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Let $R$ be a valuation ring with maximal ideal $M$ and $L$ its residue field. We let $k$ be the fractions field which we assume to be algebraically closed. Let $\mathcal{L}$ be a closed subset of $\mathbb{P}^n_R$ defined by $r+1$ linear forms with coefficients in $R$. Let $\tau: \mathbb{P}^n_R \backslash \mathcal{L} \to \mathbb{P}^r_R$ be the projection.

Let $p: \mathbb{P}^n_R \to \operatorname{Spec}R$, and $i_b: \operatorname{Spec}L \to \operatorname{Spec}R $ where the image is $b$, the closed point of $\operatorname{Spec}R$. It can be shown that the fiber of $p$ over $b$ is $\mathbb{P}^n_L$ and let $j: \mathbb{P}^n_L \to \mathbb{P}^n_R$ be the morphism we obtain from the universal property of the fiber products. It turns out that $j$ is a closed immersion and through this morphism we can view $\mathbb{P}_L^n$ inside $\mathbb{P}_R^n$. Similarly, by considering the fiber product of $p$ over the generic point, we obtain $i: \mathbb{P}^n_k \to \mathbb{P}^n_R$.

Let $Z$ be a closed subset of $\mathbb{P}^n_k$ and $\mathcal{Z}$ the closure of $i(Z)$ inside $\mathbb{P}^n_R$. Assume $\mathcal{Z}$ is disjoint from $\mathcal{L}$. Then it is proved that $\tau: Z \to \mathbb{P}_R^r$ is a finite surjective morphism.

I understand the proof until the end of page 181 (in the old version), where he deduces that $\tilde{x} \in Z$. He then claims that $\tilde{x}$ is a closed point of $Z$, because $\tau_k(\tilde{x})$, where $\tau_k: Z \to \mathbb{P}^r_k$, is a closed point. I am not seeing how this is deduced knowing that $\tau_k(\tilde{x})$ is a closed point. I would appreciate if some could clarify this. Thank you!

red_trumpet
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Johnny T.
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  • Can you provide an example of such $R$? I'm having trouble imagining the hypothesis that $k = Q(R)$ is algebraically closed. – red_trumpet Mar 16 '20 at 12:19
  • Just to be clear: Is $\mathcal{L}$ parametrized by $r+1$ linear forms? At first reading I thought $\mathcal{L}$ was cut out by $r+1$ linear forms, but then the projection doesn't make sense in general. – Servaes Mar 16 '20 at 12:37
  • According to the book $\mathcal{L}$ is cut out by $r+1$ linear forms, but the linear forms are chosen in a way that $\mathcal{L}$ is disjoint from $\mathcal{Z}$ and the dimension of the irreducible closed subset $Z \subseteq \mathbb{P}^n_k$ is $r$. – Johnny T. Mar 16 '20 at 12:47
  • @red_trumpet To be precise in the book he says: Let $k$ be an algebraically closed field and let $R \subset k$ be a valuation ring. Maybe I misinterpreted his statement? – Johnny T. Mar 16 '20 at 12:51
  • @TakeshiGouda I think you got it right. If $Q(R) \neq k$, I don't see any reason why $L$ should be algebraically closed. – red_trumpet Mar 17 '20 at 07:55

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$\tau_k^{-1}(\tau_k(\widetilde{x}))$ is closed inside $Z$, as it is the preimage of a closed subset. On the other hand, the fiber of a finite morphism over any point is finite discrete - thus every point of the fiber is closed in the fiber. As the underlying topological space of the scheme-theoretic fiber is the same as the topological fiber (ref), we have that $\{\widetilde{x}\}$ is a closed subset of $\tau_k^{-1}(\tau_k(\widetilde{x}))$ which is again a closed subset of $Z$. As a closed subset of a closed subset is again closed, we have the result.

KReiser
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  • When you say the underlying topological space of the scheme theoretic fiber is the same as the topological fiber, are you putting the subspace topology (as a subspace of $Z$) on $\tau_k^{-1}(\tau_k(\widetilde{x}))$? – Johnny T. Mar 23 '20 at 14:56
  • @JohnnyT. yes, that's correct. – KReiser Mar 23 '20 at 17:13