On his book "Introduction to transcendental numbers", page 99-100, Lang wrote
"Finally, we note that Lindemann actually proves something slightly stronger than the algebraic independenceof $e^{\alpha\_1},\cdots,e^{\alpha\_n}$ if $\alpha\_1,\cdots,\alpha\_n$ are linearly independent over $\mathbb Q$. He proved that if $\beta\_1,\cdots,\beta\_m$ are non-zero distinct algebraic numbers, then $e^{\beta\_1},\cdots,e^{\beta\_m}$ are linearly independent over $\mathbb Q$. This does not come out of the Siegel method"


I do not understand the latter is stronger than the former. Even, I thought it was the converse. I always thought about this kind of proof. Assume that the values $E(\beta\_1),\cdots,E(\beta\_m)$ of an $E$-function $E$ are $\overline{\mathbb Q}$-linearly dependant. Then there exists $\gamma\_1,\cdots,\gamma\_m$ in $\overline{\mathbb Q}$ not all zero such that $\gamma\_1E(\beta\_1)+\cdots+\gamma\_mE(\beta\_m)=0$. Denote by $G$ the Galois group of $\mathbb Q(\gamma\_1,\cdots,\gamma\_m)/\mathbb Q$. Then $\prod\_{\sigma\in G}\left(\sigma(\gamma\_1)E(\beta\_1)+\cdots+\sigma(\gamma\_m)E(\beta\_m)\right)=0$. The polynomial $\prod\_{\sigma\in G}(\sigma(\gamma\_1)X\_1+\cdots+\sigma(\gamma\_m)X\_m)$ has coefficients in $\mathbb Q$, is not zero and vanishes at $E(\gamma\_1),\cdots,E(\gamma\_m)$. It is impossible since the $E(\gamma\_i)$ are assumed to be algebraically independant over $\mathbb Q$.


I do not see where is my mistake. Any enlightenment would be welcome



Lang's remark on Lindemann-Weierstrass theorem
For brevity of notation, if $I = (i\_1,\ldots,i\_n) \in \mathbf N^n$, write $x^I = x\_1^{i\_1}\cdots x\_n^{i\_n}$. Write $\boldsymbol \alpha$ for $(\alpha\_1,\ldots,\alpha\_n)$, and set $I \cdot \boldsymbol \alpha = i\_1\alpha\_1 + \ldots + i\_n\alpha\_n$.


If $e^{\alpha\_1},\ldots,e^{\alpha\_n}$ are algebraically dependent over $\mathbf Q$, then there exists a polynomial
$$f = \sum\_{I \in \mathbf N^n} a\_I x^I \in \mathbf Q[x\_1,\ldots,x\_n]$$
such that $f(e^{\alpha\_1},\ldots,e^{\alpha\_n}) = 0$. This means that
$$\sum\_{I \in \mathbf N^n} a\_I e^{I \cdot \boldsymbol \alpha} = 0.$$
By Lindemann's result, this is only possible if there are $I, J \in \mathbf N^n$ with $I \neq J$ such that $a\_I, a\_J \neq 0$ and
$$i\_1\alpha\_1 + \ldots + i\_n \alpha\_n = j\_1\alpha\_1 + \ldots + j\_n \alpha\_n.$$
In particular the $\alpha\_1,\ldots,\alpha\_n$ are linearly dependent over $\mathbf Q$.


What's different in your argument is that you replaced $\mathbf Q$-linear dependence by $\bar{\mathbf Q}$-linear dependence.


