Algebraic Numbers of the form αT with α Algebraic and T Transcendental

Let α≠1 be a positive real number and let P(x) be a non-constant rational function with algebraic coefficients. In this paper, in particular, we prove that the set of algebraic numbers of the form αP(T), with T transcendental, is dense in some open interval of R.


Introduction
In 1900, in the International Congress of Mathematicians (in Paris), Hilbert provided a list of 23 problems and his seventh problem was about the arithmetic nature of the power α β of two algebraic numbers α and β. In 1934-1935, Gelfond and Schneider [1] (p. 9) (independently) completely solved this problem : if α and β are algebraic numbers, with α = 0 or 1, and β irrational. Then α β is transcendental. This outstanding result is called the Gelfond-Schneider theorem.
As immediate consequences, we have the transcendence of the numbers 2 and e π = i −2i . The Gelfond-Schneider theorem classifies the (arithmetic) nature of x y , for algebraic numbers x and y (because x y is algebraic if y is rational). However, when {x, y} Q, one may has all possibilities (see Table 1). The case x = y may be of more interest: is it possible T T to be algebraic, for some transcendental number T? We point out that a negative answer for this question is expected (but still unproved), since the numbers e e , π π , and (log 2) log 2 are expected to be transcendental. However, related to the previous question, Marques and Sondow [2] showed that its answer is Yes. In fact, they proved that the set of algebraic numbers of the form T T (where T runs over all transcendental numbers) is dense in the interval [e −1/e , ∞).
In this direction, it may be interesting to consider the following generalization of the previous question: given arbitrary non-constant polynomials P, Q ∈ Q[x], is there always a transcendental number T, such that P(T) Q(T) is algebraic? The answer for this question is also Yes and it was given by Marques [3]: the set of algebraic numbers of the form P(T) Q(T) , with T transcendental, is dense in some connected subset of R or C. See [4,5] for generalizations of this result and some new results in [6,7].
We still point out that, in 2020, Trojovský [8] proved some results related to the transcendence of numbers of the form n γ for positive integers n and complex numbers γ.
In this paper, we continue this program by studying the existence of algebraic numbers of the form α T , where α is algebraic and T is transcendental. More precisely, our main result is the following: Theorem 1. Let α = 1 be a positive algebraic number and let P(x) ∈ Q(x) be a non-constant rational function. Then the set of algebraic numbers of the form α P(T) , with T transcendental, is dense in some open interval of R. Example 1. In particular, for α = 2 and P(x) = x, we have that the previous result implies in the existence of an open interval I ⊆ R such that the set of algebraic numbers of the form 2 T with T transcendental, is dense in I. As we shall see in the next section, it is possible to make this interval explicit (if α and P(x) are previously given). In the case of α = 2 and P(x) = x, we infer that I = (0, ∞).

Example 2.
We remark that the transcendence of 2 π still remains as an open problem. However, as an application of the previous example, we have the existence of a sequence (T n ) n of transcendental numbers, such that 2 T n ∈ Q, for all n ≥ 1, and 2 T n tends to 2 π , as n → ∞.

Remark 1.
We recall that a rational function f (x) is the quotient of two polynomials, say P(x)/Q(x) (with no common zeros). This function is defined in the whole complex plane but the set of zeroes of Q(x) (which are poles of f (x)). The coefficients of f (x) are defined as the coefficients of P(x) and Q(x).
A major open problem in transcendental number theory is a conjecture raised by Schanuel (in the 1960s) during a course at Yale given by Lang [9] (pp. 30-31).
We remark that Schanuel's conjecture is proved only for n = 1: Lindemann's theorem asserts that e α is a transcendental number, for any non-zero algebraic number α. As an immediate consequence, we deduce that log α is transcendental for any α ∈ Q\{0, 1} (since e log α = α is algebraic).
There are many research topics on the deep consequences of the veracity of the Schanuel's conjecture (for some of them, we refer the reader to [10,11] and references therein). In the opposite direction of Theorem 1, here, we still prove the following conditional result: Theorem 2. Assume Schanuel's conjecture (SC) and let P(x) ∈ Q(x) be a non-constant rational function. If α ∈ {0, 1} is an algebraic number, then α P(e) is transcendental. Example 3. We point out that even the transcendence of 2 e is still unproved. Thus, the previous result gives (conditionally) the transcendence of this number (for the choice of α = 2 and P(x) = x).

We point out that the above fact is an open problem if (SC)
is not assumed to be true (even for a particular rational function f (x)). The number e can be replaced by some other known transcendental numbers, as π, log 2, etc.

Proof of the Theorem 1
Before proceeding further, we shall provide a field-theoretical result which will be an essential ingredient in the proof.
We know that the set of the real algebraic numbers is dense in R (because Q ⊆ Q ∩ R). Indeed, we know that the set Q m of the m-degree real algebraic numbers is dense in R, for all m ≥ 1, see [12] (Theorem 7.4). Before the proof of the theorem, we shall need a slightly stronger fact. More precisely, Lemma 1. For any m ≥ 1, the set A m = Q m \{β ∈ Q m : β n ∈ Q, for some n ∈ Z >0 } is dense in R.
Proof. Let p be a prime number. It suffices to prove that the set is an m-degree algebraic number, for any non-zero rational number Q. In fact, 1 + m √ p is a root of P(x) = (x − 1) m − p which is irreducible over Q (since P(x + 1) is irreducible by the Eisenstein's irreducibility criterion). Now, we must prove that P m ∩ {β ∈ Q m : β n ∈ Q, for some n > 0} = ∅.
Towards a contradiction, we suppose the existence of an integer number n > 0 such that (Q(1 + m √ p)) n is a rational number. Thus, so is (1 + m √ p) n . We know that B = {1, m √ p, . . . , ( m √ p) m−1 } is a basis of the field extension Q( m √ p)/Q. By the Binomial Theorem, (1 + m √ p) n = ∑ n k=0 ( n k )( m √ p) k .
If n < m, then (1 + m √ p) n / ∈ Q, by the Q-linear independence of B. When n ≥ m, we get We then rewrite the second summatory as If n − m < m, the result follows again by the Q-linear independence of B. Otherwise, we have n − m ≥ m and so Repeating this procedure = n/m times, we arrive at where a j ∈ Z >0 and we used that 0 ≤ n − m < m. Thus (1 + m √ p) n / ∈ Q, again because B is linearly independent over Q. Now, we are ready to deal with the proof of theorem.
Proof. Set P(x) = P 1 (x)/P 2 (x) and let us consider an open interval J ⊆ R such that P 2 (x) = 0, for all x ∈ J . Now, the function f : J → R given by f (x) := α P(x) is well-defined and analytical in J . We claim that f (x) is a non-constant function. Indeed, on the contrary, f (x) = 0, for all x ∈ J . However, this would imply that P(x) is constant (contradicting the hypothesis) since f (x) = α P(x) P (x) log α. Thus, there exists x 0 ∈ J with f (x 0 ) = 0. Therefore, by the Inverse Function Theorem, there are open connected intervals I and I 0 (with x 0 ∈ I) such that f : I → I 0 is a diffeomorphism (i.e., a differentiable function which has a differentiable inverse). In particular, By Lemma 1, the set A is dense in R and therefore A ∩ I 0 is dense in the interval I 0 . Now, it suffices to prove that every element of this intersection has the desired form. In fact, if A ∈ A ∩ I 0 , then A = α P(T) , for some T ∈ I (since I 0 = f (I)). We claim that T is a transcendental number. Indeed, assume that T is algebraic. Since α / ∈ {0, 1}, we use the Gelfond-Schneider theorem to conclude that P(T) is a rational number, say m/n with n > 0. Thus, we can write A n = α m . Hence deg(A n ) divides deg(α) (since Q(α n ) ⊆ Q(α)). Since deg(A n ) | deg(A) and gcd(deg(A), deg(α)) = 1, we conclude that deg(A n ) = 1 yielding that A n is a rational number which contradicts the choice of A ∈ A. This contradiction implies that T is transcendental and the proof is complete.
Since =trdeg Q Q(log α, e, α P(e) ) ≤ 3, we infer that = 3 and so log α, e, α P(e) are algebraically independent. This yields, in particular, the transcendence of α P(e) . The proof is then complete.