Topology on $p$-adic period ring in an article by Fontaine

Question

Fix a $p$-adic field $K$ with perfect residue field $k.$ Let $\mathcal{C}$ be the completion of the algebraic closure of $K,$ and let $$R = \varprojlim \mathcal{C}/p,$$ where the transition maps in the inverse limit are given by the Frobenius. One can then show that $R$ is a perfect valuation ring (of characteristic $p.$) We can then form $W(R),$ the ring of Witt vectors of $R.$

There is a natural topology on $W(R),$ making it into a topological ring, a basis of neighborhoods is given by $p^N W(R)+ W(I)$ for $N\geq 0$ and $I$ a non-zero ideal of $R.$

In this article, pg. 536, Fontaine uses the topology I just defined to define a topology on $W(R)[1/p] = K \otimes_{W(k)} W(R),$ by using what he calls the "tensor product topology". He then claims that the topology one get from the tensor product topology is the "same" (up to identifications under isomorphisms) as the one obtained by taking the topology coming from the inductive limit $$ \cdots \rightarrow W(R) \rightarrow W(R) \rightarrow \cdots$$ where transfer maps are multiplication by $p.$

My questions are the following:
1) What is precisely this tensor product topology? Naively, I would say that it is the topology on $K \otimes_{W(k)} W(R) \cong W(R)[1/p]$ where a basis of neighborhoods are given by $$(p^N W(R)+ W(I)) \otimes_{W(k)} K + W(R) \otimes_{W(k)} p^n \mathcal{O}_K$$ (where $\mathcal {O}_K$ is the valuation ring of $K).$ What makes me think this can not be a basis of neighborhoods comes from the fact that it seems to me (maybe erroneously) that $p^NW(R) \otimes_{W(k)} K \cong W(R)[1/p].$ Thus, it seems to me that Fontaine must have some other sort of topology in mind for this tensor product, or I am making a silly mistake. For example, what is a basis of neighborhoods for the tensor product topology? Is it part of a more general construction?
2. Why does the topology of the tensor product and inductive limit coincide?

Answer 1

Inductive limit topologies are always nasty to describe: Basically you can't do better than their very definition. Namely, a subset $U\subset W(R)[1/p]$ is an open neighborhood of $0$ if and only if for all $n$ the intersection $U\cap p^{-n} W(R)$ is open. In particular, this means that it must contain $p^N W(R)$ for some $N$, but also $p^{-n} W(\mathfrak a_n)$ for some open ideal $\mathfrak a_n\subset R$. So a basis of open neighborhoods of $0$ is given by the subsets

$$p^N W(R) + \sum_{n\geq 0} p^{-n} W(\mathfrak a_n)$$

for varying $N\geq 0$ and open ideals $\mathfrak a_n\subset R$, $n\geq 0$. Note that such subsets are never contained in $p^{-n} W(R)$ for any $n$...

As Dustin notes in the other question, passing to the condensed world is helpful here, as then inductive limits are completely naive (they are just inductive limits on $S$-valued points for any profinite $S$), and in fact there is no question about which condensed structure to put on $W(R)$ or $W(R)[1/p]$, and it is clear that $W(R)\otimes_{W(k)} K = W(R)[1/p]$ as condensed rings. (Also, as everything lies in the essential image of the fully faithful functor from compactly generated topological spaces to condensed sets, there is no loss of information in passing to the condensed world.)

Answer 2

What you are missing is that it is using the topology of K as a W(k) module (not as a topological field), so basis for K is like $p^{-n}W(k)$ for $n\in\mathbb{Z}$. So the basis of W(R)[1/p] is rather consisted of things like $p^{-n}$-scaled $p^{N}W(R)+W(I)$. Topology of inductive limit coincides with tensor product because inductive limit and tensor product commute and the inductive limit mentioned above is just W(R) tensored with the inductive limit W(k)->W(k)->... which is precisely expressing the basis of neighborhoods of 0 of K we chose, $p^{-n}W(k)$.

Answer 3

I think a good reference is Schneider's book Nonarchimedean Funtional Analysis, but my answer is simply an expanded version of GTA's comments.

In Chapter IV, §14, A and B you find a description of two possible "tensor product topologies" with which you can endow the tensor product of two locally convex vector spaces: in your setting, $W(R)$ is only a $W(k)$-module, but this makes almost no difference if you follow the proofs. The projective limit topology has as basis of neighborhoods the products $p^{-n}W(k)\otimes p^NW(R)\otimes W(I)$ by definition, because $K$ has, as a basis of neighborhoods, the $p^{-n}W(k)$. That the two coincide is Proposition 17.6, which holds in your case because your spaces are complete (for $W(R)$ this is again on page 536 of the paper by Fontaine); you need to use the definition of "locally convex final topology" given in Schneider's book, Chapter I, § 5, E to realise that the inductive limit topology given by multiplication by $p$ is the inductive tensor product topology.