The Stacks project

Lemma 66.17.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\mathcal{P} \in \{ (\beta ), decent, reasonable\} $. Assume

  1. $f$ is quasi-compact,

  2. $f$ is étale,

  3. $|f| : |X| \to |Y|$ is surjective, and

  4. the algebraic space $X$ has property $\mathcal{P}$.

Then $Y$ has property $\mathcal{P}$.

Proof. Let us prove this in case $\mathcal{P} = (\beta )$. Let $y \in |Y|$ be a point. We have to show that $y$ can be represented by a monomorphism from a field. Choose a point $x \in |X|$ with $f(x) = y$. By assumption we may represent $x$ by a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$, with $k$ a field. By Lemma 66.4.3 it suffices to show that the projections $\mathop{\mathrm{Spec}}(k) \times _ Y \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ are étale and quasi-compact. We can factor the first projection as

\[ \mathop{\mathrm{Spec}}(k) \times _ Y \mathop{\mathrm{Spec}}(k) \longrightarrow \mathop{\mathrm{Spec}}(k) \times _ Y X \longrightarrow \mathop{\mathrm{Spec}}(k) \]

The first morphism is a monomorphism, and the second is étale and quasi-compact. By Properties of Spaces, Lemma 64.16.8 we see that $\mathop{\mathrm{Spec}}(k) \times _ Y X$ is a scheme. Hence it is a finite disjoint union of spectra of finite separable field extensions of $k$. By Schemes, Lemma 26.23.11 we see that the first arrow identifies $\mathop{\mathrm{Spec}}(k) \times _ Y \mathop{\mathrm{Spec}}(k)$ with a finite disjoint union of spectra of finite separable field extensions of $k$. Hence the projection morphism is étale and quasi-compact.

Let us prove this in case $\mathcal{P} = decent$. We have already seen in the first paragraph of the proof that this implies that every $y \in |Y|$ can be represented by a monomorphism $y : \mathop{\mathrm{Spec}}(k) \to Y$. Pick such a $y$. Pick an affine scheme $U$ and an étale morphism $U \to X$ such that the image of $|U| \to |Y|$ contains $y$. By Lemma 66.4.5 it suffices to show that $U_ y$ is a finite scheme over $k$. The fibre product $X_ y = \mathop{\mathrm{Spec}}(k) \times _ Y X$ is a quasi-compact étale algebraic space over $k$. Hence by Properties of Spaces, Lemma 64.16.8 it is a scheme. So it is a finite disjoint union of spectra of finite separable extensions of $k$. Say $X_ y = \{ x_1, \ldots , x_ n\} $ so $x_ i$ is given by $x_ i : \mathop{\mathrm{Spec}}(k_ i) \to X$ with $[k_ i : k] < \infty $. By assumption $X$ is decent, so the schemes $U_{x_ i} = \mathop{\mathrm{Spec}}(k_ i) \times _ X U$ are finite over $k_ i$. Finally, we note that $U_ y = \coprod U_{x_ i}$ as a scheme and we conclude that $U_ y$ is finite over $k$ as desired.

Let us prove this in case $\mathcal{P} = reasonable$. Pick an affine scheme $V$ and an étale morphism $V \to Y$. We have the show the fibres of $V \to Y$ are universally bounded. The algebraic space $V \times _ Y X$ is quasi-compact. Thus we can find an affine scheme $W$ and a surjective étale morphism $W \to V \times _ Y X$, see Properties of Spaces, Lemma 64.6.3. Here is a picture (solid diagram)

\[ \xymatrix{ W \ar[r] \ar[rd] & V \times _ Y X \ar[r] \ar[d] & X \ar[d]_ f & \mathop{\mathrm{Spec}}(k) \ar@{..>}[l]^ x \ar@{..>}[ld]^ y \\ & V \ar[r] & Y } \]

The morphism $W \to X$ is universally bounded by our assumption that the space $X$ is reasonable. Let $n$ be an integer bounding the degrees of the fibres of $W \to X$. We claim that the same integer works for bounding the fibres of $V \to Y$. Namely, suppose $y \in |Y|$ is a point. Then there exists a $x \in |X|$ with $f(x) = y$ (see above). This means we can find a field $k$ and morphisms $x, y$ given as dotted arrows in the diagram above. In particular we get a surjective étale morphism

\[ \mathop{\mathrm{Spec}}(k) \times _{x, X} W \to \mathop{\mathrm{Spec}}(k) \times _{x, X} (V \times _ Y X) = \mathop{\mathrm{Spec}}(k) \times _{y, Y} V \]

which shows that the degree of $\mathop{\mathrm{Spec}}(k) \times _{y, Y} V$ over $k$ is less than or equal to the degree of $\mathop{\mathrm{Spec}}(k) \times _{x, X} W$ over $k$, i.e., $\leq n$, and we win. (This last part of the argument is the same as the argument in the proof of Lemma 66.3.4. Unfortunately that lemma is not general enough because it only applies to representable morphisms.) $\square$


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