Efficient Computation of Class Numbers of Real ... - Semantic Scholar
Efficient Computation of Class Numbers of Real Abelian Number Fields St´ephane R. Louboutin Institut de Math´ematiques de Luminy, UPR 906 163, avenue de Luminy, Case 907 13288 Marseille Cedex 9, FRANCE [email protected]
Abstract. Let {Km } be a parametrized family of real abelian number fields of known regulators, e.g. the simplest cubic fields associated with the Q-irreducible cubic polynomials Pm (x) = x3 − mx2 − (m + 3)x − 1. We develop two methods for computing the class numbers of these Km ’s. As a byproduct of our computation, we found 32 cyclotomic fields Q(ζp ) of prime conductors p < 1010 for which some prime q ≥ p divides the + class numbers h+ p of their maximal real subfields Q(ζp ) (but we did not find any conterexample to Vandiver’s conjecture!).
1
Introduction
This paper is an abridged version of [Lou5] in which the reader will find the proofs we omit here, and in which he will also find various supplementary examples (families of real cyclic quartic, sextic and octic fields). Our aim is to explain how one can generalize the technique developed in [Lou1] not only to compute efficiently class numbers of real abelian number fields of known regulators, but also to compute efficiently exact values of Gauss sums and roots numbers associated with primitive Dirichlet characters of large conductors. In [Bye], [Lou4], [LP], [Sha] and [Wa], various authors dealt with the so called simplest cubic fields, the real cyclic cubic number fields Km associated with the cubic polynomials Pm (x) = x3 − mx2 − (m + 3)x − 1 of discriminants dm = ∆2m , where ∆m := m2 + 3m + 9, and roots θm , σ(θm ) = −1/(θm + 1) and σ 2 (θm ) = −(θm + 1)/θm . Since −x3 Pm (1/x) = P−m−3 (x), we may assume that m ≥ −1. In this paper, we assume that ∆m is square-free. In that case, the conductor of Km is equal to ∆m , its discriminant is equal to ∆2m , the set {−1, θm , σ(θm )} generates the full group of algebraic units of Km and the regulator of Km is RegKm = log2 θm − (log θm )(log(1 + θm )) + log2 (1 + θm ), with θm
√ 1 27 1 2 ∆m cos = arctan( ) +m . 3 3 2m + 3
C. Fieker and D.R. Kohel (Eds.): ANTS 2002, LNCS 2369, pp. 134–147, 2002. c Springer-Verlag Berlin Heidelberg 2002
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In [Jean] and [SW], S. Jeanin, R. Schoof and L. C. Washington dealt with the so called simplest quintic fields, the real cyclic quintic number fields Km associated with the quintic polynomials Pm (x) = x5 + m2 x4 − (2m3 + 6m2 + 10m + 10)x3 +(m4 + 5m3 + 11m2 + 15m + 5)x2 + (m3 + 4m2 + 10m + 10)x + 1 of discriminants dm = (m3 + 5m2 + 10m + 7)2 ∆4m , ∆m = m4 + 5m3 + 15m2 + 2 )/(1 + (m + 2)θm ), σ 2 (θm ), 25m + 25 and roots θm , σ(θm ) = ((m + 2) + mθm − θm 3 4 σ (θm ) and σ (θm ). In this paper, we assume that ∆m is square-free. In that case, the conductor of Km is equal to ∆m , its discriminant is equal to ∆4m , the set {−1, θm , σ(θm ), σ 2 (θm ), σ 3 (θm )} generates the full group of algebraic units of Km and the regulator of Km is 1 ij RegKm = ζ5 log |σ j (θm )| . 5 1≤i≤4 0≤j≤4
Since Pm (m + 1)Pm (m + 2) = −(m3 + 5m2 + 10m + 7)2 < 0 we can use Newton’s method for computing efficiently as good as desired numerical approximations of a root θm ∈ (m + 1, m + 2) of Pm (x). Then, the four other roots are computed inductively by the transformation θ → σ(θ) := ((m+2)+mθ−θ2 )/(1+(m+2)θ). One of the motivation for computing class numbers of simplest cubic and quintic fields stems from Vandiver’s conjecture according to which p never divides + the class number h+ p of the maximal real subfield Q(ζp ) = Q(cos(2π/p)) of a cyclotomic field Q(ζp ) of prime conductor p. However, as the computation of h+ p is impossible to perform (except for very small values of p, say p ≤ 67 (see [Wa, Tables, pages 420-423])), the idea is to compute class numbers hK of subfields K of Q(ζp )+ of small degrees: Theorem 1. (i). Let p ≡ 1 (mod 12) be a prime and let h2 , h3 and h6 denote the class numbers of the real quadratic, cubic and sextic subfields of the cyclotomic field Q(ζp ). Then, h2 h3 divides h6 and h6 divides the class number h+ p of the maximal real subfield Q(ζp )+ = Q(cos(2π/p)) of Q(ζp ) (see [CW, Lemmas 1 and 2]). However, all the prime factors q of h6 are less than p (see [Mos]). (ii). Let p ≡ 1 (mod 10) be a prime and let h5 denote the class number of the real quintic subfield of the cyclotomic field Q(ζp ). Then, h5 divides h+ p. Since the simplest cubic and quintic fields have small regulators we might expect to find some of them of prime conductors and large class numbers. Therefore, by using simplest cubic fields we might expect to find examples of cyclotomic fields of prime conductors p for which h+ p ≥ p but for which, unfortunately, all could be less than p. Up to now, only one such example the prime factors q of h+ p had been found (see [CW] and [SWW]), and we will find three more examples (see Table 3). In the same way, by using simplest quintic fields we might expect to find examples of cyclotomic fields of prime conductors p for which some prime factor q of h+ p satisfies q ≥ p. Up to now, only one such example had been found (see [SW] and [Jean]), and we will find 31 more examples (see Table 2).
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First Method for Computing Class Numbers
Let K be a real abelian number field of degree q > 1, discriminant dK and conductor fK associated with a Q-irreducible unitary polynomial PK (X) = X q + aq−1 X q−1 + · · · + a0 ∈ Z[X]. Let XK denote the group (of order q) of primitive even Dirichlet characters associated with K and let RegK denote the regulator of K. According to the analytic class number formula (see [Lan, Chapter XIII, section 3, Th. 2 page 259]), s → FK (s) = (dK /π q )s/2 Γ q (s/2)ζK (s) has a simple pole at s = 0 of residue Ress=0 (FK (s)) = −2q−1 hK RegK = 2q lim sq−1 ζK (s). s→0
Since ζK (s) = χ∈XK L(s, χ) and L(0, χ) = −1/2 if χ = 1 but L(0, χ) = 0 for 1 = χ ∈ XK , we obtain hK RegK = L (0, χ). (1) 1=χ∈XK
Lemma 1. (See [Sta]). If χ is a (non-necessarily primitive) non-trivial even Dirichlet character modulo f > 1, then L (0, χ) = − χ(k) log sin(kπ/f ). 1≤k 1 and known regulators (it will practically require only O(fK ) elementary operations to compute hK , whereas our previous techniques based on (1) 1+ and (2) requires O(fK ) elementary operations to compute hK ). The idea is to generalize [WB, Section 3] to compute efficiently good enough numerical approximations to L (0, χ) for 1 = χ ∈ XK , and to use (1). Let χ be a primitive even Dirichlet character modulo f > 1. Set χ(n) exp(2nπi/f ) (Gauss sum), (3) τ (χ) = 1≤n≤f
W (χ) = τ (χ)/ and
θ(x, χ) =
f
χ(n)e−πn
(root number) 2
x/f
(4)
(x > 0).
(5)
(x > 0),
(6)
n≥1
Then, |W (χ)| = 1 and using √ θ(1/x, χ) = W (χ) xθ(x, χ) ¯
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we obtain (f /π)s/2 Γ (s/2)L(s, χ) ∞ ∞ ∞ dx dx dx = + W (χ) θ(x, χ)xs/2 θ(x, χ)xs/2 θ(x, χ)x ¯ (1−s)/2 , = x x x 0 1 1 L(0, χ) = 0 and the following result which enables us to compute numerical approximations to L (0, χ) to any prescribed accuracy: Theorem 2. Let χ be a primitive even Dirichlet character modulo f > 1. Then,
∞ ∞ 2 f χ(n) ¯ 1 −t dt L (0, χ) = χ(n) e e−t dt. (7) + W (χ) √ 2 t π n πn2 /f πn2 /f n≥1
Hence, setting
n≥1
E1 (x) :=
∞
x
e−t
(−1)k dt = − log x − γ − xk t k · k! k≥1 1 1 1 2 2 3 3 ··· = e−x z+ 1+ z+ 1+ z+ 1+ z+
(where γ = 0.577 215 664 901 532 · · · denotes Euler’s constant),
∞ 2 x (−1)k 2 E2 (x) := √ √ e−t dt = 1 − 2 xk π (2k + 1) · k! π x k≥0
1 1 2 3 4 5 1 2 2 2 2 2 = √ e−x ··· , z+ z+ z+ z+ z+ z+ π and LN (0, χ)
√ 1 W (χ) f 2 = χ(n)E1 (πn /f ) + 2 2 1≤n≤N
χ(n) ¯ E2 (πn2 /f ) n
1≤n≤N
(N ≥ 1 a positive integer), it holds that |L (0, χ) − LN (0, χ)| ≤
1 √
1
f 2 −t
2M t πt3 log3/2 (M f )
for
N ≥ B(t, f, M ) :=
tf log(M f ). π
(8)
Corollary 1. (See [Lou3, Proof of Theorem 7]). Let q ≥ 2 be a given prime. Fix t > (q − 1)/2 and M > 0, and let K range over a family of real abelian numbers fields K of degree q for which all the root numbers W (χ), 1 = χ ∈ −→ ∞ and for N ≥ B(t, fK , M ), the limit XK , are known. Then, as f1K 1 | Reg L (0, χ) − 1=χ∈XK 1=χ∈XK LN (0, χ)| is equal to zero. Reg K
K
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Efficient Computation of Root Numbers
According to Corollary 1, (1) and Theorem 2 could be used to compute efficiently numerical approximations LN (0, χ) to L (0, χ) for primitive even Dirichlet characters of order nχ > 1 and class numbers of real abelian number fields. However, as there is no known general formula for Gauss sums (see [BE]), we will now explain how to compute efficiently these root numbers W (χ) (notice that since the use of (4) to compute the exact value of W (χ) requires fχ elementary operations, it would be much simpler to use Theorem 1 than to use (4) and Theorem 2). We point out that we are going to end up with a method for computing class numbers of real abelian number fields which is more satisfactory than the one previously used (see [SW] and [SWW]): we compute exact values of root numbers, whereas in [SWW] they had three choices for W (χ) for simplest cubic fields of a given prime conductor and in [SW, Top of page 553] they had twenty choices for W (χ) for simplest quintic fields of a given prime conductor. In their computations they were lucky enough for in all cases considered only one of their possible choices gave rise to an approximation of the class number sufficiently close to an integer. To begin with, let us fix some notation. Throughout this fourth section, we let χ denote a primitive even Dirichlet character or order nχ > 1 and conductor fχ > 1. We set ω(χ) = (τ (χ))nχ , ζχ = exp(2πi/nχ ) and Q(χ) = Q(ζχ ). We let φχ = φ(nχ ) and Z[χ] = Z[ζχ ] denote the degree and the ring of algebraic integers of the cyclotomic field Q(χ), respectively. Finally, for any l relatively prime to nχ , we let σl denote the Q-automorphism of Q(χ) which is defined by σl (ζχ ) = ζχl . Notice that if gcd(l, nχ ) = 1, then χl is also a primitive Dirichlet character of order nχ and conductor fχ and that χl is even (respectively odd) if χ is even (respectively odd). Theorem 3. Let χ be a primitive Dirichlet character of conductor fχ > 1 and order nχ > 1. Then ω(χ) := (τ (χ))nχ = fχnχ /2 (W (χ))nχ ∈ Z[χ]
(9)
and σl (ω(χ)) = ω(χl ) for gcd(l, nχ ) = 1. Moreover, if nχ is prime and fχ is square-free, then ω(χ) ∈ fχ Z[χ]. 4.1
Exact Computation of ω(χ)
Fix a Z-basis B = {/1 , · · · , /φχ } and write b(k, χ)/k ∈ Z[χ]. ω(χ) =
(10)
1≤k≤φχ
with b(k, χ) ∈ Z, 1 ≤ k ≤ φχ . Let B ⊥ = {η1 , · · · , ηφχ } be the dual basis of B, relative to the trace form. Hence, TrQ(χ)/Q (/k ηk ) = 1 but TrQ(χ)/Q (/k ηl ) = 0 for k = l, and b(k, χ) = TrQ(χ)/Q (ηk ω(χ)) = fχnχ /2 σl (ηk )(W (χl ))nχ (11) 1≤l≤nχ gcd(l,nχ )=1
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(for σl (ω(χ)) = ω(χl )), and these coordinates b(k, χ) are rational integers of n /2 reasonable size: |b(k, χ)| ≤ M (B ⊥ )φχ fχ χ where M (B ⊥ ) = max{|σl (ηj )|; 1 ≤ l ≤ nχ , gcd(l, nχ ) = 1, 1 ≤ j ≤ φχ }.
(12)
For example, if nχ = q ≥ 3 is prime, then B ⊥ = {ηl := (ζq−l −1)/q; 1 ≤ l ≤ q −1} is the dual basis of the Z-basis B = {ζqk ; 1 ≤ k ≤ q − 1} of the ring of algebraic integers Z[ζq ] of Q(ζq ), and M (B ⊥ ) ≤ 2/q ≤ 1. Now assuming that 2 Hypothesis: θ(χl ) := χl (n)e−πn /fχ = 0 (13) n≥1
for 1 ≤ l ≤ nχ and gcd(l, nχ ) = 1, we explain how one can compute efficiently as good as desired numerical approximations bN (k, χ) to these coordinates b(k, χ) ∈ Z of ω(χ), hence how one can compute their exact values. The key point is that θ(χl ) = 0 implies W (χl ) = θ(χl )/θ(χl ), by (6). According to Section 4.4 below, this Hypothesis should always be satisfied. The following Lemma 2 will enable us to compute as good as desired numerical approximations θN (χl ) to θ(χl ). These approximations will then enable us to check the Hypothesis (13) prior to using Lemma 3 for computing as good as desired numerical approximations bN (k, χ) to the rational integers b(k, χ) defined in (10), whose exact values can therefore be deduced. Lemma 2. Let χ be a Dirichlet character modulo f > 1. Set 2 χ(n)e−πn /f θN (χ) = 1≤n≤N
(N ≥ 1 a positive integer). If N ≥ B(t, f, M ) (as in (8)), then |θ(χ) − θN (χ)| ≤
1 √
2M t πt
1
f 2 −t
log(M f )
.
(14)
Lemma 3. Let χ be a primitive even Dirichlet character of order nχ > 1 and conductor fχ > 1. Assume that θN (χl ) = 0 for gcd(l, nχ ) = 1, set WN (χl ) = θN (χl )/θN (χl ) and bN (k, χ) = fχnχ /2
σl (ηk )(WN (χl ))nχ
(1 ≤ k ≤ φχ ),
(15)
1≤l≤nχ gcd(l,nχ )=1
and fix / such that 0 ≤ / ≤ 1. Assume that |θ(χl ) − θN (χl )| ≤ /|θN (χl )|/nχ for 1 ≤ l ≤ nχ and gcd(l, nχ ) = 1. Then, |bN (k, χ) − b(k, χ)| ≤
27(e − 1) M (B ⊥ )φχ fχnχ /2 /. 4
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Proof. Let us simplify the notation: we set n = nχ , θ = θ(χl ), θN = θN (χl ), W = W (χl ) = θ/θ¯ (notice that θN = 0 and |θ − θN | ≤ /|θN |/n imply θ = 0), WN = WN (χl ) = θN /θ¯N and write θ = θN + /N θN with |/N | ≤ //n. Then, n | k=1 nk (/kN + /¯kN )| |(1 + /N )n − (1 + /¯N )n | n n = |W − WN | = |1 + /¯N |n |1 − /N |n yields |W n − WNn | ≤ 2
n
n
|/N |k (1 + |/N |)n − 1 (1 + //n)n − 1 =2 ≤2 . n n (1 − |/N |) (1 − |/N |) (1 − 1/n)n k=1
k
Since (1 − 1/n)n ≥ (1 − 1/3)3 = 8/27 for n ≥ 3 and since (1 + //n)n − 1 ≤ e − 1 ≤ (e − 1)/ for 0 ≤ / ≤ 1, we obtain |W (χl )n − WN (χl )n | ≤ 27(e − 1)//4 for 1 ≤ l ≤ nχ and gcd(l, nχ ) = 1, and the desired results, by (11), (12) and (15). 4.2
Exact Computation of W (χ) and τ (χ)
Now that we know how to compute the exact value of ω(χ) := (τ (χ))nχ , let us explain how one can determine which of its nχ th root is equal to τ (χ): Lemma 4. Fix / ∈ (0, 1]. Let χ be a primitive even Dirichlet character of order nχ > 2 and conductor fχ > 1. Assume that ω(χ) is known and that N is such that θN (χ) = 0 and |θ(χ) − θN (χ)| ≤ /|θN (χ)|/nχ . Fix W a nχ th root of (W (χ))nχ = ω(χ)/f nχ /2 . Then, W (χ) = ζχk0 W where k0 in the unique integer k ∈ {0, 1, · · · , n − 1} such that |WN (χ) − ζχk W | < 2//nχ (and it holds that |WN (χ) − ζχk W | > (4 − 2/)/nχ ≥ 2//nχ for k = k0 ). Proof. Since |θ − θN | ≤ /|θN |/nχ , we have θ = 0, θN = 0 and ¯ + θ(θ¯N − θ)| ¯ |θ − θN | θ θN |θ(θ¯N − θ) 2/ |W (χ) − WN (χ)| = | ¯ − ¯ | = ≤2 . ≤ |θN | nχ |θ¯θ¯N | θ θN There exists a unique k0 ∈ {0, 1, · · · , n − 1} such that W (χ) = ζχk0 W . Since for a = b we have |ζχa W − ζχb W | = |ζχa − ζχa | ≥ 2 sin(π/nχ ) > 4/nχ , we have |WN (χ) − ζχk0 W | = |WN (χ) − W (χ)| < 2//nχ and |WN (χ) − ζnk W | = |(WN (χ) − W (χ))+(ζχk0 W −ζχk W )| ≥ 2 sin(π/nχ )−|WN (χ)−W (χ)| > 4/n−2//nχ ≥ 2//nχ for k = k0 . 4.3
Computation of Class Numbers of Simplest Quintic Fields
First, we checked our present method by recomputing Table 1. Second, we used it to compute the class numbers of all the simplest quintic fields Km ’s of conductors ∆m = m4 + 5m3 + 15m2 + 25m + 25 ≤ 1010 a prime. We obtain the following consequence: there are 32 simplest quintic fields Km of prime conductors p ≤ 1010 whose class numbers are divisible by some prime q ≥ p (see Table 2). Third,
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St´ephane R. Louboutin
we used it to compute the class numbers of all the simplest cubic fields Km ’s with −1 ≤ m ≤ 554869 and ∆m = m2 + 3m + 9 ≡ 1 (mod 12) a prime. We obtain the following consequence: in the range −1 ≤ m ≤ 554869, there are only 4 simplest cubic fields Km of prime conductors ∆m = m2 + 3m + 9 ≡ 1 (mod 12) for √which the product h2 h3 of the class number h2 of the real quadratic field Q( ∆m ) and of the class number h3 of the simplest cubic field Km of conductor ∆m is greater than or equal to ∆m (see Table 3). 4.4
A Conjecture
According (i) to our numerical evidence (the computation of approximations to θ(χ) for the 10582203 primitive even Dirichlet characters χ of prime conductors p ≤ 20000 and for numerous examples of cubic, quartic and quintic primitive even Dirichlet characters of (non necessarily prime) large conductors associated with simplest cubic, quartic and quintic fields), and (ii) to the fact that as p ≥ 5 ranges over the odd primes it holds that χ∈Xp+ |θ(χ)|2 is asymptotic to √ p3/2 /(4 2) and that θ(χ) = 0 for at least p/ log p of the (p − 3)/2 characters 1 = χ ∈ Xp+ (adapt the proof of [Lou2, Theorem 1]), we put forward the following conjecture: Conjecture 1. (i) (See Hypothesis (13)). For any primitive even Dirichlet character of conductor fχ > 1 it holds that θ(χ) = 0. (ii) Let p ≥ 5 denote an odd prime and let Xp+ denote the set of order (p−3)/2 of the primitive even Dirichlet characters modulo p. For a ≥ 0 real, the limit g+ (a) = lim
p→∞
2 #{χ ∈ Xp+ ; |θ(χ)| ≤ ap1/4 } p−3
exits, a → g+ (a) is continuous, strictly increasing, g+ (0) = 0 and g+ (∞) = 1. Now, fix t0 < 1/4. Then, at least for real cyclic fields K of a given prime t0 degree q and of large prime conductors fK , we might expect to have |θ(χ)| ≥ fK for all the 1 = χ ∈ XK . In that case, for t > 1/2 − t0 we can use (14) with N ≥ B(t, fK , M ) to check numerically that θ(χ) = 0 for all the 1 = χ ∈ XK . Then, for t > (q + 1)/2 − t0 we can use Lemma 3 with N ≥ B(t, fK , M ) to compute the exact value of ω(χ) for all the 1 = χ ∈ XK . Finally, for t > 1/2 − t0 we can use Lemma 4 to compute the exact value of W (χ) for all the 1 = χ ∈ XK . Hence, according to Corollary 1, we might expect that our second method for computing class numbers of real abelian number fields K of a given degree and 0.5+ known regulators requires only O(fK ) elementary operations. In practice, it is indeed amazingly efficient and of the conjectured complexity.
5
Explicit Formulae for ω(χ)
Finally, we explain how we can dispense with Subsection 4.1 when dealing with simplest cubic and quintic fields: we know beforehand ω(χKm ) and we can use Lemma 4 for computing the root number W (χKm ), making in these two cases our method for computing class numbers simpler and faster.
Efficient Computation of Class Numbers of Real Abelian Number Fields
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5.1
Simplest Cubic Fields √ Set ω = (−1+i 3)/2. The units in Z[ω] are {±1, ±ω, ±ω 2 }. An algebraic integer α = a + bω ∈ Z[ω] is primary if α ≡ −1 (mod 3Z[ω]), i.e. if a ≡ −1 (mod 3) and b ≡ 0 (mod 3). The order of the multiplicative group (Z[ω]/3Z[ω])∗ is equal to 6 and the six units in Z[ω] form a set of representatives of this group. Therefore, if α ∈ Z[ω] and 3 does not divide its norm N (α) = αα ¯ , then exactly one of its six associates is primary. If follows that if 0 = α ∈ Z[ω] is a nonunit element such that α ≡ (−1)t (mod 3Z[ω]), where t denotes the number of irreducible factors of α (counted with multiplicity), then α can be written in a unique way as a product of primary irreducibles. Now, let π ∈ Z[ω] be a primary irreducible element of norm a rational prime p ≡ 1 (mod 3). For α ∈ Z[ω] coprime with π, let χπ (α) ∈ {1, ω, ω 2 } be the cubic residue symbol defined by α(p−1)/3 ≡ χπ (α) (mod π). Then, τ (χπ )3 = pπ (see [IR, Corollary page 115]). It follows: Theorem 4. Assume that ∆m = m2 + 3m + 9 is square-free, write ∆m = t k=1 pk where the pk ’s are distinct odd primes and set m 2m + 3 + 3i√3 ≡ (−1)t (mod 3Z[ω]). δm := (−1)t 3 2 Then, δm can be written in a unique way as a product δm = 1≤k≤t πk of t primary irreducibles elements πk ∈ Z[ω] with pk = |πk |2 . Set χδm = 1≤k≤t χπk . Then, χδm is a primitive cubic character modulo ∆m , and (16) ω(χδm ) := τ (χδm )3 = ∆m δm . t m Hence, setting /m = (1 − (−1) 3 )/2 ∈ {0, 1}, there exists km ∈ {0, 1, 2} such that √ 3 3 2km + /m 1 + π (mod 2π), (17) arg(W (χδm )) ≡ arctan 3 2m + 3 3 and, if θ(χδm ) = 0, then km can be efficiently computed by using Lemma 4. Moreover, χδm is associated with the simplest cubic field Km , i.e. the character χKm associated with Km obtained by using the technique developed in Section ¯ δm . 2 is equal either to χδm or to its conjugate character χ Since Pm (x) = x3 − mx2 − (m + 3)x − 1 has no root in the finite field with two elements, we have 1 = χKm (2) ∈ {ω, ω 2 } where χKm is the cubic character associated with Km obtained by using the technique developed in Section 2. According to the law of cubic reciprocity (see [IR, Theorem 1, page 114]), we have
2 ω (mod 2Z[ω]) if m ≡ 0 (mod 2) χδm (2) = χ2 (δm ) ≡ δm ≡ ω (mod 2Z[ω]) if m ≡ 1 (mod 2), hence
χδm (2) =
ω2 ω
if m ≡ 0 if m ≡ 1
(mod 2) (mod 2).
Hence, by computing χKm (2) and by changing χKm into its conjugate if necessary, we may assume that χKm (2) = χδm (2), which implies χKm = χδm .
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Simplest Quintic Fields
In the same way, we have: Theorem 5. Assume that ∆m = m4 + 5m3 + 15m2 + 25m + 25 is square-free. Since Pm (x) = x5 + m2 x4 − (2m3 + 6m2 + 10m + 10)x3 + (m4 + 5m3 + 11m2 + 15m + 5)x2 + (m3 + 4m2 + 10m + 10)x + 1 has no root in the finite field with two elements, we may assume that the quintic character χKm associated with the simplest quintic field Km (obtained by using the technique developed in Section 2) satisfies
3 ζ5 if m ≡ 0 (mod 2) χKm (2) = ζ54 if m ≡ 1 (mod 2). In that case, it holds that ω(χKm ) := τ (χKm )5 = (−1)t−1 ( m 5 )∆m δm where δm = (m6 + 5m5 + 5m4 + 25m2 + 125m + 125)ζ5 +(m6 + 5m5 − 5m4 − 75m3 − 175m2 − 125m)ζ52
+(m6 + 10m5 + 25m4 − 100m2 − 125m)ζ53 +(m6 + 10m5 + 40m4 + 75m3 + 50m2 )ζ54 ∈ Z[ζ5 ].
6
Simplest Cubic Fields and Class Numbers of the Maximal Real Subfields of Some Cyclotomic Fields
2 Lm = Assume √ that ∆m = m + 3m + 9 ≡ 1 (mod 12) is square-free. Let √ Km ( ∆m ) denote the compositum of the real quadratic field km = Q( ∆m ) and of the simplest cubic field Km , both of conductor ∆m . Then, Lm is a so-called simplest sextic field of conductor ∆m associated with the sextic polynomial
Pm (x) = x6 − 2mx5 − 5(m + 3)x4 − 20x3 + 5mx2 + 2(m + 3)x + 1, and a subgroup of finite index QLm (dividing 12) of the group of algebraic units of Lm is known (see [Gra]). Using this subgroup, and following the proof of [Lou4, Theorem 4], we obtain: Theorem 6. (See [Lou5]). Assume that ∆m = m2 + 3m + 9 ≡ 1 square-free. Then, ∆2m hLm ≥ . 15e log6 (4∆m )
(mod 12) is (18)
In particular, for m ≥ 105 it holds that hLm > ∆m . Notice that, in the special case that ∆m = m2 + 3m + 9 ≡ 1 (mod 12) is prime, we have hkm = h2 , hKm = h3 and hLm = h6 , with the notation of Theorem 1. With this notation, we have seen that the simplest cubic fields Km for which ∆m = m2 + 3m + 9 ≡ 1 (mod 12) is prime and such that h2 h3 ≥ ∆m are few and far between (see Table 3). However, according to the previous lower bound for hLm , as soon as m is large enough we have h6 ≥ ∆m . Moreover, using this lower bound for hLm and following the proof of [CW, Theorem 2] (see [CW, Page 269]), we obtain:
Efficient Computation of Class Numbers of Real Abelian Number Fields
145
Corollary 2. (See [CW, Theorem 2] for a worse and non-effective result). Let / > 0 be given. Set c = 13 p≡1 (mod 3) (1 − 2p−2 ) = 0.311 · · · . For at least (c + o(1))x1/2 positive odd square-free integers f ≤ x is holds that the class + numbers h+ f of the maximal real subfields Q(ζf ) of the cyclotomic fields Q(ζf ) + 2− , and the constants involved in these o(1) and are explicit. satisfy hf f Proof. Let f range over the positive square-free integers of the form f = ∆m := m2 + 3m + 6 ≡ 1 (mod 12).
7
Tables Table 1. class numbers hKm of the simplest quintic fields Km of square-free conductors ∆m < 107 m −1, −2 −3 1 −4 2 3 −6 4 −7 −8 6 −9 7 8 −11 9 −12 −13 11 −14 12 13 −17 −18 16 −19 17 18
∆m 11 31 71 101 191 451 631 941 1271 2321 3091 3931 5051 7841 9951 11671 13981 19811 23411 27311 31861 42431 62891 80251 90281 100991 112871 139471
hKm 1 1 1 1 11 5 11 16 55 305 80 256 1451 421 541 655 1375 4705 2000 7255 9680 9455 9455 37631 19301 203305 83275 32605
m −21 19 −22 21 −24 22 23 −26 24 −27 −28 26 −29 27 28 −31 29 −32 −33 31 −34 32 33 −36 34 −37 −38 36
∆m 154291 170531 187751 247951 270721 295331 349211 378611 410161 443311 515981 555671 597251 641491 736901 788231 842591 899321 1021771 1087691 1156331 1228601 1382791 1464901 1551071 1640531 1831511 1933261
hKm 108691 44605 76901 308605 153005 478775 186091 189305 591775 289025 2372005 721151 540905 1566401 1764400 1217821 760055 798256 4680055 1386275 1402000 4822625 2148080 4628591 2160455 1636721 11812625 3869525
m −39 37 38 −41 39 −42 −43 41 −44 42 43 −46 −47 −48 46 −49 47 48 −51 49 −53 51 −54 52 53 −56 54 −57
∆m 2038711 2148911 2382131 2505371 2633851 2766691 3047951 3196631 3350141 3509671 3845171 4021391 4392551 4788841 4997051 5211371 5433131 5897161 6139711 6390311 7186931 7468771 7758151 8056541 8678351 9002081 9335491 9677371
hKm 4521505 27105755 6728105 6340275 7503505 20599841 24153305 8088176 6495280 61395955 17264525 21321025 12722855 49860400 42769375 56285605 88151275 17478875 21966025 74338555 155197205 28850896 37142851 118690480 44646025 106111555 54898055 73297775
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St´ephane R. Louboutin
Table 2. the simplest quintic fields Km of prime conductors ∆m = p < 1010 for which some prime q ≥ p (in bold face letters) divides their class numbers hKm m 27 −61 66 73 77 −84 −88 −99 −102 −121 122 128 129 139 −147 −163 162 178 −187 −237 238 242 −249 −263 −264 268 271 282 291 293 −303 −312
∆m 641491 12765251 20479231 30425111 37526591 46927381 56676161 91352671 103090711 205717691 230839031 279170201 287909191 387022451 451386751 684652511 710402911 1032554351 1190654831 3089232931 3276804731 3501489071 3767856571 4694424311 4766572561 5256015221 5494201451 6437395351 7295360131 7497114671 8291171431 9325450081
hKm 1566401 66431941 182277211 335434451 3233114891 2068985771 5912208301 3144379001 3626779141 11420513591 60390377311 24178878281 32215474121 42590939281 155312785456 785372557471 421336924016 320881058831 259187494511 1634411025661 3314877124271 4793050096976 2253716261071 9653048507861 3419567237581 4240933367591 6532834598131 18156246542621 5988407760191 10748665628261 25938285252521 15721799752591
= 1566401 = 66431941 = 61 · 2988151 = 11 · 30494041 = 3233114891 = 2068985771 = 5912208301 = 3144379001 = 3626779141 = 11420513591 = 11 · 5490034301 = 24178878281 = 32215474121 = 11 · 3871903571 = 24 · 9707049091 = 41 · 19155428231 = 24 · 26333557751 = 320881058831 = 11 · 23562499501 = 1634411025661 = 71 · 46688410201 = 24 · 299565631061 = 11 · 204883296461 = 11 · 887549864351 = 112 · 2860886261 = 151 · 28085651441 = 6532834598131 = 11 · 1650567867511 = 5988407760191 = 112 · 88831947341 = 2311 · 11223836111 = 41 · 383458530551
Table 3. least values of m ≥ −1 for which ∆m is prime and h2 h3 ≥ ∆m m 102496 106253 319760 554869
∆m 10505737513 11290018777 102247416889 307881271777
|θ(χδm )| arg W (χδm )√ 3 3 20.268 · · · 13 arctan( 2m+3 )+ √ 1 3 3 34.364 · · · 3 arctan( 2m+3 ) √ 3 3 202.162 · · · 13 arctan( 2m+3 ) √ 1 3 3 88.861 · · · 3 arctan( 2m+3 ) +
π 3
π 3
h2 891 2685 1887 7983
h3 13152913 6209212 57772549 93739324
h2 h3 /∆m 1.115 · · · 1.476 · · · 1.066 · · · 2.430 · · ·
Efficient Computation of Class Numbers of Real Abelian Number Fields
147
References Ba.
E. Bach. Explicit bounds for primality testing and related problems. Math. Comp. 55 (1990), 355–380. BE. B.C. Berndt and R.J. Evans. The determination of Gauss sums. Bull. Amer. Math. Soc. 5 (2) (1981), 107–129. Corrigendum im 7 (2) (1982), 411. Bye. D. Byeon. Class number 3 problem for the simplest cubic fields. Proc. Amer. Math. Soc. 128 (2000), 1319–1323. CW. G. Cornell and L. C. Washington. Class numbers of cyclotomic fields. J. Number Theory 21 (1985), 260–274. Gra. M. N. Gras. Special units in real cyclic sextic fields. Math. Comp. 48 (1988), 543–556. IR. K. Ireland and M. Rosen. A classical introduction to modern number theory. Second edition. Graduate Texts in Mathematics, 84. Springer-Verlag, New York, 1990. Jean. S. Jeannin. Nombre de classes et unit´es des corps de nombres cycliques quintiques d’E. Lehmer. J. Th´eor. Nombres Bordeaux 8 (1996), no. 1, 75–92. Lan. S. Lang. Algebraic Number Theory. Second edition. Graduate Texts in Mathematics, 110. Springer-Verlag, New York, 1994. Lou1. S. Louboutin. Computation of relative class numbers of imaginary abelian number fields. Experimental Math. 7 (1998), 293–303. Lou2. S. Louboutin. Sur le calcul num´erique des constantes des ´equation fonctionnelles des fonctions L associ´ees aux caract`eres impairs. C. R. Acad. Sci. Paris 329 (1999), 347–350. Lou3. S. Louboutin. Computation of relative class numbers of CM-fields by using Hecke L-functions. Math. Comp. 69 (2000), 371–393. Lou4. S. Louboutin. The exponent three class group problem for some real cyclic cubic number fields. Proc. Amer. Math. Soc. 130 (2002), 353–361. Lou5. Efficient computation of root numbers, Gauss sums, and class numbers of real abelian number fields. In preparation. LP. F. Lemmermeyer and A. Peth¨ o. Simplest cubic fields. Manuscripta Math. 88 (1995), 53–58. Mos. C. Moser. Nombre de classes d’une extension cyclique r´eelle de Q de degr´e 4 ou 6 et de conducteur premier. Math. Nachr. 102 (1981), 45-52. Sha. D. Shanks. The simplest cubic fields. Math. Comp. 28 (1974), 1137–1152. Sta. H. M. Stark. Dirichlet’s class-number formula revisited. Contemp. Math. 143 (1993), 571–577. SW. R. Schoof and L.C. Washington. Quintic polynomials and real cyclotomic fields with large class numbers. Math. Comp. 50 (1987), 179–182. SWW. E. Seah, L.C. Washington and H.C. Williams. The calculation of a large cubic class number with an application to real cyclotomic fields. Math. Comp. 41 (1983), 303–305. Wa. L. C. Washington. Class numbers of the simplest cubic fields. Math. Comp. 48 (1987), 371–384. WB. H. C. Williams and J. Broere. A computational technique for evaluating L(1, χ) and the class number of a real quadratic field. Math. Comp. 30 (1976), 887-893.
Abstract. Let {Km } be a parametrized family of real abelian number fields of known regulators, e.g. the simplest cubic fields associated with the Q-irreducible cubic polynomials Pm (x) = x3 − mx2 − (m + 3)x − 1. We develop two methods for computing the class numbers of these Km ’s. As a byproduct of our computation, we found 32 cyclotomic fields Q(ζp ) of prime conductors p < 1010 for which some prime q ≥ p divides the + class numbers h+ p of their maximal real subfields Q(ζp ) (but we did not find any conterexample to Vandiver’s conjecture!).
1
Introduction
This paper is an abridged version of [Lou5] in which the reader will find the proofs we omit here, and in which he will also find various supplementary examples (families of real cyclic quartic, sextic and octic fields). Our aim is to explain how one can generalize the technique developed in [Lou1] not only to compute efficiently class numbers of real abelian number fields of known regulators, but also to compute efficiently exact values of Gauss sums and roots numbers associated with primitive Dirichlet characters of large conductors. In [Bye], [Lou4], [LP], [Sha] and [Wa], various authors dealt with the so called simplest cubic fields, the real cyclic cubic number fields Km associated with the cubic polynomials Pm (x) = x3 − mx2 − (m + 3)x − 1 of discriminants dm = ∆2m , where ∆m := m2 + 3m + 9, and roots θm , σ(θm ) = −1/(θm + 1) and σ 2 (θm ) = −(θm + 1)/θm . Since −x3 Pm (1/x) = P−m−3 (x), we may assume that m ≥ −1. In this paper, we assume that ∆m is square-free. In that case, the conductor of Km is equal to ∆m , its discriminant is equal to ∆2m , the set {−1, θm , σ(θm )} generates the full group of algebraic units of Km and the regulator of Km is RegKm = log2 θm − (log θm )(log(1 + θm )) + log2 (1 + θm ), with θm
√ 1 27 1 2 ∆m cos = arctan( ) +m . 3 3 2m + 3
C. Fieker and D.R. Kohel (Eds.): ANTS 2002, LNCS 2369, pp. 134–147, 2002. c Springer-Verlag Berlin Heidelberg 2002
Efficient Computation of Class Numbers of Real Abelian Number Fields
135
In [Jean] and [SW], S. Jeanin, R. Schoof and L. C. Washington dealt with the so called simplest quintic fields, the real cyclic quintic number fields Km associated with the quintic polynomials Pm (x) = x5 + m2 x4 − (2m3 + 6m2 + 10m + 10)x3 +(m4 + 5m3 + 11m2 + 15m + 5)x2 + (m3 + 4m2 + 10m + 10)x + 1 of discriminants dm = (m3 + 5m2 + 10m + 7)2 ∆4m , ∆m = m4 + 5m3 + 15m2 + 2 )/(1 + (m + 2)θm ), σ 2 (θm ), 25m + 25 and roots θm , σ(θm ) = ((m + 2) + mθm − θm 3 4 σ (θm ) and σ (θm ). In this paper, we assume that ∆m is square-free. In that case, the conductor of Km is equal to ∆m , its discriminant is equal to ∆4m , the set {−1, θm , σ(θm ), σ 2 (θm ), σ 3 (θm )} generates the full group of algebraic units of Km and the regulator of Km is 1 ij RegKm = ζ5 log |σ j (θm )| . 5 1≤i≤4 0≤j≤4
Since Pm (m + 1)Pm (m + 2) = −(m3 + 5m2 + 10m + 7)2 < 0 we can use Newton’s method for computing efficiently as good as desired numerical approximations of a root θm ∈ (m + 1, m + 2) of Pm (x). Then, the four other roots are computed inductively by the transformation θ → σ(θ) := ((m+2)+mθ−θ2 )/(1+(m+2)θ). One of the motivation for computing class numbers of simplest cubic and quintic fields stems from Vandiver’s conjecture according to which p never divides + the class number h+ p of the maximal real subfield Q(ζp ) = Q(cos(2π/p)) of a cyclotomic field Q(ζp ) of prime conductor p. However, as the computation of h+ p is impossible to perform (except for very small values of p, say p ≤ 67 (see [Wa, Tables, pages 420-423])), the idea is to compute class numbers hK of subfields K of Q(ζp )+ of small degrees: Theorem 1. (i). Let p ≡ 1 (mod 12) be a prime and let h2 , h3 and h6 denote the class numbers of the real quadratic, cubic and sextic subfields of the cyclotomic field Q(ζp ). Then, h2 h3 divides h6 and h6 divides the class number h+ p of the maximal real subfield Q(ζp )+ = Q(cos(2π/p)) of Q(ζp ) (see [CW, Lemmas 1 and 2]). However, all the prime factors q of h6 are less than p (see [Mos]). (ii). Let p ≡ 1 (mod 10) be a prime and let h5 denote the class number of the real quintic subfield of the cyclotomic field Q(ζp ). Then, h5 divides h+ p. Since the simplest cubic and quintic fields have small regulators we might expect to find some of them of prime conductors and large class numbers. Therefore, by using simplest cubic fields we might expect to find examples of cyclotomic fields of prime conductors p for which h+ p ≥ p but for which, unfortunately, all could be less than p. Up to now, only one such example the prime factors q of h+ p had been found (see [CW] and [SWW]), and we will find three more examples (see Table 3). In the same way, by using simplest quintic fields we might expect to find examples of cyclotomic fields of prime conductors p for which some prime factor q of h+ p satisfies q ≥ p. Up to now, only one such example had been found (see [SW] and [Jean]), and we will find 31 more examples (see Table 2).
136
2
St´ephane R. Louboutin
First Method for Computing Class Numbers
Let K be a real abelian number field of degree q > 1, discriminant dK and conductor fK associated with a Q-irreducible unitary polynomial PK (X) = X q + aq−1 X q−1 + · · · + a0 ∈ Z[X]. Let XK denote the group (of order q) of primitive even Dirichlet characters associated with K and let RegK denote the regulator of K. According to the analytic class number formula (see [Lan, Chapter XIII, section 3, Th. 2 page 259]), s → FK (s) = (dK /π q )s/2 Γ q (s/2)ζK (s) has a simple pole at s = 0 of residue Ress=0 (FK (s)) = −2q−1 hK RegK = 2q lim sq−1 ζK (s). s→0
Since ζK (s) = χ∈XK L(s, χ) and L(0, χ) = −1/2 if χ = 1 but L(0, χ) = 0 for 1 = χ ∈ XK , we obtain hK RegK = L (0, χ). (1) 1=χ∈XK
Lemma 1. (See [Sta]). If χ is a (non-necessarily primitive) non-trivial even Dirichlet character modulo f > 1, then L (0, χ) = − χ(k) log sin(kπ/f ). 1≤k 1 and known regulators (it will practically require only O(fK ) elementary operations to compute hK , whereas our previous techniques based on (1) 1+ and (2) requires O(fK ) elementary operations to compute hK ). The idea is to generalize [WB, Section 3] to compute efficiently good enough numerical approximations to L (0, χ) for 1 = χ ∈ XK , and to use (1). Let χ be a primitive even Dirichlet character modulo f > 1. Set χ(n) exp(2nπi/f ) (Gauss sum), (3) τ (χ) = 1≤n≤f
W (χ) = τ (χ)/ and
θ(x, χ) =
f
χ(n)e−πn
(root number) 2
x/f
(4)
(x > 0).
(5)
(x > 0),
(6)
n≥1
Then, |W (χ)| = 1 and using √ θ(1/x, χ) = W (χ) xθ(x, χ) ¯
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St´ephane R. Louboutin
we obtain (f /π)s/2 Γ (s/2)L(s, χ) ∞ ∞ ∞ dx dx dx = + W (χ) θ(x, χ)xs/2 θ(x, χ)xs/2 θ(x, χ)x ¯ (1−s)/2 , = x x x 0 1 1 L(0, χ) = 0 and the following result which enables us to compute numerical approximations to L (0, χ) to any prescribed accuracy: Theorem 2. Let χ be a primitive even Dirichlet character modulo f > 1. Then,
∞ ∞ 2 f χ(n) ¯ 1 −t dt L (0, χ) = χ(n) e e−t dt. (7) + W (χ) √ 2 t π n πn2 /f πn2 /f n≥1
Hence, setting
n≥1
E1 (x) :=
∞
x
e−t
(−1)k dt = − log x − γ − xk t k · k! k≥1 1 1 1 2 2 3 3 ··· = e−x z+ 1+ z+ 1+ z+ 1+ z+
(where γ = 0.577 215 664 901 532 · · · denotes Euler’s constant),
∞ 2 x (−1)k 2 E2 (x) := √ √ e−t dt = 1 − 2 xk π (2k + 1) · k! π x k≥0
1 1 2 3 4 5 1 2 2 2 2 2 = √ e−x ··· , z+ z+ z+ z+ z+ z+ π and LN (0, χ)
√ 1 W (χ) f 2 = χ(n)E1 (πn /f ) + 2 2 1≤n≤N
χ(n) ¯ E2 (πn2 /f ) n
1≤n≤N
(N ≥ 1 a positive integer), it holds that |L (0, χ) − LN (0, χ)| ≤
1 √
1
f 2 −t
2M t πt3 log3/2 (M f )
for
N ≥ B(t, f, M ) :=
tf log(M f ). π
(8)
Corollary 1. (See [Lou3, Proof of Theorem 7]). Let q ≥ 2 be a given prime. Fix t > (q − 1)/2 and M > 0, and let K range over a family of real abelian numbers fields K of degree q for which all the root numbers W (χ), 1 = χ ∈ −→ ∞ and for N ≥ B(t, fK , M ), the limit XK , are known. Then, as f1K 1 | Reg L (0, χ) − 1=χ∈XK 1=χ∈XK LN (0, χ)| is equal to zero. Reg K
K
Efficient Computation of Class Numbers of Real Abelian Number Fields
4
139
Efficient Computation of Root Numbers
According to Corollary 1, (1) and Theorem 2 could be used to compute efficiently numerical approximations LN (0, χ) to L (0, χ) for primitive even Dirichlet characters of order nχ > 1 and class numbers of real abelian number fields. However, as there is no known general formula for Gauss sums (see [BE]), we will now explain how to compute efficiently these root numbers W (χ) (notice that since the use of (4) to compute the exact value of W (χ) requires fχ elementary operations, it would be much simpler to use Theorem 1 than to use (4) and Theorem 2). We point out that we are going to end up with a method for computing class numbers of real abelian number fields which is more satisfactory than the one previously used (see [SW] and [SWW]): we compute exact values of root numbers, whereas in [SWW] they had three choices for W (χ) for simplest cubic fields of a given prime conductor and in [SW, Top of page 553] they had twenty choices for W (χ) for simplest quintic fields of a given prime conductor. In their computations they were lucky enough for in all cases considered only one of their possible choices gave rise to an approximation of the class number sufficiently close to an integer. To begin with, let us fix some notation. Throughout this fourth section, we let χ denote a primitive even Dirichlet character or order nχ > 1 and conductor fχ > 1. We set ω(χ) = (τ (χ))nχ , ζχ = exp(2πi/nχ ) and Q(χ) = Q(ζχ ). We let φχ = φ(nχ ) and Z[χ] = Z[ζχ ] denote the degree and the ring of algebraic integers of the cyclotomic field Q(χ), respectively. Finally, for any l relatively prime to nχ , we let σl denote the Q-automorphism of Q(χ) which is defined by σl (ζχ ) = ζχl . Notice that if gcd(l, nχ ) = 1, then χl is also a primitive Dirichlet character of order nχ and conductor fχ and that χl is even (respectively odd) if χ is even (respectively odd). Theorem 3. Let χ be a primitive Dirichlet character of conductor fχ > 1 and order nχ > 1. Then ω(χ) := (τ (χ))nχ = fχnχ /2 (W (χ))nχ ∈ Z[χ]
(9)
and σl (ω(χ)) = ω(χl ) for gcd(l, nχ ) = 1. Moreover, if nχ is prime and fχ is square-free, then ω(χ) ∈ fχ Z[χ]. 4.1
Exact Computation of ω(χ)
Fix a Z-basis B = {/1 , · · · , /φχ } and write b(k, χ)/k ∈ Z[χ]. ω(χ) =
(10)
1≤k≤φχ
with b(k, χ) ∈ Z, 1 ≤ k ≤ φχ . Let B ⊥ = {η1 , · · · , ηφχ } be the dual basis of B, relative to the trace form. Hence, TrQ(χ)/Q (/k ηk ) = 1 but TrQ(χ)/Q (/k ηl ) = 0 for k = l, and b(k, χ) = TrQ(χ)/Q (ηk ω(χ)) = fχnχ /2 σl (ηk )(W (χl ))nχ (11) 1≤l≤nχ gcd(l,nχ )=1
140
St´ephane R. Louboutin
(for σl (ω(χ)) = ω(χl )), and these coordinates b(k, χ) are rational integers of n /2 reasonable size: |b(k, χ)| ≤ M (B ⊥ )φχ fχ χ where M (B ⊥ ) = max{|σl (ηj )|; 1 ≤ l ≤ nχ , gcd(l, nχ ) = 1, 1 ≤ j ≤ φχ }.
(12)
For example, if nχ = q ≥ 3 is prime, then B ⊥ = {ηl := (ζq−l −1)/q; 1 ≤ l ≤ q −1} is the dual basis of the Z-basis B = {ζqk ; 1 ≤ k ≤ q − 1} of the ring of algebraic integers Z[ζq ] of Q(ζq ), and M (B ⊥ ) ≤ 2/q ≤ 1. Now assuming that 2 Hypothesis: θ(χl ) := χl (n)e−πn /fχ = 0 (13) n≥1
for 1 ≤ l ≤ nχ and gcd(l, nχ ) = 1, we explain how one can compute efficiently as good as desired numerical approximations bN (k, χ) to these coordinates b(k, χ) ∈ Z of ω(χ), hence how one can compute their exact values. The key point is that θ(χl ) = 0 implies W (χl ) = θ(χl )/θ(χl ), by (6). According to Section 4.4 below, this Hypothesis should always be satisfied. The following Lemma 2 will enable us to compute as good as desired numerical approximations θN (χl ) to θ(χl ). These approximations will then enable us to check the Hypothesis (13) prior to using Lemma 3 for computing as good as desired numerical approximations bN (k, χ) to the rational integers b(k, χ) defined in (10), whose exact values can therefore be deduced. Lemma 2. Let χ be a Dirichlet character modulo f > 1. Set 2 χ(n)e−πn /f θN (χ) = 1≤n≤N
(N ≥ 1 a positive integer). If N ≥ B(t, f, M ) (as in (8)), then |θ(χ) − θN (χ)| ≤
1 √
2M t πt
1
f 2 −t
log(M f )
.
(14)
Lemma 3. Let χ be a primitive even Dirichlet character of order nχ > 1 and conductor fχ > 1. Assume that θN (χl ) = 0 for gcd(l, nχ ) = 1, set WN (χl ) = θN (χl )/θN (χl ) and bN (k, χ) = fχnχ /2
σl (ηk )(WN (χl ))nχ
(1 ≤ k ≤ φχ ),
(15)
1≤l≤nχ gcd(l,nχ )=1
and fix / such that 0 ≤ / ≤ 1. Assume that |θ(χl ) − θN (χl )| ≤ /|θN (χl )|/nχ for 1 ≤ l ≤ nχ and gcd(l, nχ ) = 1. Then, |bN (k, χ) − b(k, χ)| ≤
27(e − 1) M (B ⊥ )φχ fχnχ /2 /. 4
Efficient Computation of Class Numbers of Real Abelian Number Fields
141
Proof. Let us simplify the notation: we set n = nχ , θ = θ(χl ), θN = θN (χl ), W = W (χl ) = θ/θ¯ (notice that θN = 0 and |θ − θN | ≤ /|θN |/n imply θ = 0), WN = WN (χl ) = θN /θ¯N and write θ = θN + /N θN with |/N | ≤ //n. Then, n | k=1 nk (/kN + /¯kN )| |(1 + /N )n − (1 + /¯N )n | n n = |W − WN | = |1 + /¯N |n |1 − /N |n yields |W n − WNn | ≤ 2
n
n
|/N |k (1 + |/N |)n − 1 (1 + //n)n − 1 =2 ≤2 . n n (1 − |/N |) (1 − |/N |) (1 − 1/n)n k=1
k
Since (1 − 1/n)n ≥ (1 − 1/3)3 = 8/27 for n ≥ 3 and since (1 + //n)n − 1 ≤ e − 1 ≤ (e − 1)/ for 0 ≤ / ≤ 1, we obtain |W (χl )n − WN (χl )n | ≤ 27(e − 1)//4 for 1 ≤ l ≤ nχ and gcd(l, nχ ) = 1, and the desired results, by (11), (12) and (15). 4.2
Exact Computation of W (χ) and τ (χ)
Now that we know how to compute the exact value of ω(χ) := (τ (χ))nχ , let us explain how one can determine which of its nχ th root is equal to τ (χ): Lemma 4. Fix / ∈ (0, 1]. Let χ be a primitive even Dirichlet character of order nχ > 2 and conductor fχ > 1. Assume that ω(χ) is known and that N is such that θN (χ) = 0 and |θ(χ) − θN (χ)| ≤ /|θN (χ)|/nχ . Fix W a nχ th root of (W (χ))nχ = ω(χ)/f nχ /2 . Then, W (χ) = ζχk0 W where k0 in the unique integer k ∈ {0, 1, · · · , n − 1} such that |WN (χ) − ζχk W | < 2//nχ (and it holds that |WN (χ) − ζχk W | > (4 − 2/)/nχ ≥ 2//nχ for k = k0 ). Proof. Since |θ − θN | ≤ /|θN |/nχ , we have θ = 0, θN = 0 and ¯ + θ(θ¯N − θ)| ¯ |θ − θN | θ θN |θ(θ¯N − θ) 2/ |W (χ) − WN (χ)| = | ¯ − ¯ | = ≤2 . ≤ |θN | nχ |θ¯θ¯N | θ θN There exists a unique k0 ∈ {0, 1, · · · , n − 1} such that W (χ) = ζχk0 W . Since for a = b we have |ζχa W − ζχb W | = |ζχa − ζχa | ≥ 2 sin(π/nχ ) > 4/nχ , we have |WN (χ) − ζχk0 W | = |WN (χ) − W (χ)| < 2//nχ and |WN (χ) − ζnk W | = |(WN (χ) − W (χ))+(ζχk0 W −ζχk W )| ≥ 2 sin(π/nχ )−|WN (χ)−W (χ)| > 4/n−2//nχ ≥ 2//nχ for k = k0 . 4.3
Computation of Class Numbers of Simplest Quintic Fields
First, we checked our present method by recomputing Table 1. Second, we used it to compute the class numbers of all the simplest quintic fields Km ’s of conductors ∆m = m4 + 5m3 + 15m2 + 25m + 25 ≤ 1010 a prime. We obtain the following consequence: there are 32 simplest quintic fields Km of prime conductors p ≤ 1010 whose class numbers are divisible by some prime q ≥ p (see Table 2). Third,
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we used it to compute the class numbers of all the simplest cubic fields Km ’s with −1 ≤ m ≤ 554869 and ∆m = m2 + 3m + 9 ≡ 1 (mod 12) a prime. We obtain the following consequence: in the range −1 ≤ m ≤ 554869, there are only 4 simplest cubic fields Km of prime conductors ∆m = m2 + 3m + 9 ≡ 1 (mod 12) for √which the product h2 h3 of the class number h2 of the real quadratic field Q( ∆m ) and of the class number h3 of the simplest cubic field Km of conductor ∆m is greater than or equal to ∆m (see Table 3). 4.4
A Conjecture
According (i) to our numerical evidence (the computation of approximations to θ(χ) for the 10582203 primitive even Dirichlet characters χ of prime conductors p ≤ 20000 and for numerous examples of cubic, quartic and quintic primitive even Dirichlet characters of (non necessarily prime) large conductors associated with simplest cubic, quartic and quintic fields), and (ii) to the fact that as p ≥ 5 ranges over the odd primes it holds that χ∈Xp+ |θ(χ)|2 is asymptotic to √ p3/2 /(4 2) and that θ(χ) = 0 for at least p/ log p of the (p − 3)/2 characters 1 = χ ∈ Xp+ (adapt the proof of [Lou2, Theorem 1]), we put forward the following conjecture: Conjecture 1. (i) (See Hypothesis (13)). For any primitive even Dirichlet character of conductor fχ > 1 it holds that θ(χ) = 0. (ii) Let p ≥ 5 denote an odd prime and let Xp+ denote the set of order (p−3)/2 of the primitive even Dirichlet characters modulo p. For a ≥ 0 real, the limit g+ (a) = lim
p→∞
2 #{χ ∈ Xp+ ; |θ(χ)| ≤ ap1/4 } p−3
exits, a → g+ (a) is continuous, strictly increasing, g+ (0) = 0 and g+ (∞) = 1. Now, fix t0 < 1/4. Then, at least for real cyclic fields K of a given prime t0 degree q and of large prime conductors fK , we might expect to have |θ(χ)| ≥ fK for all the 1 = χ ∈ XK . In that case, for t > 1/2 − t0 we can use (14) with N ≥ B(t, fK , M ) to check numerically that θ(χ) = 0 for all the 1 = χ ∈ XK . Then, for t > (q + 1)/2 − t0 we can use Lemma 3 with N ≥ B(t, fK , M ) to compute the exact value of ω(χ) for all the 1 = χ ∈ XK . Finally, for t > 1/2 − t0 we can use Lemma 4 to compute the exact value of W (χ) for all the 1 = χ ∈ XK . Hence, according to Corollary 1, we might expect that our second method for computing class numbers of real abelian number fields K of a given degree and 0.5+ known regulators requires only O(fK ) elementary operations. In practice, it is indeed amazingly efficient and of the conjectured complexity.
5
Explicit Formulae for ω(χ)
Finally, we explain how we can dispense with Subsection 4.1 when dealing with simplest cubic and quintic fields: we know beforehand ω(χKm ) and we can use Lemma 4 for computing the root number W (χKm ), making in these two cases our method for computing class numbers simpler and faster.
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5.1
Simplest Cubic Fields √ Set ω = (−1+i 3)/2. The units in Z[ω] are {±1, ±ω, ±ω 2 }. An algebraic integer α = a + bω ∈ Z[ω] is primary if α ≡ −1 (mod 3Z[ω]), i.e. if a ≡ −1 (mod 3) and b ≡ 0 (mod 3). The order of the multiplicative group (Z[ω]/3Z[ω])∗ is equal to 6 and the six units in Z[ω] form a set of representatives of this group. Therefore, if α ∈ Z[ω] and 3 does not divide its norm N (α) = αα ¯ , then exactly one of its six associates is primary. If follows that if 0 = α ∈ Z[ω] is a nonunit element such that α ≡ (−1)t (mod 3Z[ω]), where t denotes the number of irreducible factors of α (counted with multiplicity), then α can be written in a unique way as a product of primary irreducibles. Now, let π ∈ Z[ω] be a primary irreducible element of norm a rational prime p ≡ 1 (mod 3). For α ∈ Z[ω] coprime with π, let χπ (α) ∈ {1, ω, ω 2 } be the cubic residue symbol defined by α(p−1)/3 ≡ χπ (α) (mod π). Then, τ (χπ )3 = pπ (see [IR, Corollary page 115]). It follows: Theorem 4. Assume that ∆m = m2 + 3m + 9 is square-free, write ∆m = t k=1 pk where the pk ’s are distinct odd primes and set m 2m + 3 + 3i√3 ≡ (−1)t (mod 3Z[ω]). δm := (−1)t 3 2 Then, δm can be written in a unique way as a product δm = 1≤k≤t πk of t primary irreducibles elements πk ∈ Z[ω] with pk = |πk |2 . Set χδm = 1≤k≤t χπk . Then, χδm is a primitive cubic character modulo ∆m , and (16) ω(χδm ) := τ (χδm )3 = ∆m δm . t m Hence, setting /m = (1 − (−1) 3 )/2 ∈ {0, 1}, there exists km ∈ {0, 1, 2} such that √ 3 3 2km + /m 1 + π (mod 2π), (17) arg(W (χδm )) ≡ arctan 3 2m + 3 3 and, if θ(χδm ) = 0, then km can be efficiently computed by using Lemma 4. Moreover, χδm is associated with the simplest cubic field Km , i.e. the character χKm associated with Km obtained by using the technique developed in Section ¯ δm . 2 is equal either to χδm or to its conjugate character χ Since Pm (x) = x3 − mx2 − (m + 3)x − 1 has no root in the finite field with two elements, we have 1 = χKm (2) ∈ {ω, ω 2 } where χKm is the cubic character associated with Km obtained by using the technique developed in Section 2. According to the law of cubic reciprocity (see [IR, Theorem 1, page 114]), we have
2 ω (mod 2Z[ω]) if m ≡ 0 (mod 2) χδm (2) = χ2 (δm ) ≡ δm ≡ ω (mod 2Z[ω]) if m ≡ 1 (mod 2), hence
χδm (2) =
ω2 ω
if m ≡ 0 if m ≡ 1
(mod 2) (mod 2).
Hence, by computing χKm (2) and by changing χKm into its conjugate if necessary, we may assume that χKm (2) = χδm (2), which implies χKm = χδm .
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St´ephane R. Louboutin
Simplest Quintic Fields
In the same way, we have: Theorem 5. Assume that ∆m = m4 + 5m3 + 15m2 + 25m + 25 is square-free. Since Pm (x) = x5 + m2 x4 − (2m3 + 6m2 + 10m + 10)x3 + (m4 + 5m3 + 11m2 + 15m + 5)x2 + (m3 + 4m2 + 10m + 10)x + 1 has no root in the finite field with two elements, we may assume that the quintic character χKm associated with the simplest quintic field Km (obtained by using the technique developed in Section 2) satisfies
3 ζ5 if m ≡ 0 (mod 2) χKm (2) = ζ54 if m ≡ 1 (mod 2). In that case, it holds that ω(χKm ) := τ (χKm )5 = (−1)t−1 ( m 5 )∆m δm where δm = (m6 + 5m5 + 5m4 + 25m2 + 125m + 125)ζ5 +(m6 + 5m5 − 5m4 − 75m3 − 175m2 − 125m)ζ52
+(m6 + 10m5 + 25m4 − 100m2 − 125m)ζ53 +(m6 + 10m5 + 40m4 + 75m3 + 50m2 )ζ54 ∈ Z[ζ5 ].
6
Simplest Cubic Fields and Class Numbers of the Maximal Real Subfields of Some Cyclotomic Fields
2 Lm = Assume √ that ∆m = m + 3m + 9 ≡ 1 (mod 12) is square-free. Let √ Km ( ∆m ) denote the compositum of the real quadratic field km = Q( ∆m ) and of the simplest cubic field Km , both of conductor ∆m . Then, Lm is a so-called simplest sextic field of conductor ∆m associated with the sextic polynomial
Pm (x) = x6 − 2mx5 − 5(m + 3)x4 − 20x3 + 5mx2 + 2(m + 3)x + 1, and a subgroup of finite index QLm (dividing 12) of the group of algebraic units of Lm is known (see [Gra]). Using this subgroup, and following the proof of [Lou4, Theorem 4], we obtain: Theorem 6. (See [Lou5]). Assume that ∆m = m2 + 3m + 9 ≡ 1 square-free. Then, ∆2m hLm ≥ . 15e log6 (4∆m )
(mod 12) is (18)
In particular, for m ≥ 105 it holds that hLm > ∆m . Notice that, in the special case that ∆m = m2 + 3m + 9 ≡ 1 (mod 12) is prime, we have hkm = h2 , hKm = h3 and hLm = h6 , with the notation of Theorem 1. With this notation, we have seen that the simplest cubic fields Km for which ∆m = m2 + 3m + 9 ≡ 1 (mod 12) is prime and such that h2 h3 ≥ ∆m are few and far between (see Table 3). However, according to the previous lower bound for hLm , as soon as m is large enough we have h6 ≥ ∆m . Moreover, using this lower bound for hLm and following the proof of [CW, Theorem 2] (see [CW, Page 269]), we obtain:
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145
Corollary 2. (See [CW, Theorem 2] for a worse and non-effective result). Let / > 0 be given. Set c = 13 p≡1 (mod 3) (1 − 2p−2 ) = 0.311 · · · . For at least (c + o(1))x1/2 positive odd square-free integers f ≤ x is holds that the class + numbers h+ f of the maximal real subfields Q(ζf ) of the cyclotomic fields Q(ζf ) + 2− , and the constants involved in these o(1) and are explicit. satisfy hf f Proof. Let f range over the positive square-free integers of the form f = ∆m := m2 + 3m + 6 ≡ 1 (mod 12).
7
Tables Table 1. class numbers hKm of the simplest quintic fields Km of square-free conductors ∆m < 107 m −1, −2 −3 1 −4 2 3 −6 4 −7 −8 6 −9 7 8 −11 9 −12 −13 11 −14 12 13 −17 −18 16 −19 17 18
∆m 11 31 71 101 191 451 631 941 1271 2321 3091 3931 5051 7841 9951 11671 13981 19811 23411 27311 31861 42431 62891 80251 90281 100991 112871 139471
hKm 1 1 1 1 11 5 11 16 55 305 80 256 1451 421 541 655 1375 4705 2000 7255 9680 9455 9455 37631 19301 203305 83275 32605
m −21 19 −22 21 −24 22 23 −26 24 −27 −28 26 −29 27 28 −31 29 −32 −33 31 −34 32 33 −36 34 −37 −38 36
∆m 154291 170531 187751 247951 270721 295331 349211 378611 410161 443311 515981 555671 597251 641491 736901 788231 842591 899321 1021771 1087691 1156331 1228601 1382791 1464901 1551071 1640531 1831511 1933261
hKm 108691 44605 76901 308605 153005 478775 186091 189305 591775 289025 2372005 721151 540905 1566401 1764400 1217821 760055 798256 4680055 1386275 1402000 4822625 2148080 4628591 2160455 1636721 11812625 3869525
m −39 37 38 −41 39 −42 −43 41 −44 42 43 −46 −47 −48 46 −49 47 48 −51 49 −53 51 −54 52 53 −56 54 −57
∆m 2038711 2148911 2382131 2505371 2633851 2766691 3047951 3196631 3350141 3509671 3845171 4021391 4392551 4788841 4997051 5211371 5433131 5897161 6139711 6390311 7186931 7468771 7758151 8056541 8678351 9002081 9335491 9677371
hKm 4521505 27105755 6728105 6340275 7503505 20599841 24153305 8088176 6495280 61395955 17264525 21321025 12722855 49860400 42769375 56285605 88151275 17478875 21966025 74338555 155197205 28850896 37142851 118690480 44646025 106111555 54898055 73297775
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Table 2. the simplest quintic fields Km of prime conductors ∆m = p < 1010 for which some prime q ≥ p (in bold face letters) divides their class numbers hKm m 27 −61 66 73 77 −84 −88 −99 −102 −121 122 128 129 139 −147 −163 162 178 −187 −237 238 242 −249 −263 −264 268 271 282 291 293 −303 −312
∆m 641491 12765251 20479231 30425111 37526591 46927381 56676161 91352671 103090711 205717691 230839031 279170201 287909191 387022451 451386751 684652511 710402911 1032554351 1190654831 3089232931 3276804731 3501489071 3767856571 4694424311 4766572561 5256015221 5494201451 6437395351 7295360131 7497114671 8291171431 9325450081
hKm 1566401 66431941 182277211 335434451 3233114891 2068985771 5912208301 3144379001 3626779141 11420513591 60390377311 24178878281 32215474121 42590939281 155312785456 785372557471 421336924016 320881058831 259187494511 1634411025661 3314877124271 4793050096976 2253716261071 9653048507861 3419567237581 4240933367591 6532834598131 18156246542621 5988407760191 10748665628261 25938285252521 15721799752591
= 1566401 = 66431941 = 61 · 2988151 = 11 · 30494041 = 3233114891 = 2068985771 = 5912208301 = 3144379001 = 3626779141 = 11420513591 = 11 · 5490034301 = 24178878281 = 32215474121 = 11 · 3871903571 = 24 · 9707049091 = 41 · 19155428231 = 24 · 26333557751 = 320881058831 = 11 · 23562499501 = 1634411025661 = 71 · 46688410201 = 24 · 299565631061 = 11 · 204883296461 = 11 · 887549864351 = 112 · 2860886261 = 151 · 28085651441 = 6532834598131 = 11 · 1650567867511 = 5988407760191 = 112 · 88831947341 = 2311 · 11223836111 = 41 · 383458530551
Table 3. least values of m ≥ −1 for which ∆m is prime and h2 h3 ≥ ∆m m 102496 106253 319760 554869
∆m 10505737513 11290018777 102247416889 307881271777
|θ(χδm )| arg W (χδm )√ 3 3 20.268 · · · 13 arctan( 2m+3 )+ √ 1 3 3 34.364 · · · 3 arctan( 2m+3 ) √ 3 3 202.162 · · · 13 arctan( 2m+3 ) √ 1 3 3 88.861 · · · 3 arctan( 2m+3 ) +
π 3
π 3
h2 891 2685 1887 7983
h3 13152913 6209212 57772549 93739324
h2 h3 /∆m 1.115 · · · 1.476 · · · 1.066 · · · 2.430 · · ·
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