Fermat's Last Theorem for Regular Primes

FERMAT’S LAST THEOREM FOR REGULAR PRIMES DAVID JAO

Abstract. A presentation in modern language of Kummer’s proof that Fermat’s Last Theorem is true for all regular primes.

1. Introduction Throughout this article, set ζm := e2πi/m , a primitive m–th root of unity. Definition 1.1. A prime p ∈ Z is regular if p does not divide the class number h of the number field Q(ζp ). It’s worth pointing out that if Q(ζp ) has class number 1, then p is trivially regular. The only fact we need about regular primes is that if p is regular, and a is an ideal of Q(ζp ) with ap principal, then a is principal. Indeed, if a is not principal and ap is principal, then a generates a cyclic subgroup of order p of the ideal class group of Q(ζp ), and so p divides the order of the ideal class group of Q(ζp ). The main theorem which we prove in this article is the following result. Theorem 1.2 (Fermat’s Last Theorem for Regular Primes). Let p ≥ 3 be a regular prime. Suppose there exist integers x, y, z with xp + y p = z p . Then xyz = 0. 2. Background Information We recall some facts about Q(ζp ) which we need. The polynomial xp − 1 factors as xp − 1 = (x − 1)(x − ζp ) · · · (x − ζpp−1 ) and the polynomial

xp − 1 = xp−1 + · · · + x + 1 x−1 is irreducible (Proof: Transform x 7→ y + 1 and use Eisenstein’s criterion). The minimal polynomial of ζp is thus Φp (x), and the conjugates of ζp are {ζp , ζp2 , . . . , ζpp−1 }. The field extension Q(ζp )/Q has dimension p − 1, and is Galois since Q(ζp ) is the splitting field of Φp (x) over Q. We examine the ramification behavior of Q(ζp ) over Q. The discriminant 2 ζp ··· ζpp−1 Y .. .. (ζpk − ζpj )2 = ... . . 1≤j
of Φp (x) is a well–known Vandermonde determinant, and evaluates to (−1) ramify in Q(ζp ) must divide disc(Φp (x)), and the only such prime is p. Let us now show that p does indeed ramify.

p−1 2

pp−2 . The primes in Q which

Lemma 2.1. Let O denote the ring of algebraic integers in Q(ζp ). Set λ := ζp − 1. Then (λ) is a maximal ideal in O, and (λ)p−1 = (p). 1−ζ i

Proof. The element ui := 1−ζpp = 1 + ζp + · · · + ζpi−1 is a unit in O. To prove this, we need to show that u−1 i is in O. Choose j ∈ Z with ij ≡ 1 (mod p). Then u−1 = i

1 − ζpij 1 − ζp = = 1 + ζpi + · · · + (ζpi )j−1 . i 1 − ζp 1 − ζpi

Date: October 19, 2000. 1

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which shows u−1 is in O. i We have (1)

p = Φp (1) =

p−1 Y

(1 − ζpi ) = (1 − ζp )p−1

i=1

p−1 Y

ui

i=1

and this proves that (p) = (λ)p−1 . From equation (1) we already see that λ is not a unit1 . If (λ) factored any further, the ramification index of (p) would exceed the dimension p − 1 of the field extension Q(ζp )/Q. Hence (λ) is prime (and maximal). Corollary 2.2. In Z[ζp ], the elements 1 − ζp and ζpi − ζpj are associates for all i 6= j (that is, they generate the same maximal ideal). Proof. We have ζpi uj−i (1 − ζp ) = ζpi − ζpj , and ζpi uj−i is a unit in O. It is well known that the ring of algebraic integers in Q(ζp ) is Z[ζp ]. For the sake of completeness we prove it here. Proposition 2.3. The ring of algebraic integers O in Q(ζp ) is Z[ζp ]. Proof. Clearly Z[ζp ] ⊂ O. We prove the opposite inclusion. Let x ∈ O. Choose rational numbers a0 , a1 , . . . , ap−2 ∈ Q such that x = a0 + a1 ζp + · · · + ap−2 ζpp−2 . Then, for any i = 0, 1, . . . , p − 2,     − a0 − a1 − · · · − ai−1 + (p − 1)ai − ai+1 − · · · − ap−2 − − a0 − a1 − · · · − ap−2 Tr(xζp−i − xζp ) = = pai . Since xζp−i −xζp is an algebraic integer, its trace pai is an integer. Therefore we have integers b0 , . . . , bp−2 ∈ Z such that px = b0 + b1 ζp + · · · + bp−2 ζpp−2 .

(2)

Set λ := ζp − 1 and replace ζp with λ + 1 in (2). We obtain integers c0 , . . . , cp−2 ∈ Z such that px = c0 + c1 λ + · · · + cp−2 λp−2 . Inspecting this equation shows that λ | c0 . Taking norms, we see N (λ) divides cp−1 . But N (λ) = 0 ±Φp (1) = ±p, so p | c0 . Now everything in the equation except the c1 term is a multiple of λ2 , so λ2 | c1 λ. Then λ | c1 and p | c1 . Continuing, we get p | ci for all i = 0, 1, . . . , p − 2, so c0 c1 cp−2 p−2 + λ + ··· + λ x= p p p is in Z[ζp ], as desired. One nice consequence of this proposition is that we have an integral basis for Q(ζp ): Corollary 2.4. An integral basis for Q(ζp ) is {1, ζp , . . . , ζpp−2 }. Proof. The claimed basis is clearly a vector space basis for Q(ζp ) over Q. Furthermore, it spans Z[ζp ] over Z. Thus it is an integral basis. We also need the following standard result: Proposition 2.5. Let α0 ∈ Q be an algebraic integer all of whose conjugates have complex absolute value bounded by 1. Then α0 is a root of unity. Proof. Let f (x) be the minimal polynomial of α0 over Q. Factor f (x) as f (x) =

n−1 Y

(x − αi ) = xn + an−1 xn−1 + · · · + a1 x + a0 .

i=0

1 p is never a unit in the ring of integers O of any algebraic number field K. Indeed, by Gauss’s lemma, its inverse 1/p is not an algebraic integer.

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Each coefficient ai is a sum of ni monomials in the αi , and each such monomial has norm bounded by 1.
We conclude that |ai | ≤ ni . In particular, |a0 | ≤ 1. But a0 is not zero, so a0 = ±1 and each αi has absolute value exactly 1. Let Xn denote the set of algebraic integers of degree at most n whose conjugates all have complex absolute value at most 1. We claim Xn is finite.
Let α ∈ Xn . Then, as above, the integer coefficients of the minimal polynomial of α are bounded by ni . Therefore there are only finitely many possible minimal polynomials for α, so the number of elements of Xn is finite. Let K be the splitting field of f (x) over Q. Any Galois automorphism of K over Q permutes the αi . Let k ∈ N. The monic polynomial n−1 Y g(x) = (x − αik ) i=0

is fixed under any permutation of the αi , so it is invariant under Gal(K/Q). Since α0k is a root of g(x), we learn that • The degree of α0k is at most deg(g) = n. • Every conjugate of α0k is of the form αik , and these all have absolute value 1. Therefore, for any k ∈ N, the algebraic integer α0k is in the finite set Xn . Hence {α0 , α02 , α03 , . . . } is a finite subset of the unit circle, and it follows that α0 is a root of unity. 3. Kummer’s Proof—First Case Set λ := ζp − 1 as before. We first prove a couple of technical lemmas. Lemma 3.1. Let α ∈ Z[ζp ], and let α denote the complex conjugate of α. Then αp ≡ αp (mod pZ[ζp ]). Pp−2 Proof. Using our integral basis from Corollary 2.4, write α = i=0 ai ζpi . Then !p p−2 p−2 p−2 X X p X
p i i p α = ai ζp ≡ ai ζp = api ≡ αp (mod pZ[ζp ]) i=0

i=0

i=0

Lemma 3.2. Let u ∈ Z[ζp ] be a unit. Then u/u ∈ {1, ζp , . . . , ζpp−1 }. Proof. Let σ ∈ Gal(Q(ζp )/Q). Then σ(u) = σ(u)/σ(u) = 1. |σ(u/u)| = σ(u)

By Proposition 2.5, the algebraic integer u/u ∈ Q(ζp ) is a root of unity. Suppose the multiplicative order of u is m 6= 1, p. Then the least common multiple N := lcm(p, m) of p and m is greater than p. We have Q(ζp ) = Q(ζp , ζm ) = Q(ζN ). But now dimQ Q(ζN ) = φ(N ) > dimQ Q(ζp ) = φ(p) = p − 1, and this is a contradiction. It is common practice in old–style treatments of Fermat’s Theorem (i.e., everything before Wiles) to split the discussion into two cases. We do so here as well. Theorem 3.3 (Fermat’s Last Theorem—First Case). Let p ≥ 3 be a regular prime. Let x, y, z ∈ Z with gcd(x, y, z) = 1 and xp + y p + z p = 0. Then p | xyz. Proof. We assume p - xyz and arrive at a contradiction. First, let us quickly dispose of the case p = 3. Under the assumption that x, y, z 6≡ 0 (mod 3), we find (by brute force) that there are no x, y, z ∈ Z/9Z with x3 + y 3 + z 3 ≡ 0 (mod 9). For the remainder of the proof, we assume that p ≥ 5. Our main weapon is the factorization (3)

(x + y)(x + ζp y) · · · (x + ζpp−1 y) = −z p .

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Let m ⊂ Z[ζp ] be a maximal ideal which divides x + ζpi y. Then m divides z by Equation (3). We claim m does not divide x + ζpj y for any j 6= i. Indeed, if m divided both elements, then m would divide their difference (ζpi − ζpj )y. Since m is maximal, it divides one of the two factors. We dismiss both possibilities: 1. Suppose m | (y). We already know from above that m contains z. Thus m contains y, z, and also x, since xp = −z p − y p , which contradicts our assumption that gcd(x, y, z) = 1. 2. Suppose m | (ζpi − ζpj ). Then m contains the maximal ideal (λ); accordingly, m is (λ). Hence λ | z, and taking norms we find that p | z, violating our assumption that p - xyz. Let mkp be the highest power of m dividing (z p ) (this power is clearly a multiple of p). Since m does not divide any factor on the left hand side of Equation (3) other than (x + ζpi y), we conclude that mkp divides (x + ζpi y). We have shown that, in the factorization (x + ζpi y) = me11 · · · mekk of the ideal (x + ζpi y), every exponent ej is a multiple of p. Therefore, there exists an ideal Ci ⊂ Z[ζp ] such that (x + ζpi y) = Cip . Furthermore, since Cip is principal, and p is regular, the ideal Ci is principal. Choose αi ∈ Z[ζp ] such that Ci = (αi ), and choose units ui ∈ Z[ζp ] such that x + ζpi y = ui αip . for every i = 0, 1, . . . , p − 1. Since x + ζpp−i y = x + ζpi y, we can (and do) choose αi and ui so that αp−i up−i

= αi = ui

By Lemma 3.2, there exists k such that u1 /u1 = ζpk . Using Lemma 3.1, we find that x + ζp y = u1 α1p ≡ u1 αp1 ≡ ζpk u1 αp1 = ζpk (x + yζp−1 )

(mod p).

We now show that x + ζp y − ζpk x − ζpk−1 y ≡ 0 (mod p) only if k = 1. Suppose k = 0; then p divides the ideal (y(ζp − ζp−1 )) = (y)(λ), contradicting our assumption that p - y. If k = 2, then similarly p | x(1 − ζp2 ), contradicting our assumption that p - x. If k is neither 0, 1, or 2, then we proceed as follows (keeping in mind that p ≥ 5): • Suppose 2 < k < p − 1 and x + ζp y − ζpk x − ζpk−1 y = pα where α ∈ Z[ζp ]. Using our integral basis from Pp−2 Corollary 2.4, we may write α = i=0 ai ζpi . Then x + ζp y − ζpk x − ζpk−1 y =

p−2 X

pai ζpi .

i=0

The basis coefficients on both sides must be identical. Comparing them, we see that p | x and p | y. • Suppose k = p − 1 and x + ζp y − ζpp−1 x − ζpp−2 y = pα. Multiply both sides by ζp2 to obtain ζp2 x + ζp3 y − ζp x − y = p(ζp2 α) and compare basis coefficients on both sides as before to conclude that p divides both x and y. The preceding argument shows that p divides x + ζp y − ζp x − y = (x − y) + ζp (y − x). Applying the basis coefficient argument a third time yields p | (x − y), or x ≡ y (mod p). But the equation xp + y p + z p = 0 is symmetric in x, y, and z. Applying the whole argument again, we find that y ≡ z (mod p). Therefore, x, y, and z are all congruent modulo p. So, modulo p, we get xp + y p + z p ≡ 3xp ≡ 0

(mod p).

Since p ≥ 5, we may cancel 3 from the equation to obtain p | x, contradicting our assumption that p - xyz.

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4. Kummer’s Lemma We state without proof the Artin Reciprocity Law from class field theory. Theorem 4.1 (Artin Reciprocity Law). Let L be the maximal unramified abelian extension of a number field K which has no real embeddings. Let IK denote the group of fractional ideals of the ring of integers OK in K. The Artin map ψL/K : IK −→ Gal(L/K) defined by sending a prime ideal p ∈ IK to the unique element σ ∈ Gal(L/K) such that, for every prime P ∈ IL with P ∩ OK = p, • σα ≡ α|OK /p| (mod P) for all α ∈ OL • σP = P (and extending linearly) is well defined and induces an isomorphism from the ideal class group of K to the Galois group Gal(L/K). In order to prove the second case of Fermat’s Last Theorem, we need a result which is known in the literature as Kummer’s Lemma. The lemma as proved by Kummer states that any unit in Q(ζp ) which is congruent to a rational integer modulo p is a pth power. For our purposes we only need a weaker variant of the lemma, which we state and prove here. Proposition 4.2 (Kummer’s Lemma). Let p ≥ 3 be a regular prime. Let e be a unit in Z[ζp ] which is congruent to a pth power modulo λp . Then e is the pth power of a unit in Z[ζp ]. √ √ Proof. It suffices to show that e has a pth root p e in Q(ζp ), for then p e, being integral over Z[ζp , e], is √ integral over Z, and hence in Z[ζp ]. Consider the extension K := Q(ζp , p e) of Q(ζp ). Assume e is not a pth root in Q(ζp ), so that [K : Q(ζp )] > 1. The Galois group Gal(K/Q(ζp )) has at least the p elements √ √ p e 7→ ζpk p e, for k = 0, 1, . . . , p − 1, √ so the dimension [K : Q(ζp )] is at least p. On the other hand, the polynomial xp − e has p e as a root, so the dimension [K : Q(ζp )] is at most p. We conclude that the dimension is exactly p, and that K is the splitting field of xp − e over Q(ζp ). √ √ We compute √ the discriminant of the element p e in the field extension Q(ζp )[ p e] of Q(ζp ). The minimal polynomial for p e over Q(ζp ) is f (x) := xp − e, so the discriminant is (up to sign): √ disc(f (x)) = ±N (f 0 ( p e)) = ±N (pe(p−1)/p ) = ±pp ep−1 . The primes in Q(ζp ) which ramify in K must divide disc(f (x)), and the only such prime is (λ). So the only prime in Q(ζp ) that could ramify in K is (λ). We show that (λ) in fact does not ramify in K. Choose an integer α ∈ Z[ζp ] such that αp ≡ e (mod λp ). √ p Set τ := ( e − α)/λ. Then K = Q(ζp )[τ ]. Furthermore, τ is a root of the monic polynomial (λx + α)p − e . λp Writing p = uλp−1 where u is a unit, we see that all the coefficients of g(x) are algebraic integers. Hence τ is an algebraic integer. Since we already know that the dimension of the field extension K/Q(ζp ) is p, we find that g(x) is the minimal polynomial for τ over Q(ζp ). But all the coefficients of g 0 (x) are in (λ) except for the constant term uαp which is prime to (λ). Therefore g 0 (τ ) is prime to (λ). We claim that Y disc(g(x)) = ±N (g 0 (τ )) = ± g 0 (στ ) g(x) :=

σ∈Gal(K/Q(ζp ))

0

is prime to (λ). If it is not, then λ divides g (στ ) for some σ ∈ Gal(K/Q(ζp )), upon which σ −1 λ = λ divides g 0 (τ ), a contradiction. Therefore (λ) does not contain disc(g(x)), and so it does not ramify in K. In summary, the field extension K/Q(ζp ) is an unramified extension of degree p. Let L be the maximal unramified abelian extension of Q(ζp ), that is, the composite of all finite unramified abelian extensions. The field L certainly contains K, since K is unramified and abelian (with cyclic Galois group). By the Artin Reciprocity Law (Theorem 4.1), the Galois group G of the extension L/Q(ζp ) is isomorphic to the ideal class group of Q(ζp ). Since p = [K : Q(ζp )] divides |G| = [L : Q(ζP )], it follows that p divides the class number h = |G| of Q(ζp ), contradicting the assumption that p is regular.

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5. Fermat’s Last Theorem—Second Case We now proceed to the second case of Fermat’s Last Theorem. As before, we first need a technical lemma. Lemma 5.1. Let v ∈ Z[ζp ] with (v, λ) = 1. Then there exists k ∈ Z such that ζpk v ≡ m (mod λ2 ), with m ∈ Z. Proof. Write v ≡ m + nλ (mod λ2 ) where m, n ∈ Z and (m, p) = 1. By the binomial theorem, ζpk = (1 + λ)k ≡ 1 + kλ

(mod λ2 ).

Choose k ∈ Z such that n + mk ≡ 0 (mod p). Then ζpk v ≡ m + (n + km)λ ≡ m (mod λ2 ). Theorem 5.2 (Fermat’s Last Theorem—Second Case). Let p ≥ 3 be a regular prime. Suppose x, y, z ∈ Z, with xp + y p + z p = 0 and p | xyz. Then xyz = 0. Proof. We suppose xyz 6= 0 and arrive at a contradiction. We may, without loss of generality, assume that x, y, z are pairwise relatively prime with λ | z. It follows that λ does not divide either x or y, or else it would divide all three. Write −z = λm z0 with (λ, z0 ) = 1 (we can do this because z is nonzero). Then x, y, z0 ∈ Z[ζp ] is a solution to the equation X p + Y p = λpm Z p with m ≥ 1, (xyz0 , λ) = 1, and x, y, z0 pairwise relatively prime. To prove the theorem, it suffices to show that there are no solutions x, y, z ∈ Z[ζp ] to the equation X p + Y p = uλpm Z p

(4) with • • • •

x, y, z pairwise relatively prime, u a unit in Z[ζp ], (xyz, λ) = 1, m ≥ 1.

Our strategy is as follows. Assuming a solution exists, we show that m is in fact greater than 1. Then we prove that the equation X p + Y p = uλp(m−1) Z p has a solution of the same form. The required contradiction then follows from infinite descent. New let u ∈ Z[ζp ] be a unit and assume x, y, z ∈ Z[ζp ] is a solution to equation (4) above, with (xyz, λ) = 1 and x, y, z pairwise relatively prime. By Lemma 5.1, we may assume there exist a, b ∈ Z with x ≡ a (mod λ2 ) and y ≡ b (mod λ2 ). We have p−1 Y

(5)

(x + ζpi y) = uλpm z p .

i=0

We see that λ divides one term on the left hand side. But the difference between any two such terms is a multiple of λ, so λ must divide all the terms on the left hand side. Now, x + y ≡ a + b (mod λ2 ) and λ | x + y, so λ | a + b. Consequently p | a + b and λ2 | x + y. From equation (5) we see that λp+1 divides the left hand side, so m > 1, which achieves our first goal. We have x + ζpi y = (x + y) + (ζpi − 1)y, so for i ≥ 1, the term x + ζpi y is in (λ) but is not in (λ2 ). Passing to ideals, we find that (x + y) = (x + ζpi y)

(λ)p(m−1)+1 C0

= (λ)Ci for i = 1, 2, . . . , p − 1

where the ideals Ci are principal and prime to (λ). We claim the ideals Ci are pairwise relatively prime as well. If not, then there is a maximal ideal m 6= (λ) containing x + ζpi y and x + ζpj y for i 6= j. Then m contains (ζpi − ζpj )y, so either m contains ζpi − ζpj (in which case m contains (λ) after all), or m contains y (in which case m contains x, violating our assumption that (x, y) = 1). From equation (5) it follows that C0 C1 · · · Cp−1 = (z)p . Therefore each Ci equals Dip for some ideal Di ⊂ Z[ζp ]. Since p is a regular prime, and Ci is principal, we see that Di is a principal ideal. Write

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Di = (ωi ). Returning to the level of elements, we conclude that there exist units u0 , u1 , u2 ∈ Z[ζp ] such that (6) (7)

x+y x + ζp y

= u0 λp(m−1)+1 ω0p = u1 λω1p

(8)

x + ζp2 y

= u2 λω2p

Note that we need to have p > 2 here in order to write down three equations! Solving the system of equations (7) and (8), we get x = ζp u1 ω1p − u2 ω2p y

= −ζp−1 (u1 ω1p − u2 ω2p )

Plugging these back into (6), we get λζp−1 ((1 + ζp )u1 ω1p − u2 ω2p ) = u0 λp(m−1)+1 ω0p . We cancel λ from both sides and divide by ζp−1 (1 + ζp )u1 (recall from the proof of Lemma 2.1 that 1 + ζp is a unit). We thus find that there are units e2 and e in Z[ζp ] such that ω1p + e2 ω2p = eλp(m−1) ω0p . Note that since (ω2 ) is prime to (λ), the element ω2p is invertible modulo λp . So, taking congruences modulo λp , we see that e2 is congruent to a pth power modulo λp . By Proposition 4.2 (Kummer’s Lemma), we have e2 = f p for some f ∈ Z[ζp ]. Setting x0 = ω1 , y 0 = f ω2 , z 0 = ω0 , we see that x0 , y 0 , z 0 is a solution to X p + Y p = eλp(m−1) Z p with x0 , y 0 , z 0 pairwise relatively prime and (x0 y 0 z 0 , λ) = 1. This completes the descent step, and with it the second case of Fermat’s Last Theorem.