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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.

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    $\begingroup$ 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 $\operatorname{Spv} A$? $\endgroup$
    – Takumi Murayama
    Commented Mar 5, 2019 at 4:45
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    $\begingroup$ 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 $U$. $\endgroup$
    – user30211
    Commented Mar 5, 2019 at 5:05
  • $\begingroup$ 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 $\nu$. $\endgroup$
    – user30211
    Commented Mar 7, 2019 at 19:16

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