Journal of Numbers

Volume 2014 (2014), Article ID 296828, 8 pages

http://dx.doi.org/10.1155/2014/296828

Research Article

Algebraic Numbers Satisfying Polynomials with Positive Rational Coefficients

Vichian Laohakosol¹ and Suton Tadee²

¹Department of Mathematics, Faculty of Science, Kasetsart University, Bangkok 10900, Thailand

²Department of Mathematics and Statistics, Faculty of Science and Technology, Thepsatri Rajabhat University, Lopburi 15000, Thailand

Received 2 May 2014; Accepted 30 June 2014; Published 16 July 2014

Academic Editor: Ahmed Laghribi

Copyright © 2014 Vichian Laohakosol and Suton Tadee. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

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.

1. Introduction

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 [1] 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 [2], in 2007, through the following result.

Theorem 1 (see [2]). 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 [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 is a polynomial having no nonnegative roots, then there exists such that 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 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 [7], 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 [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 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

Let

and note that are its roots. From Lemma 2 part (6) with , we have . Thus,

Consequently,

and so

Since over , where is the minimal polynomial of , there exists such that . Thus,

From Lemma 6, there exists such that

Putting

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

Part (1) of Lemma 2 is closely related to the classical Eneström-Kakeya theorem, [8, Theorem 1], which states the following.

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 .

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 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 .

Proof. (i) If , then . Since is self-reciprocal, by Proposition 13, , 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 , 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 .

Remarks. The growth factor enjoys the following basic properties:

(i) is unique;(ii) (Theorem 3);(iii) (Propositions 12 and 15).

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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