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Article Schanuel’s Conjecture and the Transcendence of Power Towers

Eva Trojovská * and Pavel Trojovský

Department of Mathematics, Faculty of , University of Hradec Králové, 50003 Hradec Králové, Czech Republic; [email protected] * Correspondence: [email protected]; Tel.: +420-49-333-2859

Abstract: We give three consequences of Schanuel’s Conjecture. The first is that P(e)Q(e) and ( ) P(π)Q π are transcendental, for any non-constant P(x), Q(x) ∈ Q[x]. The second is that π 6= αβ, for any algebraic numbers α and β. The third is the case of the Gelfond’s conjecture (about the transcendence of a finite algebraic power tower) in which all elements are equal.

Keywords: Schanuel’s Conjecture; Gelfond–Schneider ; Hermite–Lindemann Theorem; algebraic independence; ; power tower

MSC: 11J81

1. Introduction: A Brief “Transcendental” Tour In 1900, at the International Congress of Mathematicians in Paris Hilbert raised the

 question of the arithmetic nature of an algebraic power of an (as his 7th  problem). After three decades, Gelfond and Schneider, independently, solved the problem

Citation: Trojovská, E.; Trojovský, P. (see Baker [1] p. 9). Schanuel’s Conjecture and the Transcendence of Power Towers. Gelfond–Schneider Theorem. If α and β are algebraic numbers, with α 6= 0 or 1, and β irra- β Mathematics 2021, 9, 717. tional, then α is transcendental. https://doi.org/10.3390/ math9070717 The Gelfond–Schneider Theorem yields a complete classification of the arithmetic nature of xy when x, y ∈ Q. However, when at least one of x and y is transcendental, y Academic Editor: Patrick Solé the arithmetic nature of x can be completely “chaotic”. A more restrictive question is: Is there a T for which TT is Received: 1 March 2021 algebraic? If the answer were no, then one has the transcendence of the expected (but Accepted: 24 March 2021 still unproved) numbers ee and ππ. However, Sondow and Marques ([2] Proposition 2.2) Published: 26 March 2021 showed that the answer is actually yes. Indeed, they proved that: A positive T is transcendental if the number A := TT satisfies either Publisher’s Note: MDPI stays neutral • An ∈ Q \ Q for all n ∈ N, or with regard to jurisdictional claims in • A ∈ Q \{nn : n ∈ N}. published maps and institutional affil- √ T T iations. In particular, T is transcendental if T = 1 + 2 or T = 2. This result was generalized by Marques ([3] Proposition): If P(x), Q(x) ∈ Q[x] are non-constant polynomials, then the set of algebraic numbers of the form P(T)Q(T), with T real and transcendental, is dense in some non-empty connected open of R or of C. Copyright: © 2021 by the authors. A further generalization to rational functions with algebraic coefficients was given Licensee MDPI, Basel, Switzerland. by Jensen and Marques [4]. Moreover, recently, Trojovský [5] extended the result for Q1(T) Qn(T) This article is an open access article algebraic numbers of the form P1(T) ··· Pn(T) , with T transcendental (where distributed under the terms and P1, ... , Pn, Q1, ... , Qn ∈ Q[x] are polynomials under some weak technical assumptions). y conditions of the Creative Commons We still refer the reader to [6] for the case x , where x is algebraic and y is transcendental. Attribution (CC BY) license (https:// Generalizing the notion of an algebraic number, the complex numbers α1, ... , αn are creativecommons.org/licenses/by/ algebraically dependent if there exists a nonzero P(x1, ... , xn) ∈ Q[x1, ... , xn] 4.0/).

Mathematics 2021, 9, 717. https://doi.org/10.3390/math9070717 https://www.mdpi.com/journal/mathematics Mathematics 2021, 9, 717 2 of 6

such that P(α1, ... , αn) = 0. Otherwise, α1, ... , αn are algebraically independent; in particular, they are all transcendental. The following major open problem in transcendental number was stated in the 1960s in a course at Yale given by Lang ([7] pp. 30–31).

Schanuel’s Conjecture. If α1, ... , αn ∈ C are linearly independent over Q, then there are at least α αn n algebraically independent numbers among α1,..., αn, e 1 ,..., e .

For several consequences and reformulations of Schanuel’s Conjecture (SC), see Cheng et al. [8], Marques and Sondow [9], and Ribenboim ([10] Chapter 10, Sec- tion 7G). In the present paper, we give three applications of SC. The first is a complement to the Marques’s proposition in [3], and says in particular that SC implies the transcendence of ee and ππ.

Theorem 1. If Schanuel’s Conjecture is true, then for any non-constant polynomials P(x), Q(x) ∈ ( ) ( ) Q[x], the numbers P(e)Q e and P(π)Q π are both transcendental.

Our proof can be adapted to show that SC also implies the transcendence of P(log 2)Q(log 2). On the other hand, by the previous discussions, there do exist transcendental numbers T for which TT and P(T)Q(T) are algebraic. In view of the Gelfond–Schneider Theorem, it is natural to ask: Which transcendental numbers are algebraic powers of algebraic numbers? For instance, e is not. In fact, if e = αβ, with α, β ∈ Q, then e1/β = α would be algebraic, which contradicts the Hermite– Lindemann Theorem that eγ is transcendental for any algebraic number γ 6= 0 (see [1] Chapter 1).

Theorem 2. Schanuel’s Conjecture implies π 6= αβ, for any algebraic numbers α and β.

In 1934, Gelfond announced several extensions [11] of the Gelfond–Schneider Theorem without proof. For the much weaker statements, as well as proofs of them and references, see Feldman and Nesterenko ([12] pp. 260–267). Here is the last of Gelfond’s extensions, stated as a conjecture.

Gelfond’s Conjecture. If m ≥ 3 and α1, α2, ... , αm ∈ C are algebraic, with α1 6= 0, 1 and α2,..., αm irrational, then the number · αm ·· α2 α1 is transcendental.

We prove a special case, assuming SC.

Theorem 3. Assume Schanuel’s Conjecture. Then, Gelfond’s Conjecture holds when α1 = α2 = ··· = αm. More generally, if α ∈ C is algebraic but is not a rational , then the power tower of α of order m ≥ 3 ·α ·· mα := α α | {z } m is transcendental; in fact, the numbers log α,3α,4α,5α,... are algebraically independent.

In the next section, we introduce some notation and prove a simple but useful lemma. The proofs of 1–3 are in Sections3–5.

2. A Useful Lemma

Let us denote the maximal number of algebraically independent elements of a set {z1,..., zk} ⊂ C by Mathematics 2021, 9, 717 3 of 6

# {z ,..., z }. Q 1 k

We say that z1,..., zk are algebraically dependent on {w1,..., w`} ⊂ C if

Q(z1,..., zk) ⊂ Q(w1,..., w`). This condition is equivalent (see [13] Chapter VIII) to

Q(z1,..., zk) ⊂ Q(w1,..., w`).

Lemma 1. Assume SC. Suppose that α1, ... , αn ∈ C are Q-linearly independent, and the numbers α αn α1,..., αn, e 1 ,..., e are algebraically dependent on the subset

α1 αn {β1,..., βn} ⊂ {α1,..., αn, e ,..., e }.

Then, the numbers β1, ... , βn are all transcendental; in fact, they are algebraically indepen- dent.

Proof. The hypotheses imply

n ≥ # {β ,..., β } = # {α ,..., α , eα1 ,..., eαn } ≥ n. Q 1 n Q 1 n Hence, # {β ,..., β } = n, and the conclusion follows. Q 1 n It is Lemma1 rather than SC that we use in the proofs of Theorems 1–3. Thus, Theorems 1–3 remain true under the weaker in Lemma1.

3. Proof of Theorem 1 Fix non-constant polynomials P(x), Q(x) ∈ Q[x]. Proof that P(e)Q(e) is transcendental. We first consider the case P(x) = xn, where n ≥ 1. Since Q(e) is transcendental, 1 and nQ(e) are Q-linearly independent. Applying Lemma1 to the subset {e, enQ(e)} ⊂ {1, nQ(e), e, enQ(e)}, it follows that enQ(e) = P(e)Q(e) is transcendental, as claimed. Now, assume P(x) 6= xn for any n ≥ 1. We show that 1 and log P(e) are Q-linearly independent. Given a Q-linear relation a + b log P(e) = 0, by clearing the denominators if necessary, we may assume that a, b ∈ Z, with b ≥ 0. Now, P(e)bea − 1 = 0. If a ≥ 0, then, since P(x) 6= xn for any n ≥ 0 and e is not algebraic, the polynomial P(x)bxa − 1 must be   identically zero; hence, a = b = 0. If a < 0, then P(x)b − x−a = x−a P(x)bxa − 1 must be the zero polynomial, and again a = b = 0. Now, Lemma1 applied to the subset

{e, log P(e)} ⊂ {1, log P(e), e, P(e)}

implies that e and log P(e) are algebraically independent. Hence, so are Q(e) and log P(e). Therefore, the three numbers 1, log P(e), and Q(e) log P(e) are Q-linearly independent. Applying Lemma1 to the subset

{e, log P(e), P(e)Q(e)} ⊂ {1, log P(e), Q(e) log P(e), e, P(e), P(e)Q(e)}, we get that P(e)Q(e) is transcendental. ( ) The proof for P(π)Q π will not require two cases, since 1 and log P(π) are Q-linearly independent even when P(x) = xn. Mathematics 2021, 9, 717 4 of 6

( ) Proof that P(π)Q π is transcendental. Note first that iπ and log P(π) are Q-linearly in- dependent because, otherwise, there would exist a Z-relation aiπ + b log P(π) = 0 with b > 0, and then P(π)b = (−1)a would be algebraic, contradicting the transcendence of π. Applying Lemma1 to the subset

{iπ, log P(π)} ⊂ {iπ, log P(π), −1, P(π)}, we get that iπ and log P(π) are algebraically independent. Then, the set {iπ, log P(π), Q(π) log P(π)} is Q-linearly independent, and Lemma1 applied to this subset of {iπ, log P(π), Q(π) log P(π), −1, P(π), P(π)Q(π)} yields the desired result.

4. Proof of Theorem 2 Suppose on the contrary that π = αβ, where α, β ∈ Q. Then, α 6= 0, 1 and β 6∈ Q because π 6∈ Q. We claim that α is not a root of unity. To see this, assume α = e2kπi/n, with n ≥ 2 and k ∈ {1, 2, . . . , n − 1}. Taking on both sides of π = αβ gives

2βk log π − iπ = 0. n However, this contradicts the algebraic independence of iπ and log π, which we proved conditionally in Section3. It follows that iπ and log α are Q-linearly independent, for otherwise there would be a Z-relation aiπ + b log α = 0 with b > 0, and then α = (−1)a/b, contradicting the fact that α is not a root of unity. Now, we can apply Lemma1 to the subset

{iπ, log α} ⊂ {iπ, log α, −1, α}

and conclude that iπ and log α are algebraically independent. Thus, in any Q-linear rela- tion, aiπ + b log α + cβ log α = 0, one has a = 0, and so b + cβ = 0. Since β is irrational, b = c = 0. Thus, iπ, log α, and β log α are Q-linearly independent. However, then, since α, β ∈ Q, Lemma1 applied to the subset

{iπ, log α, αβ} ⊂ {iπ, log α, β log α, −1, α, αβ}

implies that π and αβ are algebraically independent, contradicting π = αβ. This completes the proof.

We leave it as an exercise to use similar arguments to show, under the assumption of SC, that each of the numbers ee, e + π, and log 2 is also not an algebraic power of an algebraic number.

5. Proof of Theorem 3 It suffices to show that, if α ∈ Q \ Z and m ≥ 3, then the numbers log α,3α,4α, ... ,mα are algebraically independent. The proof is in two cases. Case 1: α ∈ Q \ Z. In this case, αα is irrational (see [2] Lemma 2.1). Thus, 1 and αα are Q-linearly independent, and then so are log α and αα log α. Since α and αα are algebraic, Lemma1 applied to the subset

α α {log α, αα } ⊂ {log α, αα log α, α, αα }

α yields the algebraic independence of log α and αα = 3α. Mathematics 2021, 9, 717 5 of 6

Now, fix m > 3 and assume inductively that log α,3α,4α, ... ,m−1α are algebraically independent. Then, in any Q-linear relation

m−1 j ∑ aj α = 0 j=1

1 2 α we must have a3 = ··· = am−1 = 0. Since α = α ∈ Q \ Z and α = α ∈/ Q, we also have a1 = a2 = 0. This implies the Q- of

{1α log α,...,m−1α log α} = {log(2α), . . . , log(mα)}.

Then, Lemma1 yields the algebraic independence of the subset

{log(2α),3α,4α,...,mα} ⊂ {log(2α), . . . , log(mα),2α,...,mα}

and, since log(2α) = α log α, hence also that of log α,3α,4α, ... ,mα. This completes the in- duction. Case 2: α ∈ Q \ Q. By the Gelfond–Schneider Theorem, αα is transcendental. Hence, 1, α, αα are Q-linearly independent, and then so are log α, α log α, αα log α. Since {α log α, αα log α} ⊂ Q(log α, α, αα) and α is algebraic, Lemma1 applied to the subset

α α {log α, αα, αα } ⊂ {log α, α log α, αα log α, α, αα, αα } α yields the algebraic independence of {log α, αα, αα } = {log α,2α,3α}. Suppose inductively that log α,2α,3α, ... ,m−1α are algebraically independent, where m > 3. Then, any Q-linear relation

m−1 j a0 + ∑ aj α = 0 j=1

1 implies a2 = ··· = am−1 = 0. Since α = α ∈/ Q, we also get a0 = a1 = 0. That implies the Q-linear independence of

{log α,1α log α,...,m−1α log α} = {log α, log(2α), . . . , log(mα)}. Since − {1α log α,...,m 1α log α} ⊂ Q(log α,1α,2α,...,mα), we may apply Lemma1 to the subset

{log α,2α,3α,...,mα} ⊂ {log(1α), . . . , log(mα),1α,...,mα}

and conclude that log α,2α,3α,...,mα are algebraically independent. Thus, in both Cases 1 and 2, the numbers log α,3α,4α, ... ,mα are algebraically indepen- dent, as desired.

Results on the arithmetic nature of power towers of x of infinite order

· ··   ∞x := lim kx = x x e−e ≤ x ≤ e1/e k→∞ can be found in ([2] Appendix).

6. Conclusions In this paper, we prove three results about the arithmetic nature of powers of algebraic and transcendental numbers by assuming that Schanuel’s Conjecture is true. The first one is that P(ξ)Q(ξ) is transcendental, for ξ ∈ {e, π} and for any non-constant polynomials Mathematics 2021, 9, 717 6 of 6

P(x), Q(x) ∈ Q[x]. The second one is that π cannot be written in the form αβ, for any α, β ∈ Q. The third one is the power tower case of the Gelfond’s conjecture (about the transcendence of a finite algebraic power tower) in which all elements are equal.

Author Contributions: P.T. conceived the presented idea, on the conceptualization, methodology, investigation. Writing—review & editing were done by E.T. All authors have read and agreed to the published version of the manuscript. Funding: The authors were supported by the Project of Specific Research PrF UHK No. 2101/2021, University of Hradec Králové, Czech Republic. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Informed consent was obtained from all subjects involved in the study. Conflicts of Interest: The authors declare no conflict of interest.

References 1. Baker, A. Transcendental ; Cambridge Mathematical Library, Cambridge University Press: Cambridge, UK, 1990. 2. Sondow, J.; Marques, D. Algebraic and transcendental solutions of some exponential equations. Ann. Math. Inform. 2010, 37, 151–164. 3. Marques, D. Algebraic numbers of the form P(T)Q(T), with T transcendental. Elem. Math. 2010, 65, 78–80. [CrossRef][PubMed] 4. Jensen, C.U.; Marques, D. Some field theoretic properties and an application concerning transcendental numbers. J. Algebra Appl. 2010, 9, 1–8. [CrossRef] 5. Trojovský, P. Algebraic numbers as product of powers of transcendental numbers. Symmetry 2019, 11, 887. [CrossRef] 6. Hubálovský, Š.; Trojovská, E. Algebraic Numbers of the form αT with α Algebraic and T Transcendental. Mathematics 2020, 8, 1687. [CrossRef] 7. Lang, S. Introduction to Transcendental Numbers; Addison-Wesley: Reading, MA, USA, 1966. 8. Cheng, C.; Dietel, B.; Herblot, M.M.; Huang, J.; Krieger, H.; Marques, D.; Mason, J.; Mereb, M.; Wilson, S.R. Some consequences of Schanuel’s Conjecture. J. Number Theory 2009, 129, 1464–1467. doi:10.1016/j.jnt.2008.10.018 [CrossRef] 9. Marques, D.; Sondow, J. Schanuel’s Conjecture and algebraic powers zw and wz with z and w transcendental. East-West J. Math. 2010, 12, 75–84. 10. Ribenboim, P. My Numbers, My Friends: Popular Lectures on Number Theory; Springer: New York, NY, USA, 2000. 11. Gelfond, M.A. Sur quelques résultats nouveaux dans la théorie des nombres transcendants. C. R. Math. Acad. Sci. Paris 1934, 199, 259. 12. Feldman, N.I.; Nesterenko, Y.V. Transcendental Numbers in Encyclopaedia of Mathematical ; Number Theory IV; Springer: Berlin, Germany, 1998; Volume 44. 13. Lang, S. Algebra; Springer: New York, NY, USA, 2002.