(I apologize in advance for these simple questions, I am a beginner trying to go through Scholze's paper Perfectoid Spaces).

Let $(R, R^+)$ be an affinoid $k$-algebra as defined in Scholze's paper Perfectoid Spaces, Definition 2.6. In particular, there exists a subring $R_0 \subset R$ such that $aR_0$, $a \in k^\times$ forms a basis of open neighborhoods of $0$.

I'm struggling to understand the role $R_0$ plays, and I suppose my real question is Question 3 below, so if you just want to answer that one, that's fine.

Question 1: What are some good examples (by good I perhaps mean those that arise in applications of perfectoid spaces) of $R$ and especially $R_0$ to keep in mind? I know examples of $R$ are quotients $Q$ of the Tate algebra $k\langle T_1, \ldots, T_n\rangle$. The topology on $Q$ I believe is induced from the Gauss/sup norm on the Tate algebra, I think in this case we can take $R_0=\{r \in Q \mid |r|\leq 1 \}$

Question 2. For any $x \in Spa(R, R^+)$, does $x$ induce the topology on $R$? I don't think so, because for example take the trivial valuation, which I believe is an element of $Spa(\mathbb{Q}_p, \mathbb{Z}_p)$, but doesn't induce the p-adic topology on $\mathbb{Q}_p$. Correct me if I'm wrong.

Question 3: Let $w \in k$ be topologically nilpotent,i.e. $|w|<1$. and let $h \in R$. In Remark 2.8, Scholze says that for sufficiently large $N$, $w^Nh \in R^+$, as $R^+ \subseteq R$ is open. Why? I tried reasoning as follows: let $R_0$ be as above. There exists $a \in k^\times$ such that $aR_0 \subseteq R^+$. We can find $N$ sufficiently large such that $|w^N| < |a|$. Got stuck.

  • 1
    $\begingroup$ Some small comments: In regards to Q2, note that the trivial valuation is not in $Spa(\mathbb{Q}_p,\mathbb{Z}_p)$ as it is not even continuous! Also in Q3, you seem to equate $w$ being topologically nilpotent and $|w|<1$. Be careful, these are not equivalent, though one does imply the other (as long as the valuation is not trivial...). $\endgroup$ Jan 28, 2017 at 23:15
  • $\begingroup$ Treatment of these sorts of foundational questions on adic spaces and the relevant categories of topological rings can be sketchy in places. I'd recommend the Scholze-Weinstein Berkeley lectures in p-adic geometry, as well as the notes from Brian Conrad's 2014-2015 Stanford learning seminar. $\endgroup$
    – dorebell
    Jan 15, 2019 at 23:51

1 Answer 1


Firstly, I don't think you should start to learn about adic spaces by thinking about perfectoid spaces. The sensible examples of adic spaces for a beginner to think about are sane Noetherian things like the closed polydisc etc. It's quite hard/messy to do any explicit basic calculations (of the sort one would like to do when getting the hang of thingS) with a general perfectoid ring because when you complete you lose a certain amount of control. I'll try to answer your questions.

The role of $R_0$ is that it defines a topology on $R$. Have you thought about examples like closed polydiscs or more general affinoids in the original sense of Tate? If $R$ is a reduced affinoid then one good choice for $R_0$ would be the functions with supremum norm at most 1; conversely, if you know this ring then (using powers of a fixed element $\pi\in k$ with norm less than 1) you know what it means for a function to be small (it's in $\pi^NR_0$) so now you know what it means for two functions to be close (their difference is small; the topology on a reduced affinoid is defined this way, so two functions are close iff they're close everywhere) so now you know the topology on $R$. More abstractly you can define the topology on $R$ to have basis $r+\pi^NR_0$ as $r,N$ vary, but the geometric intuition is what I said above. So to give $R_0$ is to give the topology.

Q1) Good examples: forget perfectoids. Let $R$ be $\mathbb{Q}_p\langle T_1,T_2,\ldots,T_n\rangle$ and let $R_0$ be $\mathbb{Z}_p\langle T_1,\ldots,T_n\rangle$. If you're not happy with the topology on $R$ then convince yourself that $r+p^nR_0$ give a basis for the topology. Next try a reduced affinoid. The spectral seminorm is a norm on that; try examples like an annulus $\mathbb{Q}_p\langle X,Y\rangle/(XY-p)$. Figure out what a general element of that ring looks like. Figure out which elements have supremum seminorm at most 1. That's a good example for $R_0$. Try $XY-p^n$ or $XY-1$ as the relationship instead. Check that $R_0$ is what you think it is. Check $r+p^nR_0$ give a basis for the topology -- do you see why this is obvious?

Now try a non-reduced affinoid, because here there is an extra phenomenon which is less geometric -- nilpotents in the structure sheaf. The simplest example is $\mathbb{Q}_p[X]/X^2$ with its topology induced from the obvious isomorphism with $\mathbb{Q}_p^2$, and already this exhibits phenomena that you need to understand. Check that the supremum seminorm is not a norm. Check that if $S$ is the elements of seminorm at most 1 then $r+p^nS$ do not form a basis for the topology. We fix this by letting $R_0$ be $\mathbb{Z}_p[X]/X^2$. Check that $R_0$ defines the topology in the usual way. In particular your "Q" example in your question is not right in general -- only when the affinoid is reduced.

Now try perfectifying one of these (not the non-reduced one, but any or all of the other ones). If you want.

Q2) No. This sort of question just shows that you're currently blind to standard behaviour of adic spaces so shouldn't be thinking about crazy non-Noetherian perfectoid spaces yet. Think about the closed unit disc. Adic spaces have lots of strange points, points corresponding to higher rank valuations, weird points if the underlying field isn't spherically complete etc. Think about "standard" points first; if you can't understand standard points like the $\mathbb{Q}_p$-points of the closed unit discs you'll never get anywhere. Think about the point $T=0$ on the closed unit disc. Figure out what the corresponding valuation is, understand the induced topology on $R$, understand it geometrically. Two functions are close for the valuation topology corresponding to the point $T=0$ if the functions take values at $T=0$ which are close to each other. That's what the valuation is telling you about -- behaviour at 0. The topology on $R$ is telling you something much more global -- it's telling you if two functions are close everywhere. We want the valuations to be continuous -- and that has a geometric meaning, and that meaning is something you should understand and it is not saying that the topology on $R$ (a global thing, telling us about when a function is small everywhere) is determined by behaviour near one point (indeed, behaviour at a point is only telling us about the functions at this point). The valuations are continuous, which means geometrically that the collection of functions $g$ close to $f$ for the valuation topology (i.e. such that $g(0)$ is close to $f(0)$) is open (i.e. contains the functions which are close to $f$ everywhere -- because "close" means "uniformly close" -- that's the definition of the topology on a reduced affinoid). Explained this way, continuity is clear, and it's also clear that the valuation topology is very different to the norm topology. You have to understand the closed unit disc very well before you can understand a perfectoid space. Indeed, you should probably understand the closed disc rigid space, then the closed disc Berkovich space, then the closed disc adic space, before you understand the perfectoid "closed disc" obtained by throwing in all the $p$-power roots of $T$.

Q3) $R^+$ is open and contains 0. So $R^+$ contains all functions which are globally small. Now $h$ is a function on an affinoid so it's globally bounded, so clearly $w^Nh$ for some large $N$ is globally small. This is the geometric idea. If you did Q1 then translate this into algebra -- it's not too hard. If you need a hint, think about the basis $r+w^NR_0$ of the topology.

Good luck! I had read all of Bosch-Guentzer-Remmert and some notes on Berkovich spaces, so I had a good feeling for both classical rigid spaces and Berkovich spaces, before I learnt adic spaces (from Wedhorn's notes), and then after that was when I dared to take on perfectoid spaces. You need the geometric intuition to guide you through the algebra.

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    $\begingroup$ Great answer. People shouldn't put the cart before the horse. $\endgroup$
    – nfdc23
    Jan 28, 2017 at 22:23

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