime q = 41 and the solution (u, v) = (139193, 76626) obtained with the parameters (s, t) = (-11, 43); we note that for e = 22 ∈ Z/41Z the polynomial:

) and we have chosen θ = 15 for which s -15 t ≡ 0 (mod 41). Using the automorphism T , we obtain the six steps:

T 0 (e) = e = 22 of order 40

, solution of the second case, T (e) = e1 = 9 of order 4

, solution of the special case, T 2 (e) = e2 = 14 of order 8

, solution of the second case, T 3 (e) = e3 = 10 of order 5

, solution of the special case, T 4 (e) = e4 = 39 of order 20

, solution of the first case,

T 5 (e) = e5 = 5 of order 20

, solution of the special case.

As a consequence, we have: We have obtained the set of orders {40, 4, 8, 5, 20}. For instance, this implies the inertia of q ξ 40 in F ξ 40 /Q(µ 40 ) and that of q ξ 4 in F ξ 4 /Q(µ 4 ), which illustrates the incompatibility with statements like Theorem 2 for p = 3. Example 6. We have found the following numerical example to illustrate Remark 12, with m = 5 for which L = Q(µ 5 ) is principal. Consider the prime q = 48738631 for which q -1 = 2 • 3 • 5 • 163 • 9967 and κ ≡ 0 (mod 3).

Then q = (ξ 2 + ξ 3 -3 -90 (3

where ξ is a primitive 5th root of unity, is a prime ideal above q.

Since ξ 2 + ξ 3 -3 ∈ L + , this ideal satisfies the relation q 1-c = (α) Z[ξ], α ≡ 1 (mod 9), which means that q totally splits in H -

Concerning the solutions (u, v) = (s 3 + t 3 -3 st 2 , 3 st(st)), s, t ∈ Z, g.c.d. (s, t) = 1, s + t ≡ 0 (mod 3), such that Φ 5 (u, v) ≡ 0 (mod q), we try to find the smallest values of the order n of v u modulo q. It is clear that the value n = 5 is by construction impossible. There is also no solution for n = 10 since Q(µ 10 ) = Q(µ 5 ) = L with q totally split in H -

We find the values: As we have seen, the orders n = 1 and 2 are impossible.

Example 7. In another point of view, in the following example we fix the solution (u, v) = (19, 18) corresponding to (s, t) = (3, 1) and we give the order n of v u modulo q for primes q < 3.10 6 with κ ≡ 0 (mod 3), such that n < q 1 3 to limit the data. Example 8. Let q = 113 = 1 + 2 4 . 7. In the following example we fix n and use a polynomial X 3 -3e -1 X 2 -3(1e -1 ) X + 1 which splits modulo 113; for e = 83, of order n = 14 modulo 113, its roots are 5, 28, and 46 modulo 113.

Recall that for ξ of order n and e ∈ Z defining the prime ideal q = (q, ξ-e) above q, the solutions (s, t) giving q | Φ n (u, v) for the corresponding solutions (u, v) = (s 3 + t 3 -3st 2 , 3st(st)), are defined for instance via the congruence s -5t ≡ 0 (mod 113), g.c.d. (s, t) = 1, and s + t ≡ 0 (mod 3). We observe a unique case where 113 2 divides Φ n (u, v).

Example 9. We consider the prime number q = 401 = 1 + 2 4 . 5 2 and for all possible values of ρ := v u modulo q, for the general solution of the second case, we give the order of ρ modulo q. The resolution of 3st(s -t)

There are 133 = 7 . 13 distinct values of such ρ with the following repartition of the orders n: 53 for order 400; 28 for 200; 13 for 80; 12 for 100; 7 for 50 and 25; 4 for 40; 3 for 20; 2 for 10; 1 for 16, 8, 5, and 4. As we know, orders 1 and 2 cannot exist. These densities are in accordance with the expression 1 3 φ(n). The above computations for p = 3 suggest the following conjecture.

Conjecture 5. For all m > 0 and for all prime numbers q ≡ 1 (mod m), q ≡ 2 or 5 (mod 9), and q totally split in F m /Q, there exists a solution (u, v) of the SFLT equation for p = 3, for which the order of v u modulo q is ≥ m.

This conjecture (to be compared with Conjecture 2 for p > 3) is very reasonable since, in practice, the order of v u modulo q is often q -1.

Conclusion

In Subsection 9.1, we have given a justification of the fact that Theorem 2 (or any weak form) cannot exist for p = 3.

For p > 3, if the number of solutions (u, v) of the SFLT equation is finite, for any bound N the number of primes q, such that the v u are of order modulo q less than N , is finite. So for primes q ′ for which we assume that all the prime ideals above q ′ in L ′ = Q(µ m ′ ), m ′ | q ′ -1 large enough, totally split in H - L ′ [p]/L ′ , we get large values of q ′ , hence large values of the orders n ′ of the v u modulo q ′ , say n ′ > > N . So, contrary to the case p = 3, the effectiveness of the statement of a weak form of Theorem 2 is more credible.

We have justified, in Subsection 9.2, why the case p = 3 is specific for the arithmetic of the fields Q(µ n ) in relation with the abelian 3-ramification; which suggests that, for p > 3, a result like Theorem 4, on the constraints on the ρ-laws of decomposition of infinitely many primes q, gives a non trivial obstruction and is likely to lead to a proof of SFLT.

In other words, we can hope that for p > 3 any statistical analysis of the decomposition laws is legitimate.