We know that for algebraic $e^{it\pi}$, $t$ can not be algebraic irrational by Baker's theorem, but his proof is analytic; is there some algebraic understanding for such fact? If $t$ is transcendent, can $t$ be a period? We say $t$ is period, iff $t$ can be expressed as the volume of domain in $R^{n} defined by polynomial inequalities with algebraic coefficients.
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$\begingroup$ What do you mean by "can $t$ be [a] period?" $\endgroup$– Noah SchweberCommented Mar 11, 2013 at 2:09
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$\begingroup$ (I fixed some grammar/spelling; I hope you don't mind.) $\endgroup$– Noah SchweberCommented Mar 11, 2013 at 2:09
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$\begingroup$ Noah S: Presumably this means "Can t be an element of the Ring of Periods?" I think the question is clear as written. See en.wikipedia.org/wiki/Periods_%28ring%29 . $\endgroup$– Steven LandsburgCommented Mar 11, 2013 at 4:01
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2$\begingroup$ The number $\pi$, the exponential function, a period... are not objects from Algebra, but from Analysis. So an algebraic proof would require to extract from these beasts their relevant properties and use them in an algebraic manner. However, the conjectures about periods by Grothendieck and Kontsevich-Zagier give a motivic interpretation of Baker's theorem and of (almost) all conjectural statements in Transcendental Number Theory. $\endgroup$– ACLCommented Mar 11, 2013 at 6:58
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The heart of all transcendence proofs is the analytic fact that there is no integer between $0$ and $1$. This is one of the things I learned from Baker's book "Transcendental Number Theory".
answered Mar 11, 2013 at 2:49