Algebraic independence of a family of numbers

Question

I need to show that given $a_1,\cdots,a_n\in\mathbb{C}$ algebraic numbers linearly independent over $\mathbb{Q}$, then the numbers $e^{a_1},\cdots,e^{a_n}$ are algebraically independent over rationals.

I know i have to apply Lindemann-Weierstrass at some point, but i don't know where to start. Any hint will be apreciated.

Edit: My notes state that the Lindemann-Weierstrass theorem says that for $a_1,\cdots,a_n\in\mathbb{C}$ algebraic numbers different two by two, adn $c_1,\cdots,c_n\in\mathbb{C}$ are algebraic numbers not all equal to zero then $\sum_{k=0}^nc_ke^{a_k}\not=0$

Answer 1

I will need a previous lemma for the solution:

Lemma: Let $L|K$ an extension and $a_1,\cdots,a_n\in L$. Let $K_0:=K$ and $K_i:=K(a_1,\cdots,a_i)$ for $1\leq i\leq n$. Then the elements $a_1,\cdots,a_n$ are algebraically independent over $K$ and $a_i$ is trascendental over $K_{i-1}$.

Now, take $w_k=e^{a_k}$ for $1\leq k\leq n$, and suppose that $w_1,\cdots,w_n$ are algebraically dependent over rationals. Without loss of generality we can suppose, by the previous lemma, that $w_n$ is algebraic over $\mathbb{Q}(w_1,\cdots,w_{n-1})$, so there exist some polynomials $g_0,\cdots,g_m\in\mathbb{Q}[x_1,\cdots,x_n]$ such as $$g_m(w_1,\cdots,w_{n-1})w_n^m+g_{m-1}(w_1,\cdots,w_{n-1})w_n^{m-1}+\cdots+g_(w_1,\cdots,w_{n-1})=0\;(1)$$ Lets take $g_k:=\sum_{0\leq j_l\leq d_{m,l}}a_{j_1,\cdots,j_{n-1},k}\;x_1^{j_1}\cdots x_{n-1}^{j_{n-1}}$, and we evaluate those polynomials in $x_l:=w_l=e^{a_l}$. So the equality $(1)$ is rewriten as:

$$\sum_{0\leq j_l\leq d_{m,l}}a_{j_1,\cdots,j_{n-1},m}(e^{a_1})^{j_1}\cdots(e^{a_{n-1}})^{j_{n-1}}e^{ma_n}+\cdots+\sum_{0\leq j_l\leq d_{0,l}}a_{j_1,\cdots,j_{n-1},0}(e^{a_1})^{j_1}\cdots(e^{a_{n-1}})^{j_{n-1}}=0$$ In consequence: $$\sum_{0\leq j_l\leq d_{m,l}}a_{j_1,\cdots,j_{n-1},m}\;e^{j_1a_1+\cdots j_{n-1}a_{n-1}+ma_n}+\cdots+\sum_{0\leq j_l\leq d_{0,l}}a_{j_1,\cdots,j_{n-1},0}\;e^{j_1a_1+\cdots j_{n-1}a_{n-1}}=0$$ By the linear independence of $a_1,\cdots,a_n$ over the rationals, we have that the exponents $$\{j_1a_1+\cdots+j_{n-1}a_{n-1}+ka_n\;:\;0\leq k\leq m,\;0\leq j_l\leq d_{k,l}\}$$ are different two by two. Hence, by the Lindemann-Weierstrass theorem (yes, your intuition was right), each coefficient $a_{j_1,\cdots,j_{n-1},k}=0$, hence $g_0=0,\cdots,g_m=0$. Contradiction!

Answer 2

You probably prove this directly using the other version of the Lindemann-Weierstrass theorem.

Consider a polynomial in $n$ variable with integral coefficients $$P(x_1, \ldots, x_n) = \sum c_{i_1 \ldots i_n} x_1^{i_1} \cdots x_n^{i_n}$$

Then

$$P(e^{a_1}, \ldots, e^{a_n}) = \sum c_{i_1 \ldots i_n} e^{i_1 a_1 + \cdots + i_n a_n}$$

Now, the number $a_1$, $\ldots$, $a_n$ are linearly independent over $\mathbb{Q}$. It follows that the numbers $$A_I \colon = A_{i_1, \ldots, i_n} = i_1 a_1 + \cdots + i_n a_n$$ are all distinct. Moreover, since the $a_l$'s are algebraic, and $i_l$'s are integers, we get $A_I$ algebraic. So now

$$P(e^{a_1}, \ldots, a_n) = \sum_{I} c_I e^{A_I}$$ and we can apply your version of L-W.