Algebraic independence of pi and e

Importance: High ✭✭✭
Subject: Number Theory
Recomm. for undergrads: no
Posted by: porton
on: July 8th, 2008
Conjecture   $ \pi $ and $ e $ are algebraically independent


* indicates original appearance(s) of problem.

Schanuel's conjecture

Assuming 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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