Closed points of a closed subscheme of $\mathbb{P}^n$ over the residue field and the fraction field of a valuation ring $R$

Question

Let $(R, M)$ be a valuation ring with algebraically closed fraction field $k$. Let $L = R/M$ be the residue field of $R$; it follows that $L$ is algebraically closed. I would like to understand the following: Suppose $Z$ is an irreducible closed subscheme of $\mathbb{P}^n_k$ defined by the homogeneous ideal $A \subset k[X_0, ..., X_n]$. Let $B = \pi\{ A \cap R[X_0, ..., X_n] \}$ where $\pi$ is the quotient modulo $M$. Let $W \subset \mathbb{P}^n_L$ be defined by $B$. Suppose $[\bar{a_0}:...: \bar{a_n}]$ is a closed point of $W$. Then I would like to know how one can prove that there exit $b_0, ..., b_n \in R$ (not all in $M$) such that $b_j + M = \bar{a_j}$ for each $0 \leq j \leq n$ and $[b_0:...: b_n]$ is a closed point of $Z$.

I am asking this question because I would like to understand the proof of the following result. In Mumford's 'Red book of schemes', Theorem 1 Chapter II Section 8, he proves: For all closed subsets $Z \subset \mathbb{P}^n_k$, there is a unique closed subset $W \subset \mathbb{P}^n_L$ such that $$ \rho(Z(k)) = W(L), $$ where $\rho: \mathbb{P}^n(k) \to \mathbb{P}^n(L)$.

I am having similar issues with the proof as in this MathSE question. If someone could provide an alternative reference for this result or an explanation for the statement in the first paragraph, it would be appreciated. Thank you

Answer 1

Let me first reformulate this question slightly. Set $S = \text{Spec}(R)$, denote $\eta \in S$ the generic point (with residue field $k$) and denote $s \in S$ the closed point (with residue field $L$). We may assume $Z$ is not only irreducible but also reduced as the question is about points. I assume that $A \subset k[X_0, \ldots, X_n]$ is exactly the set of homogeneous polynomials vanishing on $Z$. Thus we see that $A$ is a homogeneous prime ideal. The homogeneous prime ideal $A \cap R[X_0, \ldots, X_n]$ defines an integral closed subscheme $\overline{Z} \subset \mathbf{P}^n_R$ whose generic fibre is $Z$ and whose closed fibre is $W$. We are going to think of $\overline{Z}$ as a scheme over $S$. Picture: $$ \overline{Z} \longrightarrow S \quad\text{with}\quad \overline{Z}_\eta = Z \quad\text{and}\quad \overline{Z}_s = W $$ Denote $\xi \in Z = \overline{Z}_\eta$ the generic point. Clearly, $\xi$ is also the generic point of $\overline{Z}$. Denote $w = [\overline{a}_0 : \ldots : \overline{a}_n]$ the closed point of $W = \overline{Z}_s$ given to us. Then we have a specialization $$ \xi \leadsto w $$ which maps to the specialization $\eta \leadsto s$ in $S$. What you are asking for is to show that there is a closed point $z$ of $Z = \overline{Z}_\eta$ and specializations $\xi \leadsto z \leadsto w$.

(Hints: Observe that a closed point of $Z$ always is a $k$-rational point because $k$ is algebraically closed and that a $k$-rational point of $\mathbf{P}^n_k$ is the same as an $R$-rational point of $\mathbf{P}^n_R$ by the valuative criterion for projective space, so this will give you the sequence $b_0, \ldots, b_n \in R$ not all in $M$ well defined up to a unit in $R$, which specializes to $[\overline{b}_0 : \ldots : \overline{b}_n]$ in the closed fibre.)

The existence of $z$ is a general fact about the topology of finite type morphisms of schemes! AFAIK this property in general was first discovered by Brian Osserman and Sam Payne in a paper entitled "Lifting tropical intersections". In the Stacks project you can find it here.

On the other hand, I am pretty sure that you don't need this result in order to prove the statement in Mumford's book. In fact, I just looked at his proof of this fact and it uses just a little bit of algebraic geometry (projections) and going down for finite morphisms to prove this particular case of the Osserman-Payne result (the proof is on the bottom of page 130 in my edition). So I urge you to stick to this case and read what Mumford wrote there.