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110.79 Universally submersive but not V covering

Let $A$ be a valuation ring. Let $\mathfrak p \subset A$ be a prime ideal which is neither the minimal prime nor the maximal ideal. (A good case to keep in mind is when $A$ has three prime ideals and $\mathfrak p$ is the one in the “middle”.) Consider the morphism of affine schemes

\[ \mathop{\mathrm{Spec}}(A_\mathfrak p) \amalg \mathop{\mathrm{Spec}}(A/\mathfrak p) \longrightarrow \mathop{\mathrm{Spec}}(A) \]

We claim this is universally submersive. In order to prove this, let $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ be a morphism of affine schemes given by the ring map $A \to B$. Then we have to show that

\[ \mathop{\mathrm{Spec}}(B_\mathfrak p) \amalg \mathop{\mathrm{Spec}}(B/\mathfrak pB) \to \mathop{\mathrm{Spec}}(B) \]

is submersive. First of all it is surjective. Next, suppose that $T \subset \mathop{\mathrm{Spec}}(B)$ is a subset such that $T_1 = \mathop{\mathrm{Spec}}(B_\mathfrak p) \cap T$ and $T_2 = \mathop{\mathrm{Spec}}(B/\mathfrak p B) \cap T$ are closed. Then we see that $T$ is the image of the spectrum of a $B$-algebra because both $T_1$ and $T_2$ are spectra of $B$-algebras. Hence to show that $T$ is closed it suffices to show that $T$ is stable under specialization, see Algebra, Lemma 10.41.5. To see this, suppose that $p \leadsto q$ is a specialization of points in $\mathop{\mathrm{Spec}}(B)$ with $p \in T$. Let $A'$ be a valuation ring and let $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(B)$ be a morphism such that the generic point $\eta $ of $\mathop{\mathrm{Spec}}(A')$ maps to $p$ and the closed point $s$ of $\mathop{\mathrm{Spec}}(A')$ maps to $q$, see Schemes, Lemma 26.20.4. Observe that the image of the composition $\gamma : \mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ is exactly the set of points $\xi \in \mathop{\mathrm{Spec}}(A)$ with $\gamma (\eta ) \leadsto \xi \leadsto \gamma (s)$ (details omitted). If $\mathfrak p \not\in \mathop{\mathrm{Im}}(\gamma )$, then we see that either both $p, q \in \mathop{\mathrm{Spec}}(B_\mathfrak p)$ or both $p, q \in \mathop{\mathrm{Spec}}(B/\mathfrak pB)$. In this case the fact that $T_1$, resp. $T_2$ is closed implies that $q \in T_1$, resp. $q \in T_2$ and hence $q \in T$. Finally, suppose $\mathfrak p \in \mathop{\mathrm{Im}}(\gamma )$, say $\mathfrak p = \gamma (r)$. Then we have specializations $p \leadsto r$ and $r \leadsto q$. In this case $p, r \in \mathop{\mathrm{Spec}}(B_\mathfrak p)$ and $r, q \in \mathop{\mathrm{Spec}}(B/\mathfrak pB)$. Then we fist conclude $r \in T_1 \subset T$, then $r \in T_2$ as $r$ maps to $\mathfrak p$, and then $q \in T_2 \subset T$ as desired.

On the other hand, we claim that the singleton family

\[ \{ \mathop{\mathrm{Spec}}(A_\mathfrak p) \amalg \mathop{\mathrm{Spec}}(A/\mathfrak p) \longrightarrow \mathop{\mathrm{Spec}}(A)\} \]

is not a V covering. See Topologies, Definition 34.10.7. Namely, if it where a V covering, there would be an extension of valuation ring $A \subset B$ such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ factors through $\mathop{\mathrm{Spec}}(A_\mathfrak p) \amalg \mathop{\mathrm{Spec}}(A/\mathfrak p)$. This would imply $\mathop{\mathrm{Spec}}(A')$ is disconnected which is absurd.

Lemma 110.79.1. There exists a morphism $X \to Y$ of affine schemes which is universally submersive such that $\{ X \to Y\} $ is not a V covering.

Proof. See discussion above. $\square$

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