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I know that one of the Weil conjectures says something loosely like the following.

If $X$ is a projective variety obtained from a complex projective variety $X_\mathbf{C}$ by "reduction mod $p$", then the $j$-th Betti number of $X_\mathbf{C}$ is equal to the degree of the polynomial $F_j \in \mathbf{Z}[x]$, where the $F_j$ are defined by \begin{align} \prod_{j = 0}^{2n}F_j(q^{-s})^{(-1)^{j + 1}} = \zeta_X(s) = \exp\left(\sum_{d \geq 1}\frac{N_d}{d}q^{-ds}\right). \end{align}

I have two questions, but they are both related to this statement.

  1. In this context, is the $j$-th Betti number just the rank of the singular homology group $H_j(X_{\mathbf{C}})$ of $X_\mathbf{C}$ as a topological space?
  2. We can consider $X_\mathbf{C}$ as a scheme over $\textrm{Spec} \, \mathbf{Z}$ via a unique morphism $\varphi : X_\mathbf{C} \to \textrm{Spec} \, \mathbf{Z}$. Taking the fibre of $\varphi$ over the point $(p)$, one obtains a scheme $X_\mathbf{C} \times_{\textrm{Spec} \, \mathbf{Z}} \mathbf{F}_p$ over $\mathbf{F}_p$, à la Hartshorne II.3 page 89. Do we say that $X$ is obtained from $X_\mathbf{C}$ by reduction mod $p$ because $X \cong X_\mathbf{C} \times_{\textrm{Spec} \, \mathbf{Z}} \mathbf{F}_p$?
Joseph Harrison
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1 Answers1

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  1. You want $H_j(X(\Bbb C))$, the rank of the singular homology of the $\Bbb C$-points of $X$ endowed with the analytic topology. Singular homology does not do what you want on the underlying topological space of schemes over $\Bbb C$ (because for instance every irreducible topological space is contractible), but it does what you want on their $\Bbb C$-points.

  2. This is not quite right. If the map $X\to\operatorname{Spec} \Bbb Z$ factors through $\operatorname{Spec} \Bbb C$, then the fiber over $(p)$ is empty since the image of $\operatorname{Spec} \Bbb C \to \operatorname{Spec} \Bbb Z$ is the generic point. What you want to do is to have $X_{\Bbb C}\to \operatorname{Spec} \Bbb C$ be the base extension of some projective variety $X\to\operatorname{Spec} R$ where $R$ is the ring of integers of some number field (or some slight generalization; please correct me in the comments if this is not quite precise enough). Then $X(\Bbb C)\cong X_{\Bbb C}(\Bbb C)$ and the fiber of $X\to \operatorname{Spec} \Bbb Z$ is nonempty over each point and you can actually form the fiber product you speak of.

KReiser
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