# Adjoint for Spv, the Space of Valuations

The contravariant functor $$\text{Spec} : \text{Rng} \rightarrow \text{RngSp}$$ from rings to locally ringed spaces, sending a ring to its spectrum, and the contravariant functor $$\text{Glob} : \text{RngSp} \rightarrow \text{Rng}$$ from locally ringed space to rings sending $$X$$ to $$\mathcal{O}_X (X)$$ are mutually right adjoint. That is, there is an isomorphism of hom-sets $$\text{Rng}(A, \mathcal{O}_X(X) ) \cong \text{Sch}(X, \text{Spec}(A))$$ natural in $$A$$ and $$X$$.

One reason this result is nice to me is that is shows how the functor $$\text{Spec}$$ is unique in a certain respect; once we fix the global sections functor (which does not mention prime ideals or ideals at all), considering $$\text{Spec}$$ is inevitable. Supposing we wanted to represent a ring as a sheaf of rings over some topological space, this adjoint suggests we have the right way.

What about the contravariant functor $$\text{Spv}$$, which sends a ring to its space of valuations? I would like some kind of adjoint, or something similar to the above.

• I believe this holds for the adic spectrum $\operatorname{Spa}A$; see Prop. 2.1(ii) in Huber's "A generalization of formal schemes and rigid analytic varieties". I don't know what happens for $\operatorname{Spv} A$, however. Is there a commonly accepted definition for the structure sheaf on $\operatorname{Spv} A$? – Takumi Murayama Mar 5 at 4:45
• I think we have a canonical way of putting a sheaf on $X = \text{Spv}(A)$, but I don't know if it's commonly accepted. The way I know it is done in the paper "Spectral Schemes as Ringed Lattices" by Thierry Coquand. For an element $a$ of $K = \text{Frac}(A)$, put $V(a)$ to be the set of valuations whose value at $a$ is non-negative. For an open set $U$ of $X$, let $\mathcal{O}_X (U)$ be the set of $a \in K$ such that $V(a)$ contains $U$. – Dean Young Mar 5 at 5:05
• By the way, one way of seeing how we put a sheaf structure on $\text{Spv}$ is to ask that its stalk at the point $\nu$ be the valuation ring corresponding to $\nu$. – Dean Young Mar 7 at 19:16

Main Theorem: Let $$\text{ValSp}$$ be the category of ringed spaces whose stalks are valuation rings, which I call valuation ringed spaces. We assume that all valuation-ringed spaces are irreducible here, and have $$\text{limit}_{U \subset X \text{ dense}} \mathcal{O}_X (U)$$ a field, but this can be made to work in more general situations. Consider the functor $$\text{Rat}$$, the rational functions functor, from $$\text{ValSp}$$ to $$\text{Rng}$$, which assigns to a valuation ringed space $$X$$ the ring $$\text{Rat}(X) := \text{colimit}_{U \subset X \text{ nonempty}} \mathcal{O}_X (U)$$, where the colimit is taken over dense open sets of $$X$$, and acts on morphisms in the canonical way. $$\text{Rat}$$ is right adjoint to $$\text{Spv}$$.

In the first section I develop the tools to make things precise. In the second section I prove the theorem. I actually prove the theorem for fields. The theorem for rings holds, but takes a little more work.

To anyone who would like to improve this answer, the bounty is still open. My feeling is that the bounty should not go to waste

Definition: Let $$K$$ be a field. For $$a_1, ..., a_n \in K$$, let $$V(a_1, ..., a_n) = \{ R \text{ a valuation ring in } K : a_1, ..., a_n \in R \}$$ All valuations on $$K$$ are contained in $$V(1)$$. The sets $$\{ V(a_1, ..., a_n) : a \in A \}$$ are closed under intersection, as $$V(a_1, ..., a_n) \cap V(b_1, ..., b_n) = V(a_1, ..., a_n, b_1, ..., b_n)$$. So the sets $$\{ V(a_1, ..., a_n) : a_1, ..., a_n \in A \}$$ form the base for a topology on the set of valuation rings of $$A$$. Write $$\text{Spv}(K)$$ for the topological space on the set of valuation rings of $$K$$ with basis $$\{ V(a_1, ..., a_n) : a_1, ..., a_n \in K \}$$. A reference for this construction can be found here.

Lemma: Let $$K$$ be a field, and take $$f_1, ..., f_n, g_1, ..., g_n \in K$$. Then $$V(g_1, ..., g_n) \subset V(f_1, ..., f_n)$$ if and only if $$E(f_1, ..., f_n) \subset E(g_1, ..., g_n)$$, where we write $$E(f_1, ..., f_n)$$ for the integral closure of the ring generated by $$f_1, ..., f_n$$ in $$K$$.

Proof: First suppose that $$V(g_1, ..., g_n) \subset V(f_1, ..., f_n)$$. Then any valuation ring containing $$g_1, ..., g_n$$ contains $$f_1, ..., f_n$$. Hence the intersection $$E(g_1, ..., g_n)$$ of all valuation rings containing $$g_1, ..., g_n$$ contains $$f_1, ..., f_n$$, so that it contains $$E(f_1, ..., f_n)$$.

Conversely, suppose $$V(g_1, ..., g_n) \not\subset V(f_1, ..., f_n)$$. Then there is some valuation ring $$R$$ of $$K$$ containing $$g_1, ..., g_n$$ and not containing one of $$f_i$$. Hence the intersection $$E(g_1, ..., g_n)$$ of all valuation rings containing $$g_1, ..., g_n$$ does not contain $$f_i$$, and in particular does not contain the integral closure $$E(f_1, ..., f_n)$$ of the ring generated by the $$f_i$$.

Definition: Let $$K$$ be a field. We make $$\text{Spv}(K)$$ into a presheaf, for each open set $$U \subset \text{Spec}(K)$$, $$\text{Spv}(K)^\# (U) = \{ a \in K : a \in R \forall R \in U \}$$.

In fact, $$\text{Spv}(K)$$ is a sheaf. $$\text{Spv}(K)^\# (V(f_1, ..., f_n))$$ can be identified with $$E(f_1, ..., f_n)$$. The stalk of $$\text{Spv}(K)$$ at a valuation $$R \subset K$$ is $$R$$, which is a local ring. In particular, $$\text{Spec}(K)$$ is a locally ringed space.

Definition: (The Valuation Spectrum Functor) We make $$\text{Spv}$$ into a contravariant functor from $$\text{Fld}$$ to $$\text{ValSp}$$ as follows: let $$f : K \rightarrow L$$ be a morphism of fields. As a map of topological spaces, we define $$\text{Spv}(f) : \text{Spv}(L) \rightarrow \text{Spv}(K)$$ to send a valuation ring $$R \in \text{Spv}(L)$$ to $$f^{-1} (R)$$ in $$\text{Spv}(K)$$. For each open set $$V(a_1, ..., a_n)$$ of $$\text{Spv}(K)$$, $$\mathcal{O}_{ \text{Spv}(K)}(V(a_1, ..., a_n)) = E(a_1, ..., a_n)$$, and $$\text{Spf}(f)^{-1} (V(a_1, ..., a_n)) = V(f(a_1), ..., f(a_n))$$, so that \begin{align*} &\ \mathcal{O}_{\text{Spv}(L)} ( \text{Spf}(f)^{-1} (V(a_1, ..., a_n))) \\ =&\ \mathcal{O}_{\text{Spv}(L)}(V(f(a_1), ..., f(a_n))) \\ =&\ E(f(a_1), ..., f(a_n)) \end{align*} We define $$\text{Spv}(f)^\# (V(a_1, ..., a_n)) : E(a_1, ..., a_n) \rightarrow E(f(a_1), ..., f(a_n))$$ to be the map induced by the map $$f: K \rightarrow L$$.

Definition: We define a contravariant functor $$\text{Rat} : \text{ValSp} \rightarrow \text{Fld}$$, called the rational functions functor. For a valuation ringed space $$X$$, we set $$\text{Rat} (X) = \text{colimit}_{U \text{ dense in } X } \mathcal{O}_X (U)$$. For a map $$f : X \rightarrow Y$$ in $$\text{RngSp}$$, there is a cone $$\mathcal{O}_Y (U) \rightarrow \text{Rat}(X)$$, where $$U$$ ranges over open sets of $$Y$$ dense in $$Y$$. By the universal property of colimit, there is a map $$\text{Rat}(Y) \rightarrow \text{Rat}(X)$$, and we set $$\text{Rat}(f)$$ to be this map.

Now I estalish the claimed adjunction.

Definition: (The first unit of the adjunction) For each field $$K$$, we may identify $$\text{Rat}(\text{Spv}(K))$$ with $$K$$ itself. Hence $$\text{Rat} \circ \text{Spv}$$ is the identity functor. In particular, we set $$\eta : 1_{\text{Fld}} \rightarrow \text{Rat} \circ \text{Spv}$$ to be the identity natural transformation.

Definition: (The second unit of the adjunction) We define a natural transformation $$\theta : 1_{\text{FldSp}} \rightarrow \text{Spv} \circ \text{Rat}$$ as follows. Take a locally ringed space $$X$$ such that $$\text{Rat}(X)$$ is a field, and let $$Y = \text{Spv}(\text{Rat}(X))$$. We define $$\theta_X : X \rightarrow Y$$ as follows. For a point $$x \in X$$, let $$\theta_X (x)$$ be the integral closure of the stalk $$\mathcal{O}_{X, x}$$ in $$\text{Rat}(X)$$, which is necessarily a valuation ring and hence a point in $$Y$$. For an open set $$U$$ in $$\text{Rat}(X)$$, define $$\theta_X^\#(U) : \mathcal{O}_{Y}(U) \rightarrow \mathcal{O}_X(\theta_X^{-1} (U))$$ to be the unique map making the following diagram commute:

Proposition:

(i) $$\theta * \text{Spv} : \text{Spv} \rightarrow \text{Spv} \circ \text{Rat} \circ \text{Spv}$$ is the identity natural transformation.

(ii) $$\text{Rat} * \theta : \text{Rat} \circ \text{Spv} \circ \text{Rat} \rightarrow \text{Rat}$$ is the identity natural transformation.

Proof: (i) Let $$K$$ be a field and put $$X = \text{Spv}(K)$$. Now $$\text{Rat}(X) = K$$ and $$\text{Spv}(\text{Rat}(X)) = \text{Spv}(K) = X$$. As a map of topological spaces, $$\theta_X : X \rightarrow X$$ sends a point $$R \in X$$ to the integral closure of $$R = \mathcal{O}_{X, x}$$ in $$K$$. But this is identically $$\mathcal{O}_{X, x}$$, as valuation rings are integrally closed. So $$\theta_X : X \rightarrow X$$ is the identity map. Hence, for each open set $$U$$ in $$X$$, $$\mathcal{O}_X (\theta_X^{-1} (U)) = \mathcal{O}_X (U)$$. For each open set $$D(f)$$ in $$X$$, $$\theta_X^\#$$ is the unique map making the diagram below commute:

And so $$\theta_X^\#$$ must be the identity map.

(ii) Let $$X$$ be a valuation ringed space such that $$\text{Rat}(X)$$ is a field. Put $$Y = \text{Spv}(\text{Rat}(X))$$. Then $$\text{Spv}(X) = \text{Spv}(Y)$$. The following diagram commutes for each open set $$U$$ of $$Y$$:

This commutativity shows that $$\text{Rat}(Y) \rightarrow \text{Rat}(X)$$ is the canonical map in the colimit; $$\theta_X^\# : \mathcal{O}_Y \rightarrow (\theta_X)_* \mathcal{O}_X$$ induces the identity map $$\text{Spv}(X)$$ after applying $$\text{Rat}$$.

Theorem: The contravariant functors $$\text{Spv}: \text{Fld} \rightarrow \text{FldSp}$$ and $$\text{Rat} : \text{FldSp} \rightarrow \text{Fld}$$ are right adjoint to each other.

Proof: With $$\eta : 1_C \rightarrow \text{Rat} \circ \text{Spv}$$ the identity natural transformation, $$\text{Spv} * \eta : \text{Spv} \rightarrow \text{Spv} \circ \text{Rat} \circ \text{Spv}$$ and $$\eta * \text{Rat} : \text{Rat} \rightarrow \text{Rat} \circ \text{Spv} \circ \text{Rat}$$ are the identity natural transformations. $$\text{Rat} * \theta$$ and $$\theta * \text{Spv}$$ are the identity natural transformations. So we have the triangle identities: $$\text{Spv} * \eta \circ \theta * \text{Spv} = 1_{\text{Spv}}$$ and $$\text{Rat} * \theta \circ \eta * \text{Rat} = 1_{\text{Rat}}$$. We can conclude that $$\text{Spv}$$ and $$\text{Rat}$$ are right adjoint to each other.

If anyone wants to complete writeups of the smaller gaps, I will also give them the bounty for this question.

• I haven't tried to follow the whole answer, but how is Rat functorial with respect to, say, the inclusion of a proper closed subspace? To define Rat on $f : X \to Y$, don't you need to assume that the preimage of each dense open subset of $Y$ is dense in $X$? – Reid Barton Mar 12 at 13:51
• You're right. I want to assume both are irreducible spaces. – Dean Young Mar 13 at 0:49
• I don't think irreducibility is the issue: if you take a closed inclusion as @ReidBarton suggests, the preimage of the complement of $X$ in $Y$ is the empty set. I think you instead need to restrict to the subcategory of injective ring maps (equivalently, dominant rational maps of varieties). – Devlin Mallory Mar 13 at 2:33
• Oh, hey Devlin. I think you're right. Thanks, I might not have figured that out otherwise. – Dean Young Mar 13 at 3:01