In this Berkovich's paper, the following kind of algebra is studied: $$ A=A_{m,n}=k^\circ \langle T_1,\dots,T_m \rangle [[S_1,\dots,S_n]] $$ where $k$ is some non-archimedean field with non-trivial valuation and $k^\circ$ is the associated ring. If my understanding was right, it is stated (page370, the second last paragraph) that the formal scheme $\mathfrak X=\mathrm{Spf}A$ has generic fiber: $$ \mathfrak X_\eta = E^m(0,1) \times D^n(0,1) $$ where $E^m(0,1)$ and $D^n(0,1)$ are the closed and open polydisks of radius one at $0$.

Question: I hardly encounter (not necessarily convergent) formal power series in the literature. I was wondering if there is a good theory about it. For example, shall we simply consider $\mathcal M(k\langle T_1,\dots, T_m\rangle[[S_1,\dots,S_n]])$ or even $\mathcal M(k[[S_1,\dots,S_n]])$? Are they still analytic spaces?

And, I will really appreciate it if you may mention some reference.


In Berkovich's paper that you cite, he constructs a $k$-analytic space $\mathfrak{X}_{\eta}$ associated to what he calls a ``special formal $k^{\circ}$-scheme'' $\mathfrak{X}$. Such a formal $k^{\circ}$-scheme is defined to be locally of the form \begin{equation} k^{\circ}\langle T_1,\ldots,T_n\rangle[[S_1,\ldots,S_m]]/I, \end{equation} where $I$ is a finitely-presented ideal. This is the space you're interested in, and it's called the analytic generic fibre of $\mathfrak{X}$.

When there are no formal power series involved (i.e. $\mathfrak{X}$ is locally of the form $k^{\circ}\langle T_1,\ldots,T_n\rangle / I$), we say $\mathfrak{X}$ is topologically of finite presentation over $k^{\circ}$, and $\mathfrak{X}_{\eta}$ is simply (locally) the spectrum $$ \mathcal{M}(k\langle T_1,\ldots,T_n\rangle / I), $$ as you guessed.

However, when there are power series involved, the story becomes a little more complicated. In this case, $\mathfrak{X}_{\eta}$ is (to first approximation) locally the subset of $$ \mathcal{M}(k\langle T_1,\ldots,T_n,S_1\ldots,S_m\rangle / I) $$ where $|S_i| < 1$ for all $i=1,\ldots,m$. This explains the example of the product of the closed and open polydiscs that are cited in the question. Of course, the details of this are a little more complicated and are explained in Berkovich's paper, but this is the usual way that one constructs an analytic space associated to a formal power series ring.

  • $\begingroup$ Thanks a lot for your answer. So, do you mean we may anyway construct analytic spaces from formal power series rings? I am a student not an expert, but my guess is that we construct it as a union, just like the open disk is a union of closed disks. Maybe I was wrong, or? $\endgroup$
    – Hang
    Aug 27 '18 at 21:50
  • $\begingroup$ That's correct, this gives you a way to construct an analytic space from a formal power series ring: for example, the formal scheme $\mathrm{Spf}(k^{\circ}[[S]])$ has generic fibre equal to the open unit disc $D^1(0,1)$. In fact, this idea goes back to Raynaud, and in his construction it is really clear that the construction of $D^1(0,1)$ from $k^{\circ}[[S]]$ is by exhausting it by closed subdiscs, as you say. $\endgroup$
    – msteve
    Aug 28 '18 at 3:14
  • $\begingroup$ Thanks. I still feel confused about a point. What if $I$ contains some non-convergent power series? In this case, the algebra $k\langle T,S\rangle/I$ in your answer doesn't make sense since $I\not\subset k\langle T,S\rangle$. Maybe I miss something. $\endgroup$
    – Hang
    Aug 29 '18 at 17:37
  • $\begingroup$ This was an abuse of notation that I made, and it is hidden in the "to first approximation" comment. The correct thing to do is to pick a map $B = k^{\circ}\langle T \rangle[[S]] \to k^{\circ}\langle T \rangle [[S]]/I = A$, and you take the closed analytic subspace of $(\mathrm{Spf}(B))_{\eta}$ consisting of those seminorms vanishing on the elements of $I$. This is explained on pages 370-371 of Berkovich's paper. $\endgroup$
    – msteve
    Aug 31 '18 at 20:38

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