The product of two diamonds

Let {\mathcal{P}\mathrm{erf}} be the category of perfectoid spaces in characteristic {p}, equipped with its pro-étale topology. Any {X\in\mathcal{P}\mathrm{erf}} gives rise to a sheaf {h_{X}=\mathrm{Hom}_{\mathrm{Adic}}(-,X)} on {\mathcal{P\mathrm{erf}}}. By definition, a diamond {\mathcal{D}} is a sheaf on {\mathcal{P}\mathrm{erf}} which admits a relatively representable surjection {h_{X}\rightarrow\mathcal{D}} which pulls back to a pro-étale cover of {Y} under any map {h_{Y}\rightarrow\mathcal{D}}; we say {\mathcal{D}} admits a pro-étale cover by the representable sheaf {h_{X}}. We can also view {\mathcal{D}} as the quotient of {h_{X}} by the pro-étale equivalence relation {h_{X}\times_{\mathcal{D}}h_{X}\rightrightarrows h_{X}} (note that {h_{X}\times_{\mathcal{D}}h_{X}=h_{Z}} is representable, and that the maps {Z\rightarrow X} are pro-étale). Morphisms of diamonds are simply sheaf maps {\mathcal{E}\rightarrow\mathcal{D}}. Let {\mathcal{D}\mathrm{ia}} denote the category of diamonds.

Since the category of perfectoid spaces (in whatever characteristic) admits fiber products, it’s somewhat formal to check that for any maps of diamonds {\mathcal{E}\rightarrow\mathcal{D},\mathcal{F}\rightarrow\mathcal{D}}, the fiber product {\mathcal{E}\times_{\mathcal{D}}\mathcal{F}} exists as a diamond. More strangely, however, we also have “baseless” products in {\mathcal{D}\mathrm{ia}}:

Theorem. For any two diamonds {\mathcal{D},\mathcal{E}}, the product {\mathcal{D}\times\mathcal{E}} is canonically a diamond, where the product is taken in the category of sheaves of sets on {\mathcal{P}\mathrm{erf}}.

This is another sense in which diamonds are a “topological” sort of gadget: one finds in practice that most categories of ringed spaces (e.g. schemes, complex analytic spaces, rigid analytic spaces, etc.) have a final object, so “baseless” fiber products can’t exist. But diamonds, like topological spaces, don’t have any sort of structure sheaf, and they don’t have a final object.

Curiously, the above theorem is never stated explicitly in the Berkeley notes; it’s certainly implicit, however, since e.g. the diamond {(\mathrm{Spd}\,\mathbf{Q}_{p})^{n}} is mentioned repeatedly. Anyway, the key point in the proof of this theorem is the following claim:

Proposition. Let {(A,A^{+})}, {(B,B^{+})} be two perfectoid Tate-Huber pairs in characteristic {p}. Then {h_{\mathrm{Spa}(A,A^{+})}\times h_{\mathrm{Spa}(B,B^{+})}\cong h_{X}} for a certain perfectoid space {X} whose formation is bi-functorial in the data of {(A,A^{+})} and {(B,B^{+})}.

Proof. We begin by defining {X}. (Hat tip to Jared, who told me the definition of {X}, from which I reverse-engineered everything else below.) Set {C=A\otimes_{\mathbf{F}_{p}}B}, and let {C^{+}} be the integral closure of {A^{+}\otimes_{\mathbf{F}_{p}}B^{+}} in {C}. Choose pseudouniformizers {\varpi_{A}\in A,\varpi_{B}\in B}, and set {I=(\varpi_{A},\varpi_{B})\subset C^{+}} (here we abbreviate {\varpi_{A}\otimes1\in C} by {\varpi_{A}}, and likewise for {\varpi_{B}}; we continue to use this abbreviation in what follows). Let {D^{+}} be the {I}-adic completion of {C^{+}}. Then we define

\displaystyle X=\mathrm{Spa}(D^{+},D^{+})\smallsetminus\left\{ x\mid|\varpi_{A}\varpi_{B}|_{x}=0\right\} .

To see that {X} is perfectoid, consider the open subsets

\displaystyle \begin{array}{rcl} U_{n} & = & \left\{ x\mid|\varpi_{A}|_{x}^{n}\leq|\varpi_{B}|_{x}\neq0,|\varpi_{B}|_{x}^{n}\leq|\varpi_{A}|_{x}\neq0\right\} \end{array}

of {\mathrm{Spa}(D^{+},D^{+})}. It’s easy to see that { U_n \subset U_{n+1}} and {X=\bigcup_{n\geq1}U_{n}}. Furthermore, {U_{n}=U\left(\frac{\{\varpi_{A}^{n+1},\varpi_{B}^{n+1}\}}{\varpi_{A}\varpi_{B}}\right)} is a rational subset of {\mathrm{Spa}(D^{+},D^{+})}, so we may describe it explicitly as an affinoid adic space. Precisely, let {C_{n}=A\otimes_{\mathbf{F}_{p}}B} and let {C_{n}^{+}} be the integral closure of {(A^{+}\otimes_{\mathbf{F}_{p}}B^{+})[\frac{\varpi_{A}^{n}}{\varpi_{B}},\frac{\varpi_{B}^{n}}{\varpi_{A}}]} in {C_{n}}; give {C_{n}^{+}} the {I\cdot C_{n}^{+}}-adic topology, and give {C_{n}} the topology making {C_{n}^{+}} an open subring. (Note that the {I\cdot C_{n}^{+}}-adic, {\varpi_{A}\cdot C_{n}^{+}}-adic, and {\varpi_{B}\cdot C_{n}^{+}}-adic topologies on {C_{n}^{+}} all coincide.) Let {D_{n}} and {D_{n}^{+}} be the completions of {C_{n}} and {C_{n}^{+}} for these topologies. Then

\displaystyle \left(\mathcal{O}(U_{n}),\mathcal{O}(U_{n})^{+}\right)=(D_{n},D_{n}^{+}).

One checks directly that {D_{n}} is a complete perfect uniform Tate ring, which exactly characterizes the perfectoid rings in characteristic {p}. Therefore each {U_{n}} is perfectoid, so {X} is perfectoid.

Next we verify that {X} has the claimed property at the level of sheaves. Note that we have natural continuous ring maps {A^{+}\rightarrow D^{+},B^{+}\rightarrow D^{+}} inducing morphisms {\mathrm{Spa}(D^{+},D^{+})\rightarrow\mathrm{Spa}(A^{+},A^{+})} and {\mathrm{Spa}(D^{+},D^{+})\rightarrow\mathrm{Spa}(B^{+},B^{+})}, and the latter morphisms restricted to {X} factor through natural morphisms {\mathrm{pr}_{A}:X\rightarrow\mathrm{Spa}(A,A^{+})} and {\mathrm{pr}_{B}:X\rightarrow\mathrm{Spa}(B,B^{+})}, respectively.

Choose any {Y\in\mathcal{P}\mathrm{erf}}, and suppose we’re given a morphism {Y\rightarrow X}, i.e. an element of {h_{X}(Y)}. Composing this morphism with the morphisms {\mathrm{pr}_{A},\mathrm{pr}_{B}}, we get morphisms {Y\rightarrow\mathrm{Spa}(A,A^{+})}, {Y\rightarrow\mathrm{Spa}(B,B^{+})}, i.e. an element of {h_{\mathrm{Spa}(A,A^{+})}(Y)\times h_{\mathrm{Spa}(B,B^{+})}(Y)}. This association is clearly functorial in {Y}, and so defines a map of sheaves

\displaystyle F:h_{X}\rightarrow h_{\mathrm{Spa}(A,A^{+})}\times h_{\mathrm{Spa}(B,B^{+})}.

To go the other way, assume that {Y} is affinoid perfectoid, say {Y=\mathrm{Spa}(R,R^{+})} for {(R,R^{+})} some perfectoid Tate-Huber pair in characteristic {p}. Suppose we’re given an element of {h_{\mathrm{Spa}(A,A^{+})}(Y)\times h_{\mathrm{Spa}(B,B^{+})}(Y)}. This is equivalent to the data of a pair of continuous ring maps {f:A\rightarrow R}, {g:B\rightarrow R} such that {f(A^{+}),g(B^{+})\subset R^{+}}. Consider the evident ring map {f\otimes g:C\rightarrow R}. One checks directly that {(f\otimes g)(C^{+})\subset R^{+}}, using the obvious inclusion

\displaystyle (f\otimes g)(A^{+}\otimes_{\mathbf{F}_{p}}B^{+})\subset R^{+}

together with the fact that {R^{+}} is integrally closed in {R}. Choose a pseudouniformizer {\varpi_{R}\in R}; replacing {\varpi_{R}} by {\varpi_{R}^{1/p^{j}}} if necessary, we may assume that {\varpi_{R}} divides both {f(\varpi_{A})} and {g(\varpi_{B})} in {R^{+}}. This immediately implies that the ring map {f\otimes g:C^{+}\rightarrow R^{+}} induces compatible ring maps {C^{+}/I^{n}\rightarrow R^{+}/\varpi_{R}^{n}}, so passing to the inverse limit we get a continuous ring map {f\hat{\otimes}g: D^{+} \to R^{+}}. Passing to Spa’s, we may consider the composite

\displaystyle i:\mathrm{Spa}(R,R^{+})\rightarrow\mathrm{Spa}(R^{+},R^{+})\overset{\mathrm{Spa}(f\hat{\otimes}g)}{\longrightarrow}\mathrm{Spa}(D^{+},D^{+})

where the lefthand arrow is the evident open immersion. Since {(f\hat{\otimes}g)} carries {\varpi_{A}\varpi_{B}} to a unit in {R}, {i} factors through the open subset {X\subset\mathrm{Spa}(D^{+},D^{+})}, so we get a map {i:\mathrm{Spa}(R,R^{+})\rightarrow X}, i.e. an element of {h_{X}(Y)}. Summarizing our efforts in this paragraph so far, we’ve described a natural map of sets

\displaystyle G:h_{\mathrm{Spa}(A,A^{+})}(Y)\times h_{\mathrm{Spa}(B,B^{+})}(Y)\rightarrow h_{X}(Y)

for any affinoid perfectoid {Y}. One checks directly that for any map {Y\rightarrow Y'} of affinoid perfectoids, the associated diagram

\displaystyle \begin{array}{rcl} h_{\mathrm{Spa}(A,A^{+})}(Y')\times h_{\mathrm{Spa}(B,B^{+})}(Y') & \rightarrow & h_{X}(Y')\\ \downarrow\quad\quad\quad\quad\quad\quad\,\, & & \downarrow\\ h_{\mathrm{Spa}(A,A^{+})}(Y)\times h_{\mathrm{Spa}(B,B^{+})}(Y) & \rightarrow & h_{X}(Y)\end{array}

commutes, so {G} extends to a morphism of sheaves. Finally, one checks that {F} and {G} are naturally inverse by staring hard at the above constructions. {\square}

Though it’s not strictly necessary for the above proof, let me note that {D^{+}} is a perfectoid Huber ring in Gabber-Ramero’s sense, and hence is sheafy.

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