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 $(W(k),(p))$.
 A: 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.
