I'm studying field theory. When I study the proof of Abel–Ruffini theorem, I wonder if the following proposition holds.
Proposition
Let
$L_0/K$ be a field extension and $f(x)\in L_0[x]$. And suppose the set of coefficients of $f(x)$ are algebraically independent over $K$ and $L_1\supset L_0$ is a splitting field of $f(x)$
Then
the set of solutions of $f(x)$ is algebraically independent over K.
Does this proposition hold?
No, this proposition does not hold.
Let $K=\mathbb{Q}, L_0=\mathbb{Q}(\pi)$, $f(x)=x^2-\pi\in L_0[x]$.
The splitting field $L_1$ of $f(x)$ over $L_0$ is $\mathbb{Q}(\sqrt{\pi})$ and the set of solutions $S$ of f(x) is $\{\pm\sqrt \pi\}$.
Let $h(x_1, x_2)= x_1 +x_2\in \mathbb{Q}[x_1, x_2]$. Since $h(\sqrt{\pi}, -\sqrt{\pi})=0$, S is not algebraically independent over $\mathbb{Q}$.
The set of coefficients of $f(x)$ is $\{1, -\pi\}$, which is algebraically independent over $\mathbb{Q}$ as $\pi$ is not algebraic over $\mathbb{Q}$.