I have a few basic questions about Liu's [1, Section 10.1.3] description of the reduction map from the closed points of the generic fibre of a proper scheme over a complete DVR to its special fibre. Here's (my slight paraphrasing of) the section that's bothering me:

Let $R$ be a complete DVR and let $S = \operatorname{Spec}R$. Let $s \in S$ be the closed point. Let $\mathcal{X}\rightarrow S$ be a proper scheme over $S$ with generic fibre $X$ and special fibre $\mathcal{X}_s$. For any closed point $x \in X$, the Zariski closure $\overline{\{x\}}$ is an irreducible finite scheme over $S$ and is therefore a local scheme with closed point $\overline{\{x\}} \cap \mathcal{X}_s$.

One then defines the reduction map $r\colon X^0 \rightarrow \mathcal{X}_s$, where $X^0$ is the set of closed points of $X$, by $r(x) = \overline{\{x\}} \cap \mathcal{X}_s$. Here are my questions:

  1. What is the point of considering the Zariski closure of a closed point? By definition $x$ is closed precisely when $\{x\} = \overline{\{x\}}$, so this part of the definition seems redundant. Am I missing something obvious?

  2. More importantly, I don't understand how to interpret the intersection $\overline{\{x\}} \cap \mathcal{X}_s$ since the two sets in question don't share a common "parent" space. What is the space in which the intersection is taking place?

Thanks in advance.

[1] Liu, Q., Algebraic Geometry and Arithmetic Curves, Oxford Graduate Texts in Mathematics, 6. Oxford University Press, 2002

  1. The point $x$ is closed as a subset of the generic fiber $X$, but not as a subspace of the big scheme $\mathcal{X}$ over $S$ ; so it makes sense to take the closure of $x$ in the big scheme $\mathcal{X}$.

  2. Then you can take the intersection of this closure with the special fibre $\mathcal X_s$ : this intersection takes place in the big scheme $\mathcal{X}$, which is your "parent" scheme.

Maybe the source of the confusion is the simultaneous use of $X$ and $\mathcal{X}$ , which are typographically not sufficiently distinguishable?

  • $\begingroup$ The typography is definitely not the problem. I think the problem is that I'm having trouble seeing how to consider $X$ and $\mathcal{X}_s$ as being "in" $\mathcal{X}$. They are both $\mathcal{X}$-schemes having been formed by fibre products, so should the expression $\overline{\{x\}} \cap \mathcal{X}_s$ be interpreted as $\pi_X(\overline{\{x\}}) \cap \pi_{\mathcal{X}_s}(\mathcal{X}_s)$ where $\pi_X\colon X \rightarrow \mathcal{X}$ and $\pi_{\mathcal{X}_s}\colon \mathcal{X}_s \rightarrow \mathcal{X}$ are the canonical projections? $\endgroup$ – Hamish Apr 23 '11 at 14:46
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    $\begingroup$ Call $f:\mathcal X \to S$ your morphism of schemes.Then topologically $X$ is just the inverse image $X=f^{-1}(\eta )$ of the generic point $\eta$ of $S$ and $\mathcal X_s=f^{-1}(s)$ the inverse image of the closed point $s$ of $S$. So $\mathcal X$ is just the disjoint union of its open subset $X$ and of its closed subset $\mathcal X_s$. This business of fibre products (=base change) is useful for giving a scheme structure to fibres of morphisms but obscures a little the fact that you just deal with inverse images. $\endgroup$ – Georges Elencwajg Apr 23 '11 at 17:56
  • $\begingroup$ Excellent, your comment clears it up. Thanks! $\endgroup$ – Hamish Apr 23 '11 at 18:05

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