## 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.