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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:

  1. 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?
  2. How do we prove that $\Psi(C) = \prod \Psi(P)^{n_P}$ is a rational number?

Thank you very much for any help.

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If I were using the term, I'd use it to mean a $0$-cycle (i.e. algebraic cycle of dimension $0$ subvarieties) up to rational equivalence (see en.wikipedia.org/wiki/Chow_ring). The point of confusion is whether an author means dimension $0$ or codimension $0$ subvarieties by "$0$-cycle" but "rational" should mean the group of these things up to rational equivalence. –  Matt Oct 28 '12 at 0:24
    
@Matt I see, in that case, I can assume that the coefficients $n_P$ are integers then, so the statement that the product $\prod \Psi(P)^{n_P}$ is a rational number is immediate? –  Adrián Barquero Oct 28 '12 at 0:33
    
Considering the author uses "$P$" I would think these are dimension zero. Could rational possibly mean rational points, i.e. they are defined over $\Bbb Q$? –  Andrew Oct 28 '12 at 2:56
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2 Answers 2

up vote 2 down vote accepted

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.

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Dear Georges, thank you very much for your answer. Do you know a reference where I can find these definitions? –  Adrián Barquero Oct 28 '12 at 5:26
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Dear Adrián, besides the first chapter of Fulton's Intersection Theory there is a book project by Eisenbud and Harris, amusingly called 3264 and All That, a draft of which the authors generously put online‌​. –  Georges Elencwajg Oct 28 '12 at 6:06
    
Dear Georges, thanks for the reference. –  Adrián Barquero Oct 28 '12 at 14:43
    
Now that this has popped back up to the top (I must have missed this answer back when it was given), I really have to stick with my original comment and say this would be a highly non-standard definition. A $0$-cycle should mean a formal sum of closed points and the modifier rational should mean up to rational equivalence. The term "rational" shouldn't refer to the points being rational. For example, even Fulton in Chapter 13 says that the degree of a $0$-cycle is $\sum n_P[K(P):K]$ which doesn't make sense if you restrict to rational points. –  Matt Mar 29 '13 at 20:27
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Thanks for the reference, it is a treasure !

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This is better left as a comment (not sure you have enough reputation) as this area is intended for answers. Regards. –  Amzoti Mar 29 '13 at 19:41
    
Please don't add "thank you" as an answer. Once you have sufficient reputation, you will be able to vote up questions and answers that you found helpful. –  Amzoti Mar 29 '13 at 19:41
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