What is a rational $0$-cycle on an algebraic variety $X$ over $\mathbb{Q}$? I found the following assertion in a paper about Hilbert modular forms that I'm trying to read.

Let $X$ be an algebraic variety over $\mathbb{Q}$, and let $\Psi$ be a rational function on $X$ and $C = \sum n_P P$ be a rational $0$-cycle on $X$. Then $\Psi(C) = \prod \Psi(P)^{n_P}$ is a rational number.

By searching online I found some definitions of an algebraic cycle, but I haven't found what a rational $0$-cycle is. So the questions I have are:


*

*What does it mean that $C = \sum n_P P$ is a rational $0$-cycle? Does it mean that the coefficients $n_P \in \mathbb{Q}$ and that $\sum n_P = 0$? And what would be a good reference for these basic definitions?

*How do we prove that $\Psi(C) = \prod \Psi(P)^{n_P}$ is a rational number? 


Thank you very much for any help.

To say that $C = \sum n_P P$ is a rational $0$-cycle means that $n_P\in \mathbb Z$ and that $P\in X$ is a rational point i.e. a closed point with residue field $\kappa (P)=\mathbb Q$.
If the rational function $\Psi$ is defined at $P$ its value at $P$ is a rational number $\Psi(P) \in \mathbb Q$ and if $\Psi$ is defined at all $P$'s with $n_P\neq 0$ we have $\Psi(C) = \prod \Psi(P)^{n_P}\in \mathbb Q$.  
As an illustration of what it means for $P$ to be rational, take the simplest example $X =\mathbb A^1_\mathbb Q=Spec ( \mathbb Q[T])$.
Then the point $P$ corresponding to the prime ideal $J_P= \langle T-1/2\rangle\subset \mathbb Q[T]$ is rational since $\mathbb Q[T]/ \langle T-1/2\rangle=\mathbb Q$.
However the closed point $Q$  corresponding to the prime ideal $J_Q= \langle T^3-2\rangle\subset \mathbb Q[T]$ is not rational since the canonical morphism $\mathbb Q \to \mathbb Q[T]/ \langle T^3-2 \rangle $ is not an isomorphism.
