Explicit base change for a curve Let $X$ be a geometrically integral smooth curve over a field $K$. Let's assume that $K$ is a number field.
Consider the base change $X_{\mathbb C}$ of $X$ by some embedding $K\subset \mathbb C$. In other words we look at $X$ as if it was a curve over the complex numbers. By the general theory of schemes we know that we have a projection map $p:X_{\mathbb C}\to X$. I don't understand the explicit description of $p$. 


*

*If $x\in X_{\mathbb C}$, what is $p(x)$?

*If $y\in X$, what is $p^{-1}(y)$?


All sources I've checked so far, study the properties of $X_{\mathbb C}$ and $p$ but I don't find anything about the explicit description of $p$.
 A: Here's the simplest possible case that will hopefully help you get a feel for the more general case.
Let $X = \mathbf{A}^1_{\mathbf{Q}} = \operatorname{Spec}(\mathbf{Q}[x])$. Then $X_{\mathbf C} = \operatorname{Spec}(\mathbf{C}[x])$ and the map $p:X_{\mathbf{C}} \to X$ is dual to the map $\iota:\mathbf{Q}[x] \to \mathbf{C}[x]$.
In $X_{\mathbf{C}}$ there are two kinds of points: closed points $(x - a)$ and $(0)$. By definition, we have
$$ p((x - a)_{\mathbf C}) = \iota^{-1}((x -a)_{\mathbf C}) = \{ f \in \mathbf{Q}[x] : f_{\mathbf C} \in (x - a)_{\mathbf C}\}. $$
This will depend on what $a$ is.


*

*If $a$ is algebraic, then $f \in (x - a)$—meaning $(x - a)$ divides $f$ in $\mathbf{C}$ if and only if the minimal polynomial $\mu_a$ of $a$ over $\mathbf{Q}$ is a divisor of $f$ in $\mathbf{Q}$. Thus $p((x - a)_{\mathbf C}) = (\mu_a)_{\mathbf{Q}}$.

*If $a$ is transcendental, then $(x - a) \mid f$ if and only if $f = 0$.
So some examples, $p((x - i)_{\mathbf C}) = (x^2 + 1)_{\mathbf Q}$, $p((x - 3)_{\mathbf C}) = (x - 3)_{\mathbf Q}$, $p((x - \pi)_{\mathbf C}) = (0)_{\mathbf Q}$.
Conversely, we see that if $(\mu)_\mathbf{Q}$ is a prime ideal (let's say non-zero), then
$$ p^{-1}((\mu)_{\mathbf{Q}}) = \{ (x - a)_\mathbf{C} : p((x - a)_{\mathbf C}) = (\mu)_{\mathbf{Q}}\} = \{ (x - a)_{\mathbf C} : (x - a) \mid \mu \}. $$
So for example, $p^{-1}((x^2 + 1)_{\mathbf Q}) = \{ (x - i)_{\mathbf C}, (x + i)_{\mathbf C}\}$.
In the general case, the idea is that if I have a point on a curve$/\mathbf C$ with algebraic coordinates, then the minimal polynomials of those coordinates will give me the corresponding point over $\mathbf Q$.
A: I'm going to take an explicit example that maybe can make the situation clearer.
Let's start with $X=\mathbb{A}^1_K$ , in other words $Spec(K[t])$. As you can easily check, $X_{\mathbb{C}} \cong \mathbb{A}^1_{\mathbb{C}}$ in such a way that the projection map corresponds to the map induced by the inclusion $K[t] \subset \mathbb{C}[t]$. Now, points in $\mathbb{A}^1_{\mathbb{C}}$ (except from the generic one) corresponds to maximal ideal of the form $\mathfrak{m}_z=(x-z)$ with $z \in \mathbb{C}$. Now, the situation depends heavily on $z$: if $z$ is algebraic over $\mathbb{Q}$ you could try to check that $p(\mathfrak{m}_z)=(\mu_z)$ where $\mu_z$ is the minimum polynomial of $z$ over $K$.
However, if you take a point which does not come from an algebraic number ,e.g $\mathfrak{m}_{\pi}$, you can try to check that $p(\mathfrak{m}_z)$ corresponds to the generic  point of $X$. This depends on the fact that the extension $K \subset \mathbb{C}$ is not algebraic and so not finitely generated.
This is probably the most basic example, but the situation is fairly similar in every affine case. In the general case,you can apply the tecnique of  of reduction to an affine cover.
Hope it helped.
