I have a question about an argument used in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (or look up at page 127):
In the proof of Belyi we beginn with the reduction step to the case that $S$ consists of only $\mathbb{Q}$-rational points.
We take a point $P$ of degree $n=[\kappa(P):\mathbb{Q}]$ and consider the minimal polynomial $f$ of the extension $\kappa(P) /\mathbb{Q}$.
This map gets rise for the map $\phi_f: \mathbb{P}^1 \to \mathbb{P}^1, [x:1] \mapsto [f(x):1]$
Denote by $S_f$ the set of points $Q$ such that $f'(Q)=0$. In language of schemes this equation means that $f' \in \mathbb{Q}[t]$ lies in the prime ideal $m_Q$ of $ \mathbb{Q}[t]$ corresponding to point $Q$.
My question is why do all points $Q \in S_f$ induce field extensions $\kappa(Q)/\mathbb{Q}$ of degree at most $n-1$? Therefore $[\kappa(Q):\mathbb{Q}] \le n-1$?
Futhermore why their images $\phi_f(S_f)$ inherit this property?
By the way: $0$ has degree $1$,right?
Because 1) $f'$ has degree $n-1$, and 2) $f'(Q)=0$, and so any generator for $m_Q$ (an irreducible poly $g\in {\mathbb Q}[t]$) divides $f'$, and therefore has degree $\le n-1$. Therefore, $Q$ has degree $\le n-1$, by definition.
Next, if $t$ belongs to some extension $L$ of a field $k$, then $h(t)\in k(t) \subset L $, for any $h(x) \in k[x]$, the polynomial ring over $k$. So, the degree of $h(t)$ over $k$ is no more than that of $t$.
And, finally, yes: $0$ (or any rational point) has degree $1$.