Q: Sum of algebraicly independent transcendentals is transcendental? We say the transcendental numbers $\xi_1,...,\xi_n$ are algebraicly independent if  an algebraic combination of them satisfies:
$$
\sum_{1\leq i\leq n}\psi_i\xi_i= 0 \iff \psi_i=0, \forall i=1,...,n.
$$
My question is, if $\xi,\gamma$ are algebraicly independent transcendental numbers, so is $\xi+\gamma$?

A: If we take $e$ and $e-1$, we get that $\xi_1 e+\xi_2 (e-1)=0$ implies that $e=\xi_2/(\xi_1+\xi_2)$, so that we must have $\xi_1+\xi_2=0$, but then we have $-\xi_2=0$, so indeed they are independent. But we have that $e+(e-1)=1$ is not transcendental.
