Algebraic Numbers Satisfying Polynomials with Positive Rational Coefficients

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.

Lemma 2. Let r > 0 and $$. Then

• (1)

  f(x) ∈ Q(r)⇔c₀ ≥ 0, c_j ≤ rc_j+1  (0 ≤ j < J);

•  

  f(x) ∈ Q⁺(r)⇔c₀ > 0, c_j ≤ rc_j+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 c_j ≠ 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)

  let Y ∈ {P, Q, Q(r), Q⁺(r)}. Then f(x) ∈ Y⇒f(x^k) ∈ Y for all $$;

• (6)

  ([7, Lemma 2]) a₁, a₂, …, a_s are positive real numbers, and $$, 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|.

Theorem 3. Let γ be a nonzero algebraic number of degree n (over $$), let γ₁( = γ), …, γ_n be all its conjugates, and let

Then all conjugates of γ are in $$ if and only if there exists $$ such that p(γ) = 0.

Lemma 4. Let λ = |λ|e^2πiθ ≠ 0 with $$. Then, for any real r > |λ|, there exists F(x) ∈ Q⁺(r) such that F(λ) = 0.

Proof. Let u = r/|λ| > 1. 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 f₁ are >1/u > 0. Again, using (4) with Lemma 2 part (2), we get Thus, (x − u^m)f₁(x) ∈ Q and so, by Lemma 2 part (5), (x^m − u^m)f₁(x^m) ∈ Q. Let so that (x − u)f₂(x) ∈ Q. Consequently, Putting we see that and λ and $$ are roots of F(x).

Lemma 5. Let λ = |λ|e^2πiθ ≠ 0 with $$. Then there exists G(x) ∈ Q⁺(|λ|) such that G(λ) = 0.

Proof. Since $$, there is an $$ such that (λ/|λ|)^M = 1. Putting

we clearly see that (x − 1)g(x) ∈ Q, and so The result follows by taking noting that λ and $$ are two roots of G(x).

Lemma 6. Let A(x), $$. If $$, for some r > 0, then there exists a polynomial $$ such that A(x)B(x) ∈ Q⁺(r).

Proof. Writing

we have c_j > 0  (0 ≤ j ≤ k), and, by Lemma 2 part (1), For each j ∈ {0,1, …, k}, since the jth coefficient c_j depends on the first j coefficients of A(x) and $$ and since any real number is a limit of a sequence of rational numbers from both sides, we can successively choose rational numbers b₀, b₁, …, b_m so close to the corresponding coefficients of $$ in such a way that when the polynomial is multiplied to A(x) yielding these new coefficients still satisfy Lemma 2 part (1) shows then that A(x)B(x) ∈ Q⁺(r).

Lemma 7. Let γ be a nonzero algebraic number and let γ₁( = γ), …, γ_n be all its conjugates. Assume that

where r ∈ (0,1]. Then all conjugates of γ are in $$ if and only if there exists $$ such that p(γ) = 0.

Proof. Assume that the nonzero algebraic number γ and all its conjugates γ₁( = γ), …, γ_n are in $$. Arrange these numbers in such a way that

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 |γ_t| < r, for each 1 ≤ t ≤ n, there exists an odd positive integer N such that $$, for each 1 ≤ t ≤ n, and $$, for each 1 ≤ j ≤ J. Thus, for 1 ≤ j ≤ J, we have For 1 ≤ k ≤ K, by Lemma 5, there exits a polynomial For $$, by Lemma 4 and $$, there exits a polynomial Let and note that $$ are its roots. From Lemma 2 part (6) with J + K + L = n, we have E(x) ∈ Q⁺(r). Thus, Consequently, and so Since M(x)∣E(x^N) over $$, where $$ is the minimal polynomial of γ, there exists $$ such that $$. Thus, From Lemma 6, there exists $$ such that Putting we see that $$ and p(γ) = 0.

Conversely, assume that there exists $$ such that p(γ) = 0. Since $$, 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.

Proof of Theorem 3. Assume that all conjugates of $$. Since $$ and $$ is dense in $$, there exists $$ such that $$. Let

Then |λ_t| < 1 and $$, for all 1 ≤ t ≤ n. Let where $$ is the minimal polynomial of γ. Then F(x) is the minimal polynomial of λ, and λ₁( = λ), …, λ_n are conjugates of λ. By Lemma 7, there exists $$ such that p₁(λ) = 0. Then F(x)∣p₁(x); that is, M(x)∣p₁(x/r₁). Since $$ and $$, we have $$. Let Then $$ and p(γ) = 0. Since 0 < r₁ < r, by Lemma 2 part (2), we have $$.

On the other hand, assume that there exists $$ such that p(γ) = 0. Since $$, 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.

Corollary 8. Let γ be a nonzero algebraic number, let γ₁( = γ), γ₂, …, γ_n be all its conjugates, and let $$. Then the following statements are equivalent:

• (i)

  all conjugates of γ are in $$;

• (ii)

  there exists $$ such that p(γ) = 0;

• (iii)

  there exists $$ such that g(γ) = 0;

• (iv)

  there exist $$ and $$ 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

such that g(γ) = 0, and so g(γ₁) = ⋯ = g(γ_n) = 0. Since g_h > 0, for all h ∈ {1, …, m}, none of γ₁, …, γ_n can be real and positive. If γ_j ≠ 0 for some j ∈ {1,2, …, n}, choose Since g(x) is a factor of q(x), we have q(γ₁) = ⋯ = q(γ_n) = 0. From γ_j ≠ 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, …, γ_n, as desired.

That assertion (iv) implies that assertion (i) is again an immediate consequence of Lemma 2 part (4).

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 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).

Proposition 11. Let

where J ≥ 1, and let $$. If then f(x)d(x) ∉ Q(r) for all $$.

Proof. Let

If f(x)d(x) ∈ Q(r), then Lemma 2 part (1) gives where we adopt the convention that c_j = 0, for all j > J, and d_k = 0, for all k > K. From c₀ > 0, and (42), we get d₀ ≥ 0. From (43) and (40), we have From (44) and (40), we get which together with previous results yield d₂ ≥ 0. Continuing in the same manner up to (46), we get d₃ ≥ 0, d₄ ≥ 0, …, d_K ≥ 0. Thus, Since c_j > rc_j+1  (j = 0,1, 2, …, J − 1), the left-hand expression in (50) can be 0 only when d₀ = d₁ = ⋯ = d_K = 0; that is, d(x) ≡ 0, which is not possible.

Proposition 12. Let $$.

• (i)

  The upper Eneström-Kakeya quotient β[f] is the smallest r > 0 such that f(x) ∈ Q⁺(r).

• (ii)

  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 $$.

Proof. Part (i) follows from Lemma 2 part (1). Part (ii) is immediate from Proposition 11.

Proposition 13. Let $$. 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 $$, we have

and so From Proposition 12 part (i), we have If f(x) is self-reciprocal, then clearly β[f] = 1/α[f] and f(x) ∈ Q⁺(1/α[f]).

Corollary 14. If $$ is self-reciprocal, then

• (i)

  α[f] ≤ 1, and

• (ii)

  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.

Proposition 15. Let γ be an algebraic number and let

be its minimal polynomial. If M(x) is positively minimal, then, for any r ∈ (0, α[M]) and any $$, we have p(γ) ≠ 0.

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 r_γ, to be the infimum of the set of nonnegative real numbers with the following property: for any r > r_γ, there exists $$ such that p(γ) = 0.

Theorem 17. Let γ be an algebraic number and assume that its minimal polynomial

is positively minimal.

• (i)

  If α[M] = β[M], then α[M] = r_γ = β[M] and $$.

• (ii)

  We have α[M] = β[M], if and only if

  $$ ()

for some $$.

Proof. Assertion (i) follows at once from the preceding remarks.

(ii) If α[M] = β[M], then

which gives The converse is trivial.

Proposition 18. Let γ be an algebraic number of degree n over $$. Assume that its minimal polynomial M(x) is positively minimal. If n ≥ 2 and all the conjugates of γ₁ = γ, …, γ_n are negative real numbers, then

and $$.

Proof. Since all conjugates of γ are negative real numbers, we have

Lemma 2 part (6) shows that and so Proposition 12 (i) implies that $$. Since n ≥ 2 and $$, we have $$. Consequently, By Proposition 12 (i) and the definition of the upper Eneström-Kakeya quotient, we conclude that $$.

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 $$.

• (1)

  If deg⁡γ = 1, then $$.

• (2)

  If deg⁡γ = 2, let $$ and M(x) = x² + a₁x + a₀; then,

  • (2.1)

    for b = 0, we have $$,

  • (2.2)

    for b ≠ 0 and 3a² ≠ b², we have $$,

  • (2.3)

    for b ≠ 0 and 3a² = b², we have $$.

• (3)

  If deg⁡γ = 3, then $$.

Proof. Part (1) is trivial. We now consider part (2), that is, deg⁡γ = 2. If b = 0, then $$, and so

If b ≠ 0, without loss of generality, assume that γ = a + bi with $$. Then $$ and Since M(x) is positively minimal, we have a < 0 and a² + b² > 0. From we deduce that, if 3a² ≠ b², then $$ and that, if 3a² = b², then $$.

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

Since the minimal polynomial, is positively minimal, we have A − 2a > 0, a² + b² − 2Aa > 0 and A(a² + b²) > 0, which gives 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 $$.

We split our consideration into three cases depending on the maximum absolute value of the conjugates.

• (i)

  Case $$.

•  

  If A < A − 2a, or if A < A(a² + b²)/(a² + b² − 2Aa), then $$. Otherwise, we have A ≥ A − 2a and A ≥ A(a² + b²)/(a² + b² − 2Aa), which gives a = 0, a contradiction.

• (ii)

  Case $$.

•  

  If $$, 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 $$, then $$.

•  

  If $$, then

  $$ ()

•  

  If a > 0, by (71), we get

  $$ ()

•  

  yielding $$; that is, $$.

•  

  Consider now a < 0.

•  

  If $$, 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 $$, then $$.

•  

  If $$, then $$, yielding $$.

Proposition 20. Let γ be an algebraic number, and let

be its minimal polynomial. For $$, if $$, then q(γ) ≠ 0.

Proof. Assume on the contrary that there is $$ such that q(γ) = 0. Thus, M(x)∣q(x), so that

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 $$ 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 n + t − 1 ≥ t, we have c_t > 0 and Thus, a < c_t + a ≤ rc_t+1 ≤ rⁿ < a, which is a contradiction. The proofs for the cases m = n − 1 and n are similar but simpler.