The Stacks project

66.17 Relative conditions

This is a (yet another) technical section dealing with conditions on algebraic spaces having to do with points. It is probably a good idea to skip this section.

Definition 66.17.1. Let $S$ be a scheme. We say an algebraic space $X$ over $S$ has property $(\beta )$ if $X$ has the corresponding property of Lemma 66.5.1. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.

  1. We say $f$ has property $(\beta )$ if for any scheme $T$ and morphism $T \to Y$ the fibre product $T \times _ Y X$ has property $(\beta )$.

  2. We say $f$ is decent if for any scheme $T$ and morphism $T \to Y$ the fibre product $T \times _ Y X$ is a decent algebraic space.

  3. We say $f$ is reasonable if for any scheme $T$ and morphism $T \to Y$ the fibre product $T \times _ Y X$ is a reasonable algebraic space.

  4. We say $f$ is very reasonable if for any scheme $T$ and morphism $T \to Y$ the fibre product $T \times _ Y X$ is a very reasonable algebraic space.

We refer to Remark 66.17.10 for an informal discussion. It will turn out that the class of very reasonable morphisms is not so useful, but that the classes of decent and reasonable morphisms are useful.

Lemma 66.17.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. We have the following implications among the conditions on $f$:

\[ \xymatrix{ \text{representable} \ar@{=>}[rd] & & & & \\ & \text{very reasonable} \ar@{=>}[r] & \text{reasonable} \ar@{=>}[r] & \text{decent} \ar@{=>}[r] & (\beta ) \\ \text{quasi-separated} \ar@{=>}[ru] & & & & } \]

Proof. This is clear from the definitions, Lemma 66.5.1 and Morphisms of Spaces, Lemma 65.4.12. $\square$

Here is another sanity check.

Lemma 66.17.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $X$ is decent (resp. is reasonable, resp. has property $(\beta )$ of Lemma 66.5.1), then $f$ is decent (resp. reasonable, resp. has property $(\beta )$).

Proof. Let $T$ be a scheme and let $T \to Y$ be a morphism. Then $T \to Y$ is representable, hence the base change $T \times _ Y X \to X$ is representable. Hence if $X$ is decent (or reasonable), then so is $T \times _ Y X$, see Lemma 66.6.5. Similarly, for property $(\beta )$, see Lemma 66.5.3. $\square$

Lemma 66.17.4. Having property $(\beta )$, being decent, or being reasonable is preserved under arbitrary base change.

Proof. This is immediate from the definition. $\square$

Lemma 66.17.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\omega \in \{ \beta , decent, reasonable\} $. Suppose that $Y$ has property $(\omega )$ and $f : X \to Y$ has $(\omega )$. Then $X$ has $(\omega )$.

Proof. Let us prove the lemma in case $\omega = \beta $. In this case we have to show that any $x \in |X|$ is represented by a monomorphism from the spectrum of a field into $X$. Let $y = f(x) \in |Y|$. By assumption there exists a field $k$ and a monomorphism $\mathop{\mathrm{Spec}}(k) \to Y$ representing $y$. Then $x$ corresponds to a point $x'$ of $\mathop{\mathrm{Spec}}(k) \times _ Y X$. By assumption $x'$ is represented by a monomorphism $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k) \times _ Y X$. Clearly the composition $\mathop{\mathrm{Spec}}(k') \to X$ is a monomorphism representing $x$.

Let us prove the lemma in case $\omega = decent$. Let $x \in |X|$ and $y = f(x) \in |Y|$. By the result of the preceding paragraph we can choose a diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(k') \ar[r]_ x \ar[d] & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(k) \ar[r]^ y & Y } \]

whose horizontal arrows monomorphisms. As $Y$ is decent the morphism $y$ is quasi-compact. As $f$ is decent the algebraic space $\mathop{\mathrm{Spec}}(k) \times _ Y X$ is decent. Hence the monomorphism $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k) \times _ Y X$ is quasi-compact. Then the monomorphism $x : \mathop{\mathrm{Spec}}(k') \to X$ is quasi-compact as a composition of quasi-compact morphisms (use Morphisms of Spaces, Lemmas 65.8.4 and 65.8.5). As the point $x$ was arbitrary this implies $X$ is decent.

Let us prove the lemma in case $\omega = reasonable$. Choose $V \to Y$ étale with $V$ an affine scheme. Choose $U \to V \times _ Y X$ étale with $U$ an affine scheme. By assumption $V \to Y$ has universally bounded fibres. By Lemma 66.3.3 the morphism $V \times _ Y X \to X$ has universally bounded fibres. By assumption on $f$ we see that $U \to V \times _ Y X$ has universally bounded fibres. By Lemma 66.3.2 the composition $U \to X$ has universally bounded fibres. Hence there exists sufficiently many étale morphisms $U \to X$ from schemes with universally bounded fibres, and we conclude that $X$ is reasonable. $\square$

Lemma 66.17.6. Having property $(\beta )$, being decent, or being reasonable is preserved under compositions.

Proof. Let $\omega \in \{ \beta , decent, reasonable\} $. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over the scheme $S$. Assume $f$ and $g$ both have property $(\omega )$. Then we have to show that for any scheme $T$ and morphism $T \to Z$ the space $T \times _ Z X$ has $(\omega )$. By Lemma 66.17.4 this reduces us to the following claim: Suppose that $Y$ is an algebraic space having property $(\omega )$, and that $f : X \to Y$ is a morphism with $(\omega )$. Then $X$ has $(\omega )$. This is the content of Lemma 66.17.5. $\square$

Lemma 66.17.7. Let $S$ be a scheme. Let $f : X \to Y$, $g : Z \to Y$ be morphisms of algebraic spaces over $S$. If $X$ and $Z$ are decent (resp. reasonable, resp. have property $(\beta )$ of Lemma 66.5.1), then so does $X \times _ Y Z$.

Proof. Namely, by Lemma 66.17.3 the morphism $X \to Y$ has the property. Then the base change $X \times _ Y Z \to Z$ has the property by Lemma 66.17.4. And finally this implies $X \times _ Y Z$ has the property by Lemma 66.17.5. $\square$

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.) $\square$

Lemma 66.17.9. 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, very\ reasonable\} $. The following are equivalent

  1. $f$ is $\mathcal{P}$,

  2. for every affine scheme $Z$ and every morphism $Z \to Y$ the base change $Z \times _ Y X \to Z$ of $f$ is $\mathcal{P}$,

  3. for every affine scheme $Z$ and every morphism $Z \to Y$ the algebraic space $Z \times _ Y X$ is $\mathcal{P}$, and

  4. there exists a Zariski covering $Y = \bigcup Y_ i$ such that each morphism $f^{-1}(Y_ i) \to Y_ i$ has $\mathcal{P}$.

If $\mathcal{P} \in \{ (\beta ), decent, reasonable\} $, then this is also equivalent to

  1. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that the base change $V \times _ Y X \to V$ has $\mathcal{P}$.

Proof. The implications (1) $\Rightarrow $ (2) $\Rightarrow $ (3) $\Rightarrow $ (4) are trivial. The implication (3) $\Rightarrow $ (1) can be seen as follows. Let $Z \to Y$ be a morphism whose source is a scheme over $S$. Consider the algebraic space $Z \times _ Y X$. If we assume (3), then for any affine open $W \subset Z$, the open subspace $W \times _ Y X$ of $Z \times _ Y X$ has property $\mathcal{P}$. Hence by Lemma 66.5.2 the space $Z \times _ Y X$ has property $\mathcal{P}$, i.e., (1) holds. A similar argument (omitted) shows that (4) implies (1).

The implication (1) $\Rightarrow $ (5) is trivial. Let $V \to Y$ be an étale morphism from a scheme as in (5). Let $Z$ be an affine scheme, and let $Z \to Y$ be a morphism. Consider the diagram

\[ \xymatrix{ Z \times _ Y V \ar[r]_ q \ar[d]_ p & V \ar[d] \\ Z \ar[r] & Y } \]

Since $p$ is étale, and hence open, we can choose finitely many affine open subschemes $W_ i \subset Z \times _ Y V$ such that $Z = \bigcup p(W_ i)$. Consider the commutative diagram

\[ \xymatrix{ V \times _ Y X \ar[d] & (\coprod W_ i) \times _ Y X \ar[l] \ar[d] \ar[r] & Z \times _ Y X \ar[d] \\ V & \coprod W_ i \ar[l] \ar[r] & Z } \]

We know $V \times _ Y X$ has property $\mathcal{P}$. By Lemma 66.5.3 we see that $(\coprod W_ i) \times _ Y X$ has property $\mathcal{P}$. Note that the morphism $(\coprod W_ i) \times _ Y X \to Z \times _ Y X$ is étale and quasi-compact as the base change of $\coprod W_ i \to Z$. Hence by Lemma 66.17.8 we conclude that $Z \times _ Y X$ has property $\mathcal{P}$. $\square$

Remark 66.17.10. An informal description of the properties $(\beta )$, decent, reasonable, very reasonable was given in Section 66.6. A morphism has one of these properties if (very) loosely speaking the fibres of the morphism have the corresponding properties. Being decent is useful to prove things about specializations of points on $|X|$. Being reasonable is a bit stronger and technically quite easy to work with.

Here is a lemma we promised earlier which uses decent morphisms.

Lemma 66.17.11. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is quasi-compact and decent. (For example if $f$ is representable, or quasi-separated, see Lemma 66.17.2.) Then $f$ is universally closed if and only if the existence part of the valuative criterion holds.

Proof. In Morphisms of Spaces, Lemma 65.42.1 we proved that any quasi-compact morphism which satisfies the existence part of the valuative criterion is universally closed. To prove the other, assume that $f$ is universally closed. In the proof of Proposition 66.16.1 we have seen that it suffices to show, for any valuation ring $A$, and any morphism $\mathop{\mathrm{Spec}}(A) \to Y$, that the base change $f_ A : X_ A \to \mathop{\mathrm{Spec}}(A)$ satisfies the existence part of the valuative criterion. By definition the algebraic space $X_ A$ has property $(\gamma )$ and hence Proposition 66.16.1 applies to the morphism $f_ A$ and we win. $\square$


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