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.

## 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

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

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

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.

**Proof.**
See discussion above.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)