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If you search for algebraic independence, you will find the following 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.

https://en.wikipedia.org/wiki/Algebraic_independence

What does that mean in simpler terms? I am especially interested in the Brownawell-Waldschmidt theorem (here), but I dont see how it follows that for example at least one of $e\pi$ and $e^{\pi^2}$ is transcendental.

Kinheadpump
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  • As written in the referenced post, if we take $\alpha=\beta=\frac i\pi$, $\gamma=\pi^2$, theh B-W says that, if $e^{i\pi}=-1$ and $e^{\pi^2}$ are algebraic, then at least two of $\frac i\pi$, $\pi^2$, $-1$, $e^{-1}$ are algebraically independent over $\Bbb Q$. As the first three are algebraically dependent, it can only be the case that they (and henec $\pi$) and $e^{-1}$ (and hence $e$) are algebraically independent (and hence any non-trivial polynomial in them, in partticular $e\pi$, is transcendental), or the premises do not hold, i.e., $e^{\pi^2}$ is not algebraic. – Hagen von Eitzen Jul 21 '19 at 16:22
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    In my experience, wikipedia is a very bad resource to understand something. Checking wiki and then asking the MathSE is better. – peterh May 30 '21 at 15:47

3 Answers3

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It means that if $n\in\mathbb N$, if $p(x_1,\ldots,x_n)$ is a non-zero polynomial with coefficients in $K$ and is $s_1,\ldots,s_n\in D$, then $p(s_1,\ldots,s_n)\neq0$.

Now, suppose that $e\pi$ and $e^{\pi^2}$ are both algebraic. Then $\{e,\pi\}$ would be algebraically dependent over $\mathbb Q$, since $e\pi$ being algebraic means that there are rational numbers $a_0,\ldots,a_n\in\mathbb Q$, not all of which are $0$, such that$$a_0+a_1e\pi+a_2(e\pi)^2+\cdots+a_n(e\pi)^n=0.$$So, if $p(x,y)=a_0+a_1xy+a_2(xy)^2+\cdots+a_n(xy)^n$, $p(e,\pi)=0$.

José Carlos Santos
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I'll explain what it means if $S$ is a finite set. Let's write $S=\{a_1,...,a_n\}$. It is algebraically independent if there is no non-zero polynomial $f\in K[x_1,...,x_n]$ such that $f(a_1,...,a_n)=0$. You can try to prove as an easy exercise that this definition is equivalent to the following:

$1.$ Each $a_i$ is transcendental over $K(a_1,...,a_{i-1},a_{i+1},...,a_n)$

$2.$ $a_1$ is transcendental over $K$ and each $a_i$ is transcendental over $K(a_1,...,a_{i-1})$

Now, if $S$ is infinite then it is called algebraically independent if every finite subset is algebraically independent.

Mark
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Reading the wiki article a little bit further, we can see a very easily comprehensible explanation:

The numbers $x$, $y$ are algebraic independent if there is no 2-variable polynome with rational coefficients for which $P(x,y)$ would be zero.

For example, $\sqrt{\pi}$ and $2\pi+1$ are not algebraically independent, because substituting $x=\sqrt{\pi}$ and $y=2\pi+1$ into the polynome $P(x,y)=2x^2-y+1$ we get zero.

peterh
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