Algebraic Numbers Satisfying Polynomials with Positive Rational Coefficients
A theorem of Dubickas, affirming a conjecture of Kuba, states that a nonzero algebraic number is a root of a polynomial 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 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.
A nonzero complex number is called positively algebraic if it is a root of a polynomial all of whose coefficients are positive rational numbers. In 2005, Kuba  conjectured that a necessary condition for an algebraic number to be positively algebraic is that none of its conjugates is a positive real number. This conjecture was confirmed affirmatively by Dubickas , in 2007, through the following result.
Theorem 1 (see ). A nonzero complex number is a root of a polynomial with positive rational coefficients if and only if is an algebraic number such that none of its conjugates is a positive real number.
In 2009, Brunotte  gave an elementary proof of Dubickas-Kuba theorem based on the following lemma ([3, Lemma 2]), which is originated from  (see also ): if is a polynomial having no nonnegative roots, then there exists such that has only positive coefficients. See also , 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 the set of positive real numbers. For , let Some basic properties, which are needed in our work here, modified with the same proofs for from , are in the following lemma.
Lemma 2. Let and . Then (1), ; , ;(2), where ;(3);(4) and for some has a unique positive zero ; this zero is simple and all other roots have absolute values ;(5)let . Then for all ;(6)([7, Lemma 2]) are positive real numbers, and , where ;(7)([7, Lemma 3]) is a complex number which is not real positive is a root of a polynomial in for any .
Our first main result is a certain quantitative improvement of Dubickas-Kuba theorem (Theorem 1 above).
Theorem 3. Let be a nonzero algebraic number of degree (over ), let be all its conjugates, and let Then all conjugates of are in if and only if there exists such that .
The nontrivial half of Theorem 3 is its necessity part, and its main difficulty is to show the existence of a polynomial in 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 which depends not only on the conjugates of 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 for which Theorem 3 holds. In the last section, we show that the bound 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 , 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 with . Then, for any real , there exists such that .
Proof. Let . Since , by Kronecker’s theorem ([9, Theorem 439, page 376]), the set is dense in the unit circle and so there exists such that This yields Let Thus, (4) shows that all the coefficients of are . Again, using (4) with Lemma 2 part (2), we get Thus, and so, by Lemma 2 part (5), . Let so that . Consequently, Putting we see that and and are roots of .
Lemma 5. Let with . Then there exists such that .
Proof. Since , there is an such that . Putting we clearly see that , and so The result follows by taking noting that and are two roots of .
Observe that the proof of this last lemma remains true when is a negative real number (with ). The followimg result indicates how to obtain a multiple in of a given element of with a factor in .
Lemma 6. Let , . If , for some , then there exists a polynomial such that .
Proof. Writing we have , and, by Lemma 2 part (1), For each , since the th coefficient depends on the first coefficients of and and since any real number is a limit of a sequence of rational numbers from both sides, we can successively choose rational numbers so close to the corresponding coefficients of in such a way that when the polynomial is multiplied to yielding these new coefficients still satisfy Lemma 2 part (1) shows then that .
The next lemma proves Theorem 3 for the case when the algebraic number and all its conjugates lie inside the unit circle.
Lemma 7. Let be a nonzero algebraic number and let be all its conjugates. Assume that where . Then all conjugates of are in if and only if there exists such that .
Proof. Assume that the nonzero algebraic number and all its conjugates are in . Arrange these numbers in such a way that
with . (In the above arrangement, it is tacitly assumed that , , and are positive integers; indeed, should any one of the three sets be empty, the corresponding value(s) of , , or could be zero. Yet the remaining arguments below still work with only slight adjustments.) Since , for each , there exists an odd positive integer such that , for each , and , for each . Thus, for , we have
For , by Lemma 5, there exits a polynomial
For , by Lemma 4 and , there exits a polynomial
and note that are its roots. From Lemma 2 part (6) with , we have . Thus,
Since over , where is the minimal polynomial of , there exists such that . Thus,
From Lemma 6, there exists such that
we see that and .
Conversely, assume that there exists such that . Since , all conjugates of are zeros of and, since all coefficients of are positive, no real conjugate of can be positive.
We are now ready to prove Theorem 3.
Proof of Theorem 3. Assume that all conjugates of . Since and is dense in , there exists such that . Let
Then and , for all . Let
where is the minimal polynomial of . Then is the minimal polynomial of , and are conjugates of . By Lemma 7, there exists such that . Then ; that is, . Since and , we have . Let
Then and . Since , by Lemma 2 part (2), we have .
On the other hand, assume that there exists such that . Since , all conjugates of are zeros of and, since all coefficients of are positive, no real conjugate of can be positive.
It is worth noting that in [7, Lemma 4], the bound for reduces to if the polynomial mentioned in Theorem 3 is the minimal polynomial of . The next corollary contains the Dubickas-Kuba theorem and some equivalent assertions.
Corollary 8. Let be a nonzero algebraic number, let be all its conjugates, and let . Then the following statements are equivalent: (i)all conjugates of are in ;(ii)there exists such that ;(iii)there exists such that ;(iv)there exist and such that .
Proof. Assertions (i) and (ii) are equivalent by Theorem 3. Clearly, (ii) implies (iii). To show (iii) implies (iv), assume that there exists
such that , and so . Since , for all , none of can be real and positive. If for some , choose
Since is a factor of , we have . From and Lemma 2 part (4), we know that has a unique positive zero, say , which must then be distinct from all , 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 its lower and upper Eneström-Kakeya quotients are defined, respectively, by
Theorem 10. Let all of whose coefficients are positive. Then all the zeros of are contained in the annulus , where and are, respectively, the lower and upper Eneström-Kakeya quotients of .
In this section, we derive a proposition yielding conditions which are necessary for a product of two polynomials to be in .
Proposition 11. Let where , and let . If then for all .
Proof. Let If , then Lemma 2 part (1) gives where we adopt the convention that , for all , and , for all . From , and (42), we get . From (43) and (40), we have From (44) and (40), we get which together with previous results yield . Continuing in the same manner up to (46), we get , . Thus, Since , the left-hand expression in (50) can be only when ; that is, , which is not possible.
The next proposition indicates the significance of the upper and lower Eneström-Kakeya quotients.
Proposition 12. Let .(i)The upper Eneström-Kakeya quotient is the smallest such that .(ii)The lower Eneström-Kakeya quotient has the property that if with , then over .
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 be a polynomial. The reciprocal polynomial of is defined as and we say that is self-reciprocal if .
Proposition 13. Let . Then , and . Moreover, if is self-reciprocal, then and .
Proof. Writing , we have and so From Proposition 12 part (i), we have If is self-reciprocal, then clearly and .
Corollary 14. If is self-reciprocal, then (i), and(ii), if and only if .
4. Minimal Polynomials with Positive Coefficients
In this section, we aim to show that the bound , 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 be a nonzero algebraic number, and its minimal polynomial. We say that is positively minimal if all its coefficients are positive. From Proposition 12, we have the following.
Proposition 15. Let be an algebraic number and let be its minimal polynomial. If is positively minimal, then, for any and any , we have .
Proposition 15 leads naturally to the following definition.
Definition 16. Let be a nonzero algebraic number, whose minimal polynomial is positively minimal. Define the growth factor of an algebraic number with respect to , denoted by , to be the infimum of the set of nonnegative real numbers with the following property: for any , there exists such that .
For a nonzero algebraic number whose minimal polynomial is positively minimal and whose growth factor is , let A class of algebraic numbers for which the bound in Theorem 3 cannot be improved is given in the next theorem.
Theorem 17. Let be an algebraic number and assume that its minimal polynomial is positively minimal.(i)If , then and .(ii)We have , if and only if for some .
Proof. Assertion (i) follows at once from the preceding remarks.
(ii) If , then which gives The converse is trivial.
Theorem 17 poses a natural question whether a positively minimal polynomial must necessarily belong to . A negative answer is provided by the next proposition.
Proposition 18. Let be an algebraic number of degree over . Assume that its minimal polynomial is positively minimal. If and all the conjugates of are negative real numbers, then and .
Proof. Since all conjugates of are negative real numbers, we have Lemma 2 part (6) shows that and so Proposition 12 (i) implies that . Since and , we have . Consequently, By Proposition 12 (i) and the definition of the upper Eneström-Kakeya quotient, we conclude that .
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 be an algebraic number. Assume that its minimal polynomial is positively minimal and that all its conjugates are in . (1)If , then .(2)If , let and ; then,(2.1)for , we have ,(2.2)for and , we have ,(2.3)for and , we have .(3)If , then .
Proof. Part (1) is trivial. We now consider part (2), that is, . If , then , and so
If , without loss of generality, assume that with . Then and
Since is positively minimal, we have and . From
we deduce that, if , then and that, if , then .
Finally consider part (3), that is, . If and all its conjugates , , and are negative real numbers, the conclusion follows at once from Proposition 18. Assume then that Since the minimal polynomial, is positively minimal, we have , and , which gives Since , is positive rational, and is positive irrational, we must have . By the remarks after Proposition 15, there remains only the verification that .
We split our consideration into three cases depending on the maximum absolute value of the conjugates. (i)Case . If , or if , then . Otherwise, we have and , which gives , a contradiction.(ii)Case . If , since is positive rational, we deduce that must be positive rational. Since is positive rational, we conclude that is positive rational, which is a contradiction. If , then . If , then If , by (71), we get yielding ; that is, . Consider now . If , since is positive rational, we deduce that is positive rational. Since is positive rational, we conclude that is positive rational, which is a contradiction. If , then . If , then , yielding .
We end this paper with another class of algebraic numbers for which the bound in Theorem 3 is optimal.
Proposition 20. Let be an algebraic number, and let be its minimal polynomial. For , if , then .
Proof. Assume on the contrary that there is such that . Thus, , so that
If , by Lemma 2 part (1) we have , a contradiction.
If , let for some and the relation (74) becomes Invoking upon Lemma 2 part (1), we get the following chain of inequalities: Proceeding in the same manner, we have Since , we have and Thus, , which is a contradiction. The proofs for the cases and 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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