Z -basis for the orders generated by the conjugates of algebraic - - PDF document

Z-basis for the orders generated by the conjugates of algebraic integers

St´ ephane R. LOUBOUTIN Aix Marseille Universit´ e, CNRS, Centrale Marseille, I2M, Marseille, FRANCE stephane.louboutin@univ-amu.fr June 26, 2018

Contents

1 Abstract 2 2 A Z-generating system of Mα 5 3 A Z-basis of Mα in the worst case 6 4 Remark on the Z-basis of Mα 7 5 A Z-basis of Mα in the cyclic cubic case 8 6 When is the order Z[ε] Galois-invariant? 10 7 On the orders Z[αn] and Mαn, n 1 13 7.1 The case that α is an algebraic unit . . . . . . . . . . . . . . . 13 7.2 The non-normal cubic case . . . . . . . . . . . . . . . . . . . . 14 7.3 The cyclic cubic case . . . . . . . . . . . . . . . . . . . . . . . 15 1

Z-basis for the orders generated by the conjugates of algebraic integers 2

1 Abstract Let Dα := Y

1i<jn

(αj αi)2 2 Z \ {0} be the discriminant of the minimal polynomial Πα(X) = Xn an1Xn1 + · · · + (1)na0 2 Z[X]

• f an algebraic integer α of degree n, where α1, · · · , αn

are the n distinct complex roots of Πα(X). We consider Mα = Z[α1, · · · , αn], and order of Lα = Q(α1, · · · , αn) be the normal closure of Q(α). It is a free Z-module of rank r = (Lα : Q) (Q(α) : Q) = n. If M is an order of a number field, let DM 2 Z denote its

• discriminant. Notice that DZ[α] = Dα.

The goal is to determine a Z-basis and the discrimi- nant DMα of Mα and to give various applications of these determinations.

Z-basis for the orders generated by the conjugates of algebraic integers 3

Let us explain how one usually constructs parametrized families of number fields of known discriminants and reg-

• ulators. One usually starts from explicit parametrized

families of monic polynomials with integral coefficients and constant coefficient equal to ±1, so that their com- plex roots are algebraic units. Let us for example consider the simplest cubic fields Q(α), where Πα(X) = X3 aX2 +(a3)X +1, a 2. Since Dα = (a2 3a + 9)2 is a square, Q(α)/Q is a Galois cyclic cubic extension and since Πα(X) = (X α) (X (α2 + (a 1)α + 2))(X (α2 aα + a 2)), the order Z[α] is Galois invariant. Moreover, the three conjugates α, α0 and α00 of α are algebraic units. Since Q(α)/Q is of prime degree, any 2 of these conjugates are multiplicatively independent in the group of units Z[α]⇥

• f the order Z[α]. In fact, for the simplest cubic fields

we have Z[α]⇥ = h1, α, α0i. Hence, in the cases that Z[α] is equal to the ring of algebraic integers ZK of the number field K, we end up with cyclic cubic fields of known discriminants and regulators. Now, since Dα = (ZK : Z[α])2dK and dK > 1, it follows that if a23a+9 = p is prime then ZK = Z[α]. Since the class number of K divides the class number h+

p of the real cyclotomic field

Q(ζp)+, we easily end up with with examples of prime numbers p > 3 for which h+

p > 1 (see [CW]).

However, still assuming that α is an algebraic unit such that Q(α)/Q Galois cyclic of prime degree p 3,

Z-basis for the orders generated by the conjugates of algebraic integers 4

whereas the order Z[α] is not generally Galois invari- ant, the order Mα is always Galois invariant. Hence it would be much more satisfactory to have families of parametrized polynomials for which DMα would be known and for which any p 1 of the p conjugates of α would form a system of fundamental units of the order Mα. In this respect we proved: Theorem 1 (See [LL14, Theorem 1.2]). Let ε, ε0 and ε00 be the three real roots of any one of the follow- ing parametrized families of Q-irreducible cubic poly- nomials of discriminants a square:

Φn(X) = X3 n(n2 + n + 3)(n2 + 2)X2 (n3 + 2n2 + 3n + 3)X 1 n 2 Z, Ξn(X) = X3 (n3 2n2 + 3n 3)X2 n2X 1 1, 2 6= n 2 Z, Ψn(X) = X3+(n8+2n63n5+3n44n3+5n23n+3)X2(n32)n2X1 n 2 Z.

Then, {1, ε, ε2ε0} is a Z-basis of the totally real cubic

• rder Z[ε, ε0] and {ε, ε0} is a system of fundamental

units of this cubic order Z[ε, ε0]. Indeed, in these three cases 3ac b2 divides Dα and 3b a2. Hence the results on the Z-basis follow from Theorem 7.

Z-basis for the orders generated by the conjugates of algebraic integers 5

2 A Z-generating system of Mα Proposition 2 (See [Lou16b]). The set Ωα := {αe1

1 · · · een n ; 0  ek  n k}

is a Z-generating system (with n! elements) of Mα.

• Proof. Let us prove Proposition 2 for n = 3.

We must show that Ωα = {1, α, α2, α0, αα0, α2α0} is a Z-generating system of the order Mα = Z[α, α0, α00], where α, α0 and α00 are the complex roots of the minimal polynomial Πα(X) = X3 aX2 + bX c 2 Z[X]. Since a = α + α0 + α00 2 Z, we have Mα = Z[α, α0]. Now, α0 being a root of Πα(X) X α = X2 (a α)X + (α2 aα + b) 2 Z[α][X], it is integral over Z[α] and we have Mα = Z[α, α0] = Z[α][α0] = Z[α] + α0Z[α]. Since Z[α] = Z + Zα + Zα2, the desired result follows. •

Z-basis for the orders generated by the conjugates of algebraic integers 6

3 A Z-basis of Mα in the worst case Theorem 3 (See [Lou16b]). Assume that Gal(Lα/Q) is isomorphic to the symmetric group Sn. Then Mα is a free Z-module of rank n!, Ωα := {αe1

1 · · · αen n ; 0  ek  n k}

is a Z-basis of Mα and the discriminant of Mα is DMα = Dn!/2

α

. Now, of particular interest is the case where Q(α) is a normal number field. In that case Mα is a Galois in- variant order of Q(α) and r = n. The matrix Mα of the coordinates of the n! elements of Ωα in the canonical Q-basis Bα = {1, α, · · · , αn1} of Q(α) is in Mn,n!(Q). Consequently, it is rather easy to develop and algorithm for constructing a Z-basis of Mα and to compute DMα. However, from a theorytical point of view, in the Galois case, we present the only cases where a Z-basis of Mα has been obtained: the quadratic and cubic cases.

Z-basis for the orders generated by the conjugates of algebraic integers 7

4 Remark on the Z-basis of Mα Lemma 4 Let {ω1, · · · , ωr} be a Z-basis of a free Z- module M of rank r 1. There exists a Z-basis of M containing a given ω = a1ω1 + · · · + arωr 2 M if and

• nly if gcd(a1, · · · , ar) = 1.

Consequently, if 1 2 M and M \ Q = Z, e.g. if M is an order of a number field of degree r, then there exists a Z-basis of M of the form {1, ω2, · · · , ωr}.

• Proof. Clearly, the condition is necessary.

Conversely, assume that gcd(a1, · · · , ar) = 1. Let u1, · · · , ur 2 Z be such that a1u1 + · · · + arur = 1 (B´ ezout). Define a Z-linear map φ : M ! Z by x = x1ω1+· · ·+xrωr 2 M 7! φ(x) = u1x1+· · ·+urxr 2 Z. Since x = φ(x)ω + (x φ(x)ω) for x 2 M, we have M = Zω ker φ. Hence, there exist ω0

2, · · · , ω0 r 2 M such that {ω, ω0 2, · · · , ω0 r}

is a Z-basis of M. • Proposition 5 There exists a Z-basis of Mα of the form {1, α, ω3, · · · , ωr}.

Z-basis for the orders generated by the conjugates of algebraic integers 8

5 A Z-basis of Mα in the cyclic cubic case Corollary 6 Let α ba a cubic algebraic integer. As- sume that Q(α)/Q is Galois, i.e. assume that Dα = D2 is a square. Then DMα = (D/(Mα : Z[α])2 and the index (Mα : Z[α]) is equal to the least d 1 such that dMα ✓ Z[α], i.e. is equal to the least common multiple of the denominators of the entries

• f the matrix Mα of the coordinates in the Q-basis

Bα = {1, α, α2} of any Z-generating system of Mα. We may assume that Ωα = {1, α, α2, α0, αα0, α2α0} with α0 such that its coordinates in the canonical Q-basis Bα = {1, α, α2} of Q(α) are given by the fourth column

• f the following matrix Mα (see [Lou12, Proposition 10]):

B @ 1 0 0

a2b+3ac4b2+Da 2D (2a26b)c 2D (ab9cD)c 2D

0 1 0

2a3+7ab9cD 2D a2b+3ac+2b2+aD 2D 2a2cab2+3bc+bD 2D

0 0 1

2a26b 2D ab9cD 2D 6ac+2b2 2D

1 C A Letting ni,j denote the numerator of the coefficient (i, j)

• f Mα, we have d := (Mα : Z[α]) = 2D/ gcd(ni,j) and

DMα = (D/d)2 = ✓1 2 gcd

4j6

(n3,j) ◆2 , DMα = gcd

• Dα, (a2 3b)2, (b2 3ac)2

.

Z-basis for the orders generated by the conjugates of algebraic integers 9

Theorem 7 (See [LL16]). Assume that the Q(α)/Q is Galois, i.e. assume that Dα = D2 is a square in Z. Set ∆ = gcd(D, 3b a2, 3ac b2), where Πα(X) = X3 aX2 + bX c 2 Z[X]. Let x, y, z 2 Z be such that ∆ = xD + y(3b a2) + z(3ac b2). Then {1, α1, η = xα2

1 + yα2 + zα2α2 1}

is a Z-basis of the order Mα = Z[α1, α2, α3] and DMα = ∆2 = gcd(Dα, (3b a2)2, (3ac b2)2). Corollary 8 Under the assumptions of Theorem 7, the cubic order Z[α] is Gal(Q(α)/Q)-invariant if and

• nly if D divides 3b a2 and 3ac b2.

Conjecture 9 Under the assumptions of Theorem 7, if Z[α] is Gal(Q(α)/Q)-invariant then c is odd. If one could prove this conjecture, then using [Coh, Theorem 6.4.6] one could obtain cyclic cubic number fields with non-monogenic rings of algebraic integers.

Z-basis for the orders generated by the conjugates of algebraic integers 10

6 When is the order Z[ε] Galois-invariant? Theorem 10 Let ε be an algebraic cubic unit. Assume that Πε(X) = X3 aX2 + bX c 2 Z[X], c 2 {±1}, is reduced, i.e. that |b|  a. Then Q(ε) is Galois and the order Z[ε] is Galois in- variant if and only if we are in one of the following cases:

• 1. Πε(X) = X3 4X2 + 3X + 1, X3 6X2 + 5X 1
• r X3 20X2 9X 1, in which cases dε = 72,
• r Πε(X) = X3 9X2 + 6X 1, in which case

dε = 92.

• 2. Πε(X) = X3 aX2 + (a 3)X + 1, a 2, i.e.

if Q(ε) is a so-called simplest cubic field, in which case dε = (a2 3a + 9)2. Question 11 Let α be a cubic algebraic integer for which Q(α)/Q is Galois. If ZQ(α) = Z[α] then the

• rder Z[α] is Galois invariant. This seldom happens

(Theorem 10 and Table (1)). Does anyone know a neccessary and sufficient condition on Πα(X) for having ZQ(α) = Z[α]? In contrast, the order Mα = Z[α1, α2, α3] is always Galois and we often have ZQ(α) = Mα (Table (1)). It would be nice to have a necessary and sufficient condition on Πα(X) for having ZQ(α) = Z[α1, α2, α3].

Z-basis for the orders generated by the conjugates of algebraic integers 11

For a given bound B we computed the number N(B)

• f Q-irreducible cubic polynomials Π(X) = X3 aX2 +

bX c 2 Z[X] with 0  a, |b|, |c|  B and whose discriminants are squares in Z. Let α denote any root of Π(X) and α1, α2, α3 denote its three real roots. We com- puted the number N(α) of these polynomials for which Z[α] = ZQ(α), i.e. for which Dα = DK, the number Ninv(α) of these polynomials for which the order Z[α] is Galois invariant, i.e. for which D diivides 3b a2 and 3ac b2 (Corollary 8), and the number N(α1, α2, α3) of these polynomials for which Z[α1, α2, α3] = ZQ(α), i.e. for which ∆2 = DK (Theorem 7).

B N(B) N(α) Ninv(α) N(α1, α2, α3) 10 62 30 (48.4%) 36 (58.1%) 44 (71.0%) 20 190 64 (33.7%) 77 (40.5%) 137 (72.1%) 30 387 97 (25.1%) 116 (30.0%) 280 (72.4%) 40 613 136 (22.2%) 161 (26.3%) 431 (70.3%) 50 853 168 (19.7%) 202 (23.7%) 592 (69.4%) 100 2506 351 (14.0%) 414 (16.5%) 1686 (67.3%) 200 7125 713 (10.0%) 840 (11.8%) 4663 (65.4%) 300 12762 1071 (8.4%) 1261 (9.9%) 8263 (64.7%) 500 26349 1794 (6.8%) 2117 (8.0%) 16991 (64.5%) 1000 69696 3603 (5.2%) 4266 (6.1%) 44005 (63.1%) (1)

Z-basis for the orders generated by the conjugates of algebraic integers 12

Conjecture 12 Set α = 2 cos(2π/11). Let ε be an algebraic quintic unit. Assume that Πε(X) = X5aX4+bX3cX2+dXe 2 Z[X], e 2 {±1}, is reduced, i.e. that |d|  a. Then Q(ε) is a cyclic quintic number field and the

• rder Z[ε] is Galois invariant if and only if we are in
• ne of the following 8 cases:
• 1. Πε(X) = X5 3X4 3X3 + 4X2 + X 1 = Π −1

α (X).

• 2. Πε(X) = X5 4X4 + 2X3 + 5X2 2X 1 = Π1+α(X).
• 3. Πε(X) = X5 6X4 X3 + 10X2 6X + 1 = Π

1 1−α(X).

• 4. Πε(X) = X5 6X4 + 10X3 X2 6X + 1 = Π1α(X).
• 5. Πε(X) = X5 7X4 + 13X3 5X2 2X + 1 = Π

α α+1(X).

• 6. Πε(X) = X5 8X4 + 19X3 15X2 + X + 1 = Π α−1

α (X).

• 7. Πε(X) = X5 10X4 15X3 3X2 + 3X + 1 = Π −α−1

α+2 (X).

• 8. Πε(X) = X5 15X4 + 35X3 28X2 + 9X 1 = Π

1 α+2(X).

In these 8 cases we have dε = 114 and Z[ε] = ZQ(ζ11)+.

Z-basis for the orders generated by the conjugates of algebraic integers 13

7 On the orders Z[αn] and Mαn, n 1 We would like to understand the behaviors of Dαk and Z[αk], or of DZ[αk

1,··· ,αk n] and Z[αk

1, · · · , αk n]

as k 1 varies. Hence, Z[αm] ✓ Z[αn] and Mm ✓ Mn if m is a multiple of n. Can it happen that for some α’s we have Z[αn] = Z[α] or Mn = M1 for infinitely many n’s? Or even for a positive proportion of n’s?

7.1 The case that α is an algebraic unit

Theorem 13 Let ε be an algebraic unit. Assume that ε is not a complex root of unity. Then dεk goes to infinity as k goes to infinity.

• Proof. (J. Oesterl´

e, personal communication). Since Q(ε) has only finitely many subfields, by consider- ing subsequences if necessary, we may assume that Q(εk) = K for all k 1, where K is a given number field. Set m = (K : Q). Since ε is not a complex root of unity, the εk’s are pairwise distinct elements of Z⇥

• K. Therefore, it

suffices to show that for any given A > 0 the set X = {η 2 Z⇥

K; Q(η) = K and dη  A} is finite. Let K be

the normal closure of K. Let σi, 1  i  m, be the com- plex embeddings of K. Hence, σi(K) ✓ K. Let S be the set of places of K above the rational primes less than or equal to A. Set Y = {η 2 Z⇥

K; η 1 is a S-unit of ZK}.

Z-basis for the orders generated by the conjugates of algebraic integers 14

By Siegel’s theorem, Y is finite. Now, let η 2 X. Then σi(η) η divides dη in ZK. Hence, σi(η) η is a S-unit and so is each σi(η)/η. Therefore, φ : η 2 X ! φ(η) = ✓σ1(η) η , · · · , σn(η) η ◆ 2 Y m is well defined and φ(η) = φ(η0) if and only if η0/η is invariant under the action of all the σi’s, hence if and only if η0 = ±η. Therefore, X is finite and #X  2(#Y )m.

• Question 14 Can anyone prove the same result with-
• ut assuming that the algebraic integer α is a unit?

7.2 The non-normal cubic case

For non-totally real cubic units, we proved: Theorem 15 (See [Lou10, Theorem 1]). Let α be a real cubic algebraic unit of negative discriminant Dα < 0. Then |Dα| max(|α|, |α|1)3/2/2. Consequently, Dαk and DZ[αk

1,αk 2,αk 3] go exponentially to

infinity as k goes to infinity. Proof. Since Q(αk)/Q = Q(α)/Q is not Galois, the group Gal(Q(αk

1, αk 2, αk 3)/Q) is isomorphic to S3 and

DMk = DZ[αk

1,αk 2,αk 3] = D3

αk (Theorem 3). •

Z-basis for the orders generated by the conjugates of algebraic integers 15

For totally real cubic units, we proved: Theorem 16 (See [Lou12] and [Lou15, Theorem 33]). Let α1, α2, α3 be the 3 real conjugates of a totally real cubic algebraic unit α. Then Dα max(|α1|, |α1|1, |α2|, |α2|1, |α3|, |α3|1)3/2/2. Consequently, Dαk goes exponentially to infinity as k goes to infinity and if Q(α)/Q is not Galois, i.e. if Dα is not a square in Z, then DZ[αk

1,αk 2,αk 3] goes expo-

nentially to infinity as k goes to infinity.

7.3 The cyclic cubic case

Now, let us come back to the problem considered in [Lou16b]. We did some computation to determine N(X) := #{n 2 [1, X]; Mn = M1} for various cubic algebraic integers α such that Q(α)/Q is Galois (we used Theorem 7 to compute DM1 and the DMn’s, and we notice that since Mn ✓ M1, we have Mn = M1 if an only if DMn = DM1). According to these computations, and contrary to the non-Galois cubic case, it seems reasonable to conjecture that for any such α there is a positive proportion of n’s for which Mn = M1:

Z-basis for the orders generated by the conjugates of algebraic integers 16 Πα(X) fα X N(X)/X X3 3X2 4X 1 7 102 0.49 103 0.479 104 0.4645 105 0.45166 X3 X2 4X 1 13 102 0.48 103 0.464 104 0.4553 105 0.44489 X3 10X2 + 7X + 1 79 102 0.47 103 0.467 104 0.4651 105 0.45369 X3 43X2 + 40X + 1 7 · 13 · 19 102 0.55 103 0.492 104 0.4817 105 0.47455 X3 31X2 25X 1 23 · 7 · 13 102 0.30 103 0.305 104 0.2968 105 0.28942 X3 54X2 + 69X 1 32 · 5 · 7 · 11 102 0.18 103 0.161 104 0.1547 105 0.15172 X3 15X2 + 14X 3 61 102 0.64 103 0.587 104 0.5655 105 0.55305 X3 24X2 + 23X 5 132 102 0.65 103 0.621 104 0.5938 105 0.58007 X3 33X2 + 32X 7 331 102 0.72 103 0.624 104 0.5974 105 0.57832

Z-basis for the orders generated by the conjugates of algebraic integers 17

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