# Finite non-empty coproduct in the absolute prismatic site

Let $$(R/A)_\Delta$$ be the prismatic site over $$R$$ relative to a prism $$(A, I)$$, then it is known that $$(R/A)_\Delta$$ admits finite non-empty coproduct, for instance, by Cor. 5.2 in Bhatt's lecture notes V on prismatic cohomology.

My question is, if we consider the absolute prismatic site $$R_\Delta$$, does $$R_\Delta$$ still always have finite non-empty coproduct?

If we restrict to the existence of finite non-empty self coproducts, it seems to me that the answer is yes, as it has been suggested (implicitly) in work:

Arthur-César Le Bras and Johannes Anschütz: Prismatic Dieudonné theory.

Or an upcoming work of Bhatt and Scholze, see the notes of Scholze on this topic in the RAMpAGe Seminar.

A more concrete question is the following:

Let $$R=O_K$$ be a complete discrete valuation ring of mixed characteristic with perfect residue field $$k$$, and let $$(\mathfrak{S},(E))$$ be a Breuil-Kisin prism in $$R_\Delta$$, i.e., $$\mathfrak S=W(k)[\![u]\!]$$ and $$E \in W(k)[u]$$ is the Eisenstein polynomial of a uniformizer of $$O_K$$. In the above two works, they claim the self product of $$(\mathfrak{S},(E))$$ in $$R_\Delta$$ exists and is equal to a prismatic envelop of $$\mathfrak{S}\otimes_{W(k)}\mathfrak{S}$$. Why is the tensor product over $$W(k)$$? I know for $$(R/A)_\Delta$$, the similar construction takes tensors over $$A$$, but it is clear that $$(\mathfrak{S},(E))$$ is not a prism over $$(W(k),(p))$$.

Yes, it does admit nonempty finite coproducts. If you have two prisms $$(A_1,I_1)$$ and $$(A_2,I_2)$$ with maps $$R\to A_i/I_i$$, you need to find the initial prism $$(A,I)$$ with maps from both $$(A_i,I_i)$$ such that the two induced maps $$R\to A_i/I_i\to A/I$$ agree. For this, start with $$A_0=A_1\hat{\otimes}_{\mathbb Z_p} A_2$$ (where the tensor product is $$(p,I_1,I_2)$$-adically completed, say) and then take a suitable prismatic envelope to ensure the required conditions.
You raise an interesting point, that if $$R=\mathcal O_K$$, then one can take the tensor product over $$W(k)$$ instead of $$\mathbb Z_p$$, where $$k$$ is the residue field of $$\mathcal O_K$$. The reason is that for any prism $$(A,I)$$ with a map $$\mathcal O_K\to A/I$$, you get in particular a map $$W(k)\to A/I$$, and this lifts uniquely to a map $$W(k)\to A$$: This is using that the $$p$$-completed cotangent complex of $$W(k)/\mathbb Z_p$$ vanishes, or some more concrete assertion involving Teichmüller lifts. The map $$W(k)\to A$$ is also necessarily a map of $$\delta$$-rings. So all those $$A$$'s live canonically over $$W(k)$$, so one can as well directly take the tensor product there.
• Dear Peter, thank you very much for the answer! And for the first part, when you say take a suitable prismatic envelope, I guess you mean that we should first regard $A_0$ (together with a certain ideal) as a $\delta$-pair over $(A_1, I_1)$ (or $(A_2, I_2)$), and then take the prismatic envelope relative to it, right? Or do we know that the "prismatic envelope" also exists in this absolute setting? Jan 28 at 1:18