J. Ramanujan Math. Soc. 23, No.2 (2008) 191–210
The fundamental unit of some quadratic, cubic or
quartic orders
Stéphane R. Louboutin
Institut de Mathématiques de Luminy, UMR 6206, 163, avenue de Luminy, Case 907,
13288 Marseille Cedex 9, France
e-mail: [email protected]
Communicated by: V. Kumar Murty
Dedicated to Florence F.
Received: March 24, 2008
Abstract.
We explain why it is reasonable to conjecture that if is a
totally imaginary quartic unit, then is in general a fundamental unit of
the quartic order Z[], order whose group of units is of rank equal to one.
We partially prove this conjecture. This generalizes a result of T. Nagell,
who proved in 1930 a similar result for real cubic units with two non real
conjugates.
2000 Mathematics Subject Classification: Primary 11R16, 11R27.
Secondary 11R09.
1. Introduction
Let be an algebraic unit. Assume that the group of units of the order Z[] is
of rank equal to 1. It is a natural question to ask whether is a generator of this
unit group. We must assume that is not a complex root of unity, otherwise,
being of finite order, it cannot be a generator of the infinite group of units of
the order Z[].
Clearly, K = Q() must be a real quadratic number field, in which case
we may assume that > 1, or must be a cubic number field with one real
embedding and two non-real complex embeddings, in which case we may
assume that is real and that > 1, or must be a totally imaginary quartic
number field, in which case we may assume that || > 1 ([8, Lemma 1.6]).
In these quadratic and cubic situations, the only complex roots of unity
contained in Z[] are +1 and −1, and there exists only one unit η > 1 of
Z[] such that the unit group of the order Z[] is {±η n ; n ∈ Z}, this unit
191
192
Stéphane R. Louboutin
being called the fundamental unit of this order. In both these situations, there
exists some n ≥ 1 such that = η n , which implies Z[] = Z[η n ] = Z[η] and
d = dη (absolute values of discriminants of minimal monic polynomials).
In Theorem 1, we will prove that, in the quadratic case, is always
a fundamental unit of the quadratic order Z[], with one exception. In
Theorem 3, we will prove that, in the cubic case, is always a fundamental
unit of the cubic order Z[], with a one parameter infinite family of exception
where is a square of a fundamental unit, together with eight sporadic exceptions. In Conjecture 8, we will conjecture that, in the quartic case, is always
a fundamental unit of the quartic order Z[], with a one parameter infinite
family of exceptions where is a square of a fundamental unit, together with
fourteen sporadic exceptions. We will not be able to fully prove this conjecture, but at least we will prove in Theorem 10 that it holds true if either
(i) the quartic field Q() contains an imaginary quadratic subfield, or (ii) if
the quartic order Z[] contains non trivial (i.e. = ±1) complex roots of unity.
We will also add some compelling arguments making this conjecture very
reasonable. Finally, we will explain what we lack to prove this conjecture:
a result on quartic polynomials (see Conjecture 22) which we can prove for
quadratic polynomials (see (1) in Theorem 1) and for cubic polynomials (see
(3) in Theorem 4) and that we used in Theorems 1 and 3 to settle our problem
in the quadratic and cubic cases. In the quartic case, we can only prove the
easy part of this Conjecture 22 (see (8)).
2. The quadratic case
The quadratic case is easy:
Theorem 1. Let > 1 be a real quadratic unit of discriminant d > 0.
Then,√ is the fundamental unit of the quadratic order
√ Z[], except if =
(3+ 5)/2, in which case = η 2 , where 1 < η = (1+ 5)/2 = −1 ∈ Z[]
is the fundamental unit of Z[] = Z[η], and d = dη = 5. Moreover, we have
2 /3 ≤ d ≤ 22 .
(1)
Proof. To begin with, let α > 1 be a quadratic algebraic unit, let α = ±1/α,
Πα (X) = X 2 − aX ± 1 ∈ Z[X] and dα = a2 ∓ 4 > 0 be its conjugate,√its minimal monic
√ quadratic polynomial and its discriminant. Then, α =
(a+ dα )/2 ≥ (1+ 5)/2 and (α−1/α)2 ≤ dα = (α−α )2 ≤ (α+1/α)2 .
Now, write = η n and assume that n ≥ 2. Then,
(η 2 − η −2 )2 ≤ (η n − η −n )2 = ( − −1 )2 ≤ d = dη ≤ (η + η −1 )2 .
√
√
Hence, 0 < η − η −1 ≤ 1 , 1 < η ≤ (1 + 5)/2 and η = η0 = (1 + √5)/2.
Then, (η0n − η0−n )2 ≤ (η0 + η0−1 )2 implies n = 2 and = η02 = (3 + 5)/2.
The fundamental unit of some quadratic, cubic or quartic orders
193
√
For (1), we use 3−2 5 2 = (1 − 1/η02 )2 2 ≤ (1 − 1/2 )2 2 ≤ d ≤
√
(1 + 1/2 )2 2 ≤ (1 + 1/η02 )2 2 = 5 3−2 5 2 .
2
The idea used in this proof will be used to prove Proposition 12.
3. The cubic case
Definition 2. A cubic polynomial of type (T) is a monic cubic polynomial
P (X) = X 3 − aX 2 + bX − 1 ∈ Z[X] which is Q-irreducible (⇔ b = a
and b = −a − 2), of negative discriminant −dP (X) < 0, with dP (X) =
4(a3 + b3 ) − a2 b2 − 18ab + 27 > 0, and whose only real root P satisfies P > 1 (⇔ P (1) < 0 ⇔ b ≤ a − 1). In that case, we have (see
[1, Lemma 1]):
√
|b| < 1 + 2 a + 2 and 0 ≤ a ≤ P + 2.
(2)
Hence, we can sort the cubic polynomials of type (T) by increasing values of
a ≥ 0.
A non totally real algebraic cubic unit > 1 is the real root of a monic
cubic polynomial Π (X) = X 3 − aX 2 + bX − 1 ∈ Z[X] of type (T), its
minimal monic polynomial. Set d = |dΠ (X) | > 0. If = η n for some n ≥ 2
and some algebraic unit η ∈ Z[] satisfying η > 1, then Z[] = Z[η], hence
d = dη , and Π (X) and Πη (X) are two cubic polynomials of type (T) of the
same discriminant. In 1930, T. Nagell proved the following result:
Theorem 3 (See [2]). Let > 1 be a real cubic algebraic unit of negative
discriminant −d < 0. Let η > 1 be the fundamental unit of the cubic order
Z[]. Then, = η , except in the following cases:
1. The infinite family of exceptions for which is the only real root of X 3 −
M 2 X 2 − 2M X − 1, M ≥ 1, in which case = η 2 where η = 2 −
M 2 − M ∈ Z[] is the real root of X 3 − M X 2 − 1, Z[η] = Z[] and
d = dη = 4M 3 + 27.
2. The 8 following sporadic exceptions:
(a)
i. Π (X) = X 3 − 2X 2 + X − 1, in which case = η 2
η = 2 − ∈ Z[].
ii. Π (X) = X 3 − 3X 2 + 2X − 1, in which case = η 3
η = − 1 ∈ Z[].
iii. Π (X) = X 3 − 2X 2 − 3X − 1, in which case = η 4
η = 2 − 2 − 2 ∈ Z[],
iv. Π (X) = X 3 − 5X 2 + 4X − 1, in which case = η 5
η = 2 − 4 + 1 ∈ Z[].
where
where
where
where
194
Stéphane R. Louboutin
v. Π (X) = X 3 − 12X 2 − 7X − 1, in which case = η 9 where
η = −32 + 37 + 10 ∈ Z[].1
In these five cases, η > 1 is the real root of Πη (X) = X 3 − X − 1,
Z[η] = Z[] and dη = d = 23.
i. Π (X) = X 3 − 4X 2 + 3X − 1, in which case = η 3 where
η = 2 − 3 + 1 ∈ Z[].
ii. Π (X) = X 3 − 6X 2 − 5X − 1, in which case = η 5 where
η = −22 + 13 + 5 ∈ Z[].
In these two cases, η > 1 is the real root of Πη (X) = X 3 −X 2 −1,
Z[η] = Z[] and dη = d = 31.
(c) Π (X) = X 3 − 7X 2 + 5X − 1, in which case = η 3 , where 1 < η =
−2 + 7 − 3 ∈ Z[] is the real root of Πη (X) = X 3 − X 2 − X − 1,
Z[η] = Z[] and dη = d = 44.
(b)
In 2006 we gave a new proof of Theorem 3 by using the following
Theorem 4, the following Lemma 5, and numerical computations and the
following Lemma 6 to deal with the finitely many cases where d < 712.
Theorem 4 (See [1, Lemma 2 and Theorem 2]). Let > 1 be a real cubic
algebraic unit of negative discriminant −d < 0. Then (compare with (1)),
3/2 /2 ≤ d ≤ 643 .
(3)
Hence, by (2), there are only finitely many such cubic units of bounded discriminant. Finally, if d ≥ 712, then is either the fundamental unit of the
cubic order Z[] or its square. In any case, is at most an eleventh power of
the fundamental unit of this order.
Proof. In fact, we have the better upper bound d ≤ 4(3/4 + −3/4 )4 (see
[1, (2)]). Assume that = η n with 1 < η ∈ Z[] and n ≥ 3. Then,
3/2 /2 ≤ d = dη ≤ 4(η 3/4 + η −3/4 )4 ≤ 4(1/4 + −1/4 )4 .
Hence, < 126.51 and d ≤ 4(1/4 + −1/4 )4 < 712.
For the last assertion, we first check using (2) that we always have ≥
0 = 1.32471 . . . , where 0 > 1 is the only real root of X 3 − X − 1 of type
(T). Hence, if = η n with n ≥ 2 and η ∈ Z[], we have
η 3n/2 = 3/2 ≤ 2d = 2dη ≤ 8(η 3/4 + η −3/4 )4 ,
which in using η ≥ 0 yields n ≤ 11.
1
2
7
Note the misprint in [1, Theorem 4(2)] where we wrote = η instead of
= η 9 . Note also the misprint in the displayed formula in [1, Proof of Theorem 2,
3/2
−3/2
case (i)] which should be dP ≥ (2P − 4P + 1)(1 − P )4 ≥ 2P /2 for P ≥ 65.
The fundamental unit of some quadratic, cubic or quartic orders
195
Lemma 5 (Compare with Lemma 16). Let > 1 be a real cubic algebraic
unit of negative discriminant −d < 0. Then, is a square in Z[] if and only
if we are in one of the following three cases:
1. Π (X) = X 3 − M 2 X 2 − 2M X − 1 with M ≥ 1, in which case = η 2
where η = 2 − M 2 − M ∈ Z[], Πη (X) = X 3 − M X 2 − 1 and
dη = d = 4M 3 + 27.
2. Π (X) = X 3 − 2X 2 − 3X − 1, in which case = η 2 where η =
−2 + 3 + 2 ∈ Z[], Πη (X) = X 3 − 2X 2 + X − 1 and dη = d = 23.
3. Π (X) = X 3 − 2X 2 + X − 1, in which case = η 2 where η = 2 − ∈
Z[], Πη (X) = X 3 − X − 1 and dη = d = 23.
Proof. Assume that = η 2 , with 1 < η ∈ Z[]. Then, Z[] = Z[η] and
d = dη . The index (Z[η] : Z[η 2 ]) is equal to |ab − 1|, where Πη (X) =
X 3 − aX 2 + bX − 1. Hence, we must have |ab − 1| = 1, and we will
have η = (2 − (a2 − b) − a)/(1 − ab) and Π (X) = Πη2 (X) = X 3 −
(a2 − 2b)X 2 + (b2 − 2a)X − 1.
First, assume that ab = 2. Then, a = 2 and b = 1 (for a ≥ 0 and
b ≤ a−1), Πη (X) = X 3 −2X 2 +X +1 and Π (X) = X 3 −2X 2 −3X −1.
Second, assume that ab = 0. If a = 0, then b ≤ a − 1 = −1 and
dη = 4b3 + 27 > 0 yields b = −1, Πη (X) = X 3 − X − 1, and Π (X) =
X 3 − 2X 2 + X − 1. If a = 0, then b = 0, Πη (X) = X 3 − aX 2 − 1,
dη = 4a3 + 27 and Π (X) = X 3 − a2 X 2 − 2aX − 1.
2
Lemma 6 (Compare with Lemma 18). 2 Let > 1 and 1 < η ∈ Z[]
be two real algebraic cubic units of negative discriminant. Write Π (X) =
X 3 −a X 2 +b X −1 ∈ Z[X] and Πη (X) = X 3 −aη X 2 +bη X −1 ∈ Z[X].
If = η n with n ≥ 3, then a > aη and d = dη . Hence, if is neither the
fundamental unit of the order Z[] nor its square, then there exists P (X) =
X 3 − aX 2 + bX + 1 ∈ Z[X] of type (T) such that 0 ≤ a < a and dP = d .
Proof. If P > 1 is the real root of P (X) = X 3 − aX 2 + bX − 1 of type
(T) and if P and ¯
P are its two complex conjugate roots, then P |P |2 = 1.
Hence,
√
√
P − 2/ P ≤ a = P + 2(P ) ≤ P + 2/ P .
Suppose that 0 ≤ a ≤ aη . We have
√
√
η 3 − 2/η 3/2 ≤ − 2/ ≤ a ≤ aη ≤ η + 2/ η,
which implies η < 1.61.
2
Notice that [1, Lemma 3] should have been stated as follows: Let P (X) =
X − aP X 2 + bP X − 1 and Q(X) = X 3 − aQ X 2 + bQ X − 1 be two cubic
polynomials of type (T). Then, P ≥ 2Q implies aP > aQ except in the following
two cases: (i) Q(X) = X 3 −2X 2 +X −1 and P (X) = X 3 −2X 2 −3X −1 where
P = 2Q , aP = aQ = 2 and dP = dQ = 23, and (ii) Q(X) = X 3 − X 2 − 1 and
P (X) = X 3 − X 2 − 2X − 1 where P = 2Q , aP = aQ = 1 and dP = dQ = 31.
3
196
Stéphane R. Louboutin
Now, by (2), if η < 1.61, then either (i) Πη (X) = X 3 − X − 1 with dη =
23 and aη = 0, but by (2) there is no cubic polynomial Π (X) = Πη (X)
of type (T) with a = 0 and d = 23, or (ii) Πη (X) = X 3 − X 2 − 1 with
dη = 31 and aη = 31, but by (2) Π (X) = X 3 − X 2 − 2X − 1 is the
only cubic polynomial Π (X) = Πη (X) of type (T) with 0 ≤ a ≤ 1 and
d = 31 and this yields = η n with n = 2.
2
4. The quartic case
We would like to prove similar results for quartic orders of unit groups of
rank 1. So, let be a totally imaginary quartic algebraic unit. Let Π (X) =
X 4 − aX 3 + bX 2 − cX + 1 ∈ Z[X] be its minimal monic polynomial. It is
Q-irreducible and of positive discriminant d . Since Z[] = Z[−] =
Z[1/] = Z[−1/], we may assume that 0 ≤ |c| ≤ a. Let , ¯, and ¯
denote the four complex conjugate of . Since, | | = 1/||, we may also
assume that || ≥ 1. If || = 1, then || = | | = 1, is a complex root of
unity (see [8, Lemma 1.6]), and being of finite order, it is not a fundamental unit of the totally imaginary quartic order Z[]. We exclude this case by
defining:
Definition 7 (Compare with Definition 2). A quartic polynomial of type
(T) is a monic quartic polynomial P (X) = X 4 − aX 3 + bX 2 − cX + 1 ∈
Z[X] which is Q-irreducible, satisfying 0 ≤ |c| ≤ a, of positive discriminant
dP = −4a3 c3 + a2 b2 c2 + 18a3 bc + 18abc3 − 4a2 b3 − 4b3 c2 − 27a4 − 27c4 −
6a2 c2 − 80ab2 c + 16b4 + 144a2 b + 144bc2 − 192ac − 128b2 + 256 > 0,
with no real root and with at least one complex root of absolute value greater
than one. In that case, for any complex root P of P (X), we have
√
|c| ≤ a ≤ 4b + 24 and − 1 ≤ b ≤ |P |2 + 1/|P |2 + 4.
(4)
Hence, we can sort the quartic polynomials of type (T) by increasing values
of b ≥ −1.
Proof. Let , ¯, and ¯ be the four complex roots of P (X). Then, a =
2() + 2( ) and b = ||2 + | |2 + 4()( ). Using | | = 1/|| and
|| > 1, we obtain a ≤ 2(|| + 1/||), b > −2 and b ≥ ||2 + 1/||2 − 4 ≥
2
(a/2)2 − 6.
From now on, is a totally imaginary quartic algebraic unit satisfying
|| > 1 and whose minimal monic quartic polynomial Π (X) is of type (T).
We want to prove that, in general, is a fundamental unit of the totally imaginary quartic order Z[] (whose group of units is of rank one).
Let μ() denote the finite cyclic group of complex roots of unity in
Z[]. Hence, {±1} ⊆ μ(). Let 2m ≥ 2 denote the order of μ() and let
The fundamental unit of some quadratic, cubic or quartic orders
197
ζ2m = exp(iπ/m) be a generator of this cyclic group μ() = ζ2m . Let
η ∈ Z[] denote a fundamental unit of the quartic order Z[], i.e. assume that
the group of units of this order Z[] is equal to {ζη n ; ζ ∈ μ(), n ∈ Z}.
(Notice that, in the case that μ() = {±1}, η could be a real quadratic
unit or a totally real quartic unit. However, if μ() = {±1}, then η is also
a totally imaginary quartic unit, and by changing η into −η , 1/η or −1/η if
necessary, we can assume that its minimal monic polynomial Πη (X) is also of
type (T).)
Then, = ζη n with ζ ∈ μ() and n ∈ Z \ {0} and, if ζ ∈ μ(η), hence in
particular if μ() = {±1}, then Z[] = Z[η n ] = Z[η] and d = dη .
We will explain why it is reasonable to conjecture the following:
Conjecture 8 (Compare with Theorems 1 and 3). Let be a totally imaginary quartic algebraic unit whose minimal monic quartic polynomial Π (X)
is of type (T). Let η denote a fundamental unit of the totally imaginary quartic
order Z[]. Then, we can choose η = , except in the following cases:
1. The infinite family of exceptions for which is a root of Π (X) = X 4 −
2bX 3 + (b2 + 2)X 2 − (2b − 1)X + 1, b ≥ 3, in which cases =
−1/η 2 where η = 3 − 2b2 + (b2 + 1) − (b − 1) ∈ Z[] is a root of
Πη (X) = X 4 − X 3 + bX 2 + 1 of type (T), Z[η] = Z[] and dη = d =
16b4 − 4b3 − 128b2 + 144b + 229.
2. The 14 following sporadic exceptions:
(a)
i. Π (X) = X 4 − 3X 3 + 2X 2 + 1, in which case ζ3 = 3 −
22 + − 1 ∈ Z[] and = −ζ3 η 2 where η = 2 − ∈ Z[].
ii. Π (X) = X 4 − 3X 3 + 5X 2 − 3X + 1, in which case in which
case ζ3 = 3 − 32 + 4 − 2 ∈ Z[] and = ζ32 η 2 where
η = −3 + 22 − 3 + 1 ∈ Z[].
iii. Π (X) = X 4 − 5X 3 + 8X 2 − 4X + 1, in which case ζ3 =
−3 + 42 − 5 + 1 ∈ Z[] and = ζ32 η 3 where η = −2 +
3 − 1 ∈ Z[].
In these three cases, η is a root of Πη (X) = X 4 − 2X 3 + 2X 2 −
X + 1 of type (T), ζ3 = η 2 − η ∈ Z[η], Z[η] = Z[] and dη =
d = 117.
(b) Π (X) = X 4 − 5X 3 + 9X 2 − 5X + 1, in which case = −η 2
where η = −3 + 42 − 6 + 2 ∈ Z[] is a root of Πη (X) = X 4 −
X 3 + 3X 2 − X + 1 of type (T), Z[η] = Z[] and dη = d = 189.
(c)
i. Π (X) = X 4 − X 3 + 2X 2 + 1, in which case = η 2 where
η = −3 + 2 − ∈ Z[].
ii. Π (X) = X 4 − 3X 3 + 3X 2 − X + 1, in which case = 1/η 3
where η = − + 1 ∈ Z[].
198
Stéphane R. Louboutin
iii. Π (X) = X 4 −4X 3 +6X 2 −3X+1, in which case = −1/η 4
where η = 3 − 32 + 3 ∈ Z[].
iv. Π (X) = X 4 − 5X 3 + 5X 2 + 3X + 1, in which case = −η 6
where η = 2 − 2 − 1 ∈ Z[].
v. Π (X) = X 4 − 7X 3 + 14X 2 − 6X + 1, in which case =
−1/η 7 where η = 2 − 4 + 2 ∈ Z[].
In these five cases, η is a root of Πη (X) = X 4 − X 3 + 1 of type
(T), Z[η] = Z[] and dη = d = 229.
(d)
i. Π (X) = X 4 −2X 2 +3X 2 −X +1, in which case = −1/η 2
where η = 3 − 22 + 2 ∈ Z[].
ii. Π (X) = X 4 − 3X 3 + X 2 + 2X + 1, in which case = 1/η 3
where η = 2 − 2 ∈ Z[].
iii. Π (X) = X 4 − 5X 3 + 7X 2 − 2X + 1, in which case = −η 4
where η = 2 − 2 ∈ Z[].
In these three cases, η is a root of Πη (X) = X 4 − X 3 + X 2 + 1
of type (T), Z[η] = Z[] and dη = d = 257.
(e) Π (X) = X 4 −4X 3 +7X 2 −4X +1, in which case ζ4 = 3 −42 +
6 − 2 ∈ Z[] and = ζ43 η 2 , where η = −3 + 32 − 4 + 1 ∈ Z[]
is a root of Πη (X) = X 4 − 2X 3 + X 2 + 1 of type (T), Z[η] = Z[]
and dη = d = 272.
(f) Π (X) = X 4 − 13X 3 + 43X 2 − 5X + 1, in which case = −η 3
where η = −3 + 62 + 3 ∈ Z[] is a root of Πη (X) = X 4 −
2X 3 + 4X 2 − X + 1 of type (T), Z[η] = Z[] and dη = d = 1229.
We will partially prove this conjecture. First, we will prove that in all these
fifteen cases the given η is indeed a fundamental unit of the quartic order Z[].
Conversely, we will determine a fundamental unit of the quartic order Z[]
(i) in the case that μ() = {±1}, and (ii) in the case that the totally imaginary
quartic number field K = Q() contains an imaginary quadratic subfield,
a special situation of this latter case being:
Theorem 9. Let be a quartic unit whose minimal monic polynomial is of
the form Π (X) = X 4 − aX 3 + bX 2 − aX + 1 ∈ Z[X], with a ≥ 0 (if not,
consider −). Then, is totally imaginary
√ and is not a complex root of unity
if and only if b ≥ 3 and 0 ≤ a ≤ 4b − 11. In that case, Q() contains
the imaginary quadratic field Q( −(4b − 8 − a2 )) and d = ((b + 2)2 −
4a2 )(4b − 8 − a2 )2 . Let η denote a fundamental unit of the quartic order Z[].
By Theorem 10, we can choose η = , except in 3 cases:
1. Π (X) = X 4 − 3X 3 + 5X 2 − 3X + 1, in which case = η 2 where
η = −3 +22 −2 ∈ Z[] is a root of Πη (X) = X 4 −X 3 −X 2 +X +1
of type (T), Z[η] = Z[] and dη = d = 117
The fundamental unit of some quadratic, cubic or quartic orders
199
2. Π (X) = X 4 −5X 3 +9X 2 −5X +1, in which case = −η 2 where η =
−3 +42 −6+2 ∈ Z[] is a root of Πη (X) = X 4 −X 3 +3X 2 −X +1
of type (T), Z[η] = Z[] and dη = d = 189.
3. Π (X) = X 4 − 4X 3 + 7X 2 − 4X + 1, in which case ζ4 = 3 − 42 +
6 − 2 ∈ Z[] and = ζ43 η 2 , where η = −3 + 32 − 4 + 1 ∈ Z[] is
a root of Πη (X) = X 4 − 2X 3 + X 2 + 1 of type (T), Z[η] = Z[] and
dη = d = 272.
Proof. Here, + 1/ and ¯
+ 1/¯
are roots of P (X) = X 2 − aX + b − 2,
with dP = a2 − 4b + 8. If is totally imaginary and is not a complex root
of unity, then || = 1 and + 1/ and ¯
+ 1/¯
aredistinct, hence not real,
and dP < 0, i.e., dP ≤ −3, and Q( + 1/) = Q( −(4b − 8 − a2 )) is an
imaginary quadratic subfield of Q(). Conversely, if dP < 0, then its roots
+ 1/ and ¯ + 1/¯
are not real and distinct, hence is totally imaginary and
|| = 1, i.e. is not a complex root of unity.
2
5. Partial proof of the conjecture
The aim of this section is to partially settle Conjecture 8:
Theorem 10. Let be a totally imaginary quartic algebraic unit, with
Π (X) of type (T). Assume either (i) that Z[] contains non trivial (i.e. = ±1)
complex roots of unity, or (ii) that Q() contains an imaginary quadratic subfield. Let η denote a fundamental unit of the order Z[]. Then, we can choose
η = , except in the following five cases:
1. Π (X) = X 4 − 3X 3 + 2X 2 + 1, in which case ζ3 = 3 − 22 + − 1 ∈
Z[] and = −η −2 , where η = −2 + + 1 ∈ Z[] is a root of
Πη (X) = X 4 − 2X 3 + 2X 2 − X + 1 of type (T), Z[η] = Z[] and
d = dη = 117 (case 2a.i of Conjecture 8).
2. Π (X) = X 4 − 3X 3 + 5X 2 − 3X + 1, in which case ζ3 = −3 +
32 − 4 + 1 ∈ Z[] and = η 2 , where η = −3 + 22 − 2 ∈ Z[] is
a root of Πη (X) = X 4 − X 3 − X 2 + X + 1 of type (T), Z[η] = Z[]
and d = dη = 117 (case 2a.ii of Conjecture 8).
3. Π (X) = X 4 − 5X 3 + 8X 2 − 4X + 1, in which case ζ3 = −3 + 42 −
5 + 1 ∈ Z[] and = ζ32 η 3 , where η = −2 + 3 − 1 ∈ Z[] is a root
of Πη (X) = X 4 − 2X 3 + 2X 2 − X + 1 of type (T), Z[η] = Z[] and
d = dη = 117 (case 2a.iii of Conjecture 8).
4. Π (X) = X 4 − 5X 3 + 9X 2 − 5X + 1, in which case ζ3 = −3 + 52 −
8 + 2 ∈ Z[] and = −η 2 , where η = −3 + 42 − 6 + 2 ∈ Z[] is
a root of Πη (X) = X 4 − X 3 + 3X 2 − X + 1 of type (T), Z[η] = Z[]
and d = dη = 189 (case 2b of Conjecture 8).
200
Stéphane R. Louboutin
5. Π (X) = X 4 − 4X 3 + 7X 2 − 4X + 1, in which case ζ4 = 3 − 42 +
6 − 2 ∈ Z[] and = ζ43 η 2 , where η = −3 + 32 − 4 + 1 ∈ Z[] is
a root of Πη (X) = X 4 − 2X 3 + X 2 + 1 of type (T), Z[η] = Z[] and
d = dη = 272 (case 2e of Conjecture 8).
Let ζ2m ∈ Z[] be a generator of order 2m ≥ 2 of the finite cyclic group
μ() = ζ2m of the complex roots of unity in Z[]. Then, φ(2m) divides 4.
Hence, #μ() = 2m ∈ {2, 4, 6, 8, 10, 12}, and there are three cases:
1. If 2m ∈ {8, 10, 12}, then Q() contains no imaginary quadratic subfield
k for which μ() ⊆ k. This case is dealt with in Section 5.1.
2. If 2m ∈ {4, 6}, then Q() contains an imaginary quadratic subfield k for
which μ() ⊆ k, which may also happen in the case that 2m = 2. These
cases are dealt with in Section 5.2, the key point being Proposition 12.
3. μ() = {±1} and Q() contains no imaginary quadratic subfield.
5.1 The case that μ() is of order 8, 10 or 12
In that case, Q() = Q(ζ2m ) (for Q(ζ2m ) ⊆ Q() and both fields have
degree 4), hence Z[ζ2m ] ⊆ Z[] ⊆ Z[ζ2m ] (for is an algebraic integer and
Z[ζ2m ] is the ring of algebraic integers of Q(ζ2m ) = Q()) and Z[ζ2m ] =
Z[], which implies d = dζ2m ∈ {256, 125, 144}.
Proposition 11. Assume that 2m ∈ {8, 10, 12} and that Z[ζ2m ] = Z[],
i.e. that the ring of algebraic integers Z[ζ2m ] of the cyclotomic number field
Q(ζ2m ) is generated by a totally imaginary
quartic unit with || > 1. Then,
√
either (i) 2m = 10, = ±ζ5a 1+2 5 with a ∈ {1, 2, 3, 4} and is a funda√
a (1+ζ4 )(1+ 3)
with
mental unit of Z[] = Z[ζ10 ] or (ii) 2m = 12, = ζ12
2
a ∈ {1, 2, 4, 5, 7, 8, 10, 11} and is a fundamental unit of Z[] = Z[ζ12 ].
Proof. The result follows from [3, Théorème 8] or [5, Theorem 1.1], [3,
Théorème 7] or [4, Theorem 5.1], and [3, Théorème 9]. We give a proof which
does not use T. Nagell and L. Robertson’s
results.
√
If 2m = 8, then η = 1 + 2 is a fundamental unit of Z[ζ8 ], we must
have = ζ8a η b with b ≥ 1 and a ∈ {1, 2, 3, 5, 6, 7} (for || > 1 and is not
quadratic over Q), and
d = 16 sin2 (πa/4) sin2 (3πa/4)η 8b
4 4
× 1 − (−1)b ζ82a /η 2b 1 − (−1)b ζ84a /η 2b .
Hence, d → ∞ as b → ∞, and it easily follows that d is never equal to 256
for b ≥ 1.
The fundamental unit of some quadratic, cubic or quartic orders
201
√
If 2m = 10, then η = (1 + 5)/2 > 1 is a fundamental unit of Z[ζ5 ], we
must have = ±ζ5a η b with b ≥ 1 and a ∈ {1, 2, 3, 4} (for || > 1 and is
not quadratic over Q), and
4 4
d = 5η 8b 1 − (−1)b ζ5a /η 2b 1 − (−1)b ζ52a /η 2b .
Hence, d → ∞ as b → ∞, and it easily follows that d = 125 if and only if
b = 1 and a ∈ {1, 2, 3, 4}.
√
4
If 2m = 12, then 1+ζ
η is a fundamental unit of Z[ζ12 ], where η = 1+ 3,
2
1+ζ4 b
a
η with b ≥ 1, and
we must have = ζ12
2
π(2a + 9b)
π(10a + 9b) η 8b
2
2
d = sin
sin
12
12
16b
b 4 b 4
2
2 4a+6b
6a+3b
× 1 − ζ12
1 − ζ12
.
2
η
η2 Hence, assuming that a and b ≥ 1 range over rational integers for which this
discriminant is not equal to zero, d → ∞ as b → ∞, and it easily follows
that d = 144 if and only if b = 1 and a ∈ {1, 2, 4, 5, 7, 8, 10, 11}.
2
5.2 The case that Q() contains an imaginary quadratic field
If the quartic number field K = Q() contains an imaginary quadratic subfield k, then d2k divides dK , hence d2k divides d = (AK : Z[])2 dK , where
AK is the ring of algebraic integers of K .
Proposition 12. Let be a totally imaginary quartic unit, with || > 1.
Assume that K = Q() contains an imaginary quadratic subfield k and that
μ() ⊆ k. Assume that is not a fundamental unit of the order Z[]. Let η
be a fundamental unit of this order. Write Πη (X) = X 4 − aX 3 + bX 2 −
cX + 1 ∈ Z[X] and = ζη n , with ζ ∈ μ(). Then, 2 ≤ |n| ≤ 5 and −1 ≤
b ≤ 7.
Proof. Let be the conjugate of over k and Ak = Z[ω] be the ring of
algebraic integers of k Then, NK /k () = is an algebraic unit of k. hence
a complex root of unity. Therefore, | | = 1/|| < 1. We may assume
that |η| > 1. Hence, n ≥ 2. Notice that η and η n are quadratic over k,
otherwise η n would be in k, hence would be a complex root of unity and
so would be = ζη n . Then, Ak [η] = Ak [η n ] (for η ∈ Z[] and ζ ∈ Ak
yield Ak [η n ] ⊆ Ak [η] ⊆ Ak [] = Ak [ζη n ] = Ak [η n ]). If η n = αη − β
with α, β ∈ Ak , then 1 = (Ak [η] : Ak [η n ]) = |α|2 = |η n − η n |2 /
202
Stéphane R. Louboutin
|η − η |2 . Indeed, {1, ω, η, ωη} and {1, ω, η n , ωη n } being two Z-basis of the
free Z-modules Ak [η] and Ak [η n ], in writing α = a + bω and β = c + dω ,
we have
⎛
⎞
1 0 −c
dNk/Q (ω)
⎜0 1 −d −c − dT r (ω)⎟
k/Q
⎜
⎟
(Ak [η] : Ak [η n ]) = det ⎜
⎟
⎝0 0 a
−bNk/Q (ω) ⎠
0 0 b a + bT r (ω) k/Q
= Nk/Q (a + bω).
Hence,
| − |2 = |η n − η n |2 = |η − η |2
(5)
and
|η|n − 1/|η|n ≤ |η n − η n | = |η − η | ≤ |η| + 1/|η|
(6)
√
As in the proof of Theorem 1, using (6), we obtain |η| ≤ (1 + 5)/2. Hence,
|b| ≤ |η|2 + 1/|η|2 + 4 ≤ 7 (express b in terms of the complex roots of
Πη (X)). Using (6) and Lemma 13, we obtain n ≤ 5.
2
Lemma 13. Let η be a totally imaginary quartic unit. Assume that |η| > 1.
Then, |η| ≥ |η0 | = 1.18375 . . . , where Πη0 (X) = X 4 − X 3 + 1.
4
− xi ) = X 4 − σ1 X 3 + σ2 X 2 − σ3 X + σ4 , then
6
5
4
2
1≤i<j≤4 (X − xi xj ) = X − σ2 X + (σ1 σ3 − σ4 )X − (σ1 σ4 −
2σ2 σ4 + σ32 )X 3 + (σ1 σ3 σ4 − σ42 )X 2 −σ2 σ42 X + σ43 . In our situation,
x1 = η , x2 = η̄ , x3 = η and x4 = η̄ are the complex roots of
Πη (X) = X 4 − aX 3 + bX 2 − cX + 1 ∈ Z[X]. Hence, |η|2 > 1 and
1/|η|2 < 1 are real roots of
Proof. If
i=1 (X
Qη (X) = X 6 − bX 5 + (ac − 1)X 4 − (a2 − 2b + c2 )X 3
+ (ac − 1)X 2 − bX + 1.
The absolute values of its four other complex roots ηη , η η̄ , η̄η and η̄ η̄ are
equal to 1.
If c = a, then Qη (0) = 1 > 0 and Qη (1) = −(a − c)2 < 0. Hence, using
the dichotomy method, it is easy to compute numerical approximations to the
only real root 1/|η|2 in [0, 1] of Qη (X).
If c = a, then Qη (X) = (X − 1)2 Rη (X) with
Rη (X) = X 4 − (b − 2)X 3 + (a2 − 2b + 2)X 2 − (b − 2)X + 1.
The fundamental unit of some quadratic, cubic or quartic orders
203
We may assume that η = 1/η (for 1/η is also a root of Πη (X) =
X 4 Πη (1/X)). Therefore, |η|2 > 1, 1/|η|2 < 1, η/η̄ and η̄/η are the four
roots of Rη (X). Hence, |η|2 + 1/|η|2 > 2 and η/η̄ + η̄/η ∈ [−2, 2]
are roots of S(X) = X 2 − (b − 2)X + a2 − 2b. Hence, we must have
2
2
2
2
2
dS = (b +
2) − 4a > 0, S(2) < 0 ⇔ a − 4b + 8 < 0, |η| + 1/|η| =
2
2
(b − 2 + (b + 2) − 4a ))/2 and
|η|2 = b − 2 + (b + 2)2 − 4a2
+
2b2 − 4a2 − 8 + 2(b − 2) (b + 2)2 − 4a2 /4.
Finally, we adapt the proof of [1, Lemma 2]: if 1 < |η| ≤ 1.2, then −1 ≤
b ≤ 6. Using (4), we make the list of all the possible monic quartic polynomials Πη (X) = X 4 − aX 3 + bX 2 − cX + 1 of type (T) in this range. For
each Πη (X) in this list, we compute Qη (X) and numerical approximations
to 1/|η|2 if c = a, whereas we use our formula above for |η|2 in the case that
c = a. We obtain the desired result: if 1 < |η| ≤ 1.2, then |η| ≥ |η0 |.
2
5.2.1 The case that μ() is of order 4
Lemma 14. Let be a totally imaginary quartic unit whose minimal monic
polynomial Π (X) is of type (T). Assume that μ() = {±1, ±ζ4 }. If = ζ4a η 2
for some a ∈ {0, 1, 2, 3} and some unit η ∈ Z[], then Π (X) = X 4 −
4X 3 + 7X 2 − 4X + 1, d = 272, ζ4 = 3 − 42 + 6 − 2 ∈ Z[], = ζ43 η 2
and η = −3 +32 −4+1 ∈ Z[] is a root of Πη (X) = X 4 −2X 3 +X 2 +1
of type (T), = −η 3 + η 2 + 1 ∈ Z[η], Z[η] = Z[], dη = d = 272 and
μ() = {±1, ±ζ4 } (case 5 of of Theorem 10).
Proof. We may assume that || > 1. First, η is quadratic over k = Q(ζ4 )
(if η ∈ Q(ζ4 ), then η is a complex root of unity, and so is = ζ4a η 2 , a
contradiction), and η 2 − αη + β = 0 for some α and β in Z[ζ4 ]. Then β
is a root of unity in k. Moreover, |α|2 = |η + η |2 = 1, by (5), and α is
also a complex root of unity in k. Moreover, since η = ±αη satisfies η 2 −
±α2 η +α2 β = 0, we may assume that η 2 −η +β = 0 with β ∈ {±1, ±ζ4 }.
We cannot have β = +1, otherwise η is a complex root of unity, and so is
, a contradiction. By changing η and into
may assume that
√ η̄ and ¯, we
a 2
β ∈ {−1, −ζ4 }. If β = −1, then η = (1+ 5)/2, = ζ4 η with a ∈ {1, 3}
(for is quartic), hence η = −(2 + 2)/3 ∈ Z[], a contradiction. Therefore,
β = −ζ4 , Πη (X) = (X 2 −X +β)(X 2 −X + β̄) = X 4 −2X 3 +X 2 +1 and
ζ4 = η 2 − η ∈ Z[η] ⊆ Z[], which implies Z[] = Z[ζ4 ][] = Z[ζ4 ][ζ4a η 2 ] =
Z[ζ4 ][η 2 ] ⊆ Z[ζ4 ][η] = Z[η] ⊆ Z[]. Hence Z[] = Z[η] and d = dη .
However, we have the following Table:
204
Stéphane R. Louboutin
a
0
1
2
3
Πζ4a η2 (X)
4
3
2
X − 2X + 3X + 2X
X 4 + 4X 3 + 7X 2 + 4X
X 4 + 2X 3 + 3X 2 − 2X
X 4 − 4X 3 + 7X 2 − 4X
dζ4a η2
+1
+1
+1
+1
4352 = 42 · 272
272
4352 = 42 · 272
272
Since Π (X) must be of type (T), we have = ζ43 η 2 = (η 2 − η)3 η 2 =
−η 3 +η 2 +1 and Π (X) = X 4 −4X 3 +7X 2 −4X +1. Finally, using Πη (X),
4
j−1
= i=1 pi,j η i−1 , to
we compute (qi,j )1≤i,j≤4 = (pi,j )−1
1≤i,j≤4 , where 4
obtain η = i=1 qi,2 i−1 . Then, using Π (X), we compute ζ4 = η 2 − η in
Z[], which completes the proof (since d = 272 ∈ {256, 125, 144}, the last
assertion follows from the argument just before Proposition 11).
2
5.2.2 The case that μ() is of order 6
Lemma 15. Let be a totally imaginary quartic unit whose minimal monic
polynomial Π (X) is of type (T). Assume that μ() = {±1, ±ζ3 , ±ζ32 }. If
= ±ζ3a η 3 for some a ∈ {0, 1, 2} and some unit η ∈ Z[], then Π (X) =
X 4 − 5X 3 + 8X 2 − 4X + 1, d = 117, ζ3 = −3 + 42 − 5 + 1 ∈ Z[],
= ζ32 η 3 and η = −2 + 3 − 1 ∈ Z[] is a root of Πη (X) = X 4 −
2X 3 + 2X 2 − X + 1 of type (T), = −η 3 + η 2 + 1 ∈ Z[η], Z[η] = Z[],
dη = d = 117 and μ() = {±1, ±ζ3 , ±ζ32 } (case 3 of of Theorem 10).
Proof. We may assume that || > 1. Again, η 2 − αη + β = 0 for some
α ∈ Z[ζ3 ] and β a sixth root of unity. Since |(η+η )2 −ηη |2 = |α2 −β|2 = 1,
by (5), we have |α|2 ≤ 2. Hence, either α = 0, which would imply |η|2 =
| − β| = 1, a contradiction, or α ∈ {±1, ±ζ3 , ±ζ32 }. Since η = ±α2 η
satisfies η 2 − ±α3 η + α4 β = 0, we may assume that η 2 − η + β = 0 with
2
β ∈ {±1, ±ζ
we cannot have β = +1. If β = −1, then
√ 3 , ±ζ3 }. Hereaagain,
η = (1 + 5)/2, = ±ζ3 η 2 with a ∈ {1, 2}, hence η = (±3 − 5)/8 ∈
Z[], a contradiction. Hence, by changing η and into η̄ and ¯, we may assume
that β ∈ {±ζ3 }. Hence, ζ3 ∈ Z[η], and here again Z[] = Z[η] and d = dη .
We have Πη (X) = X 4 − 2X 3 + (1 + β + β̄)X 2 − (β + β̄)X + 1. If β = ζ3 ,
then |(η + η )2 − ηη |2 = |1 − ζ3 |2 = 3, which contradicts (5), Hence,
β = −ζ3 , ζ3 = η 2 − η , Πη (X) = X 4 − 2X 3 + 2X 2 − X + 1, dη = 117
and we have the following Table:
a
Π±ζ3a η3 (X)
d±ζ3a η3
0
1
2
X 4 ± X 3 + 5X 2 ∓ X + 1
X 4 ± 4X 3 − 8X 2 ± 5X + 1
X 4 ∓ 5X 3 + 8X 2 ∓ 4X + 1
9477 = 92 · 117
117
117
The fundamental unit of some quadratic, cubic or quartic orders
205
Since Π (X) must be of type (T), we have = ζ32 η 3 = (η 2 − η)2 η 3 =
2
−η 3 + η 2 + 1 and Π (X) = X 4 − 5X 3 + 8X 2 − 4X + 1.
5.2.3 The case that ± is a square in Z[]
Let be a totally imaginary quartic algebraic unit. We may assume that Π (X)
is of type (T), by changing into −, 1/ or −1/, if necessary. Let η be
a fundamental unit of the order Z[]. Assume that = ±η n for some n ∈
Z\{0}. Then, η is a totally imaginary quartic unit, Z[] = Z[η], and d = dη .
We may assume that Πη (X) is of type (T), by changing η into −η , 1/η or
−1/η , if necessary.
Lemma 16 (Compare with Lemma 5). Let be a totally imaginary quartic algebraic unit, with Π (X) of type (T). Then, ± is a square in Z[] if and
only we are in one of the seven following cases:
1. Π (X) = X 4 − 2bX 3 + (b2 + 2)X 2 − (2b − 1)X + 1, b ≥ 1, in which
cases = −1/η 2 where η = 3 − 2b2 + (b2 + 1) − (b − 1) ∈ Z[]
is a root of Πη (X) = X 4 − X 3 + bX 2 + 1 of type (T), and d = dη =
16b4 −4b3 −128b2 +114b+229 (cases 1, 2c.iii and 2d.i of Conjecture 8).
2. Π (X) = X 4 − X 3 + 2X 2 + 1, in which case = η 2 where η =
−3 + 2 − ∈ Z[] is a root of Πη (X) = X 4 − X 3 + 1 of type (T), and
d = dη = 229 (case 2c.i of Conjecture 8).
3. Π (X) = X 4 − 3X 3 + 2X 2 + 1, in which case = −η −2 where
η = −2 ++1 ∈ Z[] is a root of Πη (X) = X 4 −2X 3 +2X 2 −X +1
of type (T), and d = dη = 117 (case 1 of Theorem 10).
4. Π (X) = X 4 − 3X 3 + 5X 2 − 3X + 1, in which case = η 2 where
η = −3 +22 −2 ∈ Z[] is a root of Πη (X) = X 4 −X 3 −X 2 +X +1
of type (T), and d = dη = 117 (case 2 of of Theorem 10).
5. Π (X) = X 4 −5X 3 +9X 2 −5X +1, in which case = −η 2 where η =
−3 +42 −6+2 ∈ Z[] is a root of Πη (X) = X 4 −X 3 +3X 2 −X +1
of type (T), and d = dη = 189 (case 4 of of Theorem 10).
6. Π (X) = X 4 − 5X 3 + 5X 2 + 3X + 1, in which case = −η −2 where
η = −2 +2+2 ∈ Z[] is a root of Πη (X) = X 4 −3X 3 +3X 2 −X +1
of type (T), dη = d = 229 (case 2c.iv of Conjecture 8).
7. Π (X) = X 4 − 5X 3 + 7X 2 − 2X + 1, in which case = −η −2 where
η = 3 −42 +4 ∈ Z[] is a root of Πη (X) = X 4 −2X 3 +3X 2 −X +1
of type (T), and dη = d = 257 (case 2d.iii of Conjecture 8).
Proof. Assume that = ±η 2 or ±η −2 for some η ∈ Z[], with Πη (X) =
X 4 − aX 3 + bX 2 − cX + 1 of type (T). Hence, |c| ≤ a. The index (Z[η] :
Z[η 2 ]) is equal to |a2 + c2 − abc|. Hence, we must have |a2 + c2 − abc| = 1.
If c = 0, then 1 = |a2 + c2 − abc| = a2 , hence a = 1 and we are in the
first or the second case.
206
Stéphane R. Louboutin
√ Now, assume that c = 0. Then, 1 ≤ |c| ≤ a. If b ≥ 9, then b ≥
4b + 24 + 1 ≥ a + 1 ≥ |c| + 1 and we obtain 7 ≤ b − 2 ≤ |c|b −
2
2
2
2
(1 + c2 ) ≤ a|c|b − (a
√ + c ) ≤ |abc − a − c |. Hence, −1 ≤ b ≤ 8
and 1 ≤ |c| ≤ a ≤ 4b + 24. In this range, there are 14 triplets (a, b, c)
for which |a2 + c2 − abc| = 1. By getting rid of those for which Πη (X) =
X 4 − aX 3 + bX 2 − cX + 1 is not of type (T), we fall in one of the five
remaining last cases. Finally, by choosing between the four units ±η 2 or ±η −2
the ones whose minimal monic polynomials are of type (T), we complete the
proof of this Lemma.
2
5.2.4 The case that ± is an nth power in Z[]
Lemma 17. Let be a totally imaginary quartic algebraic unit such that
= ±η n , where η ∈ Z[] is therefore a totally imaginary quartic algebraic
unit whose minimal monic polynomial Πη (X) = X 4 − aX 3 + bX 2 − cX +
1 may be assumed to be of type (T). Then, Z[η] = Z[η n ] and dηn = dη .
Moreover, in the range −1 ≤ b ≤ 10000, we have dηn = dη for some
n ∈ {3, . . . , 10} if and only if we are in one of the following cases:
1. dη3 = dη if and only if we are in one of one of the following four cases:
(a) Πη (X) = X 4 − X 3 + 1, in which case Π1/η3 (X) = X 4 − 3X 3 +
3X 2 − X + 1 is of type (T) and dη3 = dη = 229.
(b) Πη (X) = X 4 − X 3 + 2X 2 + 1, in which case Π−η3 (X) = X 4 −
5X 3 + 5X 2 + 3X + 1 is of type (T) and dη3 = dη = 229.
(c) Πη (X) = X 4 − X 3 + X 2 + 1, in which case Π1/η3 (X) = X 4 −
3X 3 + X 2 + 2X + 1 is of type (T) and dη3 = dη = 257.
(d) Πη (X) = X 4 − 2X 3 + 4X 2 − X + 1, in which case Π−η3 (X) =
X 4 − 13X 3 + 43X 2 − 5X + 1 is of type (T) and dη3 = dη = 1229.
2. dη4 = dη if and only if we are in one of one of the following two cases:
(a) Πη (X) = X 4 − X 3 + 1, in which case Π−1/η4 (X) = X 4 − 4X 3 +
6X 2 − 3X + 1 is of type (T) and dη4 = dη = 229.
(b) Πη (X) = X 4 − X 3 + X 2 + 1, in which case Π−η4 (X) = X 4 −
5X 3 + 7X 2 − 2X + 1 is of type (T) and dη4 = dη = 257.
3. dη6 = dη if and only if Πη (X) = X 4 − X 3 + 1, in which case
Π−η6 (X) = X 4 − 5X 3 + 5X 2 + 3X + 1 and dη6 = dη = 229.
4. dη7 = dη if and only if Πη (X) = X 4 − X 3 + 1, in which case
Π−1/η7 (X) = X 4 − 7X 3 + 14X 2 − 6X + 1 and dη7 = dη = 229.
5. dη5 , dη8 , dη9 and dη10 are never equal to dη .
n−1
Proof. Noticing that Πηn (X n ) = k=0 Πη (ζnk X), we compute Πηn (X) and
dηn from the knowledge of Πη (X). For example, if n = 3, there are 6 triplets
The fundamental unit of some quadratic, cubic or quartic orders
207
√
(a, b, c) in the range |c| ≤ a ≤ 4b + 24 and −1 ≤ b ≤ 10000 (see (4))
such that dη3 = dη > 0: (a, b, c) ∈ {(0, 0, 0), (1, 0, 0), (1, 1, 0), (1, 1, 1),
(1, 2, 0), (2, 4, 1)}. Since X 4 + 1 and X 4 − X 3 + X 2 − X + 1 are not of type
(T) (their complex roots are roots of unity), we obtain the first result.
2
Lemma 18 (Compare with Lemma 6). Let be a totally imaginary algebraic quartic unit, with Π (X) = X 4 − a X 3 + b X 2 − c X + 1 ∈ Z[X] of
type (T). Assume that = ±η n with |n| ≥ 3 and η ∈ Z[] a root of Πη (X) =
X 4 − aη X 3 + bη X 2 − cη X + 1 ∈ Z[X] of type (T). Then, 0 ≤ bη < b and
dη = d , except in the case that Π (X) = X 3 − 3X 3 + X 2 + 2X + 1 where
d = 257, = η −3 with η = 2 − ∈ Z[] and Πη (X) = X 4 −X 3 +X 2 +1.
Proof. Suppose that b ≤ bη . We have |η|6 +1/|η|6 −4 ≤ ||2 +1/||2 −4 ≤
b < bη ≤ |η|2 + 1/|η|2 + 4. Hence, 0 < | log |η|| < 0.393. By (4), there
are 8 such possible quartic polynomials Πη (X) of type (T). They all satisfy
−1 ≤ b ≤ 2. Hence, we would have −1 ≤ b ≤ 2. By (4), there are 22
quartic polynomials P (X) of type (T) with −1 ≤ b ≤ 2, and the repeated
discriminants are 117, 229 and 257 and are obtained for the following polynomials:
b
−1
0
1
1
2
2
2
P (X)
X4 − X3 − X2 + X + 1
X4 − X3 + 1
X4 − X3 + X2 + 1
X 4 − 3X 3 + X 2 + 2X + 1
X 4 − 2X 3 + 2X 2 − X + 1
X 4 − 3X 3 + 2X 2 + 1
X 4 − X 3 + 2X 2 + 1
dP
| log(P )|
117
229
257
257
117
117
229
0.27176 . . .
0.16868 . . .
0.22106 . . .
0.66320 . . .
0.27176 . . .
0.54353 . . .
0.33737 . . .
This proves the Lemma.
2
5.2.5 The case that Π (X) = X 4 − X 3 + bX 2 + 1 ∈ Z[X], b ≥ 3
Proposition 19. Let be a complex root of Π (X) = X 4 −X 3 +bX 2 +1 ∈
Z[X], b ≥ 3. Then, is a totally imaginary quartic unit, d = 16b4 − 4b3 −
144b + 229 > 0√is odd, μ() = {±1}, hence we may assume that
128b2 + √
|| > 1, b − 1 ≤ || ≤ b, d is asymptotic to 16||8 as b → ∞, and is
a fundamental unit of the quartic order Z[]. It follows that ˜ = 2 which is
a root of Π˜(X) = X 4 + (2b − 1)X 3 + (b2 + 2)X 2 + 2bX 2 + 1 is such
that d˜ = d is asymptotic to 16|˜
|4 as b → ∞. Finaly, Q() contains no
imaginary quadratic subfield.
Proof. Since X 4 −X 3 +1 has no real root, we have Π (x) ≥ x4 −x3 +1 > 0
for x real, and is totally imaginary. Since Π (X) is Q-irreducible, is a
208
Stéphane R. Louboutin
totally imaginary quartic unit. We now that μ() is of order 8, 10, 12, in which
cases d = 256, 125 or 144, respectively, or of order 4 or 6, in which cases
42 or 32 divides d , respectively. Since d is odd, not divisible by 3 and never
equal to 125, we have μ() = {±1}. Since 1/||2 is the only real root in
(0, 1) of Q(X) = X 6 − bX 5 − X 4 + (2b − 1)X 3 − X 2 − bX + 1 (see
the proof of Lemma 13), and since Q(1/b) = (b4 − b3 − 2b2 + 1)/b6 > 0
and Q(1) = −1 < 0, we have ||2 < b. Since Q(1/(b − 1)) < 0, we have
||2 > b − 1. Let η be fundamental unit of Z[]. Suppose that = ±η n with
|n| ≥ 2. We have |n| ≥ 3, by Lemma 16. Hence, by (8) below, we have
d = dη ≤ 16(|η| + 1/|η|)8 ≤ 16(||1/3 + 1/||1/3 )8
≤ 16(b1/6 + 1/b1/6 )8 ,
which implies b < 5. Hence, is indeed a fundamental unit of Z[] for
b ≥ 5. For b = 3, we have || = 1.67635 . . . and for b = 4, we have
|| = 1.95375 . . . . By Lemma 13, if = ±η n for some η ∈ Z[], then
|n| ≤ 4. We then use Proposition 17 to settle these two cases. Let us prove
the last assertion. The cubic resolvent of X 4 Π (1/X) = X 4 + bX 2 − X + 1
is R(X) = X 3 − 2bX 2 + (b2 − 4)X + 1, and dR(X) = dΠ (X) = d > 0.
It is easy to see that R(X) is Q-irreducible and that dR(X) is never a perfect
square. It follows that Q() is a non-normal quartic field and that the Galois
group of its normal closure is the symmetric group S4 (see [6, Theorem 7.5.4]
or [7, Theorem 80]). In particular, Q() contains no quadratic subfield.
2
In the same spirit, we have:
Proposition 20. Let be a complex root of P√(X) = X 4 − aX 3 + bX 2 −
aX + 1 ∈ Z[X] with b ≥ 4 and 0 ≤ a ≤ 4b − 11. Then, is a totally
imaginary quartic unit, it is not a complex root of unity,
assume
√ hence we may √
that || > 1, d = ((b + 2)2 − 4a2 )(4b − a2 − 8)2 , b − 4 ≤ || ≤ b and
5||4 ≤ d = dP ≤ 17||8 .
(7)
Moreover, if b = (a2 − 11)/4, then d is asymptotic to 9||4 as a → ∞,
whereas if a is fixed, then d is asymptotic to 16||8 as b → ∞.
Proof. We use the explict formula for ||2 obtained in the proof of Lemma 13.
Since dP ≥ 9((b + 2)2 − 4(4b − 11)) = 9(b2 − 12b + 48) and
2
2
4|| ≤ b − 2 + (b + 2) + 2b2 − 8 + 2(b − 2) (b + 2)2 ≤ 4b,
the lower bound in (7) holds true for b ≥ 23, and we numerically checked it
for 3 ≤ b ≤ 22. Now, we have dP ≤ (b+2)2 (4b−8)2 = 16(b2 −4)2 ≤ 16b4 .
On the other hand, we have (b + 2)2 − 4a2 ≥ (b + 2)2 − 4(4b − 11) =
The fundamental unit of some quadratic, cubic or quartic orders
209
b2 − 12b + 48 and 2b2 − 4a2 − 8 ≥ 2b2 − 4(4b − 11) − 8 = 2b2 − 16b + 36.
Hence,
√
4||2 ≥ b − 2 + b2 − 12b + 48
√
+ 2b2 − 16b + 36 + 2(b − 2) b2 − 12b + 48
and ||2 ≥ b − 4. Since, dP ≤ 16b4 ≤ 17(b − 4)4 ≤ 17||8 for b ≥ 266, the
upper bound in (7) holds true for b ≥ 266, and we numerically checked it for
3 ≤ b ≤ 265.
2
5.3 Proof of Theorem 10
Assume that is not a fundamental unit of Z[]. We prove that we are in
one of the five considered cases. Then, = ζη n with ζ ∈ μ(), η ∈ Z[]
and |n| ≥ 2. By Proposition 11, we may assume that K = Q() contains
an imaginary quadratic subfield k such that μ() ⊆ k. In particular, μ() =
{±1}, μ() = {±1, ±ζ4 } or μ() = {±1, ±ζ3 , ±ζ32 }. By Proposition 12,
we have |n| ≤ 5. By Lemmas 14 and 16, we may assume that |n| ∈ {3, 5}.
By Lemmas 15, 17 and the last assertion of Proposition 19, we may assume
that |n| = 5. Finally, Lemma 17 settles this case. Conversely, in all these five
cases, η is the fundamental unit of Z[] (apply the same resoning to η instead
of applying it to . In the same way, we can prove that in all the fiveteen cases
of Conjecture 8, the given η is indeed a fundamental unit of the quartic order
Z[]).
6. How to prove conjecture 8?
Theorem 10 and Lemmas 16, 17 and 18 show that if is a totally imaginary
algebraic quartic unit, with Π (X) = X 4 − a X 3 + b X 2 − c X + 1 ∈ Z[X]
of type (T), and if is not a fundamental unit of the order Z[], then either we
are in one of the cases of Conjecture 8, or |b | > 10000, or ± is at least an
eleventh power of a unit η ∈ Z[] (and these ranges could be easily extended).
We now explain which result whose proof eludes us would enable us to settle
Conjecture 8.
Proposition 21. Assume P (X) is of type (T). Let be any one of the two
complex roots of P (X) satisfying || > 1. Then,
0 < d ≤ 16(|| + 1/||)8 ≤ 4096||8 .
Proof. We have dP = 16()2 ( )2 | − |4 | − ¯ |4 .
(8)
2
210
Stéphane R. Louboutin
This Proposition and Propositions 19 and 20 make it reasonable to conjecture the following result on quartic polynomials which would enable us to
prove Conjecture 8:
Conjecture 22. There exists some explicit constant C > 0 such that for
any totally complex quartic algebraic unit satisfying || > 1 we have d ≥
C||4 . In particular, there would be only finitely many totally complex quartic
algebraic units of bounded discriminant, and if || were large enough then would be either a fundamental unit of the quartic order Z[] or its square, up
to a product by a complex root of unity in Z[].
References
[1] S. Louboutin, The class-number one problem for some real cubic number fields with
negative discriminants, J. Number Theory, 121 (2006) 30–39.
[2] T. Nagell, Zur Theorie der kubischen Irrationalitäten, Acta Math., 55 (1930) 33–65.
[3] T. Nagell, Sur les discriminants des nombres algébriques, Ark. Mat., 7 (1967) 265–282.
[4] L. Robertson, Power bases for cyclotomic integer rings, J. Number Theory, 69 (1998)
98–118.
[5] L. Robertson, Power bases for 2-power cyclotomic fields, J. Number Theory, 88 (2001)
196–209.
[6] S. Roman, Field Theory, Second Edition, Grad. Texts Math., 158, Springer, New York
(2006).
[7] J. Rotman, Galois Theory, Springer-Verlag, Universitext (1990).
[8] L. C. Washington, Introduction to Cyclotomic Fields, Springer-Verlag, Grad. Texts
Math., 83, Second Edition (1997).