Let $A$ be a discrete ring, and $\mathrm {Spa}(A,A)$ the space of (continuous) valuations on $A$, up to equivalence, which take values $\le1$ on all of $A$.

In Clausen–Scholze’s Condensed Mathematics notes (page 64, just after Definition 9.5) it is claimed that one can identify $\mathrm{Spa}(A,A)$ with the the set of ring homomorphisms $A\to V$ where $V$ is a valuation ring, modulo faithfully flat maps of valuation rings (i.e. if $V\to W$ is such a map, than $A\to V$ is equivalent to $A\to V\to W$). I don’t understand why this holds. Here is an idea of an incomplete proof.

The natural way to identify the latter set with $\mathrm{Spa}(A,A)$ appears to be the following. Let $\varphi:A\to V$ be given. Then since $V$ is a valuation ring, there exists a valuation $|\cdot|_V$ on its field of fractions which takes values $\le1$ on $V$. (See e.g. Wedhorn’s notes.) Then $|\cdot|_V\circ \varphi$ (or rather its equivalence class) is an element of $\mathrm{Spa}(A,A)$.

In the other direction, let $|\cdot|\in\mathrm{Spa}(A,A)$. Its support $\mathfrak p$ is a prime ideal in $A$. Note that it only depends on the equivalence class of $|\cdot|$ (Wedhorn 1.27). Let $\mathfrak m$ be a maximal ideal containing $\mathfrak p$. Consider the local ring $(A/\mathfrak p)_{\mathfrak m/\mathfrak p}$, and let $V$ be a local ring in the field of fractions of $(A/\mathfrak p)_{\mathfrak m/\mathfrak p}$, maximal with respect to domination. Then $V$ is a valuation ring (Wedhorn 2.2), and the composition $A\to (A/\mathfrak p)_{\mathfrak m/\mathfrak p}\hookrightarrow V$ is a morphism of the desired type.

We would like to show that the two maps are inverse to one another. If we start out with $|\cdot|$, and pass to $\varphi: A\to (A/\mathfrak p)_{\mathfrak m/\mathfrak p}\hookrightarrow V$, then the valuation on $V$ agrees with the valuation on $(A/\mathfrak p)_{\mathfrak m/\mathfrak p}$ induced by $|\cdot|$, hence we get back the original $|\cdot|$.

It is easily seen that the map from Spa to the set of maps $\{A\to V\}$ is surjective. Indeed, let $\varphi:A\to V$ be such a map. As above, $|\cdot|_V\circ\varphi$ is a valuation on $A$. Then if we let $\mathfrak p$ be its support and $\mathfrak m$ any maximal ideal containing it, $\varphi$ induces a map from $(A/\mathfrak p)_{\mathfrak m/\mathfrak p}$ to $V$, which will be injective because the support was quotiented out. Thus we get that $\varphi$ factors as $A\to (A/\mathfrak p)_{\mathfrak m/\mathfrak p}\hookrightarrow V$, as claimed.

When it comes to the other direction, i.e. injectivity of $\{A\to V\}/\sim\to \mathrm{Spa}(A,A)$, I am stuck. I know that there are multiple possible choices for $\mathfrak m$ and $V$, as there may be various maximal ideals containing a prime ideal resp. local rings dominating a local ring, so I guess the equivalence by faithfully flat maps is supposed to take care of this somehow.


1 Answer 1


Be careful: $A_\mathfrak{p}$ need not be an integral domain, so you don't want to use the phrase field of fractions there. But we can instead use $k(\mathfrak p):=\operatorname{Frac}(A/\mathfrak{p})$.

On one hand, as you say, if $A\to V$ is a map to a valuation ring then you can restrict the valuation $V$ naturally has to a valuation which is $\le 1$ on $A$.

On the other hand, if you have a valuation $v$ which is $\le1$ on $A$, then it induces a valuation (let's also denote it $v$) on $k(\mathfrak p)$ for $\mathfrak p=\ker(v)$. If you let $V$ be the valuation ring of $v$ inside $k(\mathfrak p)$, i.e. $V=\{x\in k(\mathfrak p)\mid v(x)\le1\}$, then you have $A/\mathfrak p\subseteq V$, so you get a natural map $A\to V$ and this is the map to a valuation ring we want.

We need to know this respects equivalence in both directions. The following will be useful for reasoning this out:

Lemma: A map $V\to W$ of valuation rings is faithfully flat if and only if it is injective and local.

One one hand, it is simple that if $v$ and $w$ are equivalent valuations on $A$ then $\ker(v)=\ker(w)=:\mathfrak p$ and they have the same valuation ring inside $k(\mathfrak p)$, so the map $A\to V$ is unambiguous.

Conversely, if we have $A\to V\to W$ with $V\to W$ faithfully flat, then using our lemma above one can see that $W':=\operatorname{Frac}(V)\cap W$ is a valuation ring of $\operatorname{Frac}(V)$ containing $V$ and also with $\mathfrak m_V\subseteq\mathfrak m_{W'}$, and from this one can deduce that $\operatorname{Frac}(V)\cap W=V$. From this you can deduce that the valuations $v$ and $w$ induced on $A$ by $A\to V$ and $A\to W$ are equivalent.

Now we want to know our two maps are inverse to each other: as you've pretty much noted, if $v$ is a valuation on $A$ and you take the induced map to a valuation ring $A\to V$ we described, then the valuation you get back from this is the original valuation $v$.

Conversely, let's say we started with a map $A\to V$ and we get the induced valuation on $A$. For $\mathfrak p=\ker(v)$ we have the valuation ring $V'$ of $k(\mathfrak p)$ coming from $v$, and we want to show that our maps $A\to V$ and $A\to V'$ are equivalent. We have that $A/\mathfrak p\hookrightarrow V$ is injective, so we have an induced map of fields of fractions $k(\mathfrak p)\hookrightarrow\operatorname{Frac}(V)$ under which one can see that $V'\hookrightarrow V$. Also because $v$ by definition came from $V$, one can see that $\mathfrak m_{V'}\hookrightarrow\mathfrak m_V$. Thus using our lemma above we see that $V'\to V$ is faithfully flat, which is what we wanted.

  • $\begingroup$ Thanks for the answer, Alex! Do you perhaps have a reference for the Lemma? $\endgroup$ Jun 29, 2021 at 22:40
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    $\begingroup$ Hi, I know it's stated as part of Def. 1.1. here: arxiv.org/pdf/1807.04725.pdf, but I'm not actually sure I know of a place it's written down $\endgroup$ Jun 29, 2021 at 22:45
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    $\begingroup$ @HermeticallySealedHalibut Some facts you should know that can help you prove it are the following: (1) a module over a valuation ring is flat if and only if it is torsion-free (this lets you relate injectivity to flatness) and (2) a flat R-module M is faithfully flat if and only if $M/\mathfrak m M\neq0$ for all maximal ideals $\mathfrak m\subset R$ (this lets you relate "faithfully" to being local) $\endgroup$ Jun 29, 2021 at 22:55
  • $\begingroup$ This explains the lemma, thank you. $\endgroup$ Jun 30, 2021 at 9:49

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