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I am reading Lindemann–Weierstrass theorem, which is:

— if $α_1, ..., α_n$ are algebraic numbers which are linearly independent over the rational numbers ℚ, then $e^{α_1}, ..., e^{α_n}$ are algebraically independent over ℚ;

in other words the extension field ℚ($e^{α_1}, ..., e^{α_n}$) has transcendence degree n over ℚ

This is a definition about algebraically independence stated on Wikipedia:

In abstract algebra, a subset S of a field L is algebraically independent over a subfield K if the elements of S do not satisfy any non-trivial polynomial equation with coefficients in K.
In particular, a one element set {α} is algebraically independent over K if and only if α is transcendental over K. In general, all the elements of an algebraically independent set S over K are by necessity transcendental over K, and over all of the field extensions over K generated by the remaining elements of S.

So, now there is some ambiguity for me in the thm that I need them to be cleared. Here I am okay with the notion of algebraically independence, but I don't know what is meant in Lindemann-Weierstrass thm by mentioning about the linear dependence of numbers. What is meant by saying "$α_1, ..., α_n$, that are algebraic numbers, are linearly independent over the rational numbers ℚ" ? And can I conclude that since $e^{α_1}, ..., e^{α_n}$ are algebraically independent over ℚ, $e^{α_1}, ..., e^{α_n}$ are transcendental numbers?

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Linearly independent over the rational numbers means that if $$ q_1\alpha_1+\ldots + q_n\alpha_n = 0$$ for $q_1,\ldots q_n\in \mathbb Q,$ then $q_1=q_2=\ldots = q_n = 0.$

A common special case is when $n=1.$ Then, $\{\alpha_1\}$ being linearly independent over the rationals has no content other than that $\alpha_1\ne 0$ (so the only operative assumption is that $\alpha_1$ is algebraic). In this case $\{e^{\alpha_1}\}$ being algebraically independent over $\mathbb Q$ means exactly that $e^{\alpha_1}$ is transcendental. So this special case can be rephrased: if $\alpha$ is a nonzero algebraic number then $e^{\alpha}$ is transcendental.

In the case where $n$ is not necessarily equal to one, $\{e^{\alpha_1}, \ldots, e^{\alpha_n}\}$ being algebraically independent over $\mathbb Q$ implies that all the $e^{\alpha_i}$ are transcendental, but is stronger than that since it means that there is no nonzero polynomial $f(y_1,\ldots y_n)$ with coefficients in $\mathbb Q$ such that $f(e^{\alpha_1},\ldots, e^{\alpha_n}) = 0.$ That $e^{\alpha_i}$ is transcendental follows from the special case where the polynomial only depends on $y_i$ and not any of the other $y$'s.

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  • $\begingroup$ I checked the proofs of the L-W theorem on some websites and I can say I couldn't understand the intuition behind the theorem, actually they were the proofs of the reformulations of L-W theorem, not the proof of the actual theorem I wrote above. Also, I have just read this: "if $\alpha$ is a nonzero complex number and $e^\alpha$ is algebraic, then $\alpha$ must be transcendental" How does this one hold? @spaceisdarkgreen $\endgroup$ – Leyla Alkan Jan 1 '18 at 6:46
  • $\begingroup$ It is just the contrapositive of the theorem. $\endgroup$ – spaceisdarkgreen Jan 1 '18 at 7:41
  • $\begingroup$ Oh yes of course! $\endgroup$ – Leyla Alkan Jan 1 '18 at 8:04

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