Monogenous integral basis in cubic algebraic number fields

Andrea Previtali

Department of Mathematics, University of Mainz 55099 Mainz, DE

June 9, 2003

Abstract

Given a prime p, K = ℚ( ^{3} \sqrt{p}), 𝔒_{K} its ring of integers, we show that A = {p : 𝔒_{K} = ℤ[x]} and A^{c} are infinite.

1.Introduction

Suppose [K : ℚ] = n, i.e. K is an algebraic number field, 𝔒_{K} is the integral closure of ℤ in K, x ∈ 𝔒_{K} a primitive element for K over ℚ, f_{x}(w) ∈ ℤ[w] the monic polynomial of minimal degree such that f_{x}(x) = 0 and Δ(f_{x}) its discriminant. It is well known that (see [2, Ch.4]) ℤ[x] ⊆ 𝔒 ⊆ \frac{1}{d}ℤ[x], where d|Δ(f_{x}), and that 𝔒_{K} is a free ℤ–module of rank n. A ℤ–basis for 𝔒_{K} will be called an integral basis. Given a ℚ–basis ω_{1}, …, ω_{n} for K, we may define the regular representation D of K into (ℚ)_{n}, setting D(x) = (x_{ij}), where xω_{i} = ∑_{i=1}^{n} x_{ij}ω_{j}, x_{ij} ∈ ℚ. It is known that Δ_{D(x)}(w), the characteristic polynomial of D(x), is equal to f_{x}(w)^{n/ deg f_{x}}. This allows us to determine constructively 𝔒_{K}. Since ∀x ∈ K, ∃a ∈ ℕ, such that ax ∈ 𝔒_{K}, we may suppose that ω_{i} ∈ 𝔒_{K}. Let Δ(ω_{1}, …, ω_{n}) = det(T r(ω_{i}ω_{j})) denote the discriminant of ω_{1}, …, ω_{n}, then Δ(ω_{1}, …, ω_{n}) ∈ ℤ. Given a prime r, x ∈ 𝔒_{K}, (\frac{1}{r}ℤ[x] ∖ ℤ[x])∩𝔒_{K} ≠ ∅ only if r^{2}|Δ(\underline{x}), where \underline{x} = (1, x, …, x^{n−1}). This provides a constructive procedure to determine an integral basis for 𝔒_{K}. If 𝔒_{K} = ℤ[x], we say that 𝔒_{K} admits a monogenous basis. Denote with ℙ the set of prime elements in ℤ or rational primes. Let p ∈ ℙ and K = ℚ( ^{3} \sqrt{p}), 𝔒 = 𝔒_{K}, α = ^{3} \sqrt{p}, A = \{p ∈ ℙ : 𝔒 = ℤ[x]\}. We prove that {p ∈ ℙ : p ≢ ±1 (mod 9)} ⊆ A. Dirichlet’s theorem on the density of primes in arithmetic progressions implies that A is infinite. Using the law of cubic reciprocity, we show that also ℙ ∖ A is infinite.

2.Main Results

If p ∈ ℙ, p ≠ 3, set ε = ± ≡ p (mod 3).

Theorem 1 𝔒 admits the following integral basis:

\{ \begin{matrix}1, α, α^{2} & p ≢ ±1 (mod 9) \\ 1, α, β & p ≡ ±1 (mod 9) \end{matrix} ,

where β = \frac{1}{3}(1 + εα + α^{2}).

Proof. Let D denote the regular representation of K into (ℚ)_{3} with respect to the ℚ–basis 1, α, α^{2} for K. Since Δ(1, α, α^{2}) = −3^{3}p^{2}, then x = \frac{1}{r}(a + bα + cα^{2}) ∈ 𝔒 only if r = 3 or p. Now

D(rx) = ( \begin{matrix}a & b & c \\ pc & a & b \\ pb & pc & a \end{matrix} ) .

Set Δ_{D(rx)}(w) = ∑_{i=0}^{3} σ_{i}w^{3−i}, then x ∈ 𝔒 iff σ_{i} ≡ 0 (mod r^{i}), that is,

\{ \begin{matrix}a^{3} + pb^{3} + p^{2}c^{3} − 3abc & ≡ 0 (mod r^{3}) \\ 3(a^{2} − pbc) & ≡ 0 (mod r^{2}) \\ 3a & ≡ 0 (mod r) \end{matrix} .

If r = p, then σ_{3} ≡ 0 (mod p^{3}) implies that a = 0 and b^{3} + pc^{3} ≡ 0 (mod p^{2}), that is, b, c = 0. Hence p ≠ r = 3. If a = 0 then bc = 0 and σ_{3} ≡ 0 (mod 27) yealds b = c = 0. We may suppose that a = 1, so pbc ≡ 1 (mod 3). Hence

pc ≡ b, 1 + p(b^{3} + pc^{3}) ≡ 1 + 2pb (mod 3)

and

c = 1, p ≡ b (mod 3).

Substituting in the last equation of the system, we get p ≡ b (mod 9). Since 9 |̸ Δ(1, α, β), the procedure can’t be applied once again and 1, α, β is an integral basis . □

Corollary 1 There exist infinitely many primes p, for which 𝔒 admits a monogenous basis.

Proof. By Dirichlet’s theorem (see [1,pg.251]), there are infinitely many primes satisfying p ≢ ±1 (mod 9) . □

We show that ℙ ∖ A is infinite. We concentrate on the case p ≡ ±1 (mod 9) and determine under which conditions 𝔒 admits a monogenous basis.

Lemma 1 Suppose p ≡ ±1 (mod 9), then 𝔒 admits a monogenous basis iff the elliptic curve 𝒞_{p} : (3a + εb)^{3} = 9 + pb^{3} has integral points.

Proof. As it is well known, any pair of ℤ–basis for 𝔒 are unimodular equivalent. We observe that ℤ[θ] = ℤ[σ], whenever σ − θ ∈ ℤ. Infact

\underline{σ}^{t} = T \underline{θ}^{t},

where T is a unitriangular, hence unimodular, lower matrix. Without loss of generality, we may suppose θ = aα + bβ, a, b ∈ ℤ. Then

(\begin{matrix}1 \\ θ \\ θ^{2} \end{matrix} ) = T (\begin{matrix}1 \\ α \\ β \end{matrix})

and det T = \frac{(3a + εb)^{3} − pb^{3}}{9} . Hence , if 𝔒 = ℤ[θ], then (3a+εb)^{3} = pb^{3} ±9. Since p is odd, it’s enough to consider the plus sign case . □

In particular 9 has to be a cubic residue modulo p. If p ≡ −1 (mod 9), then (3, p − 1) = 1, so there exist u, v ∈ ℤ such that 3u + v(p − 1) = 1, hence ∀x ∈ 𝔽_{p}^{∗}, x = x^{3u} and any element is a third power. We show that 9 is not a cubic residue for infinitely many primes p ≡ 1 (mod 9). We need some results about the cubic reciprocity law (see [1,Ch.9]). Let ω = \frac{−1+\sqrt{−3}}{2} and R = ℤ[ω] = 𝔒_{ℚ(ω)}, then R is a unique factorization domain with units R^{∗} = {±1, ±ω, ±ω^{2}}. We say x, y ∈ R are associate, x ∼ y, if x = uy, u ∈ R^{∗}. For x = a + bω, let Nx = x\overline{x} = a^{2} − ab + b^{2} be the norm. Let π denote a prime element in R, then Nπ = q^{2}, q a rational prime q ≡ 2 (mod 3), Nπ = p, p a rational prime p ≡ 1 (mod 3) or π = 1 − ω = λ and 3 = Nλ. If π ≁ λ, then ∃!u ∈ R^{∗} such that uπ = −1 + 3(m + nω), m, n ∈ ℤ. If this holds, we say that uπ is a primary element. If π |̸ x, let

χ_{π}(x) ≡ x^{\frac{Nπ−1}{3}} (mod π), χ_{π}(x) ∈ \{1, ω, ω^{2}\}

be the cubic character with respect to π. Then χ_{π}(x) = 1 iff y^{3} ≡ x (mod π) has a solution. We are now ready to state the main result.

Theorem 2 There are infinitely many rational primes p ≡ 1 (mod 9) for which 9 is not a cubic residue.

Proof. Let p ∈ ℙ and π = −1 + 3(m + nω) primary such that Nπ = p, then

m + n ≡ 0 (mod 3) ⟺ p ≡ 1 (mod 9).

Since 9 = ωλ^{4} and χ is multiplicative, we get χ_{π}(9) = χ_{π}(ω)χ_{π}(λ). Now χ_{π}(ω) = ω^{m+n} (see [1,Ex.9.3]). Since p ≡ 1 (mod 9), then χ_{π}(ω) = 1. By a theorem of Eisenstein, χ_{π}(λ) = ω^{2m}. Therefore 9 is a cubic residue for π iff m ≡ 0 (mod 3). Let π_{1}, …, π_{k} denote primary elements satisfying the following conditions:

1. χ_{π}(9) ≠ 1,

2. Nπ_{i} = p_{i} ≡ 1 (mod 9), p_{i} prime.

Clearly 9 is not a cubic residue modulo p_{i}. Eventually for k = 0, we consider the following expression:

(μπ_{1} ⋯ π_{k})^{3} − ω = γ

and prove that for a suitable choice of μ there exist a primary divisor of γ satisfying conditions 1,2. Since γ ≡ (−1)^{k}μ^{3} + 1 (mod λ), for a suitable choice of μ, e.g. μ = ±λ, γ ≢ 0 (mod λ). So γ = u ∏_{i} ρ_{i}, with ρ_{i} primary and u ∈ R^{∗}. Set ρ_{i} = −1 + 3(m_{i} + n_{i}ω). Since χ_{ρ_{i}}(ω) = 1, by construction, then

m_{i} + n_{i} ≡ 0 (mod 3).

If ρ_{i} = q_{i} = 3m_{i} − 1 is a rational prime, then m_{i} ≡ 0 (mod 3) and q_{i} ≡ −1 (mod 9). By way of contradiction, suppose that ∀ρ_{i}, ρ_{i} not rational, m_{i} ≡ 0 (mod 3) holds, for any choice of μ. Therefore ρ_{i} ≡ −1 (mod 9) and

(μπ_{1} ⋯ π_{k})^{3} ≡ ω ± u (mod 9)

hold. Now x^{3} ≡ a (mod 3), where a ∈ ℤ, hence

(μπ_{1} ⋯ π_{k})^{3} ≡ 0, −1 (mod 9).

Since π_{i} |̸ 9 and μ = ±λ ≢ 0 (mod 9), only (μπ_{1} ⋯ π_{k})^{3} ≡ −1 (mod 9) may hold. This congruence can’t clearly be met for both μ = λ and μ = −λ, that is (±λπ_{1} ⋯ π_{k})^{3} − ω has always a primary divisor π_{k+1} such that χ_{π_{k+1}}(9) ≠ 1 for a suitable choice of sign. By construction, Nπ_{k+1} = p_{k+1} ≡ 1 (mod 9) and 9 is not a cubic residue modulo p_{k+1}. Since π_{k+1} ≁ π_{i} for i = 1, …, k, we have shown that there are infinitely many primes π_{i}, for which 9 is not a cubic residue . □

Corollary 2 There are infinitely many primes, for which 𝔒 doesn’t admit a monogenous basis.

Proof. By the preceding theorem, there are infinitely many primes for which 𝒞_{p} has no integral points . □

Aknowledgment

This paper is part of my degree thesis. I would like to dedicate this result to the memory of my advisor Prof. M.Sce.

References

1. K. Ireland and M. Rosen, “A Classical Introduction to Modern Number Theory,” 2nd ed., Springer–Verlag, Berlin–Heidelberg–New York, 1990.

2. M. Pohst and H. Zassenhaus, “Algorithmic Algebraic Number Theory,” Encyclopedia of Mathematics and its Applications, G.Rota eds., 1989.