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I just read an article of Ram Murty about transcendence of special values of L-functions, and it seems that Schanuel's conjecture plays a crucial role in it. So given a positive integer $n$, let's define the set $\mathcal{L}_{n}$ in the following way: $A$ is an element of $\mathcal{L}_{n}$ if and only if $A$ is a set of $n$ complex numbers linearly independent over $\mathbb{Q}$. Now let's denote $E(A)$ the set $\{e^{\alpha_{i}},\alpha_{i}\in A\}$.
My question is: Is it true that there exists a positive integer $k_n$ depending only on $n$ such that for all $\sigma\in\frak{S}(\mathcal{L}_{n})$ the transcendence degree of $\mathbb{Q}(\sigma(A),E(\sigma(A)))$ is $k_n$? If so, proving Schanuel's conjecture would pertain to showing that $k_n\geqslant n$.
Thanks in advance.

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I feel I must have misunderstood the question: I thought $\frak{S}(\mathcal{L}_n)$ denoted the symmetry group of the (infinite) set $\mathcal{L}_n$, where, unless I've further misunderstood, the statement is not true. Could you explain the symbolism? – Todd Trimble Aug 3 '13 at 12:48
I think you perfectly understood the question. As you say the statement is not true, would you have a counter-example that shows it? – Sylvain JULIEN Aug 3 '13 at 12:52
Well, just to make sure, let me ask this: What is the difference between this question and asking whether $\mathbb{Q}(A, E(A))$ has the same transcendence degree $k_n$ as $\mathbb{Q}(B, E(B))$, for any $A, B \in \mathcal{L}_n$? Because there is always a $\sigma$ in the symmetric group that takes one element $A$ to another $B$. – Todd Trimble Aug 3 '13 at 13:55
Anyway, what I'm leading up to can be explained for the case $\mathcal{L}_2$: it should almost always be true (in the sense of measure) that for $\alpha, \beta \in \mathbb{C}$, we have $\alpha, \beta, e^\alpha, e^\beta$ are algebraically independent (thus yielding transcendence degree 4) whereas for some $\alpha, \beta$ (say $\sqrt{2}$ and $\sqrt{3}$) the transcendence degree is at most 2. So you get different transcendence degrees for different $A, B \in \mathcal{L}_n$. – Todd Trimble Aug 3 '13 at 14:01
Ok, I got it. Thank you for explaining why my statement can't be true. – Sylvain JULIEN Aug 3 '13 at 16:18

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