Proof. Let $T$ be an affine scheme. Let $f : U \to T$ be a proper surjective morphism. Let $U = \bigcup _{j = 1, \ldots , m} U_ j$ be a finite affine open covering. We have to show that $\{ U_ j \to T\}$ is a standard V covering, see Definition 34.8.1. Let $g : \mathop{\mathrm{Spec}}(V) \to T$ be a morphism where $V$ is a valuation ring with fraction field $K$. Since $U \to T$ is surjective, we may choose a field extension $L/K$ and a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(L) \ar[rr] \ar[d] & & U \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar[r] & \mathop{\mathrm{Spec}}(V) \ar[r]^ g & T }$

By Algebra, Lemma 10.50.2 we can choose a valuation ring $W \subset L$ dominating $V$. By the valuative criterion of properness (Morphisms, Lemma 29.42.1) we can then find the morphism $h$ in the commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(L) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(W) \ar[r]_ h \ar[d] & U \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar[r] & \mathop{\mathrm{Spec}}(V) \ar[r]^ g & X }$

Since $\mathop{\mathrm{Spec}}(W)$ has a unique closed point, we see that $\mathop{\mathrm{Im}}(h)$ is contained in $U_ j$ for some $j$. Thus $h : \mathop{\mathrm{Spec}}(W) \to U_ j$ is the desired lift and we conclude $\{ U_ j \to T\}$ is a standard V covering. $\square$

There are also:

• 4 comment(s) on Section 34.10: The V topology

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).