Assuming Schanuel's conjecture, one can show that and are algebraically independent over .
After all, e to the pi i = -1, so this shows that pi and e are not always algebraically independent.
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
pi and 4-pi are both transcedential and sum to 4, so are not algebraically independant.
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.