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It is an open problem to prove that $\pi$ and $e$ are algebraically independent (over $\mathbb{Q}$).

  • What are some of the important results leading toward proving this?
  • What are the most promising theories and approaches for this problem?
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up vote 16 down vote accepted

People in model theory are currently studying the complex numbers with exponentiation. Z'ilber has an axiomatisation of an exponential field (field with exponential function) that looks like the complex numbers with exp. but satisfies Schanuel's conjecture. He proved that there is exactly one such field of the size of $\mathbb C$. I would find it odd if Z'ilber's field turned out to be different from the complex numbers.

By results of Wilkie, the reals with exponentiation are well understood, and the complex numbers with exponentiation is in some way the next step up. The model theoretic frame work (o-minimality) that works for the reals with exp. fails for the complex numbers, but there might be a similar theory that works for the complex field with exponentiation.

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But is such an approach able (theoretically) to skirt the Hilbert's 10th barrier? Namely, if you really had the complex exponential, then you'd have the rationals defined as the set of quotients $q_1 / q_2$ of numbers which satisfy $e^{q_1} = e^{q_2} = 1$ and $q_2 \neq 0$. From this, it's a short skip to defining integers and ending up with undecidability, I thought. At the very least, it's far from o-minimal. So what does Zilber do to get around this? Or does he get around this at all? – Marty Jul 30 '10 at 0:33
The structure is not o-minimal. First of all, it lacks order. Moreover, there are definable sets that are more complicated then anything made up from finitely many very simple ingredients. Your argument for instance gives countably infinite discrete definable sets. A promising theory would have to allow for definable sets that are countable unions of basic objects (or complements of such) and tell us something about the structure, even if the theory is undecidable. And it might be possible to show that Z'ilber's field is isomorphic to the complex numbers. – Stefan Geschke 0 – Stefan Geschke Jul 30 '10 at 13:09

Schanuel's conjecture would imply this result. It states that if $z_1, \ldots, z_n$ are linearly independent over $\mathbb{Q}$, then $\mathbb{Q}(z_1, \ldots, z_n, e^{z_1}, \ldots, e^{z_n})$ has transcendence degree at least $n$ over $\mathbb{Q}$. In particular, if we take $z_1 = 1$, $z_2 = \pi i$, then Schanuel's conjecture would imply that $\mathbb{Q}(1, \pi i, e, -1) = \mathbb{Q}(e, \pi i)$ has transcendence degree 2 over $\mathbb{Q}$.

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This is the conjecture I am asking about. – muad Jul 29 '10 at 18:58

There is a proof of the algebraic independence of $\pi$ and $e^\pi$ in Introduction to Algebraic Independence Theory and a detailed exposition of methods created in last the 25 years although I have not read it.

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The origin of this result is published in a shorter paper [Yu.V. Nesterenko, Modular functions and transcendence questions, Sb. Math. 187 (1996), no. 9, 1319--1348] ( What is in fact known is that $\pi$, $e^\pi$ and $\Gamma(1/4)$ are algebraically independent. – Wadim Zudilin Jul 29 '10 at 23:41

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