Algebraic independence of pi and e

Importance: High ✭✭✭
Subject: Number Theory
Recomm. for undergrads: no
Posted by: porton
on: July 8th, 2008

\begin{conjecture} $\pi$ and $e$ are \Def[algebraically independent]{Algebraic_independence} \end{conjecture}

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Schanuel's conjecture

Assuming \Def[Schanuel's conjecture]{Schanuel's conjecture}, one can show that $\pi$ and $e$ are algebraically independent over $\mathbb Q$.

in which subfield K of which field L?

After all, e to the pi i = -1, so this shows that pi and e are not always algebraically independent.

By definition?

I think any two distinct transcendental numbers must be algebraically independent, almost by definition. Since e and pi are transcendental, they must be a. i. No? - David Spector

not all transcedentials are algebraically independant

pi and 4-pi are both transcedential and sum to 4, so are not algebraically independant.

two transcendentals are not necessarily algebraically independen

e and e^2 are both transcendental but (e,e^2) makes the two-variable polynomial f(x,y)=x^2-y equal to zero


but that's not a polynomial.

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