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](http://www-personal.umich.edu/%7Ebhattb/teaching/prismatic-columbia/lecture5-prismatic-site.pdf).


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.](https://arxiv.org/abs/1907.10525)


Or an upcoming work of Bhatt and Scholze, see the [notes of Scholze](http://math.bu.edu/people/jsweinst/rampage/Scholze.pdf) 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 $(W(k),(p))$.



Finite non-empty coproduct in the absolute prismatic site
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.


