This article on Wikipedia about the Lindemann-Weierstrass theorem mentions two equivalent formulations, and says that they are equivalent by an argument about symmetric polynomials. But I cannot see what the argument should be.
More in general, I would like to know:
What is the relation between algebraic independence over $\mathbb{Q}$ and linear independence over $\overline{\mathbb{Q}}$ for a finite set $\{\alpha_1, \dots,\alpha_n\}$ of complex numbers? Does any of the two properties imply the other?
It's clear that for $n=1$ the first property (being a transcendental number) implies the second (being a nonzero complex number), but for $n>1$ I cannot see why any of them should imply the other.
Best Answer
If the set $\{\alpha_1,\dots,\alpha_n\}$ is algebraically independent over $\mathbb{Q}$ then it is also algebraically independent over $\bar{\mathbb{Q}}$.
Otherwise we can assume, without loss of generality, that $\alpha_n$ is algebraic over $\bar{\mathbb{Q}}(\alpha_1,\dots,\alpha_{n-1})$.
Let $f(X)$ be a nonzero polynomial in $\bar{\mathbb{Q}}(\alpha_1,\dots,\alpha_{n-1})[X]$, $f=p_0+p_1X+\dots+p_kX^k$, where $p_i\in\bar{\mathbb{Q}}[\alpha_1,\dots,\alpha_{n-1}]$, such that $f(\alpha_n)=0$ (take the minimal polynomial and clear the denominators). Let $S$ be the set of the coefficients of the monomials in $\alpha_1,\dots,\alpha_{n-1}$ appearing in $p_i$, for $i=1,2,\dots,k$. Then $S$ is a finite set of elements algebraic over $\mathbb{Q}$, so $\alpha_n$ is algebraic over $\mathbb{Q}(S)(\alpha_1,\dots,\alpha_{n-1})=\mathbb{Q}(\alpha_1,\dots,\alpha_{n-1})(S)$, which is finite dimensional over $\mathbb{Q}(\alpha_1,\dots,\alpha_{n-1})$. Hence $\alpha_n$ is algebraic over $\mathbb{Q}(\alpha_1,\dots,\alpha_{n-1})$, a contradiction.
Algebraic independence implies linear independence. The converse is not true: $\pi$ and $\pi^2$ are linearly independent over $\mathbb{Q}$ (or any algebraic extension of $\mathbb{Q}$), but not algebraically independent.
This generalizes verbatim to the case where we have fields $F\subseteq K\subseteq L$, with $K$ algebraic over $F$. A subset $\{\alpha_1,\dots,\alpha_n\}$ of $L$ is algebraically independent over $K$ if and only if it is algebraically independent over $F$.