Multiplicative relations with conjugate algebraic numbers
We study which algebraic numbers can be represented by a product of conjugate over a fixed number field K algebraic numbers in fixed integer powers. The problem is nontrivial if the sum of these integer powers is equal to zero. The norm over K of such number must be a root of unity. We show that t...
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nasplib_isofts_kiev_ua-123456789-1642122025-02-09T15:35:03Z Multiplicative relations with conjugate algebraic numbers Мультиплікативні співвідношення зі спряженими алгебраїчними числами Dubickas, A. Статті We study which algebraic numbers can be represented by a product of conjugate over a fixed number field K algebraic numbers in fixed integer powers. The problem is nontrivial if the sum of these integer powers is equal to zero. The norm over K of such number must be a root of unity. We show that there are infinitely many algebraic numbers whose norm over K is a root of unity and which cannot be represented by such product. Conversely, every algebraic number can be expressed by every sufficiently long product in conjugate over K algebraic numbers. We also construct nonsymmetric algebraic numbers, i.e., such that none elements of the respective Galois group acting on the full set of their conjugates form a Latin square. Досліджено, які алгебраїчні числа можуть бути зображені у вигляді добутку спряжених над фіксованим числовим полем K алгебраїчних чисел у фіксованих цілих степенях. Розглядувана задача є нетривіальною, якщо сума цих цілих степенів дорівнює нулю. Норма над K такого числа має бути коренем з одиниці. Показано, що існує нескінченно багато алгебраїчних чисел, норма над K яких є коренем з одиниці і які не можуть бути зображені згаданим добутком. Навпаки, кожне алгебраїчне число можна виразити будь-яким достатньо довгим добутком спряжених над K алгебраїчних чисел. Побудовано також несиметричні алгебраїчні числа, тобто такі, що жоден елемент відповідної групи Галуа, яка діє на повній множині їхніх спряжень, не формує Латинський квадрат. This research was partially supported by the Lithuanian State Science and Studies Foundation. 2007 Article Multiplicative relations with conjugate algebraic numbers / A. Dubickas // Український математичний журнал. — 2007. — Т. 59, № 7. — С. 890–900. — Бібліогр.: 20 назв. — англ. 1027-3190 https://nasplib.isofts.kiev.ua/handle/123456789/164212 512.5 en Український математичний журнал application/pdf Інститут математики НАН України |
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Статті Статті Dubickas, A. Multiplicative relations with conjugate algebraic numbers Український математичний журнал |
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We study which algebraic numbers can be represented by a product of conjugate over a fixed number
field K algebraic numbers in fixed integer powers. The problem is nontrivial if the sum of these integer
powers is equal to zero. The norm over K of such number must be a root of unity. We show that there
are infinitely many algebraic numbers whose norm over K is a root of unity and which cannot be
represented by such product. Conversely, every algebraic number can be expressed by every sufficiently
long product in conjugate over K algebraic numbers. We also construct nonsymmetric algebraic
numbers, i.e., such that none elements of the respective Galois group acting on the full set of their
conjugates form a Latin square. |
| format |
Article |
| author |
Dubickas, A. |
| author_facet |
Dubickas, A. |
| author_sort |
Dubickas, A. |
| title |
Multiplicative relations with conjugate algebraic numbers |
| title_short |
Multiplicative relations with conjugate algebraic numbers |
| title_full |
Multiplicative relations with conjugate algebraic numbers |
| title_fullStr |
Multiplicative relations with conjugate algebraic numbers |
| title_full_unstemmed |
Multiplicative relations with conjugate algebraic numbers |
| title_sort |
multiplicative relations with conjugate algebraic numbers |
| publisher |
Інститут математики НАН України |
| publishDate |
2007 |
| topic_facet |
Статті |
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https://nasplib.isofts.kiev.ua/handle/123456789/164212 |
| citation_txt |
Multiplicative relations with conjugate algebraic numbers / A. Dubickas // Український математичний журнал. — 2007. — Т. 59, № 7. — С. 890–900. — Бібліогр.: 20 назв. — англ. |
| series |
Український математичний журнал |
| work_keys_str_mv |
AT dubickasa multiplicativerelationswithconjugatealgebraicnumbers AT dubickasa mulʹtiplíkativníspívvídnošennâzísprâženimialgebraíčnimičislami |
| first_indexed |
2025-11-27T11:40:54Z |
| last_indexed |
2025-11-27T11:40:54Z |
| _version_ |
1849943560912109568 |
| fulltext |
UDC 512.5
A. Dubickas (Vilnius Univ. and Inst. Math. and Inform., Lithuania)
MULTIPLICATIVE RELATIONS
WITH CONJUGATE ALGEBRAIC NUMBERS*
MUL|TYPLIKATYVNI SPIVVIDNOÍENNQ
ZI SPRQÛENYMY ALHEBRA}ÇNYMY ÇYSLAMY
We study which algebraic numbers can be represented by a product of conjugate over a fixed number
field K algebraic numbers in fixed integer powers. The problem is nontrivial if the sum of these integer
powers is equal to zero. The norm over K of such number must be a root of unity. We show that there
are infinitely many algebraic numbers whose norm over K is a root of unity and which cannot be
represented by such product. Conversely, every algebraic number can be expressed by every sufficiently
long product in conjugate over K algebraic numbers. We also construct nonsymmetric algebraic
numbers, i.e., such that none elements of the respective Galois group acting on the full set of their
conjugates form a Latin square.
DoslidΩeno, qki alhebra]çni çysla moΩut\ buty zobraΩeni u vyhlqdi dobutku sprqΩenyx nad
fiksovanym çyslovym polem K alhebra]çnyx çysel u fiksovanyx cilyx stepenqx. Rozhlqdu-
vana zadaça [ netryvial\nog, qkwo suma cyx cilyx stepeniv dorivng[ nulg. Norma nad K tako-
ho çysla ma[ buty korenem z odynyci. Pokazano, wo isnu[ neskinçenno bahato alhebra]çnyx çy-
sel, norma nad K qkyx [ korenem z odynyci i qki ne moΩut\ buty zobraΩeni zhadanym dobutkom.
Navpaky, koΩne alhebra]çne çyslo moΩna vyrazyty bud\-qkym dostatn\o dovhym dobutkom
sprqΩenyx nad K alhebra]çnyx çysel. Pobudovano takoΩ nesymetryçni alhebra]çni çysla,
tobto taki, wo Ωoden element vidpovidno] hrupy Halua, qka di[ na povnij mnoΩyni ]xnix sprq-
Ωen\, ne formu[ Latyns\kyj kvadrat.
1. Introduction. Let K be a number field, i.e., a finite extension of the field of
rational numbers Q . In this paper we investigate multiplicative relations with
conjugate algebraic numbers. More precisely, given β ∈ Q and k1 , … , kn ∈ Z
*, our
main concern is to determine whether or not β can be expressed as α α1
1k
n
kn… with
some algebraic numbers α1 , … , αn conjugate over K. (Throughout, as usual, Q
denotes the set of algebraic numbers, and Z* denotes the set of non-zero integers.)
Let M ( K; k1 , … , kn ) be the set of all β expressible as α α1
1k
n
kn… . Here, we do
not assume that α1 , … , αn are all distinct, nor we assume that the degree of α = α1
over K is equal to n. Throughout, we reserve the letter d for the degree of β over
K. Also, with β1 = β , β2 , … , βd being the full set of conjugates of β over K, let
L = K ( β1 , … , βd ) be the normal closure of K ( β ) over K, and let G = Gal ( L / K ) be
the Galois group of L / K.
As in [1], it is easily seen that M ( K; k1 , … , kn ) = Q , unless k1 + … + kn = 0.
(Just take α1 = … = αn = β1 1/( +…+ )k kn .) Also,
M ( K; 1, – 1 ) ⊂ M ( K; k1 , … , kn ).
Indeed, the equality k1 + … + kn = 0 with non-zero k1 , … , kn implies that n ≥ 2.
The above inclusion now easily follows, by setting α2 = … = αn and observing that
M ( K; k k1 , … , k kn ) = M ( K; k1 , … , kn ) for k ∈ Z
* (by Theorem 1 below).
The structure of M ( K; k1 , … , kn ) is nontrivial if n ≥ 2 and k1 + … + kn = 0 (see,
for instance, Corollary 2 in Section 5). Note that if β ∈ M ( K; k1 , … , kn ) with k1 , …
*
This research was partially supported by the Lithuanian State Science and Studies Foundation.
© A. DUBICKAS, 2007
890 ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 7
MULTIPLICATIVE RELATIONS WITH CONJUGATE ALGEBRAIC NUMBERS 891
… , kn ∈ Z
* such that k1 + … + kn = 0, then its norm over K, namely, Norm ( β ) =
= β1 , … , βd must be a root of unity. Indeed, setting F for the normal closure of
L ( α ) over K and substituting β = α α1
1k
n
kn… , we deduce that
Norm Norm Norm/ / /F K F K j
k
j
n
F K
k kj n( ) = ( ) = ( ) = ( )
∈ =
+…+∏ ∏β σ β α α
σ G 1
1 = 1,
where G = Gal ( L / K ). Since Norm /F K ( )β is a natural power of Norm ( β ), the latter
number is a root of unity.
In the next section, we state the main results of this paper. Their comparison with
earlier results (in particular, with additive results) will be discussed in Section 3. In
Section 4, we prove Theorem 1 and Corollary 1. Section 5 contains the proofs of
Theorems 2 and 3 which show that the condition on the norm of β is not sufficient for
it to belong to M ( K; k1 , … , kn ). In Section 6 we prove Theorem 4 which asserts that
every β whose norm is a root of unity can be represented by every sufficiently long
multiplicative form. We also present an example showing how, for a given β, one can
find the respective α. The last section contains the construction of nonsymmetric
numbers (see the definition at the end of Section 2).
2. Main results. Below, k1 , … , kn are integers, K is an arbitrary number field,
L is the normal closure of K ( β ) over K, and G = Gal ( L / K ). Also, for r ∈ Q , the
number β βr r= { }exp log is defined by taking the principal branch of the logarithm.
Our first theorem shows that the set M ( K ; k1 , … , kn ) is invariant under
multiplication by roots of unity. This implies that the search for possible a can be
reduced to those numbers whose powers lie in the field L.
Theorem 1. Suppose that β ∈ M ( K; k1 , … , kn ), r ∈ Q , and ζ is a root of
unity. Then ζ β, βr ∈ M ( K; k1 , … , kn ).
Corollary 1. Given integers k1 , … , kn , assume that β = α α1
1k
n
kn… , where
α1 , … , αn are all conjugate to α over K. Then α can be chosen so that one of
its natural powers lies in L.
In the next two theorems we show that not all algebraic numbers whose norm is a
root of unity lie in the set M ( K; k1 , … , kn ), where k1 + … + kn = 0. The proof of
Theorem 3 is constructive and, at the same time, it is rather unusual for this kind of
proofs. It uses, for instance, some elementary properties of the Pell equation.
Theorem 2. Suppose that β ∈ M ( K; k1 , … , kn ) with k1 , … , kn ∈ Z
* such
that k1 + … + kn = 0. Then there is a subgroup H of G , generated by n – 1
(not necessarily distinct) elements, such that σ βσ ( )∈∏ H
is a root of unity.
Theorem 3. Assume that k1 , … , kn ∈ Z
* are such that k1 + … + kn = 0. Then
there exists an algebraic number β ∉ M ( K; k1 , … , kn ) of degree d = 2n over K
whose norm over K is equal to 1.
Our final theorem shows that the condition on Norm ( β ) to be a root of unity is not
only necessary, but also sufficient for β to lie in M ( K; k1 , … , kn ), provided that n
is sufficiently large.
Theorem 4. Assume that k1 , … , kn ∈ Z
*, and β is an algebraic number of
degree d over K whose norm over K is a root of unity. If n ≥ 2d – 5, then
β ∈ M ( K; k1 , … , kn ).
ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 7
892 A. DUBICKAS
Similarly to Theorem 3 in [1], the inequality n ≥ 2d – 5 can be replaced by n ≥
≥ 2 [ d / 2 ] – 1 for symmetric β. Here, [ … ] stands for the integral part, and β ∈ K
of degree d over K is called symmetric over K if there exist σ2 , … , σd ∈ G such
that the matrix σ βi j i j d
( )
= …, , ,1
, where σ1 stands for the identity, is a Latin square,
namely, each of its rows and each of its columns is a permutation of β1 , … , βd . In
Section 7 we will prove the result which was announced in [1]: the smallest possible
degree for nonsymmetric numbers to occur is equal to 6.
3. Comparison with earlier results and comments. There are several types of
problems concerning additive and multiplicative relations in conjugates of an algebraic
number. Given a field K, an algebraic number β over K, and k1 , … , kn ∈ K, one
can ask, for instance, whether β can be expressed as k1 α1 + … + kn αn with distinct
α1 , … , αn conjugate over K. Similarly, for integer k1 , … , kn , one can ask whether
β is expressible as α α1
1k
n
kn… . The cases β = 0 (and β = 1 in the multiplicative
setting, respectively) were studied earlier by V. A. Kurbatov [2], C. J. Smyth [3, 4],
K. Girstmair [5, 6], J. D. Dixon [7], M. Drmota and M. Skalba [8] (see also [9, 10]).
Similar problems were also studied by E. M. Matveev [11], the author [12] and
T. Zaimi [13 – 15].
Given a positive integer n and non-zero k1 , … , kn ∈ K, one can also ask which
algebraic numbers β over K can be written as
β = k1 α1 + … + kn αn
with algebraic numbers α1 , … , αn conjugate over K . For n = 2, the complete
answer was given in [16]: an algebraic number β can be written as a difference α1 –
– α2 of algebraic numbers α1
, α2 conjugate over a number field K if and only if
there is σ ∈ G such that σ βi
i
( )=
−∑ 0
1v
= 0. (Here, v is the order of the cyclic group
〈 σ 〉 generated by σ .) The case n ≥ 3 was the main subject of our paper [1].
Similarly, β can be written as a quotient α1 / α2 of algebraic numbers α1
, α 2
conjugate over a number field k if and only if there is σ ∈ G such that
σ βi
i
( )=
−∏ 0
1v
is a root of unity. Note that in Hilbert’s Theorem 90 (see, e.g., [17, 18] and also [19,
20] for generalizations), where both β and α are only allowed to lie in a fixed cyclic
extension of K, the answer is different.
Let k1 , … , kn ∈ Z
*. Assume that β = α α1
1k
n
kn… with α1 , … , αn conjugate to α
over a number field K. In [1] we asked whether it is true that α can be chosen so that
its natural power is equal to a a
d
adβ β1
1 … with integer a, a1 , … , ad
? This, as we
claimed, would be sufficient in order to give the additive theorems of [1] in the
multiplicative form. In the present paper, we use a much weaker version of this
statement (Corollary 1), but still attain the same goals as in [1].
There is nothing like Theorem 1 needed in the additive case, because, for r ∈ Q ,
the numbers r α and r α′ are conjugate over K if so are α and α′. This is, in
general, false in the multiplicative case: αr and α′
r need not be conjugate for α and
α′ being conjugate. Theorem 2 is a direct analogue of the respective additive theorem
in [1] both in terms of the result and in terms of the proof. The proof of Theorem 3 is
much more subtle compared to its additive analogue (see the construction before
Corollary 1 in [1]), because now we cannot use the normal basis theorem. The present
construction uses, for instance, the fact that the Pell equation X mY2 2− = 1, where m
is square-free, has infinitely many solutions in positive integers X, Y . It also involves
an extra part of combinatorics. Finally, Theorem 4 looks essentially the same as does
ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 7
MULTIPLICATIVE RELATIONS WITH CONJUGATE ALGEBRAIC NUMBERS 893
its additive analogue (Theorem 3 in [1]), although, because of what was said earlier, the
practical computation becomes more difficult (see the example in Section 6). In
particular, for d = 4, it follows that every β of degree ≤ 4 over Q can be
represented by every form α α α1 2 3
1 2 3k k k of length 3 with fixed k1 , k2 , k3 ∈ Z
* and some
algebraic numbers α1 , α2 , α3 conjugate over Q. Thus, for d = 4, the inequality n ≥
≥ 3 of Theorem 4 is sharp. It cannot be replaced by n ≥ 2, which is shown by the
example of β = 1 + 2 6+ ∉ M ( Q; 1, – 1 ) (see [16] or apply Theorem 2 with n =
= 2 combined with the fact that, for this β, G is the Klein 4-group).
4. Restrictions on algebraic numbers. Proof of Theorem 1. Write β =
= α α1
1k
n
kn… with α1 , … , αn conjugate over a number field K. Assume that m is a
positive integer and ζ1 , … , ζ n are arbitrary mth roots of unity. We will show first
that there is a positive integer a such that ζ α ζ α1
1
1
1a am
n
m
n
/ /, ,… are conjugate over
K. Here, a m1/ denotes the positive mth root of a.
Indeed, let F be the normal closure of K ( α, µm ) over K , where µm is the
primitive mth root of unity. Take a positive integer a such that the polynomial zm –
– a is irreducible over F. (This is possible, e.g., by Theorem 16 on p. 221 in Lang’s
book [18].) Then, firstly, α 1 , … , αn are all conjugate over K a m( )1/ , for otherwise
the minimal polynomial of α over K is reducible over K a m( )1/ . We thus get
D > [ ] [ ][ ]
[ ]
[ ]( ) ( ) = ( ) ( ) ( )
( )
= ( )K a K a
K a K K K
K a K
K Km m
m
m
1 1
1
1
/ /
/
/, :
, : :
:
:α α α α α = D,
where D is the degree of α over K, a contradiction. Secondly, F a m( )1/
/ F and
F / K are both Galois extensions, hence there are automorphisms τ1 , … , τn in the
Galois group of F a m( )1/
/ K fixing F and taking a m1/ to ζ ζ1
1 1a am
n
m/ /, ,… ,
respectively. Finally, F a m( )1/
/ K a m( )1/ is a Galois extension whose Galois group
isomorphic to that of F / K (see, for instance, Theorem 4 of Ch. VIII in. [18]). Thus
there are automorphisms σ 1 , … , σn in the Galois group of F a m( )1/
/ K fixing
K a m( )1/ and taking α to α 1 , … , αn
, respectively. Note that τ σ αj j
ma( )1/ =
= τ αj
m
ja( )1/ = ζ αj
m
ja1/ , where j = 1, … , n. It follows that ζ α ζ α1
1
1
1a am
n
m
n
/ /, ,…
are all conjugate over K, as claimed.
Write ζ = exp /{ π − }2 1u m with u < m coprime. Let k ′ be the greatest common
divisor of k1 , … , kn
. We can certainly assume that k ′ = 1, for otherwise the initial set
of conjugates α1 , … , αn can be replaced by the set α α1
′ ′…k
n
k, , . Clearly, there exist
nonnegative integers r1 , … , rn < m such that r1 k1 + … + rn kn is equal to u modulo
m. Take a ∈ Z
* so that δ1 = µ αm
r ma1 1
1
/ , … , δn = µ αm
r m
n
n a1/ are conjugate over K.
Using k1 + … + kn = 0 (which can be assumed without loss of generality, for
otherwise M ( K; k1 , … , kn ) = Q , and there is nothing to prove), we obtain that
δ δ µ α α µ β ζβ1 1
1 1 1 1k
n
k
m
r k r k k
n
k
m
un n n n… = … = =+…+ .
Consequently, ζ β ∈ M ( K; k1 , … , kn ) .
For every r ∈ Q, there are roots of unity ζ1 , … , ζ n such that ζ α ζ α1 1
r
n n
r, ,… are
all conjugate to αr over K. (Recall that αr = exp log{ }r α is defined by taking the
principal branch of the logarithm.) Thus β = α α1
1k
n
kn… implies that
ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 7
894 A. DUBICKAS
( ) … ( ) =ζ α ζ α ζβ1 1
1r k
n n
r k rn
with some root of unity ζ. Hence ζβr lies in M ( K; k1 , … , kn ), and, by the above,
so does ζ ζβ−1 r = βr
.
Proof of Corollary 1. Let F be the normal closure of L ( α ) over K , and let
G = = Gal ( F / K ). We have that K ⊂ L ⊂ F. By the main theorem of Galois theory, G
= = G / H, where
H = { σ ∈ G | σ ( x ) = x for all x ∈ L }.
Assuming that H = { σ1 , … , σm }, we set
ϕ ( x ) = σ1 ( x ) … σm ( x )
for every x ∈ F. Clearly, ϕ ( β ) = βm
, since β ∈ L. On applying ϕ to the equality
β = α α1
1k
n
kn… , we deduce that
βm
= ϕ α ϕ α( ) … ( )1
1k
n
kn .
Also, as H is a group, σj ( ϕ ( α ) ) = ϕ ( α ) for every j = 1, … , n. Hence
ϕ ( α ) ∈ L = { x ∈ F | σ ( x ) = x for all σ ∈ H }.
The numbers ζ ϕ α ζ ϕ α1 1
1 1( ) … ( )/ /, ,m
n n
m are conjugate over K for some m th
roots of unity ζ1 , … , ζ n
. Now, as in the proof of Theorem 1, it follows that there is a
positive integer a such that β = δ δ1
1k
n
kn… , where δ1 = ζ ϕ α1
1
1
1a m m/ /( ) , … , δn =
= ζ ϕ αn
m
n
ma1 1/ /( ) are all conjugate over K. This completes the proof, since δ1
m =
= δm
= a ϕ ( α ) ∈ L.
5. On numbers which cannot be represented. Proof of Theorem 2. Suppose
that β can be expressed as α α1
1k
n
kn… . By Corollary 1, there is a positive integer m
such that αm
∈ L. On replacing β, α 1 , … , αn by their m th powers (without
changing the notation for α ), we see that the new α lies in L. It follows that βm
=
= α σ α σ αk k
n
kn1 2
2( ) … ( ) with σ2 , … , σn ∈ G. Setting H = 〈 σ2 , … , σn 〉, we
deduce that
σ β σ α σ α σ α σ α
σ σ σ
( ) = ( ) … ( ) = ( )
∈ ∈
+…+
∈
∏ ∏ ∏( )m
H
k k
n
k
H
k k
H
n n1 2 1
2 = 1,
which implies Theorem 2.
Let K be a number field, and let p1 , … , pn be prime numbers such that p1 ∉
∉ K, p2 ∉ K p1( ), … , pn ∉ K p pn1 1, ,…( )− . Let S1 , … , Sl be all l = 2n – 1
nonempty subsets of the set { p1 , … , pn
} (in an arbitrary order). Set mi = p
p Si∈∏ .
Assume that xi
, yi are solutions of the Pell equations
X m Yi
2 2− = 1
(in positive integers), where i = 1, … , l, satisfying xi > ( )+2 1
2
xi
l for 1 ≤ i ≤ l – 1.
Consider the number
β = ( + )
=
∏ x y mi i i
i
l
1
.
ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 7
MULTIPLICATIVE RELATIONS WITH CONJUGATE ALGEBRAIC NUMBERS 895
Lemma 1. The number β is a unit of degree 2n over K such that no product
of fewer than 2n of its conjugates is a root of unity.
Proof of Lemma 1. We see at once that β is a unit, because it is a product of
units x y mi i i+ . Since
K p p Kn1 1, , :…( )[ ]− = 2n,
it follows immediately that the degree of β over K is at most 2 n. The Galois group
of K p pn1, ,…( ) / K is generated by n elements of order 2, say σ1 , … , σn , where
σj maps pj to – pj and every other pi , i ≠ j, to itself. The conjugates of β
are all of the form
β′ = ( + )
=
∏ x y mi i i i
i
l
ε
1
,
where εi ∈ { 1, – 1 }. We call ( ε1 , ε2 , … , εl ) the signature of β′. The signature of
every β′ is uniquely prescribed by the n signs εi which correspond to Si containing
exactly one prime number. Consider the table with 2n rows and 2 n – 1 = l columns,
whose first row contains 2n – 1 of plus signs, and whose other 2n – 1 rows
correspond to the signatures of different β′.
We first show that every row except for the first contains 2n
–
1 – 1 of plus signs
and 2n
–
1 of minus signs. This is, of course, the case for n = 1. Assume that this is
true with n – 1 instead of n. By adding the square root of the pn th prime with plus
sign, we increase the number of plus signs by 1 + ( 2n
–
2 – 1 ) = 2n
–
2. The total number
of plus signs will be 2n
–
2 – 1 + 2n
–
2 = 2n
–
1 – 1. Similarly, after adding the square
root of the pn th prime with minus sign, the total number of plus signs will be 2 n
–
2 –
– 1 + 2n
–
2 = 2n
–
1 – 1, unless all square roots p pn1 1, ,… − were with plus signs.
The latter situation however could be also achieved by adding the square root of the
pn – 1 th prime with minus sign which leads to the former situation. Alternatively, if just
one pn is with minus sign, then one can find the total number of minus signs by the
formulae
1 1
1
2
1
1
+ − +
−
+ … +
−
−
n
n n
n
= 2n
–
1.
Furthermore, every column of the table contains 2 n
–
1 of plus signs and 2n
–
1 of
minus signs. Indeed, if the sign of the column is determined by the sign of the product
of v signs, then it is minus in
v v v
v
v v v
1 3 2 1 2 1
2 2 1
+
+ … +
( − ) +
=[ ]
− − + −
/
n n = 2n
–
1
cases. The product of β and all different β′ is thus equal to 1. Clearly, β and all β′
are positive. Both remaining claims of the lemma will therefore follow if the product
of < 2n (not necessarily distinct) conjugates β is newer equal to 1.
Suppose, contrary to our claim, that the product of some s < 2n conjugates of β is
equal to 1. Let s1 of these be β itself, and let s – s1 be different from β. There is
no loss of generality to assume that s1 ≥ 1, since we can map arbitrary β′ to β. Now,
consider the table with s rows and 2n – 1 columns which correspond to the
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896 A. DUBICKAS
conjugates involved in the product, where every row is taken with corresponding
multiplicity. The total number of plus signs in the new (smaller) table is equal to
s s s s sn n n n
1 1
1 1
1
12 1 2 1 2 1 2( − ) + ( − )( − ) = ( − ) +− − − .
This number is greater than s n( − )2 1 2/ , so that the number of plus signs is greater
than the number of minus signs. By our construction, the product of conjugates is
equal to
( + )
=
∏ x y mi i i
e
i
l
i
1
= 1,
where ei is the difference between the number of plus signs and the number of minus
signs in the i th column. The last equality can be also written as
( + ) = ( + )∏ ∏ −x y m x y mj j j
e
j
k k k
e
k
j k ,
where j are all indices with positive ej
, and k are all indices with negative ek
.
(Here, at least one side is greater than 1, because e1 + … + el > 0.)
Assume that q is the smallest number among all j and all k. The side which
contains the index q is at least x y mq q q+ > xq
. We immediately have a
contradiction if q = l. Otherwise, since | ei | ≤ s ≤ l and x y mi i i+ < 2xi
, the other
side is at most ( ) < ( )+
( − )
+2 21 1
2
x xq
l l q
q
l < xq
, a contradiction again. The proof of
Lemma 1 is now completed.
Proof of Theorem 3. Consider β as defined before Lemma 1. Every element of
G, except for identity, is of order 2. Furthermore, G is abelian. Therefore, every
subgroup of G generated by n – 1 of its elements has the order at most 2n
–
1. By
Theorem 2, it follows that if β ∈ M ( K; k1 , … , kn ), then the product of at most 2n
–
1
of its conjugates is a root of unity. This is however not the case, by Lemma 1, a
contradiction. This completes the proof of Theorem 3, because, by Lemma 1 again, the
degree of β over K is 2n. (Norm ( β ) = 1, because every column in the table of
signatures contains equal number of plus and minus signs.)
Corollary 2. Let k1 , … , kn ∈ Z
* be such that k1 + … + kn = 0. Then M ( K;
k1 , … , kn ) is not a multiplicative semigroup.
By the results of [16], every algebraic number of prime degree whose norm is a root
of unity belongs to M ( K; 1, – 1 ). Thus, as every quadratic unit x y mi i i+ , where
xi
, yi are positive integers and mi is an integer which is not a perfect square, is a
quotient of two conjugates over a number field K, this number belongs to every M ( K;
k1 , … , kn ). For the proof of Corollary 2, note that the algebraic number β considered
in Theorem 3 (see Lemma 1) is the product of quadratic units x y mi i i+ , but β ∉
∉ M ( K; k1 , … , kn ), by Theorem 3.
6. Representation by sufficiently long forms. The next lemma is a part of
Lemma 2 proved in [1].
Lemma 2. Suppose that the d × d matrix, where d ≥ 4, with negative real
entries in the main diagonal and nonnegative real entries outside the main diagonal is
such that the sums of its elements in every row and in every column are all equal to
zero. If the first row contains at least d – 2 positive entries, then the rank of the
matrix is equal to d – 1.
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MULTIPLICATIVE RELATIONS WITH CONJUGATE ALGEBRAIC NUMBERS 897
Proof of Theorem 4. For d = 1 the theorem follows from Theorem 1, whereas
for every β of prime degree d (including d = 2 and d = 3) it follows from the fact
that already M ( K; 1, – 1 ) contains β. It therefore suffices to prove the theorem for
the case d ≥ 4 and kjj
n
=∑ 1
= 0, where k1 , … , kn ∈ Z
*. As n ≥ 2d – 5, at least d –
– 2 elements of the multiset k1 , … , kn are either positive or negative. Without loss of
generality we may assume that k2 , … , kd – 1 are all positive. On replacing of the
remaining ones k1 and kd , … , kn by k1 + kd + … + kn and n – d + 1 zeroes,
respectively, and writing k1 again for the sum k1 + kd + … + kn , we will show that
there is an m ∈ Z
* such that
β α αmd k
d
kd= …1
1 ,
where kd = 0, has a solution in conjugates of α over K.
Since ( … )β β1 d
m = 1 for some positive integer m, we see that
β β β β β1
1
2 1
( − ) − −… = =d m m
d
m md md .
Write α = β β1
1x
d
xd… with unknowns x1 , … , xd ∈ Z. Choose the automorphisms
σ2 , … , σd ∈ G such that σ βi( )1 = βi , i = 2, … , d, and let σ1 be the identity.
Setting αi = σ αi
− ( )1 , where i = 1, … , d, we deduce that
σ α σ α β β β β1
1 1
1
1
2
1− − ( − ) − −( ) … ( ) = … =k
d
k d m m
d
m mdd ,
if
M x x x d m m md
t t( … ) = ( − ) − … −( )1 2 1, , , , , ,
has a solution in x1 , … , xd ∈ Z. Here, t stands for the transpose, and M is the d × d
matrix mij i j d, , ,= …1
, where mij = kr∑ and the sum is taken over every r such that
σ βr i( ) = βj .
By Lemma 2, the rank of M is equal to d – 1. Summing the rows of the d × ( d +
+ 1 ) matrix M* which is obtained by adding the ( d + 1 ) st column
( )( − ) − … −d m m m t1 , , ,
to M, we see they are linearly dependent over Q. It follows that d – 1 = rank M ≤
≤ rank M* ≤ d – 1, thus rank M = rank M* = d – 1. By the Kronecker – Capelli
theorem, we conclude that the linear system has a non-zero rational solution. Let x′ be
the least positive integer such that x′ xi ∈ Z
* for every i = 1, … , d. On replacing
every xi by x′ xi and m by x′ m, we get the desired conclusion.
If in Lemma 2 the condition on the first row of the matrix to contain at least d – 2
positive entries is replaced by the condition to contain at least [ d / 2 ] positive entries,
and, in addition, the d × d matrix is a Latin square, then, by Lemma 2 of [1], its rank
is also equal to d – 1. Hence, if β is symmetric over K and if n ≥ 2 [ d / 2 ] – 1, then
at least [ d / 2 ] elements among k1 , … , kn are either positive or negative. Thus we can
argue as above with the automorphisms σ2 , … , σd such that σ βi j i j d
( )
= …, , ,1
is a
Latin square. This shows that, for symmetric β, in Theorem 4 the inequality n ≥ 2d –
– 5 can by replaced by the inequality n ≥ 2 [ d / 2 ] – 1.
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898 A. DUBICKAS
Example. Let K = Q , β = 1 + 2 6+ , d = 4, n = 3, k1 = k2 = 1, k3 = – 2.
Then α can be chosen as ( )( + + + )8 5 2 4 3 3 6 2 1 2/ / .
By Theorem 4, we know that the equation
β = 1 2 6 1 2 3
2+ + = −α α α
has a solution in conjugate over Q algebraic numbers α1 , α2 , α3 . We will show how
to find some solutions.
Let us choose the following indices: β1 = 1 + 2 6+ , β2 = 1 – 2 6− , β3 =
= 1 – 2 6+ and β4 = 1 + 2 6− . Now, following the proof of Theorem 4 with
m – 1, we obtain the system of linear equations
x1 + x2 – 2x3 = 3,
x1 + x2 – 2x4 = – 1,
– 2x1 + x3 + x4 = – 1,
– 2x2 + x3 + x4 = – 1.
(Of course, there is no need to put the negative elements of M on the main diagonal.
It suffices to assume that every row and every column of M contains precisely one
negative element.) One of its solutions is x1 = x2 = 0, x3 = – 3 / 2, x4 = 1 / 2. So we
can choose x′ = 2, which gives the integer solution x1 = x2 = 0, x3 = – 3, x4 = 1 for
m = 2. This choice shows that
β δ δ δ8
1 2 3
2= − ,
where δ1 = β β3
3
4
− , δ2 = β β3 4
3− , δ3 = β β1
3
2
− . On replacing δ by – δ, we compute
δ = δ1 = ( + − ) ( + ) = − + −1 2 6 7 4 3 11591 8196 2 6692 3 4732 64 3 .
The minimal polynomial of δ over Q is
Q ( z ) = z z z z4 3 246364 10950 284 1− + − + .
Since the polynomial
Q z z z z z( ) = − + − +5 625 5795500 273750 1420 18 32 24 16 8
is irreducible over Q, the equation β = α α α1 2 3
2− is solvable in conjugates of
α = ( ) = ( − + − )( )δ / // /5 11591 8196 2 6692 3 4732 6 51 8 1 8
of degree 32.
It is not the smallest possible degree for α. The polynomial Q ( z8
) is the product
of three irreducible polynomials z8 + 16z6 + 20z4 + 8z2 + 1, z8 – 16z6 + 20z4 – 8z2 +
+ 1, and z16 + 216z12 + 146z8 + 24z4 + 1. In this example, it happens that the roots of
Q ( z8
) satisfying 1 2 6 1 2 3
2+ + = −α α α are all roots of the second polynomial,
namely, z8 – 16z6 + 20z4 – 8z2 + 1 (which is not always the case). The roots are α1 =
= ( )( + + + )8 5 2 4 3 3 6 2 1 2/ / , α 2 = ( )( − + − )8 5 2 4 3 3 6 2 1 2/ / and α 3 =
= ( )( + − − )8 5 2 4 3 3 6 2 1 2/ / , giving the second solution with α of degree 8, as
claimed.
7. Nonsymmetric numbers. Clearly, every β of prime degree d over K is
symmetric, since the Galois group G contains a d-cycle. If, for d = 4, G does not
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MULTIPLICATIVE RELATIONS WITH CONJUGATE ALGEBRAIC NUMBERS 899
contain a 4-cycle, then it is the Klein 4-group, so the respective β is also symmetric.
Hence the smallest d for nonsymmetric β must be greater than or equal to 6. We
will show now that nonsymmetric numbers of degree 6 exist. For this, we first
introduce an „auxiliary” number α.
Let α be of degree 4 over K with the Galois group of K( )α α α α1 2 3 4, , , / K
isomorphic to the full symmetric group on four symbols. Assume that ± α1 ± α 2 ±
± α3 ± α4 ≠ 0. (It is clear that such α exist.) Set β = β1 = α1 + α2
. Such β is of
degree 6 over K with the remaining conjugates being β2 = α2 + α3
, β3 = α3 + α4
,
β4 = α1 + α4
, β5 = α1 + α3
, β6 = α2 + α4
, and with Galois group G of order 24.
We claim that β is nonsymmetric.
Assume that β is symmetric over K . Then some five elements of the full
symmetric group acting on α 1
, α2
, α3
, α4 can be chosen so that their action on the
row β1 , … , β6 together with identity form a Latin square. Let S be the set of these
five elements. Evidently, none of the elements in S is a transposition nor it is a
product of two transpositions. The remaining fifteen elements are the identity τ0
, and
the following fourteen elements: τ1 = (123), τ2 = (132), τ3 = (124), τ4 = (142), τ5 =
= (134), τ6 = (143), τ7 = (234), τ8 = (243), τ9 = (1234), τ10 = (1243), τ11 = (1324),
τ12 = (1342), τ13 = (1423), τ14 = (1432), five of which do form the set S. Their
action on β1 , … , β6 can be described as follows:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 2 5 6 4 2 6 5 4 2 6 3 5 3 4
2 5 1 3 5 6 1 3 6 3 4 6 4 5 1
3 4 6 5 2 4 5 6 2 4 5 1 6 1 2
4 6 3 1 6 5 3 1 5 1 2 5 2 6 3
5 1 2 2 3 3 4 4 1 6 1 2 3 4 6
6 3 4 4 1 1 2 2 3 5 3 4 1 2 5
Here, the symbol i in the first row means τi
, whereas in other five rows i stands
for βi
.
Since τ11 ( β1 ) = τ13 ( β1 ), τ10 ( β2 ) = τ12 ( β2 ), τ 9 ( β5 ) = τ14 ( β5 ), the set S
contains at most one element from the pair { τ11
, τ13 }, at most one from { τ10
, τ12 },
and at most one from { τ11
, τ13 }. If τ11 ∈ S, then τ2 ∉ S, since τ2 ( β5 ) = τ11 ( β5 )
(see the table). Similarly, τ3
, τ5
, τ8 ∉ S, because τ3 ( β5 ) = τ11 ( β5 ) and τ5 ( β2 ) =
= τ8 ( β2 ) = τ11 ( β2 ). So, if τ11 ∈ S, then the elements of S8 = { τ1 , … , τ8 } belonging
to S all must lie in the set S1 = { τ1 , τ4 , τ6 , τ7 }. Similarly, if τ13 ∈ S, or τ10 ∈ S, or
τ12 ∈ S, or τ9 ∈ S, or τ14 ∈ S, then the elements of S8 belonging to S must lie,
respectively, in S– 1 = { τ2 , τ3 , τ5 , τ8 }, S2 = { τ2 , τ4 , τ5 , τ7 }, S– 2 = { τ1 , τ3 , τ6 , τ8 },
S3 = { τ2 , τ4 , τ6 , τ8 }, S– 3 = { τ1 , τ3 , τ5 , τ7 }. Note that Si ∩ S– i = ∅ for i = 1, 2, 3
and, for any choice of signs, S S S± ± ±1 2 3∩ ∩ = 1. We can therefore conclude by
observing the following. If S ∩ { … }τ τ9 14, , = 3, then S S∩ 8 = 1, so S = 3 +
+ 1 = 4, a contradiction. If however S ∩ { … }τ τ9 14, , < 3, then, once again, S <
< 3 + 2 = 5, because S S∩ 8 ≤ 2. (Indeed, by symmetry, there is no loss of
generality to assume that τ1 ∈ S. Then τ4 , τ5 , τ8 ∉ S, and, moreover, at most one
element from { τ2 , τ3 , τ6 , τ7 } can belong to S.)
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Received 31.01.2005
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