$\mathrm{Spa}(A,A)=\{A\to V \text{ valuation ring}\}$ modulo faithfully flat maps

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.

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.

• Thanks for the answer, Alex! Do you perhaps have a reference for the Lemma? Jun 29, 2021 at 22:40
• 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 Jun 29, 2021 at 22:45
• @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) Jun 29, 2021 at 22:55
• This explains the lemma, thank you. Jun 30, 2021 at 9:49