AWSLecture3.pdf and

contents.dvi - Waldschmidt1.pdf and (duplicate?)

contents.dvi - SurveyTrdceEllipt2006.pdf -- Elliptic Functions and Transcendence by Michel Waldschmidt

He starts out by reviewing some transcendence results associated with the exponential function and its inverse, the logarithmic function.

- Hermite 1873: e is transcendental.
- Lindemann 1882: pi is transcendental.
- Hermite-Lindemann 1882: the nonzero logarithm of any nonzero algebraic number is transcendental.
- Hermite-Lindemann 1882: the exponential of any nonzero algebraic number is transcendental.
- Lindemann–Weierstrass 1885: for a1, ..., an algebraic numbers that are linearly independent over the rational numbers, e^a1, ..., e^an are algebraically independent.
- Lindemann–Weierstrass 1885: for a1, ..., an distinct algebraic numbers, e^a1, ..., e^an are linearly independent over the algebraic numbers.
- Gel'fond–Schneider 1934: for a a nonzero algebraic number and b an irrational algebraic number, then a^b = exp(b*log(a)) is transcendental.
- Gel'fond–Schneider 1934: for algebraic numbers a1, a2 with nonzero logarithms, and for log(a1)/log(a2) irrational, then that ratio is transcendental.
- Baker 1966: let log(a1), ..., log(an) be logarithms of algebraic numbers that are linearly independent over the rational numbers. Then 1, log(a1), ..., log(an) are linearly independent over the algebraic numbers.

Then the Six Exponentials Theorem.

Consider n complex numbers x1, ..., xn that are linearly independent in the rational numbers, and presumably all nonzero, and m complex numbers y1, ..., ym also with that property. If m*n > m + n, then at least one of e^(xi*yj) is transcendental.

There is a Four Exponentials Conjecture which states that that is true for m = n = 2 and thus that m*n = m + n. So let us consider m*n >= m + n.

Let us see what (m,n) values are possible. Let n >= m for convenience. Try m = 1 first. n >= n + 1 not possible. So m >= 2. That means n >= m/(m-1). The right-hand side is only an integer if m = 2, and that gives n >= 2. Thus, the only positive-integer solution of m*n = m + n is m = n = 2.

n >= m/(m-1) gives us n >= 1 + 1/(m-1) <= 2. Thus, for m*n > m + n, m = 2 gives us n >= 3 and m and n can both be >= 3.

The six comes from the smallest number that makes at least one transcendental number: n = 3, m = 2, n*m = 6.

The four comes from that conjectured number: n = m = 2, n*m = 4.

For m = 1, I will show that there is a case where all the e^(x*y) are algebraic. Set y = 1 and set the x's to natural logarithms of prime numbers: 2, 3, 5, 7, 11, 13, 17, 19... this means that their natural logarithms are linearly independent over the rational numbers, since none of them is a factor of any other of them. By Baker's theorem, 1 and their logs are linearly independent over the algebraic numbers, meaning that the logs are all transcendental numbers.

But the e^(x*y) = e^(log(p)*1) = p -- algebraic. So there is a case of all-algebraic for m = 1.