mixed characteristic

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For rings: $R$ is of mixed characteristic $(0, p)$ when $\operatorname{ch}R = 0$ but for some $I{~\trianglelefteq~}R$, $\operatorname{ch}R/I = p>0$.

For fields $K$ equipped with a valuation: the valuation ring $R$ is local with maximal ideal ${\mathfrak{m}}$ inducing a residue field, so $K$ is mixed characteristic when $\operatorname{ch}K = 0$ but $\operatorname{ch}R/{\mathfrak{m}}= p$,

$K$ of mixed characteristic $(0,p)$ means that $K$ has characteristic 0, but its residue field $\kappa$ has characteristic $p$)

Example: schemes defined over ${ {\mathbf{Z}}_{\widehat{p}} }$?

Motivations

Why non-Noetherian rings?

In mixed characteristic algebraic geometry, the basic geometric objects are smoot p-adic formal schemes over the ring of integers $\mathcal{O}_{K}$ of a complete algebraically closed nonarchimedean field $K / { {\mathbf{Q}}_p }$.These rings are often non-Noetherian, e.g. the value group of ${\mathcal{O}}_K$ is a divisible group. Replacing ${\mathcal{O}}_K$ with a DVR like ${ {\mathbf{Z}}_{\widehat{p}} }$ is not ideal. Applications of these non-Noetherian rings: perfectoid MOC geometry, descent for fine topologies( pro-etale, quasi-syntomic, v topology, arc topology), and the theory of delta rings.

Why derived/higher geometry?

Given a pair $(R, I)$ where $I{~\trianglelefteq~}R$ is a finitely generated ideal, the subcategory of $I{\hbox{-}}$adically complete $R{\hbox{-}}$modules is not abelian, while the category of derived $I{\hbox{-}}$adically complete modules is abelian.

The functor $R\mapsto \mathbf{D} { {}_{R}{\mathsf{Mod}} } { {}_{ \widehat{I} } }$ into the derived category of $I{\hbox{-}}$complete $R{\hbox{-}}$complexes is a stacks MOC for the flat topology, which does not hold at the level of triangulated categories.

Examples

${\mathbf{Z}}$ and the ideal $\left\langle{p}\right\rangle$ is $(0, p)$.
${\mathcal{O}}_K$ the ring of integers of any $K\in \mathsf{Field}_{/ {{\mathbf{Q}}}} $.
Localization: $L_{\left\langle{p}\right\rangle} {\mathbf{Z}}$ the p-local integers with the ideal $\left\langle{p}\right\rangle$ is $(0, p)$
Completion: ${ {\mathbf{Z}}_{\widehat{p}} }$ the p-adic with the ideal $\left\langle{p}\right\rangle$ is $(0, p)$