A theorem of Dubickas, affirming a conjecture of Kuba, states that a nonzero algebraic number is a root of a polynomial f with positive rational coefficients if and only if none of its conjugates is a positive real number. A certain quantitative version of this result, yielding a growth factor for the coefficients of f similar to the condition of the classical Eneström-Kakeya theorem of such polynomial, is derived. The bound for the growth factor so obtained is shown to be sharp for some particular classes of algebraic numbers.
1. Introduction
A nonzero complex number γ is called positively algebraic if it is a root of a polynomial all of whose coefficients are positive rational numbers. In 2005, Kuba [1] conjectured that a necessary condition for an algebraic number γ to be positively algebraic is that none of its conjugates is a positive real number. This conjecture was confirmed affirmatively by Dubickas [2], in 2007, through the following result.
Theorem 1 (see [2]).
A nonzero complex number γ is a root of a polynomial with positive rational coefficients if and only if γ is an algebraic number such that none of its conjugates is a positive real number.
In 2009, Brunotte [3] gave an elementary proof of Dubickas-Kuba theorem based on the following lemma ([3, Lemma 2]), which is originated from [4] (see also [5]): if f(x)∈R[x] is a polynomial having no nonnegative roots, then there exists m∈N such that (1+x)mf(x) has only positive coefficients. See also [6], where a bound for the degree of the polynomial with rational positive coefficients was given.
We fix the following notation and terminology throughout. Denote by R+ the set of positive real numbers. For r∈R+, let
(1)P∶={f(x)=∑j=0Jcjxj∈R[x];aaaaa∑j=0JcJ>0,cj≥0(0≤j≤J-1),J∈N∪{0}},Q∶={f(x)=∑j=0Jcjxj∈R[x];aaaaa∑j=0JcJ>0,cj≤0(0≤j≤J-1),J∈N∪{0}},Q(r)∶={f(x)=∑j=0Jcjxj∈R[x];aaaaa∑j=0J(x-r)f(x)∈Q,J∈N∪{0}},Q(0)∶=Q,Q+(r)∶=Q(r)∩R+[x],Q+(>r)∶=∪s>rQ+(s).
Some basic properties, which are needed in our work here, modified with the same proofs for Q+(r) from [7], are in the following lemma.
Lemma 2.
Let r>0 and f(x)=∑j=0Jcjxj∈R[x]∖{0}. Then
f(x)∈Q(r)⇔c0≥0, cj≤rcj+1(0≤j<J);
f(x)∈Q+(r)⇔c0>0, cj≤rcj+1(0≤j<J);
0<a<b⇒X(a)⊊X(b), where X∈{Q,Q+};
Q+(r)⊊Q(r)⊊P;
f(x)∈Q and cj≠0 for some j∈{0,…,J-1}⇒f(x) has a unique positive zero z0; this zero is simple and all other roots have absolute values ≤z0;
let Y∈{P,Q,Q(r),Q+(r)}. Then f(x)∈Y⇒f(xk)∈Y for all k∈N;
([7, Lemma 2]) a1,a2,…,as are positive real numbers, and fk(x)∈X(ak)(1≤k≤s)⇒∏k=1sfk(x)∈X(∑k=1sak), where X∈{Q,Q+};
([7, Lemma 3]) γ is a complex number which is not real positive ⇒γ is a root of a polynomial in Q(r) for any r>|γ|.
Our first main result is a certain quantitative improvement of Dubickas-Kuba theorem (Theorem 1 above).
Theorem 3.
Let γ be a nonzero algebraic number of degree n (over Q), let γ1(=γ),…,γn be all its conjugates, and let
(2)r>|γ|¯∶=max{|γt|;1≤t≤n}.
Then all conjugates of γ are in C∖R+ if and only if there exists p(x)∈Q+(r)∩Q[x] such that p(γ)=0.
The nontrivial half of Theorem 3 is its necessity part, and its main difficulty is to show the existence of a polynomial in Q+(r) all of whose coefficients are rational numbers. Should we need only a polynomial with nonnegative real coefficients, this necessity part follows from Roitman-Rubinstein’s result [7, Lemma 4] but with a bound for r which depends not only on the conjugates of α but also on other roots. Theorem 3 is proved in the next section.
Combining Theorem 3 with part (2) of Lemma 2, it is natural to ask for the least possible value of r for which Theorem 3 holds. In the last section, we show that the bound |γ|¯ is optimal for a class of algebraic numbers without nonnegative conjugates. The investigation in this second part leads us to some interesting connections with the classical result of Eneström-Kakeya [8], which gives upper and lower bounds for the absolute values of the roots of polynomials with positive coefficients.
2. Proof of Theorem 3
The proof of Theorem 3 is done by analyzing the nature of the roots which are complex numbers, and we simplify by treating them separately in the next two lemmas.
Lemma 4.
Let λ=|λ|e2πiθ≠0 with θ∈(0,1)∖Q. Then, for any real r>|λ|, there exists F(x)∈Q+(r) such that F(λ)=0.
Proof.
Let u=r/|λ|>1. Since θ∈R∖Q, by Kronecker’s theorem ([9, Theorem 439, page 376]), the set {e2πinθ:n∈N} is dense in the unit circle and so there exists m(=mθ)∈N such that
(3)-12<cos(2πmθ)<-12u.
This yields
(4)1<-2ucos(2πmθ)<u,1u<-2cos(2πmθ)<1<u.
Let
(5)f1(x)=(x-(λ|λ|)m)(x-(λ¯|λ¯|)m)=x2-2xcos(2πmθ)+1.
Thus, (4) shows that all the coefficients of f1 are >1/u>0. Again, using (4) with Lemma 2 part (2), we get
(6)f1(x)∈Q+(u)⊆Q+(um).
Thus, (x-um)f1(x)∈Q and so, by Lemma 2 part (5), (xm-um)f1(xm)∈Q. Let
(7)f2(x)=(xm-umx-u)f1(xm),
so that (x-u)f2(x)∈Q. Consequently,
(8)(x|λ|-r|λ|)f2(x|λ|)∈Q.
Putting
(9)F(x)=1|λ|f2(x|λ|),
we see that
(10)F(x)=1|λ|3m(xm-rmx-r)(xm-λm)(xm-λ¯m)∈Q+(r),
and λ and λ¯ are roots of F(x).
Lemma 5.
Let λ=|λ|e2πiθ≠0 with θ∈(0,1)∩Q. Then there exists G(x)∈Q+(|λ|) such that G(λ)=0.
Proof.
Since θ∈Q, there is an M(=Mθ)∈N such that (λ/|λ|)M=1. Putting
(11)g(x)=xM-(λ/|λ|)Mx-1,
we clearly see that (x-1)g(x)∈Q, and so
(12)(x|λ|-1)g(x|λ|)∈Q.
The result follows by taking
(13)G(x)=1|λ|g(x|λ|)=1|λ|M(xM-λMx-|λ|)=1|λ|{(x|λ|)M-1+(x|λ|)M-2aaaaaaa+⋯+(x|λ|)2+x|λ|+1}∈Q+(|λ|),
noting that λ and λ¯ are two roots of G(x).
Observe that the proof of this last lemma remains true when λ is a negative real number (with M=2). The followimg result indicates how to obtain a multiple in Q+(r) of a given element of R[x] with a factor in Q[x].
Lemma 6.
Let A(x), B(x)∈R[x]. If A(x)B(x)∈Q+(r), for some r>0, then there exists a polynomial B(x)∈Q[x] such that A(x)B(x)∈Q+(r).
Proof.
Writing
(14)A(x)B(x)=ckxk+ck-1xk-1+⋯+c1x+c0
we have cj>0(0≤j≤k), and, by Lemma 2 part (1),
(15)0<c0≤rc1≤r2c2≤r3c3≤⋯≤rkck.
For each j∈{0,1,…,k}, since the jth coefficient cj depends on the first j coefficients of A(x) and B(x) and since any real number is a limit of a sequence of rational numbers from both sides, we can successively choose rational numbers b0,b1,…,bm so close to the corresponding coefficients of B(x) in such a way that when the polynomial
(16)B(x)=bmxm+bm-1xm-1+⋯+b1x+b0∈Q[x]
is multiplied to A(x) yielding
(17)A(x)B(x)=c~kxk+c~k-1xk-1+⋯+c~1x+c~0,
these new coefficients still satisfy
(18)0<c~0≤rc~1≤r2c~2≤r3c~3≤⋯≤rkc~k.
Lemma 2 part (1) shows then that A(x)B(x)∈Q+(r).
The next lemma proves Theorem 3 for the case when the algebraic number γ and all its conjugates lie inside the unit circle.
Lemma 7.
Let γ be a nonzero algebraic number and let γ1(=γ),…,γn be all its conjugates. Assume that
(19)|γ|¯∶=max1≤t≤n{|γt|}<r,
where r∈(0,1]. Then all conjugates of γ are in C∖R+ if and only if there exists p(x)∈Q+(r)∩Q[x] such that p(γ)=0.
Proof.
Assume that the nonzero algebraic number γ and all its conjugates γ1(=γ),…,γn are in C∖R+. Arrange these numbers in such a way that
(20){γ1,γ2,…,γn}={γj∈R-;1≤j≤J}∪{γJ+k=|γJ+k|e2πiθJ+k;θJ+k∈Q,1≤k≤K}∪{γJ+K+l=|γJ+K+l|e2πiθJ+K+l;θJ+K+l∈R∖Q,1≤l≤L}
with J+K+L=n. (In the above arrangement, it is tacitly assumed that J, K, and L are positive integers; indeed, should any one of the three sets be empty, the corresponding value(s) of J, K, or L could be zero. Yet the remaining arguments below still work with only slight adjustments.) Since |γt|<r, for each 1≤t≤n, there exists an odd positive integer N such that |γt|N<r/n, for each 1≤t≤n, and γjN<0, for each 1≤j≤J. Thus, for 1≤j≤J, we have
(21)Hj(x)∶=(x-γjN)∈Q+(|γj|N)⊆Q+(rn)aaaaaaaaaaaaaaaaawithHj(γjN)=0.
For 1≤k≤K, by Lemma 5, there exits a polynomial
(22)Gk(x)∈Q+(|γJ+k|N)⊆Q+(rn)withGk(γJ+kN)=0.
For 1≤l≤L, by Lemma 4 and |γJ+K+l|N<r/n, there exits a polynomial
(23)Fl(x)∈Q+(rn)withFl(γJ+K+lN)=0.
Let
(24)E(x)∶=∏1≤j≤JHj(x)∏1≤k≤KGk(x)∏1≤l≤LFl(x),
and note that γ1N,…,γnN are its roots. From Lemma 2 part (6) with J+K+L=n, we have E(x)∈Q+(r). Thus,
(25)(x-1)E(x)∈Qandso(xN-1)E(xN)∈Q.
Consequently,
(26)(x-1)(xN-1+xN-2+⋯+x+1)E(xN)∈Q
and so
(27)(xN-1+xN-2+⋯+x+1)E(xN)∈Q+(1).
Since M(x)∣E(xN) over R[x], where M(x)∈Q[x] is the minimal polynomial of γ, there exists B(x)∈R[x] such that E(xN)=M(x)B(x). Thus,
(28)(xN-1+xN-2+⋯+x+1)M(x)B(x)∈Q+(r).
From Lemma 6, there exists B(x)∈Q[x] such that
(29)(xN-1+xN-2+⋯+x+1)M(x)B(x)∈Q+(r)∩Q[x].
Putting
(30)p(x)∶=(xN-1+xN-2+⋯+x+1)M(x)B(x),
we see that p(x)∈Q+(r)∩Q[x] and p(γ)=0.
Conversely, assume that there exists p(x)∈Q+(r)∩Q[x] such that p(γ)=0. Since p(x)∈Q[x], all conjugates of γ are zeros of p(x) and, since all coefficients of p(x) are positive, no real conjugate of γ can be positive.
We are now ready to prove Theorem 3.
Proof of Theorem 3.
Assume that all conjugates of γ∈C∖R+. Since r>|γ|¯ and Q is dense in R, there exists r1∈Q such that |γ|¯<r1<r. Let
(31)λt∶=γtr1(1≤t≤n),(32)λ∶=λ1.
Then |λt|<1 and λt∈C∖R+, for all 1≤t≤n. Let
(33)F(x)∶=M(r1x),
where M(x)∈Q[x] is the minimal polynomial of γ. Then F(x) is the minimal polynomial of λ, and λ1(=λ),…,λn are conjugates of λ. By Lemma 7, there exists p1(x)∈Q+(1)∩Q[x] such that p1(λ)=0. Then F(x)∣p1(x); that is, M(x)∣p1(x/r1). Since p1(x)∈Q+(1)∩Q[x] and r1∈Q, we have p1(x/r1)∈Q+(r1)∩Q[x]. Let
(34)p(x)∶=p1(xr1).
Then p(x)∈Q+(r1)∩Q[x] and p(γ)=0. Since 0<r1<r, by Lemma 2 part (2), we have p(x)∈Q+(r)∩Q[x].
On the other hand, assume that there exists p(x)∈Q+(r)∩Q[x] such that p(γ)=0. Since p(x)∈Q[x], all conjugates of γ are zeros of p(x) and, since all coefficients of p(x) are positive, no real conjugate of γ can be positive.
It is worth noting that in [7, Lemma 4], the bound for r reduces to |γ|¯ if the polynomial p(x) mentioned in Theorem 3 is the minimal polynomial of γ. The next corollary contains the Dubickas-Kuba theorem and some equivalent assertions.
Corollary 8.
Let γ be a nonzero algebraic number, let γ1(=γ),γ2,…,γn be all its conjugates, and let r>|γ|¯. Then the following statements are equivalent:
all conjugates of γ are in C∖R+;
there exists p(x)∈Q+(r)∩Q[x] such that p(γ)=0;
there exists g(x)∈R+[x]∩Q[x] such that g(γ)=0;
there exist q(x)∈Q∩Q[x] and w∈R+∖{γ1,…,γn} such that q(γ)=q(w)=0.
Proof.
Assertions (i) and (ii) are equivalent by Theorem 3. Clearly, (ii) implies (iii). To show (iii) implies (iv), assume that there exists
(35)g(x)=xm+g1xm-1+⋯+gm-1x+gm∈R+[x]∩Q[x]
such that g(γ)=0, and so g(γ1)=⋯=g(γn)=0. Since gh>0, for all h∈{1,…,m}, none of γ1,…,γn can be real and positive. If γj≠0 for some j∈{1,2,…,n}, choose
(36)q(x)=(xm)2-(g1xm-1+g2xm-2+⋯+gm-1x+gm)2∈Q.
Since g(x) is a factor of q(x), we have q(γ1)=⋯=q(γn)=0. From γj≠0 and Lemma 2 part (4), we know that q(x) has a unique positive zero, say w, which must then be distinct from all γ1,…,γn, as desired.
That assertion (iv) implies that assertion (i) is again an immediate consequence of Lemma 2 part (4).
3. Eneström-Kakeya Theorem
To proceed further, we need a new notion.
Definition 9.
For
(37)f(x)=∑j=0Jcjxj∈R+[x]∖R,
its lower and upper Eneström-Kakeya quotients are defined, respectively, by
(38)α[f]∶=min{c0c1,c1c2,…,cJ-2cJ-1,cJ-1cJ},β[f]∶=max{c0c1,c1c2,…,cJ-2cJ-1,cJ-1cJ}.
Part (1) of Lemma 2 is closely related to the classical Eneström-Kakeya theorem, [8, Theorem 1], which states the following.
Theorem 10.
Let p(x)∈R[x]∖R all of whose coefficients are positive. Then all the zeros of p(x) are contained in the annulus α[p]≤|x|≤β[p], where α[p] and β[p] are, respectively, the lower and upper Eneström-Kakeya quotients of p(x).
In this section, we derive a proposition yielding conditions which are necessary for a product of two polynomials to be in Q(r).
Proposition 11.
Let
(39)f(x)=cJxJ+cJ-1xJ-1+⋯+c1x+c0∈R+[x],
where J≥1, and let r∈R+. If
(40)cj>rcj+1(0≤j<J),
then f(x)d(x)∉Q(r) for all d(x)∈R[x]∖{0}.
Proof.
Let
(41)d(x)=∑k=0Kdkxk∈R[x]∖{0}.
If f(x)d(x)∈Q(r), then Lemma 2 part (1) gives
(42)0≤c0d0(43)c0d0≤r(c1d0+c0d1)(44)c1d0+c0d1≤r(c2d0+c1d1+c0d2)(45)⋮(46)cK-1d0+cK-2d1+⋯+c0dK-1≤r(cKd0+cK-1d1+⋯+c0dK)(47)cKd0+cK-1d1+⋯+c0dK≤r(cK+1d0+cKd1+⋯+c1dK),
where we adopt the convention that cj=0, for all j>J, and dk=0, for all k>K. From c0>0, and (42), we get d0≥0. From (43) and (40), we have
(48)(c0-rc1)d0≤rc0d1,d1≥0.
From (44) and (40), we get
(49)(c1-rc2)d0+(c0-rc1)d1≤rc0d2,
which together with previous results yield d2≥0. Continuing in the same manner up to (46), we get d3≥0, d4≥0,…,dK≥0. Thus,
(50)(cK-rcK+1)d0+(cK-1-rcK)d1+⋯+(c1-rc2)dK-1+(c0-rc1)dK≤0.
Since cj>rcj+1(j=0,1,2,…,J-1), the left-hand expression in (50) can be 0 only when d0=d1=⋯=dK=0; that is, d(x)≡0, which is not possible.
The next proposition indicates the significance of the upper and lower Eneström-Kakeya quotients.
Proposition 12.
Let f(x)∈R+[x].
The upper Eneström-Kakeya quotient β[f] is the smallest r>0 such that f(x)∈Q+(r).
The lower Eneström-Kakeya quotient α[f] has the property that if p(x)∈Q+(r) with 0<r<α[f], then f(x)∤p(x) over R[x].
Proof.
Part (i) follows from Lemma 2 part (1). Part (ii) is immediate from Proposition 11.
In passing, from the definition of Eneström-Kakeya quotients, it seems natural to ask whether one quotient can be a reciprocal of the other. This is indeed the case when the polynomial is self-reciprocal. Let g(x) be a polynomial. The reciprocal polynomial of g(x) is defined as
(51)g*(x)=xdeggg(1x),
and we say that g(x) is self-reciprocal if g(x)=g*(x).
Proposition 13.
Let f(x)∈R+[x]∖R. Then β[f*]=1/α[f], and f*(x)∈Q+(1/α[f]). Moreover, if f(x) is self-reciprocal, then β[f]=1/α[f] and f(x)∈Q+(1/α[f]).
Proof.
Writing f(x)=cJxJ+cJ-1xJ-1+⋯+c1x+c0∈R+[x]∖R, we have
(52)f*(x)=xJf(1x)=cJ+cJ-1x+⋯+c1xJ-1+c0xJ
and so
(53)β[f*]=max{c1c0,c2c1,…,cJ-1cJ-2,cJcJ-1}=1α[f].
From Proposition 12 part (i), we have
(54)f*(x)∈Q+(β[f*])=Q+(1α[f]).
If f(x) is self-reciprocal, then clearly β[f]=1/α[f] and f(x)∈Q+(1/α[f]).
Corollary 14.
If f(x)∈R+[x]∖R is self-reciprocal, then
α[f]≤1, and
α[f]=1, if and only if β[f]=1.
Proof.
(i) If α[f]>1, then 1/α[f]<1<α[f]. Since f is self-reciprocal, by Proposition 13, f(x)∈Q+(1/α[f]), which contradicts Proposition 12 part (ii). Part (ii) follows readily from Proposition 13.
4. Minimal Polynomials with Positive Coefficients
In this section, we aim to show that the bound |γ|¯, in Theorem 3, is best possible by showing that it cannot be improved for a subclass of the class of algebraic numbers whose minimal polynomials have positive coefficients. Let γ be a nonzero algebraic number, and M(x)∈Q[x] its minimal polynomial. We say that M(x) is positively minimal if all its coefficients are positive. From Proposition 12, we have the following.
Proposition 15.
Let γ be an algebraic number and let
(55)M(x)=xn+an-1xn-1+⋯+a1x+a0
be its minimal polynomial. If M(x) is positively minimal, then, for any r∈(0,α[M]) and any p(x)∈Q+(r)∩Q[x], we have p(γ)≠0.
Proposition 15 leads naturally to the following definition.
Definition 16.
Let γ be a nonzero algebraic number, whose minimal polynomial M(x) is positively minimal. Define the growth factor of an algebraic number γ with respect to M(x), denoted by rγ, to be the infimum of the set of nonnegative real numbers with the following property: for any r>rγ, there exists p(x)∈Q+(r)∩Q[x] such that p(γ)=0.
Remarks. The growth factor enjoys the following basic properties:
rγ is unique;
rγ≤|γ|¯ (Theorem 3);
α[M]≤rγ≤β[M] (Propositions 12 and 15).
For a nonzero algebraic number γ whose minimal polynomial is positively minimal and whose growth factor is rγ, let
(56)Pγ∶={p(x)∈Q+(rγ)∩Q[x];p(γ)=0}.
A class of algebraic numbers for which the bound |γ|¯ in Theorem 3 cannot be improved is given in the next theorem.
Theorem 17.
Let γ be an algebraic number and assume that its minimal polynomial
(57)M(x)=xn+an-1xn-1+⋯+a1x+a0
is positively minimal.
If α[M]=β[M], then α[M]=rγ=β[M] and M(x)∈Pγ.
We have α[M]=β[M], if and only if
(58)M(x)=xn+axn-1+a2xn-2+⋯+an-1x+an
for some a∈Q+.
Proof.
Assertion (i) follows at once from the preceding remarks.
(ii) If α[M]=β[M], then
(59)a0a1=a1a2=⋯=an-2an-1=an-1,
which gives
(60)M(x)=xn+an-1xn-1+an-12xn-2+⋯+an-1n-1x+an-1n∈Q+[x].
The converse is trivial.
Theorem 17 poses a natural question whether a positively minimal polynomial must necessarily belong to Pγ. A negative answer is provided by the next proposition.
Proposition 18.
Let γ be an algebraic number of degree n over Q. Assume that its minimal polynomial M(x) is positively minimal. If n≥2 and all the conjugates of γ1=γ,…,γn are negative real numbers, then
(61)α[M]≤rγ≤|γ|¯<β[M]=∑t=1n|γt|and M(x)∉Pγ.
Proof.
Since all conjugates of γ are negative real numbers, we have
(62)x-γt∈Q+(|γt|)(t=1,…,n).
Lemma 2 part (6) shows that
(63)M(x)=(x-γ1)(x-γ2)⋯(x-γn)∶=xn+an-1xn-1+⋯+a1x+a0∈Q+(∑t=1n|γt|)∩Q[x],
and so Proposition 12 (i) implies that β[M]≤∑t=1n|γt|. Since n≥2 and an-1=∑t=1n|γt|, we have |γ|¯<∑t=1n|γt|=an-1≤β[M]. Consequently,
(64)α[M]≤rγ≤|γ|¯<β[M]=∑t=1n|γt|.
By Proposition 12 (i) and the definition of the upper Eneström-Kakeya quotient, we conclude that M(x)∉Pγ.
It may be of interest to look at the growth factors and the Eneström-Kakeya quotients for positively minimal polynomials of small degrees.
Proposition 19.
Let γ be an algebraic number. Assume that its minimal polynomial M(x) is positively minimal and that all its conjugates are in C∖R+.
If degγ=1, then α[M]=rγ=β[M]=|γ|¯=|γ|.
If degγ=2, let γ=a+bi(a,b∈R) and M(x)=x2+a1x+a0; then,
for b=0, we have a0/a1=α[M]<|γ|¯<β[M]=a1,rγ∈[α[M],|γ|¯],
for b≠0 and 3a2≠b2, we have α[M]<|γ|¯=|γ|<β[M],rγ∈[α[M],|γ|¯],
for b≠0 and 3a2=b2, we have α[M]=rγ=β[M]=|γ|¯=|γ|.
If degγ=3, then α[M]≤rγ≤|γ|¯<β[M].
Proof.
Part (1) is trivial. We now consider part (2), that is, degγ=2. If b=0, then a12≥4a0>0, and so
(65)α[M]=a0a1<|γ|¯=a12+a124-a0<a1=β[M].
If b≠0, without loss of generality, assume that γ=a+bi with b∈R+. Then |γ|¯=a2+b2 and
(66)M(x)=(x-a-bi)(x-a+bi)=x2-2ax+a2+b2∈Q[x].
Since M(x) is positively minimal, we have a<0 and a2+b2>0. From
(67)a2+b2≥-2a⟺a2+b2≥4a2⟺b2≥3a2,
we deduce that, if 3a2≠b2, then α[M]<|γ|¯<β[M] and that, if 3a2=b2, then α[M]=rγ=β[M]=|γ|¯=|γ|.
Finally consider part (3), that is, degγ=3. If γ and all its conjugates γ1, γ2, and γ3 are negative real numbers, the conclusion follows at once from Proposition 18. Assume then that
(68)γ1=-A,γ2=a+bi,γ3=a-biaawithA∈R+∖Q,a∈R,b∈R+.
Since the minimal polynomial,
(69)M(x)=(x+A)(x-a-bi)(x-a+bi)=x3+(A-2a)x2+(a2+b2-2Aa)x+A(a2+b2)∈Q[x],
is positively minimal, we have A-2a>0, a2+b2-2Aa>0 and A(a2+b2)>0, which gives
(70)β[M]=max{A(a2+b2)a2+b2-2Aa,a2+b2-2AaA-2a,A-2a}.
Since degγ=3, A-2a is positive rational, and A is positive irrational, we must have a≠0. By the remarks after Proposition 15, there remains only the verification that |γ|¯<β[M].
We split our consideration into three cases depending on the maximum absolute value of the conjugates.
Case |γ|¯=A.
If A<A-2a, or if A<A(a2+b2)/(a2+b2-2Aa), then |γ|¯<β[M]. Otherwise, we have A≥A-2a and A≥A(a2+b2)/(a2+b2-2Aa), which gives a=0, a contradiction.
Case |γ|¯=a2+b2.
If A(a2+b2)/(a2+b2-2Aa)=a2+b2, since A(a2+b2)/(a2+b2-2Aa) is positive rational, we deduce that a2+b2 must be positive rational. Since A(a2+b2) is positive rational, we conclude that A is positive rational, which is a contradiction.
If A(a2+b2)/(a2+b2-2Aa)>a2+b2, then β[M]>|γ|¯.
If A(a2+b2)/(a2+b2-2Aa)<a2+b2, then
(71)2Aa<a2+b2(a2+b2-A).
If a>0, by (71), we get
(72)a2+b2(a2+b2-A)+2aa2+b2>2Aa
yielding (a2+b2-2Aa)/(A-2a)>a2+b2; that is, β[M]>|γ|¯.
Consider now a<0.
If A-2a=a2+b2, since A-2a is positive rational, we deduce that a2+b2 is positive rational. Since A(a2+b2) is positive rational, we conclude that A is positive rational, which is a contradiction.
If A-2a>a2+b2, then β[M]>|γ|¯.
If A-2a<a2+b2, then a2+b2(a2+b2-A+2a)>0>2Aa, yielding β[M]>|γ|¯.
We end this paper with another class of algebraic numbers for which the bound in Theorem 3 is optimal.
Proposition 20.
Let γ be an algebraic number, and let
(73)M(x)=xn+a(n∈N,a∈Q,a>0)
be its minimal polynomial. For 0<r<|γ|¯=a1/n, if q(x)∈Q+(r)∩Q[x], then q(γ)≠0.
Proof.
Assume on the contrary that there is q(x)∈Q+(r)∩Q[x] such that q(γ)=0. Thus, M(x)∣q(x), so that
(74)q(x)=(xn+a)×(xm+cm-1xm-1+⋯+c3x3+c2x2+c1x+c0)=xn+m+cm-1xn+m-1+⋯+c3xn+3+c2xn+2+c1xn+1+c0xn+axm+acm-1xm-1+⋯+ac3x3+ac2x2+ac1x+ac0.
If m<n-1, by Lemma 2 part (1) we have 0<a≤0, a contradiction.
If m>n, let m=n+t for some t∈N and the relation (74) becomes
(75)q(x)=x2n+t+cn+t-1x2n+t-1+cn+t-2x2n+t-2+⋯+ct+1xn+t+1+(ct+a)xn+t+(ct-1+acn+t-1)xn+t-1+(ct-2+acn+t-2)xn+t-2+⋯+(c4+acn+4)xn+4+(c3+acn+3)xn+3+(c2+acn+2)xn+2+(c1+acn+1)xn+1+(c0+acn)xn+acn-1xn-1+acn-2xn-2+⋯+ac3x3+ac2x2+ac1x+ac0.
Invoking upon Lemma 2 part (1), we get the following chain of inequalities:
(76)0<ac0⟹c0>0ac0≤rac1⟹c1>0,c0≤rc1ac1≤rac2⟹c2>0,c1≤rc2,c0≤r2c2⋮acn-2≤racn-1⟹cn-1>0,cn-2≤rcn-1,cn-3≤r2cn-1,…,c0≤rn-1cn-1acn-1≤r(c0+acn)≤rncn-1+racn<acn-1+racn⟹0<racn,cn>0c0+acn≤r(c1+acn+1)r(c0+acn)≤r2c1+r2acn+1acn-1≤r2c1+r2acn+1≤rncn-1+r2acn+1<acn-1+r2acn+1⟹0<r2acn+1,cn+1>0c1+acn+1≤r(c2+acn+2)r(c1+acn+1)≤r2c2+r2acn+2c0+acn≤r2c2+r2acn+2r(c0+acn)≤r3c2+r3acn+2acn-1≤r3c2+r3acn+2≤rncn-1+r3acn+2<acn-1+r3acn+2⟹0<r3acn+2,cn+2>0.
Proceeding in the same manner, we have
(77)c2+acn+2≤r(c3+acn+3)⟹cn+3>0c3+acn+3≤r(c4+acn+4)⟹cn+4>0⋮ct-2+acn+t-2≤r(ct-1+acn+t-1)⟹cn+t-1>0.
Since n+t-1≥t, we have ct>0 and
(78)cn+t-1≤r,cn+t-2≤rcn+t-1≤r2,cn+t-3≤rcn+t-2≤r3,…,ct+1≤rn-1.
Thus, a<ct+a≤rct+1≤rn<a, which is a contradiction. The proofs for the cases m=n-1 and n are similar but simpler.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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