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Article

Schanuel’s Conjecture and the Transcendence of Power Towers

Department of Mathematics, Faculty of Science, University of Hradec Králové, 50003 Hradec Králové, Czech Republic
*
Author to whom correspondence should be addressed.
Mathematics 2021, 9(7), 717; https://doi.org/10.3390/math9070717
Submission received: 1 March 2021 / Accepted: 24 March 2021 / Published: 26 March 2021
(This article belongs to the Special Issue New Developments in Number Theory)

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 polynomials P ( x ) , Q ( x ) Q ¯ [ x ] . The second is that π α β , 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.

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 algebraic number (as his 7th problem). After three decades, Gelfond and Schneider, independently, solved the problem (see Baker [1] p. 9).
Gelfond–Schneider Theorem.
If α and β are algebraic numbers, with α 0 or 1, and β irrational, then α β is transcendental.
The Gelfond–Schneider Theorem yields a complete classification of the arithmetic nature of x y when x , y Q ¯ . However, when at least one of x and y is transcendental, the arithmetic nature of x y can be completely “chaotic”.
A more restrictive question is: Is there a transcendental number T for which T T is algebraic? If the answer were no, then one has the transcendence of the expected (but still unproved) numbers e e and π π . However, Sondow and Marques ([2] Proposition 2.2) showed that the answer is actually yes. Indeed, they proved that: A positive real number T is transcendental if the number A : = T T satisfies either
  • A n Q ¯ Q for all n N , or
  • A Q { n n : n N } .
In particular, T is transcendental if T T = 1 + 2 or T 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 subset of R or of C .
A further generalization to rational functions with algebraic coefficients was given by Jensen and Marques [4]. Moreover, recently, Trojovský [5] extended the result for algebraic numbers of the form P 1 ( T ) Q 1 ( T ) P n ( T ) Q n ( T ) , with T transcendental (where P 1 , , P n , Q 1 , , Q n Q ¯ [ x ] are polynomials under some weak technical assumptions). We still refer the reader to [6] for the case x y , where x is algebraic and y is transcendental.
Generalizing the notion of an algebraic number, the complex numbers α 1 , , α n are algebraically dependent if there exists a nonzero polynomial P ( x 1 , , x n ) Q [ x 1 , , x n ] such that P ( α 1 , , α n ) = 0 . Otherwise, α 1 , , α n are algebraically independent; in particular, they are all transcendental.
The following major open problem in transcendental number theory was stated in the 1960s in a course at Yale given by Lang ([7] pp. 30–31).
Schanuel’s Conjecture.
If α 1 , , α n C are linearly independent over Q , then there are at least n algebraically independent numbers among α 1 , , α n , e α 1 , , e α n .
For several consequences and reformulations of Schanuel’s Conjecture (SC), see Cheng et al. [8], Marques and Sondow [9], and Ribenboim ([10] Chapter 10, Section 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 e e and π π .
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 T T 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 = α β , with α , β Q ¯ , then e 1 / β = α would be algebraic, which contradicts the Hermite–Lindemann Theorem that e γ is transcendental for any algebraic number γ 0 (see [1] Chapter 1).
Theorem 2.
Schanuel’s Conjecture implies π α β , 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 α 1 , α 2 , , α m C are algebraic, with α 1 0 , 1 and α 2 , , α m irrational, then the number
α 1 α 2 · · · α m
is transcendental.
We prove a special case, assuming SC.
Theorem 3.
Assume Schanuel’s Conjecture. Then, Gelfond’s Conjecture holds when α 1 = α 2 = = α m . More generally, if α C is algebraic but is not a rational integer, then the power tower of α of order m 3
m α : = α α · · · α m
is transcendental; in fact, the numbers log α , 3 α , 4 α , 5 α , are algebraically independent.
In the next section, we introduce some notation and prove a simple but useful lemma. The proofs of Theorems 1–3 are in Section 3, Section 4 and Section 5.

2. A Useful Lemma

Let us denote the maximal number of algebraically independent elements of a set { z 1 , , z k } C by
# Q ¯ { z 1 , , z k } .
We say that z 1 , , z k are algebraically dependent on { w 1 , , w } C if
Q ( z 1 , , z k ) Q ( w 1 , , w ) .
This condition is equivalent (see [13] Chapter VIII) to
Q ¯ ( z 1 , , z k ) Q ¯ ( w 1 , , w ) .
Lemma 1.
Assume SC. Suppose that α 1 , , α n C are Q -linearly independent, and the numbers α 1 , , α n , e α 1 , , e α n are algebraically dependent on the subset
{ β 1 , , β n } { α 1 , , α n , e α 1 , , e α n } .
Then, the numbers β 1 , , β n are all transcendental; in fact, they are algebraically independent.
Proof. 
The hypotheses imply
n # Q ¯ { β 1 , , β n } = # Q ¯ { α 1 , , α n , e α 1 , , e α n } n .
Hence, # Q ¯ { β 1 , , β n } = n , and the conclusion follows. □
It is Lemma 1 rather than SC that we use in the proofs of Theorems 1–3. Thus, Theorems 1–3 remain true under the weaker hypothesis in Lemma 1.

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 ) = x n , where n 1 . Since Q ( e ) is transcendental, 1 and n Q ( e ) are Q -linearly independent. Applying Lemma 1 to the subset
{ e , e n Q ( e ) } { 1 , n Q ( e ) , e , e n Q ( e ) } ,
it follows that e n Q ( e ) = P ( e ) Q ( e ) is transcendental, as claimed.
Now, assume P ( x ) x n 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 ) b e a 1 = 0 . If a 0 , then, since P ( x ) x n for any n 0 and e is not algebraic, the polynomial P ( x ) b x a 1 must be identically zero; hence, a = b = 0 . If a < 0 , then P ( x ) b x a = x a P ( x ) b x a 1 must be the zero polynomial, and again a = b = 0 .
Now, Lemma 1 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 Lemma 1 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 ) = x n .
Proof that P ( π ) Q ( π ) is transcendental. 
Note first that i π and log P ( π ) are Q -linearly independent because, otherwise, there would exist a Z -relation a i π + b log P ( π ) = 0 with b > 0 , and then P ( π ) b = ( 1 ) a would be algebraic, contradicting the transcendence of π .
Applying Lemma 1 to the subset
{ i π , log P ( π ) } { i π , log P ( π ) , 1 , P ( π ) } ,
we get that i π and log P ( π ) are algebraically independent. Then, the set { i π , log P ( π ) , Q ( π ) log P ( π ) } is Q -linearly independent, and Lemma 1 applied to this subset of { i π , 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 = α β , where α , β Q ¯ . Then, α 0 , 1 and β Q because π Q ¯ .
We claim that α is not a root of unity. To see this, assume α = e 2 k π i / n , with n 2 and k { 1 , 2 , , n 1 } . Taking logarithms on both sides of π = α β gives
log π 2 β k n i π = 0 .
However, this contradicts the algebraic independence of i π and log π , which we proved conditionally in Section 3.
It follows that i π and log α are Q -linearly independent, for otherwise there would be a Z -relation a i π + 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 Lemma 1 to the subset
{ i π , log α } { i π , log α , 1 , α }
and conclude that i π and log α are algebraically independent. Thus, in any Q -linear relation,
a i π + b log α + c β log α = 0 ,
one has a = 0 , and so b + c β = 0 . Since β is irrational, b = c = 0 . Thus, i π , log α , and β log α are Q -linearly independent.
However, then, since α , β Q ¯ , Lemma 1 applied to the subset
{ i π , log α , α β } { i π , log α , β log α , 1 , α , α β }
implies that π and α β are algebraically independent, contradicting π = α β . 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 e e , 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 , 3 α , 4 α , , m α are algebraically independent. The proof is in two cases.
Case 1: α Q Z . In this case, α α is irrational (see [2] Lemma 2.1). Thus, 1 and α α are Q -linearly independent, and then so are log α and α α log α . Since α and α α are algebraic, Lemma 1 applied to the subset
{ log α , α α α } { log α , α α log α , α , α α α }
yields the algebraic independence of log α and α α α = 3 α .
Now, fix m > 3 and assume inductively that log α , 3 α , 4 α , , m 1 α are algebraically independent. Then, in any Q -linear relation
j = 1 m 1 a j j α = 0
we must have a 3 = = a m 1 = 0 . Since 1 α = α Q Z and 2 α = α α Q , we also have a 1 = a 2 = 0 . This implies the Q -linear independence of
{ 1 α log α , , m 1 α log α } = { log ( 2 α ) , , log ( m α ) } .
Then, Lemma 1 yields the algebraic independence of the subset
{ log ( 2 α ) , 3 α , 4 α , , m α } { log ( 2 α ) , , log ( m α ) , 2 α , , m α }
and, since log ( 2 α ) = α log α , hence also that of log α , 3 α , 4 α , , m α . This completes the induction.
Case 2: α Q ¯ Q . By the Gelfond–Schneider Theorem, α α is transcendental. Hence, 1 , α , α α are Q -linearly independent, and then so are log α , α log α , α α log α . Since { α log α , α α log α } Q ¯ ( log α , α , α α ) and α is algebraic, Lemma 1 applied to the subset
{ log α , α α , α α α } { log α , α log α , α α log α , α , α α , α α α }
yields the algebraic independence of { log α , α α , α α α } = { log α , 2 α , 3 α } .
Suppose inductively that log α , 2 α , 3 α , , m 1 α are algebraically independent, where m > 3 . Then, any Q -linear relation
a 0 + j = 1 m 1 a j j α = 0
implies a 2 = = a m 1 = 0 . Since 1 α = α Q , we also get a 0 = a 1 = 0 . That implies the Q -linear independence of
{ log α , 1 α log α , , m 1 α log α } = { log α , log ( 2 α ) , , log ( m α ) } .
Since
{ 1 α log α , , m 1 α log α } Q ¯ ( log α , 1 α , 2 α , , m α ) ,
we may apply Lemma 1 to the subset
{ log α , 2 α , 3 α , , m α } { log ( 1 α ) , , log ( m α ) , 1 α , , m α }
and conclude that log α , 2 α , 3 α , , m α are algebraically independent.
Thus, in both Cases 1 and 2, the numbers log α , 3 α , 4 α , , m α are algebraically independent, as desired.
Results on the arithmetic nature of power towers of x of infinite order
x : = lim k k x = x x · · · e e x e 1 / e
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 P ( x ) , Q ( x ) Q ¯ [ x ] . The second one is that π cannot be written in the form α β , 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

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Trojovská, E.; Trojovský, P. Schanuel’s Conjecture and the Transcendence of Power Towers. Mathematics 2021, 9, 717. https://doi.org/10.3390/math9070717

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Trojovská E, Trojovský P. Schanuel’s Conjecture and the Transcendence of Power Towers. Mathematics. 2021; 9(7):717. https://doi.org/10.3390/math9070717

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Trojovská, Eva, and Pavel Trojovský. 2021. "Schanuel’s Conjecture and the Transcendence of Power Towers" Mathematics 9, no. 7: 717. https://doi.org/10.3390/math9070717

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