# Natural map to the adic spectrum

Mathematics Asked by test123 on November 7, 2020

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 $$operatorname{Spec}Asubseteq C$$ denote the function field of $$C$$. There is a natural map $$nu:Cto operatorname{Spa}(K,k)$$, where the latter is the set of all valuation rings $$V$$ with $$ksubseteq Vsubseteq K$$ and $$operatorname{Frac}(V)=K$$, defined by $$xmapsto mathcal{O}_{C,x}$$.

I can see that $$ksubseteq K$$ 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 $$xin C$$ not a generic point.

However, I don’t understand the following things:

1. Why does $$ksubseteq mathcal{O}_{C,x}subseteq K$$ 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.

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

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

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

$$Uto Xto mathrm{Spec}(R)$$

in particular gives rise to a ring map

$$Rto mathcal{O}(X)to mathcal{O}(U)$$

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 $$Vsubseteq U$$ 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

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

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

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

$$mathcal{O}_{X,x}tomathcal{O}_{X,y}$$

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 $$fin A$$. But, $$mathcal{O}(D(f))=A_f$$, the localization of $$A$$ at $$f$$, and the transition maps

$$mathcal{O}(D(f))tomathcal{O}(D(g))$$

for $$D(g)subseteq D(f)$$ are the inclusions $$A_fhookrightarrow A_g$$ (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 $$ain A$$, from where it's clear that if we've inverted $$g$$ we've inverted $$f$$ and thus there is an inclusion $$A_ghookrightarrow A_f$$). 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

$$kto mathcal{O}_{X,x}tomathcal{O}_{X,eta}cong mathrm{Frac}(A)$$

Correct answer by Alex Youcis on November 7, 2020

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