Hilbert's seventh problem. Gelfond, Schneider


Hilbert's seventh problem (after a preamble) asked for a proof that ( any value of) alpha^beta (for algebraic alpha <> 0, 1 , and irrational algebraic beta ), e.g. 2^sqrt(2) or e^Pi = i^(-2*i) , is transcendental (or at least an irrational number). Hilbert wrote: " It is certain that the solution of these and similar problems must lead us to entirely new methods and to a new insight into the nature of special irrational and transcendental numbers."

> evalc(sqrt(-1)^(-2*sqrt(-1)));


> evalc((-1)^(-sqrt(-1))); # alternatively


> evalc(I^(-2*I)); # alternatively



Although Hilbert didn't specifically refer to (what I earlier called) Euler's surmise, he must have had it in mind...


From Constance Reid's biography of Hilbert (p.164) I quote: "Siegel came to Göttingen as a student in 1919... he was always to remember a lecture on number theory which he heard from Hilbert at this time. Hilbert wanted to give his listeners examples of the characteristic problems of the theory of numbers which seem at first glance so very simple but turn out to be incredibly difficult to solve. He mentioned Riemann's hypothesis, Fermat's [Last] theorem, and the transcendence of 2^sqrt(2) as examples of this type of problem. Then he went on to say that there had recently been much progress on Riemann's hypothesis and he was very hopeful that he would live to see it proved. Fermat's problem had been around for a very long time and apparently demanded entirely new methods for its solution - perhaps the youngest members of his audience would live to see it solved. But as for establishing the transcendence of 2^sqrt(2) no one present in the lecture hall would live to see that!

The first two problems which Hilbert mentioned are still unsolved [JC comment. As everyone knows, Fermat's Last Theorem has been settled by Andrew Wiles]. But less than ten years later a young Russian mathematician named Gelfond established the transcendence of 2^sqrt(-2) . Utilising this work, Siegel himself was shortly able to establish the desired transcendence of 2^sqrt(2) .

Siegel wrote to Hilbert about the proof. He reminded him of what he had said in his 1920 lecture and emphasised that the important work was that of Gelfond. Hilbert was frequently criticized for "acting as if everything had been done in Göttingen." Now he responded with enthusiastic delight to Siegel's letter, but the made no mention of the young Russian's contribution. He wanted only to publish Siegel's solution. Siegel refused, certain that Gelfond himself would eventually solve this problem too. Hilbert immediately lost all interest in the matter."

As it happened, Gelfond didn't find a solution himself, and another Russian mathematician - Kuzmin - published a solution the following year. In 1933, K. Boehle, using Gelfond's method did manage to prove that if beta is algebraic of degree n (at least 3), then at least one of the numbers alpha^beta, alpha^(beta^2), `...`, alpha^(beta^(n-1... is transcendental, and then in 1934, with some justice, Gelfond himself published a complete solution (April 1st 1934?) to the entire seventh problem of Hilbert. Within six weeks, a PhD student of Siegel's - the great Schneider - found another independent solution, which he submitted on May 28th 1934, and was published later that year.

Gelfond's (1929) Theorem. Let alpha be any algebraic number ( alpha <> 0, 1 ) and let beta be any imaginary quadratic (i.e. a solution of a quadratic equation with negative discriminant), then ( any value of) alpha^beta is transcendental.

I believe it is possible for a novice to follow at least the gist of Gelfond's 1929 proof (if I cut some big corners, and hand wave just a little). It is helpful, I believe, if one follows the proof of the following very beautiful result (though Analysis folk would probably regard it as being rather trivial):

Theorem (Polya 1916). Let f( z ) be an entire function of z such that f(0), f(1), f(2), ... are all integers, and suppose

f( z ) grows slower than 2^z (in the sense that abs(f(z)) <= C*2^(A*abs(z)) , where A and C and are constants, and A < 1 ), then f( z ) is a polynomial [it's a trivial exercise that the polynomial f( z ) must have rational coefficients].

Recall that an entire function f( z ) of the complex variable z , is one that is defined for all z , and is differentiable everywhere (f '( z ) exists). Simple examples are

  • all polynomials (of finite degree)
    z*(z+1)/2 = z^2/2+z/2 , is an integer for z = 0, 1, 2, 3, ...

    z*(z+1)/3 = z^2/3+z/3 , is an integer for some, but not all z = 0, 1, 2, 3, ...
  • sin(z), cos(z), e^z, 2^z, 3^z, (2^z+3^z)/2, `...`
    sin(0) = 0, but none of sin(1), sin(2), sin(3), ... is an integer

(sketch of) Proof of Polya's theorem. The key is to expand f( z ) as a Newton interpolation series (which is different from the usual Taylor series) at the points of interpolation 0, 1, 2, 3, ... :

f(z) = A[0]+A[1]*(z-1)+A[2]*z*(z-1)+`...`+A[n]*z*(z...

Now, express the coefficient A[n] as a contour integral around C[n] (where C[n] is (say) any circle, containing all the points z = 0, 1, 2, ..., n ), and described in the positive sense ):

A[n] = 1/(2*Pi*i) int(f(z)/(z*(z-1)*`...`*(z-n)),z)

Then, by the residue theorem , we have

A[n] = the residue of f(z)/(z*(z-1)*`...`*(z-n)) at z = 0 + the residue of f(z)/(z*(z-1)*`...`*(z-n)) at z = 1

+ ... + the residue of f(z)/(z*(z-1)*`...`*(z-n)) at z = n

= f(0)/((0-1)*(0-2)*`...`*(0-n)) + f(1)/(1*(1-2)*(1-3)*`...`*(1-n)) + f(2)/(2*(2-1)*(2-3)*`...`*(2-n))

+ ... + f(n)/(n*(n-1)*(n-2)*`...`*(n-(n-1))) (all the poles are simple )

Since all the f(0), f(1), ... , f( n ) are integers then it is clear that n!*A[n] is an integer (think about it).

Now choose C[n] to be the circle abs(z) = 2*n and proceed to estimate the integrand on C[n] . It emerges (here I cut a corner, but I point out that one uses abs(f(z)) <= C*2^(A*2*n) ) that

abs(A[n]) <= 1/(2*Pi) . 2*Pi*2*n*C*2^(A*2*n)/(2*n*(2*n-1)*`...`*(n+1)*n) , giving

abs(n!*A[n]) <= n!*2*n*C*2^(A*2*n)/(2*n*(2*n-1)*`..... = 2*n!^2*C*2^(A*2*n)/(2*n)!

and it's an easy school exercise that %? as proc (n) options operator, arrow; infinity end proc... . (Hint. Use the ratio test, and note the relevance of the condition that A is less than 1). Thus n!*A[n] must be 0 for sufficiently large n . In short, returning to the Newton interpolation series for f( z ), one sees that f( z ) is a polynomial (a lovely proof, I hope you agree. Hardy, and others, gave refinements. E.g. it was possible to improve the growth restriction to abs(f(z)) = o(2^abs(z)) as proc (z) options operator, arrow; infinity end proc... , meaning %? as proc (z) options operator, arrow; infinity end proc... .) [end] I hope you can now follow the gist of:

Gelfond's Theorem (1929). e^Pi [understood to be a value of (-1)^(-sqrt(-1)) ] is transcendental. Actually the full version of his 1929 theorem

Sketch proof . Suppose gamma = e^Pi = (-1)^(-sqrt(-1)) is algebraic. Then (the first idea) the entire function e^(Pi*z) would have an algebraic value for every z = a+i*b , where a and b are any two (ordinary) integers `...`, -2, -1, 0, 1, 2, 3, `...`

(Those a+i*b are known as the Gaussian integers .)

For z = 3+4*i , e^(Pi*z) would have value:

e^(Pi*(3+4*i)) = (-1)^(-i*(3+4*i)) = (-1)^(-3*i)*(-1)^(-i*4*i) = gamma^3*(-1)^4 = gamma^3 , is algebraic.

For z = -4-5*i , e^(Pi*z) would have value:

e^(Pi*(-4-5*i)) = (-1)^(-i*(-4-5*i)) = (-1)^(4*i)*(-1)^(i*5*i) = gamma^(-4)*(-1)^5 = -gamma^(-4) , is also algebraic:

> evalc((-1)^(-sqrt(-1)));


> evalc(((-1)^(-sqrt(-1)))^(3 + 4*sqrt(-1)));


> evalc(((-1)^(-sqrt(-1)))^(-4 - 5*sqrt(-1)));



Arrange the Gaussian integers - namely all complex numbers of the form a+i*b , where a and b vary over the (ordinary) integers `...`, -3, -2, -1, 0, 1, 2, 3, `...` ) - in a sequence z[0], z[1], z[2], `...` according to increasing modulus , and where several have the same modulus, by increasing argument , and then expand the (entire) function e^(Pi*z) as a Newton interpolation series with the above points of interpolation:

e^(Pi*z) = A[0]+A[1]*(z-z[0])*(z-z[1])+A[2]*(z-z[0]...

Let C[n] be a circle centered at the origin containing the points z[0], z[1], `...`, z[n] , then

A[n] = 1/(2*Pi*i) int(f(z)/((z-z[0])*(z-z[1])*`...`*(z-z[n])),z)

= f(z[0])/((z[0]-z[1])*(z[0]-z[2])*`...`*(z[0]-z[n]))... + f(z[1])/((z[1]-z[0])*(z[1]-z[2])*`...`*(z[1]-z[n]))... + ... + f(z[n])/((z[n]-z[0])*`...`*(z[n]-z[n-1]))

Note that

  • all those numerators are algebraic numbers
  • all those denominators are Gaussian integers
  • thus every term in that sum is an algebraic number
  • thus A[n] is an algebraic number

By choosing the circle C[n] to be abs(z) = n

  • it can be argued that A[n] is at most e^(-n*log(n)+O(n)) , where "O( n )" means a function of n whose size is at most a constant times n , whereas
  • it can also be argued (and here I have to cut a very large corner ) that that whenever A[n] is not zero it must be at least e^(-n*log(n)/2+O(n))

Those minimum and maximum values are incompatible for large n , and it follows that A[n] is eventually zero. Thus e^(Pi*z) is a polynomial, of finite degree (as a consequence of assuming that e^Pi is algebraic). But e^(Pi*z) is most definitely not a polynomial of finite degree. Thus e^Pi is transcendental. [end]

Suffice it to say that complete proofs (Gelfond's or Schneider's) of the full version of Hilbert's seventh problem are really quite difficult, and have no place here. I will only remark that both proofs make use - as does the one above - of functions of a complex variable. An interested reader will find a very good exposition in Niven's MAA classic Irrational Numbers . Also a reader might ask: is it possible to give a proof that avoids the use of complex variable methods in the real case of Hilbert's seventh problem. In other words, if alpha ( 0 < alpha , alpha <> 1 ) and beta are real algebraic numbers, beta irrational, can alpha^beta (namely e^(beta*log[e](alpha)) ) be proved to be transcendental by using arguments not involving complex variable methods? Gelfond himself gave such a proof, which formed an entire chapter in the great Gelfond and Linnik classic, as I've already mentioned in an earlier section. Incidentally Gelfond also gave there an entirely elementary proof of the transcendence of e^alpha for non-zero real algebraic alpha .

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This page was last updated 18 February 2005 15:10:00 -0000