Valuative extension of projective coordinates

Since not all $a_j$ are zero, fix an index $i$ with $a_i\mathrm{\ne }0$. By [F8], applied to the inclusion $R\hookrightarrow K$ and the elements $a_{\mathrm{\ell }}\mathrm{/}{a}_i\in K$ for $\mathrm{\ell }\mathrm{\ne }i$, there is a unique $R$-algebra map $\psi :R[x_{\mathrm{\ell }}^{(i)}:\mathrm{\ell }\mathrm{\ne }i]\to K$ with $\psi (x_{\mathrm{\ell }}^{(i)})=a_{\mathrm{\ell }}\mathrm{/}{a}_i$; by [F3] it corresponds to a morphism $\mathrm{Spec}K\to U_i^R\subseteq {\mathbb{P}}_R^n$, which is the $K$-point $[a_0:\cdots :a_n]$ of the tuple and is a morphism over $\mathrm{Spec}R$ because the composite $R\to R[x_{\mathrm{\ell }}^{(i)}]\stackrel{\psi }{\to }K$ is the structure map $R\hookrightarrow K$ of [F2]. Replacing the tuple by $(\lambda a_j)$ with $\lambda \in K^{\times }$ multiplies each ratio $a_{\mathrm{\ell }}\mathrm{/}{a}_i$ by $\lambda {\lambda }^{-1}=1$, hence gives the same $\psi$ and the same point.

There is an index $m$ with $a_m\mathrm{\ne }0$ and $a_j\mathrm{/}{a}_m\in R$ for every $j$. Start with $m:=i$, so that $a_m=a_i\mathrm{\ne }0$; process the indices $j\mathrm{\ne }i$ one at a time, maintaining the invariant that $a_m\mathrm{\ne }0$ and $a_{j^{\mathrm{\prime }}}\mathrm{/}{a}_m\in R$ for every already processed $j^{\mathrm{\prime }}$. If $a_j=0$, then $a_j\mathrm{/}{a}_m=0\in R$ and $m$ is kept. If $a_j\mathrm{\ne }0$, apply [F2] to $x=a_j\mathrm{/}{a}_m\in K^{\times }$: either $a_j\mathrm{/}{a}_m\in R$ and $m$ is kept, or $a_m\mathrm{/}{a}_j\in R$ and we replace $m$ by $j$; in the second case $a_{j^{\mathrm{\prime }}}\mathrm{/}{a}_j=(a_{j^{\mathrm{\prime }}}\mathrm{/}{a}_m)(a_m\mathrm{/}{a}_j)\in R$ for every processed $j^{\mathrm{\prime }}$, while $a_j\mathrm{/}{a}_j=1$, so the invariant is preserved. After the finitely many indices have been processed, $a_j\mathrm{/}{a}_m\in R$ for all $j$ and $a_m\mathrm{\ne }0$.

Conversely every morphism $p:\mathrm{Spec}K\to {\mathbb{P}}_R^n$ over $\mathrm{Spec}R$ arises from such a tuple: by [F1] the charts cover ${\mathbb{P}}_R^n$, so the image of the unique point of $\mathrm{Spec}K$ lies in some chart $U_i^R$, and $p$ factors through the open immersion $U_i^R\hookrightarrow {\mathbb{P}}_R^n$ by [F4]. The factorisation corresponds by [F3] to an $R$-algebra map $\psi :R[x_{\mathrm{\ell }}^{(i)}:\mathrm{\ell }\mathrm{\ne }i]\to K$, and putting $a_{\mathrm{\ell }}:=\psi (x_{\mathrm{\ell }}^{(i)})$ for $\mathrm{\ell }\mathrm{\ne }i$ and $a_i:=1$ gives a tuple with $a_i\mathrm{\ne }0$ that determines $p$ by step 1.1.

Since $\psi (x_m^{(i)})=a_m\mathrm{/}{a}_i\mathrm{\ne }0$ by step 1.2, the image point of $p$ lies in the distinguished open $D(x_m^{(i)})=U_i^R\cap U_m^R$ of [F1]; hence $p$ factors through the open subscheme $U_m^R$ by [F4]. By the transition formula of [F1] the factorisation corresponds by [F3] to the $R$-algebra map $R[x_{\mathrm{\ell }}^{(m)}:\mathrm{\ell }\mathrm{\ne }m]\to K$ sending $x_{\mathrm{\ell }}^{(m)}$ to $c_{\mathrm{\ell }}:=a_{\mathrm{\ell }}\mathrm{/}{a}_m$ for $\mathrm{\ell }\mathrm{\ne }m$, and $c_{\mathrm{\ell }}\in R$ for every $\mathrm{\ell }\mathrm{\ne }m$ by step 1.2.

By [F8], applied to the inclusion $R\hookrightarrow R$ and the elements $c_{\mathrm{\ell }}\in R$ of step 2.2, there is a unique $R$-algebra map $\phi :R[x_{\mathrm{\ell }}^{(m)}:\mathrm{\ell }\mathrm{\ne }m]\to R$ with $\phi (x_{\mathrm{\ell }}^{(m)})=c_{\mathrm{\ell }}$; by [F3] it corresponds to a morphism $q:\mathrm{Spec}R\to U_m^R\subseteq {\mathbb{P}}_R^n$.

The composite $\pi \circ q$ corresponds by [F3] to the ring map $R\to R[x_{\mathrm{\ell }}^{(m)}]\stackrel{\phi }{\to }R$, which is the identity of $R$; by the affine anti-equivalence [F3], this identity of ring maps gives $\pi \circ q={\mathrm{id}}_{\mathrm{Spec}R}$.

The composite $q\circ j$ corresponds by [F3] to the ring map $R[x_{\mathrm{\ell }}^{(m)}]\stackrel{\phi }{\to }R\hookrightarrow K$, which sends $x_{\mathrm{\ell }}^{(m)}$ to $c_{\mathrm{\ell }}$; by step 2.2 this is the same $R$-algebra map $R[x_{\mathrm{\ell }}^{(m)}]\to K$ that describes the factorisation of $p$ through $U_m^R$, so $q\circ j=p$.

Let $q^{\mathrm{\prime }}:\mathrm{Spec}R\to {\mathbb{P}}_R^n$ be any morphism over $\mathrm{Spec}R$ with $q^{\mathrm{\prime }}\circ j=p$. Then $q$ and $q^{\mathrm{\prime }}$ are both lifts of the valuative diagram of [F6] consisting of $j$, the generic map $p$ and the base morphism ${\mathrm{id}}_{\mathrm{Spec}R}$: indeed $\pi q=\mathrm{id}$ and $\pi q^{\mathrm{\prime }}=\mathrm{id}$ and $qj=p=q^{\mathrm{\prime }}j$. Since $\pi$ is separated by [F5], [F7] gives $q^{\mathrm{\prime }}=q$. Moreover the condition of being over $\mathrm{Spec}R$ is automatic for a morphism $q^{\mathrm{\prime }}$ with $q^{\mathrm{\prime }}\circ j=p$: let $u:R\to R$ be the ring map corresponding to $\pi \circ q^{\mathrm{\prime }}$ by [F3]. Since $(\pi q^{\mathrm{\prime }})j=\pi p=j$, the composite $R\stackrel{u}{\to }R\hookrightarrow K$ equals the given inclusion $R\hookrightarrow K$. The inclusion is injective, hence $u={\mathrm{id}}_R$ and $\pi q^{\mathrm{\prime }}={\mathrm{id}}_{\mathrm{Spec}R}$ by [F3]. Thus no extension of $p$ to $\mathrm{Spec}R$ other than $q$ exists, and the extension is unique.

This completes the example: the index $i$ of the statement is the index constructed in step 1.2, the ratios $a_j\mathrm{/}{a}_i$ are the elements $c_{\mathrm{\ell }}$ of step 2.2, and step 2.2, step 3.1, step 4.1, step 4.2 and step 5.1 exhibit them as the unique extension $\mathrm{Spec}R\to {\mathbb{P}}_R^n$ of the point. The construction is explicit and uses no choice principle: only finitely many coordinates are inspected and the index is updated by the dichotomy [F2], so the case of zero coordinates is included, the case $n=0$ has the single chart $U_0^R=\mathrm{Spec}R$ with no variables and the structure maps as $\psi$ and $\phi$, and the case $R=K$ is included as a valuation ring that is a field.