Natural map to the adic spectrum

Question

I'm struggling to understand the following proposition regarding the adic spectrum corresponding to a normal curve over a field $k$:

Proposition: Let $C$ be a normal curve over a field $k$ and $K:=\operatorname{Frac}(A)$ for an arbitrary $\operatorname{Spec}A\subseteq C$ denote the function field of $C$. There is a natural map $\nu:C\to \operatorname{Spa}(K,k)$, where the latter is the set of all valuation rings $V$ with $k\subseteq V\subseteq K$ and $\operatorname{Frac}(V)=K$, defined by $x\mapsto \mathcal{O}_{C,x}$.

I can see that $k\subseteq K$ since $C$ is of finite type over $k$ and I also know that $C$ being a normal curve over $k$ implies that $\mathcal{O}_{C,x}$ is a discrete valuation ring for every $x\in C$ not a generic point.

However, I don't understand the following things:

1. Why does $k\subseteq \mathcal{O}_{C,x}\subseteq K$ hold?
2. Why is $\operatorname{Frac}(A)=\mathcal{O}_{C,\eta}\,,\,$ with $\eta$ being the generic point of $C$? (I suppose this is connected to the first question)

Thank you very much in advance.

Answer 1

If you want to google this further, the adic spectrum $\mathrm{Spa}(K,k)$ is more classically known as the Riemann--Zariski space $\mathrm{RZ}(K,k)$.

Your other questions then have nothing to do with adic spaces, or even Riemann--Zariski spaces, or even curves--they just pertain to the study of integral schemes.

So, assume that $X$ is an integral(=irreducible+reduced) scheme over a ring $R$. Note then for each $x$ in $X$ one has, by definition, that

$$\mathcal{O}_{X,x}=\varinjlim\mathcal{O}(U)$$

where $U$ ranges over the open neighborhoods $U$ of $x$. Note though that for each $U$ the composition

$$U\to X\to \mathrm{Spec}(R)$$

in particular gives rise to a ring map

$$R\to \mathcal{O}(X)\to \mathcal{O}(U)$$

which gives each $\mathcal{O}(U)$ the structure of an $R$-algebra. Clearly, by construction, the tranisition maps

$$\mathcal{O}(V)\to \mathcal{O}(U)$$

for $V\subseteq U$ open neighborhoods of $x$ are maps of $R$-algebars and thus we see, by passing to the colimit, that $\mathcal{O}_{X,x}$ is an $R$-algebra. Moreover, for any open $U$ we have a factorization

$$R\to \mathcal{O}(U)\to \mathcal{O}_{X,x}$$

so that the maps $\mathcal{O}(U)\to\mathcal{O}_{X,x}$ are maps of $R$-algebras.

Let us moreover note that if $y$ is a point generalizing $x$ then the natural map

$$\mathcal{O}_{X,x}\to\mathcal{O}_{X,y}$$

comes from the fact that the neighborhoods containing $x$ all contain $y$ and thus in particular, this map is clearly also a map of $R$-algebras.

Finally, we observe that if $X=\mathrm{Spec}(A)$, then $A$ is an integral domain and if $\eta$ denotes the generic point of $X$ then the natural map

$$A=\mathcal{O}(X)\to \mathcal{O}_{X,\eta}$$

induces an isomorphism

$$\mathrm{Frac}(A)\xrightarrow{\approx}\mathcal{O}_{X,\eta}$$

The reason is simple. Namely, since every open subset of $X$ contains $\eta$ we hav, by definition, that

$$\mathcal{O}_{X,\eta}=\varinjlim_U \mathcal{O}(U)$$

where $U$ travels over all open subsets of $X$. But, it suffices to take this limit over a cofinal system of opens which can be taken to be the basic opens $D(f)$ for $f\in A$. But, $\mathcal{O}(D(f))=A_f$, the localization of $A$ at $f$, and the transition maps

$$\mathcal{O}(D(f))\to\mathcal{O}(D(g))$$

for $D(g)\subseteq D(f)$ are the inclusions $A_f\hookrightarrow A_g$ (note that $D(g)\subseteq D(f)$ means that $V(g)\supseteq V(f)$ which means that $\sqrt{(f)}\subseteq \sqrt{(g)}$ which means that $f=g^n a$ for some $a\in A$, from where it's clear that if we've inverted $g$ we've inverted $f$ and thus there is an inclusion $A_g\hookrightarrow A_f$). But, it's then clear that

$$\mathcal{O}_{X,\eta}=\varinjlim_U \mathcal{O}(U)=\varinjlim_{D(f)}\mathcal{O}(D(f))=\varinjlim_f A_f=\mathrm{Frac}(A)$$

as desired.

So then, if we take $A=k$ then the above discussion shows that we have a sequence of maps

$$k\to \mathcal{O}_{X,x}\to\mathcal{O}_{X,\eta}\cong \mathrm{Frac}(A)$$