Lemma 65.50.1. Let $S$ be a scheme. Consider a commutative diagram

\[ \xymatrix{ V' \ar[r] \ar[rd] & T' \times _ T X \ar[r] \ar[d] & X \ar[d] \\ & T' \ar[r] & T } \]

of algebraic spaces over $S$. Assume

$T' \to T$ is an étale morphism of affine schemes,

$X \to T$ is a separated, locally quasi-finite morphism,

$V'$ is an open subspace of $T' \times _ T X$, and

$V' \to T'$ is quasi-affine.

In this situation the image $U$ of $V'$ in $X$ is a quasi-compact open subspace of $X$ which is representable.

**Proof.**
We first make some trivial observations. Note that $V'$ is representable by Lemma 65.21.3. It is also quasi-compact (as a quasi-affine scheme over an affine scheme, see Morphisms, Lemma 29.12.2). Since $T' \times _ T X \to X$ is étale (Properties of Spaces, Lemma 64.16.5) the map $|T' \times _ T X| \to |X|$ is open, see Properties of Spaces, Lemma 64.16.7. Let $U \subset X$ be the open subspace corresponding to the image of $|V'|$, see Properties of Spaces, Lemma 64.4.8. As $|V'|$ is quasi-compact we see that $|U|$ is quasi-compact, hence $U$ is a quasi-compact algebraic space, by Properties of Spaces, Lemma 64.5.2.

By Morphisms, Lemma 29.54.10 the morphism $T' \to T$ is universally bounded. Hence we can do induction on the integer $n$ bounding the degree of the fibres of $T' \to T$, see Morphisms, Lemma 29.54.9 for a description of this integer in the case of an étale morphism. If $n = 1$, then $T' \to T$ is an open immersion (see Étale Morphisms, Theorem 41.14.1), and the result is clear. Assume $n > 1$.

Consider the affine scheme $T'' = T' \times _ T T'$. As $T' \to T$ is étale we have a decomposition (into open and closed affine subschemes) $T'' = \Delta (T') \amalg T^*$. Namely $\Delta = \Delta _{T'/T}$ is open by Morphisms, Lemma 29.33.13 and closed because $T' \to T$ is separated as a morphism of affines. As a base change the degrees of the fibres of the second projection $\text{pr}_1 : T' \times _ T T' \to T'$ are bounded by $n$, see Morphisms, Lemma 29.54.6. On the other hand, $\text{pr}_1|_{\Delta (T')} : \Delta (T') \to T'$ is an isomorphism and every fibre has exactly one point. Thus, on applying Morphisms, Lemma 29.54.9 we conclude the degrees of the fibres of the restriction $\text{pr}_1|_{T^*} : T^* \to T'$ are bounded by $n - 1$. Hence the induction hypothesis applied to the diagram

\[ \xymatrix{ p_0^{-1}(V') \cap X^* \ar[r] \ar[rd] & X^* \ar[r]_{p_1|_{X^*}} \ar[d] & X' \ar[d] \\ & T^* \ar[r]^{\text{pr}_1|_{T^*}} & T' } \]

gives that $p_1(p_0^{-1}(V') \cap X^*)$ is a quasi-compact scheme. Here we set $X'' = T'' \times _ T X$, $X^* = T^* \times _ T X$, and $X' = T' \times _ T X$, and $p_0, p_1 : X'' \to X'$ are the base changes of $\text{pr}_0, \text{pr}_1$. Most of the hypotheses of the lemma imply by base change the corresponding hypothesis for the diagram above. For example $p_0^{-1}(V') = T'' \times _{T'} V'$ is a scheme quasi-affine over $T''$ as a base change. Some verifications omitted.

By Properties of Spaces, Lemma 64.13.1 we conclude that

\[ p_1(p_0^{-1}(V')) = V' \cup p_1(p_0^{-1}(V') \cap X^*) \]

is a quasi-compact scheme. Moreover, it is clear that $p_1(p_0^{-1}(V'))$ is the inverse image of the quasi-compact open subspace $U \subset X$ discussed in the first paragraph of the proof. In other words, $T' \times _ T U$ is a scheme! Note that $T' \times _ T U$ is quasi-compact and separated and locally quasi-finite over $T'$, as $T' \times _ T X \to T'$ is locally quasi-finite and separated being a base change of the original morphism $X \to T$ (see Lemmas 65.4.4 and 65.27.4). This implies by More on Morphisms, Lemma 37.38.2 that $T' \times _ T U \to T'$ is quasi-affine.

By Descent, Lemma 35.36.1 this gives a descent datum on $T' \times _ T U / T'$ relative to the étale covering $\{ T' \to W\} $, where $W \subset T$ is the image of the morphism $T' \to T$. Because $U'$ is quasi-affine over $T'$ we see from Descent, Lemma 35.35.1 that this datum is effective, and by the last part of Descent, Lemma 35.36.1 this implies that $U$ is a scheme as desired. Some minor details omitted.
$\square$

## Comments (0)