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This article on Wikipedia about the Lindemann-Weierstrass theorem mentions two equivalent formulations, and says that they are equivalent by an argument about symmetric polynomials. But I cannot see what the argument should be.

More in general, I would like to know:

What is the relation between algebraic independence over $\mathbb{Q}$ and linear independence over $\overline{\mathbb{Q}}$ for a finite set $\{\alpha_1, \dots,\alpha_n\}$ of complex numbers? Does any of the two properties imply the other?

It's clear that for $n=1$ the first property (being a transcendental number) implies the second (being a nonzero complex number), but for $n>1$ I cannot see why any of them should imply the other.

57Jimmy
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1 Answers1

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If the set $\{\alpha_1,\dots,\alpha_n\}$ is algebraically independent over $\mathbb{Q}$ then it is also algebraically independent over $\bar{\mathbb{Q}}$.

Otherwise we can assume, without loss of generality, that $\alpha_n$ is algebraic over $\bar{\mathbb{Q}}(\alpha_1,\dots,\alpha_{n-1})$.

Let $f(X)$ be a nonzero polynomial in $\bar{\mathbb{Q}}(\alpha_1,\dots,\alpha_{n-1})[X]$, $f=p_0+p_1X+\dots+p_kX^k$, where $p_i\in\bar{\mathbb{Q}}[\alpha_1,\dots,\alpha_{n-1}]$, such that $f(\alpha_n)=0$ (take the minimal polynomial and clear the denominators). Let $S$ be the set of the coefficients of the monomials in $\alpha_1,\dots,\alpha_{n-1}$ appearing in $p_i$, for $i=1,2,\dots,k$. Then $S$ is a finite set of elements algebraic over $\mathbb{Q}$, so $\alpha_n$ is algebraic over $\mathbb{Q}(S)(\alpha_1,\dots,\alpha_{n-1})=\mathbb{Q}(\alpha_1,\dots,\alpha_{n-1})(S)$, which is finite dimensional over $\mathbb{Q}(\alpha_1,\dots,\alpha_{n-1})$. Hence $\alpha_n$ is algebraic over $\mathbb{Q}(\alpha_1,\dots,\alpha_{n-1})$, a contradiction.

Algebraic independence implies linear independence. The converse is not true: $\pi$ and $\pi^2$ are linearly independent over $\mathbb{Q}$ (or any algebraic extension of $\mathbb{Q}$), but not algebraically independent.

This generalizes verbatim to the case where we have fields $F\subseteq K\subseteq L$, with $K$ algebraic over $F$. A subset $\{\alpha_1,\dots,\alpha_n\}$ of $L$ is algebraically independent over $K$ if and only if it is algebraically independent over $F$.

egreg
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  • Thanks for your answer. So in general if $M/L/K$ is a tower of field extensions with $L/K$ algebraic, the same argument shows that for a subset $S\subseteq M$ to be algebraically independent over $L$ is the same as being algebraically independent over $K$, right? Which explains why the transcendence degree is invariant under taking algebraic extensions, right? – 57Jimmy Feb 05 '19 at 16:41
  • @57Jimmy Yes, that's it. – egreg Feb 05 '19 at 16:49
  • @egreg hello why is $\mathbb{Q}(a_1,...,a_n)(S)$ finite dimensional over $\mathbb{Q}(a_1,...,a_n)$. I know that if the $p_i$ are algebraic than it would be true but how do you show that. Im assuming $S={p_1,p_2,...,p_k}$ – Jhon Doe Feb 13 '22 at 16:44
  • @JhonDoe $S$ is a finite set of algebraic elements over $F=\mathbb{Q}(\alpha_1,\dots,\alpha_{n-1})$, so $F(S)$ is finite dimensional over $F$. – egreg Feb 13 '22 at 16:46
  • @egreg Thanks but how do you show that the elements of S are algebraic? – Jhon Doe Feb 13 '22 at 16:58
  • @JhonDoe They are algebraic by construction. We have $p_i\in\bar{\mathbb{Q}}(\alpha_1,\dots,\alpha_{n-1})$; then we can express $p_i$ as a polynomial with coefficients in $\bar{\mathbb{Q}}$ and $S$ consists of these coefficients. – egreg Feb 13 '22 at 17:00
  • @egreg Aren't the $p_i=\frac{f(a_1,...a_{n-1})}{g(a_1,...a_{n-1})}$ with $f,g\in \mathbb{\bar Q}[X_1,...X_{n-1}]$. How does that how does that show that they are algebraic over F? Apologies for the many questions – Jhon Doe Feb 13 '22 at 17:05
  • @JhonDoe Yes, I should have said “quotient of polynomials in the indeterminates $\alpha_1,\dots,\alpha_{n-1}$ with coefficients in $\bar{\mathbb{Q}}$”, but this doesn't change much, does it? These coefficients are algebraic by definition. I'm not saying that $f$ and $g$ are algebraic, but their coefficients. – egreg Feb 13 '22 at 17:07
  • @egreg Ok thanks. I think I am confused as to what the definition of S is. Is it the set ${p_1,....p_n}$? Or is it the set of coefficients of $f$ and $g$ for each $p_i$. Thanks. – Jhon Doe Feb 13 '22 at 17:10
  • @JhonDoe As I've said a few times, it's the set of the coefficients. – egreg Feb 13 '22 at 17:12
  • @egreg I see thanks. When you mentioned coefficients i thought you were referring to the coefficients of the minimal polynomial of $\alpha_n$. apologies once again. – Jhon Doe Feb 13 '22 at 17:14
  • @JhonDoe You're welcome. I might reword the presentation, in the light of your comments. – egreg Feb 13 '22 at 17:20