$\mathbb{Q}$-rational Point obtain Preimage in $X_{\overline{k}}$ 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 132):

REMARK: The setting of 4.8.6 (e.g. the definitions of $\pi(n)$ and $\Pi(n)$ and the assumption that $P_i$ are $\mathbb{Q}$-rational) is given here:
S

Now the QUESTION:
The author says 

Since $P_i$ is rational over $\mathbb{Q}$, there is a unique point $\overline{Q_i}$ 
  of $X_{\overline{k}}= X \times Spec(\overline{k})$ lying above $P_i$ .

By assumption on previous pages $k$ is a field of characteristic $0$.
My questions are:
1.: What is $X$ here? Seems to be assumed as an integral proper normal curve defined over
$k$ of characteristic $0$ but it's just my guess.
2.: why $P_i$ MUST here be a rational $\mathbb{Q}$-point? What could fail if $P_i$ is for example a $K$-point where $K$ is an algebraic extension of $\mathbb{Q}$? Nevertheless there exist is a point $\overline{Q_i}$ 
of $X_{\overline{k}}$ lying above $P_i$ by pullback structure of $X_{\overline{k}}$. What problem occure here if $K \neq \mathbb{Q}$? 
 A: Let $X$ be a finite type $k$-scheme. For any closed point $x\in X$ let us denote by $\mathrm{Spec}(k(x))\to X$ the map picking out that point. Note then that, by definition, you have a fibered diagram
$$\begin{matrix} X_{\overline{k}} & \to & X\\ \downarrow & & \downarrow\\ \mathrm{Spec}(\overline{k}) & \to & \mathrm{Spec}(k)\end{matrix}$$
Let $f:X_{\overline{k}}\to X$ be the natural map above. Note then that we have a natural fibered diagram
$$\begin{matrix}f^{-1}(x) & \to & \mathrm{Spec}(k(x))\\ \downarrow & & \downarrow\\ X_{\overline{k}} & \to & X\end{matrix}$$
Combining these we get a fibered diagram
$$\begin{matrix}f^{-1}(x) & \to & \mathrm{Spec}(k(x))\\ \downarrow & & \downarrow\\ \mathrm{Spec}(\overline{k}) & \to & \mathrm{Spec}(k)\end{matrix}$$
But, this means that 
$$f^{-1}(x)=\mathrm{Spec}(k(x))\times_{\mathrm{Spec}(k)}\mathrm{Spec}(\overline{k})=\mathrm{Spec}(k(x)\otimes_{k}\overline{k})$$
Note then that if $k(x)/k$ is separable of degree $n$ (which is the case you're in) then $k(x)\otimes_k \overline{k}\cong \overline{k}^n$. Thus, 
$$\mathrm{Spec}(k(x)\otimes_k \overline{k})\cong \mathrm{Spec}(\overline{k}^n)=\bigsqcup_{i=1}^n \mathrm{Spec}(\overline{k})$$
Thus, $f^{-1}(x)$ is a singleton if and only if $x$ is rational. 
More generally, $f^{-1}(x)$ will be a singleton if and only if $k(x)/k$ is totally inseparable. Since you're in characteristic $0$, the point is that's equivalent to being trivial.
