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?


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.

| cite | improve this answer | |
  • $\begingroup$ 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 $\endgroup$ – nojdar 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! $\endgroup$ – nojdar Jan 1 '18 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.