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.

One Answer

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


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


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


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


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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