What it means to say $\pi$ and $e^\pi$ are algebraically independent $?$

Question

$\pi$ and $e^\pi$ are algebraically independent. But what does that mean?

Is that if $a_0, a_1, \cdots, a_n$ and $b_0, b_1, \cdots, b_n$ are algebraic numbers then

$$a_0\pi + a_1\pi + \cdots + a_n\pi + b_0e^\pi+ b_1e^\pi + \cdots + b_ne^\pi$$

is transcendental or $0$ $?$ If the above is wrong, what is the correct form?

Answer 1

It means that if $P(x,y)$ is a non-null polynomial with rational coefficients, then $P(\pi,e^\pi)\neq0$.

Answer 2

Let $L$ be a field and $K\subseteq L$ a subfield. The elements of a subset $I\subseteq L$ are algebraically independent over $K$ if for every polynomial $p\in K[x]$ one has $$(\forall x\in I\,p(x) = 0) \Rightarrow p=0. $$ Thus, if we choose $p\neq 0$ (likely over $\mathbb Q$) then $p(\pi) \neq 0$ or $p(e^\pi)\neq 0$.