# Algebraic Numbers Satisfying Polynomials with Positive Rational Coefficients

Authors: Vichian Laohakosol; Suton Tadee
Journal: Journal of Numbers
 (2014)

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## Abstract

A theorem of Dubickas, affirming a conjecture of Kuba, states that a nonzero algebraic number is a root of a polynomialf 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.


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## Body

## 1. Introduction

A nonzero complex numberγ is calledpositively 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 numberγ 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]):iff(x)∈R[x]is a polynomial having no nonnegative roots, then there existsm∈Nsuch 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 byR+ 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.
Letr>0 and f(x)=∑j=0J‍cjxj∈R[x]∖{0}. Then (1)
f(x)∈Q(r)⇔c0≥0, cj≤rcj+1(0≤j<J);f(x)∈Q+(r)⇔c0>0, cj≤rcj+1(0≤j<J);(2)
0<a<b⇒X(a)⊊X(b), where X∈{Q,Q+};(3)
Q+(r)⊊Q(r)⊊P;(4)
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;(5)
letY∈{P,Q,Q(r),Q+(r)}. Then f(x)∈Y⇒f(xk)∈Y for all k∈N;(6)
([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)
([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 (Theorem1 above).Theorem 3.
Letγ 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 Theorem3 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 Theorem3 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 Theorem3

The proof of Theorem3 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.
Letλ=|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.
Letu=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.
Letλ=|λ|e2πiθ≠0 with θ∈(0,1)∩Q. Then there exists G(x)∈Q+(|λ|) such that G(λ)=0.Proof.
Sinceθ∈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 whenλ 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.
LetA(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 Theorem3 for the case when the algebraic number γ and all its conjugates lie inside the unit circle.Lemma 7.
Letγ be a nonzero algebraic number and let γ1(=γ),…,γn be all its conjugates. Assume that
(19)|γ|¯∶=max⁡1≤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 numberγ 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 existsp(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 Theorem3.Proof of Theorem3.
Assume that all conjugates ofγ∈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 existsp(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.
Letγ be a nonzero algebraic number, let γ1(=γ),γ2,…,γn be all its conjugates, and let r>|γ|¯. Then the following statements are equivalent: (i)
all conjugates ofγ are in C∖R+;(ii)
there existsp(x)∈Q+(r)∩Q[x] such that p(γ)=0;(iii)
there existsg(x)∈R+[x]∩Q[x] such that g(γ)=0;(iv)
there existq(x)∈Q∩Q[x] and w∈R+∖{γ1,…,γn} such that q(γ)=q(w)=0.Proof.
Assertions (i) and (ii) are equivalent by Theorem3. 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 Lemma2 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 Lemma2 is closely related to the classical Eneström-Kakeya theorem, [8, Theorem 1], which states the following.Theorem 10.
Letp(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 inQ(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.
Letf(x)∈R+[x].(i)
The upper Eneström-Kakeya quotientβ[f] is the smallest r>0 such that f(x)∈Q+(r).(ii)
The lower Eneström-Kakeya quotientα[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 Lemma2 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. Letg(x) be a polynomial. The reciprocal polynomial of g(x) is defined as
(51)g*(x)=xdeg⁡gg(1x),
and we say that g(x) isself-reciprocal if g(x)=g*(x).Proposition 13.
Letf(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.
Writingf(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.
Iff(x)∈R+[x]∖R is self-reciprocal, then (i)
α[f]≤1, and(ii)
α[f]=1, if and only if β[f]=1.Proof.
(i) Ifα[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) ispositively minimal if all its coefficients are positive. From Proposition 12, we have the following.Proposition 15.
Letγ 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.Proposition15 leads naturally to the following definition.Definition 16.
Letγ be a nonzero algebraic number, whose minimal polynomial M(x) is positively minimal. Define thegrowth 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:(i)
rγ is unique;(ii)
rγ≤|γ|¯ (Theorem 3);(iii)
α[M]≤rγ≤β[M] (Propositions 12 and 15).For a nonzero algebraic numberγ 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.
Letγ be an algebraic number and assume that its minimal polynomial
(57)M(x)=xn+an-1xn-1+⋯+a1x+a0
is positively minimal.(i)
Ifα[M]=β[M], then α[M]=rγ=β[M] and M(x)∈Pγ.(ii)
We haveα[M]=β[M], if and only if
(58)M(x)=xn+axn-1+a2xn-2+⋯+an-1x+anfor some a∈Q+.Proof.
Assertion (i) follows at once from the preceding remarks.
(ii) Ifα[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.Theorem17 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.
Letγ 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 arenegative real numbers, then
(61)α[M]≤rγ≤|γ|¯<β[M]=∑t=1n‍|γt|andM(x)∉Pγ.Proof.
Since all conjugates ofγ 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.
Letγ be an algebraic number. Assume that its minimal polynomial M(x) is positively minimal and that all its conjugates are in C∖R+. (1)
Ifdeg⁡γ=1, then α[M]=rγ=β[M]=|γ|¯=|γ|.(2)
Ifdeg⁡γ=2, let γ=a+bi(a,b∈R) and M(x)=x2+a1x+a0; then,
(2.1)
forb=0, we have a0/a1=α[M]<|γ|¯<β[M]=a1,rγ∈[α[M],|γ|¯],(2.2)
forb≠0 and 3a2≠b2, we have α[M]<|γ|¯=|γ|<β[M],rγ∈[α[M],|γ|¯],(2.3)
forb≠0 and 3a2=b2, we have α[M]=rγ=β[M]=|γ|¯=|γ|.(3)
Ifdeg⁡γ=3, then α[M]≤rγ≤|γ|¯<β[M].Proof.
Part (1) is trivial. We now consider part (2), that is,deg⁡γ=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,deg⁡γ=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 deg⁡γ=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.(i)
Case|γ|¯=A.IfA<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.(ii)
Case|γ|¯=a2+b2.IfA(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.IfA(a2+b2)/(a2+b2-2Aa)>a2+b2, then β[M]>|γ|¯.IfA(a2+b2)/(a2+b2-2Aa)<a2+b2, then
(71)2Aa<a2+b2(a2+b2-A).Ifa>0, by (71), we get
(72)a2+b2(a2+b2-A)+2aa2+b2>2Aayielding(a2+b2-2Aa)/(A-2a)>a2+b2; that is, β[M]>|γ|¯.Consider nowa<0.IfA-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.IfA-2a>a2+b2, then β[M]>|γ|¯.IfA-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 Theorem3 is optimal.Proposition 20.
Letγ 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 isq(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.
Ifm>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.



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*Source: 296828-2014-07-16.xml*