10
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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 $\operatorname{Spv} A$? $\endgroup$ Mar 5, 2019 at 4:45
  • 1
    $\begingroup$ 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 $U$. $\endgroup$ 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 $\nu$. $\endgroup$ Mar 7, 2019 at 19:16

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.