The Stacks project

Proof. A base change of a quasi-compact morphism is quasi-compact, see Schemes, Lemma 26.19.3. Being quasi-compact is Zariski local on the base, see Schemes, Lemma 26.19.2. Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is quasi-compact. Then $X'$ is quasi-compact, and $X' \to X$ is surjective. Hence $X$ is quasi-compact. This implies that $f$ is quasi-compact. Therefore Lemma 35.22.4 applies and we win. $\square$

Proof. Any base change of a quasi-separated morphism is quasi-separated, see Schemes, Lemma 26.21.12. Being quasi-separated is Zariski local on the base (from the definition or by Schemes, Lemma 26.21.6). Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is quasi-separated. This means that $\Delta ' : X' \to X'\times _{S'} X'$ is quasi-compact. Note that $\Delta '$ is the base change of $\Delta : X \to X \times _ S X$ via $S' \to S$. By Lemma 35.23.1 this implies $\Delta $ is quasi-compact, and hence $f$ is quasi-separated. Therefore Lemma 35.22.4 applies and we win. $\square$

Proof. A base change of a universally closed morphism is universally closed by definition. Being universally closed is Zariski local on the base (from the definition or by Morphisms, Lemma 29.41.2). Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is universally closed. Let $T \to S$ be any morphism. Consider the diagram

in which both squares are cartesian. Thus the assumption implies that the middle vertical arrow is closed. The right horizontal arrows are flat, quasi-compact and surjective (as base changes of $S' \to S$). Hence a subset of $T$ is closed if and only if its inverse image in $S' \times _ S T$ is closed, see Morphisms, Lemma 29.25.12. An easy diagram chase shows that the right vertical arrow is closed too, and we conclude $X \to S$ is universally closed. Therefore Lemma 35.22.4 applies and we win. $\square$

Proof. The proof is the same as the proof of Lemma 35.23.3. $\square$

Proof. The proof is the same as the proof of Lemma 35.23.3 using that a quasi-compact flat surjective morphism is universally submersive by Morphisms, Lemma 29.25.12. $\square$

Proof. A base change of a separated morphism is separated, see Schemes, Lemma 26.21.12. Being separated is Zariski local on the base (from the definition or by Schemes, Lemma 26.21.7). Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is separated. This means that $\Delta ' : X' \to X'\times _{S'} X'$ is a closed immersion, hence universally closed. Note that $\Delta '$ is the base change of $\Delta : X \to X \times _ S X$ via $S' \to S$. By Lemma 35.23.3 this implies $\Delta $ is universally closed. Since it is an immersion (Schemes, Lemma 26.21.2) we conclude $\Delta $ is a closed immersion. Hence $f$ is separated. Therefore Lemma 35.22.4 applies and we win. $\square$

Proof. This is clear. $\square$

Proof. A base change of a universally injective morphism is universally injective (this is formal). Being universally injective is Zariski local on the base; this is clear from the definition. Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is universally injective. Let $K$ be a field, and let $a, b : \mathop{\mathrm{Spec}}(K) \to X$ be two morphisms such that $f \circ a = f \circ b$. As $S' \to S$ is surjective and by the discussion in Schemes, Section 26.13 there exists a field extension $K'/K$ and a morphism $\mathop{\mathrm{Spec}}(K') \to S'$ such that the following solid diagram commutes

As the square is cartesian we get the two dotted arrows $a'$, $b'$ making the diagram commute. Since $X' \to S'$ is universally injective we get $a' = b'$, by Morphisms, Lemma 29.10.2. Clearly this forces $a = b$ (by the discussion in Schemes, Section 26.13). Therefore Lemma 35.22.4 applies and we win.

An alternative proof would be to use the characterization of a universally injective morphism as one whose diagonal is surjective, see Morphisms, Lemma 29.10.2. The lemma then follows from the fact that the property of being surjective is fpqc local on the base, see Lemma 35.23.7. (Hint: use that the base change of the diagonal is the diagonal of the base change.) $\square$

Proof. This can be proved in exactly the same manner as Lemma 35.23.3. Alternatively, one can use that a map of topological spaces is a homeomorphism if and only if it is injective, surjective, and open. Thus a universal homeomorphism is the same thing as a surjective, universally injective, and universally open morphism. Thus the lemma follows from Lemmas 35.23.7, 35.23.8, and 35.23.4. $\square$

Proof. Being locally of finite type is preserved under base change, see Morphisms, Lemma 29.15.4. Being locally of finite type is Zariski local on the base, see Morphisms, Lemma 29.15.2. Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is locally of finite type. Let $U \subset X$ be an affine open. Then $U' = S' \times _ S U$ is affine and of finite type over $S'$. Write $S = \mathop{\mathrm{Spec}}(R)$, $S' = \mathop{\mathrm{Spec}}(R')$, $U = \mathop{\mathrm{Spec}}(A)$, and $U' = \mathop{\mathrm{Spec}}(A')$. We know that $R \to R'$ is faithfully flat, $A' = R' \otimes _ R A$ and $R' \to A'$ is of finite type. We have to show that $R \to A$ is of finite type. This is the result of Algebra, Lemma 10.126.1. It follows that $f$ is locally of finite type. Therefore Lemma 35.22.4 applies and we win. $\square$

Proof. Being locally of finite presentation is preserved under base change, see Morphisms, Lemma 29.21.4. Being locally of finite type is Zariski local on the base, see Morphisms, Lemma 29.21.2. Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is locally of finite presentation. Let $U \subset X$ be an affine open. Then $U' = S' \times _ S U$ is affine and of finite type over $S'$. Write $S = \mathop{\mathrm{Spec}}(R)$, $S' = \mathop{\mathrm{Spec}}(R')$, $U = \mathop{\mathrm{Spec}}(A)$, and $U' = \mathop{\mathrm{Spec}}(A')$. We know that $R \to R'$ is faithfully flat, $A' = R' \otimes _ R A$ and $R' \to A'$ is of finite presentation. We have to show that $R \to A$ is of finite presentation. This is the result of Algebra, Lemma 10.126.2. It follows that $f$ is locally of finite presentation. Therefore Lemma 35.22.4 applies and we win. $\square$

Proof. Combine Lemmas 35.23.1 and 35.23.10. $\square$

Proof. Combine Lemmas 35.23.1, 35.23.2 and 35.23.11. $\square$

Proof. The lemma follows by combining Lemmas 35.23.3, 35.23.6 and 35.23.12. $\square$

Proof. Being flat is preserved under arbitrary base change, see Morphisms, Lemma 29.25.8. Being flat is Zariski local on the base by definition. Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is flat. Let $U \subset X$ be an affine open. Then $U' = S' \times _ S U$ is affine. Write $S = \mathop{\mathrm{Spec}}(R)$, $S' = \mathop{\mathrm{Spec}}(R')$, $U = \mathop{\mathrm{Spec}}(A)$, and $U' = \mathop{\mathrm{Spec}}(A')$. We know that $R \to R'$ is faithfully flat, $A' = R' \otimes _ R A$ and $R' \to A'$ is flat. Goal: Show that $R \to A$ is flat. This follows immediately from Algebra, Lemma 10.39.8. Hence $f$ is flat. Therefore Lemma 35.22.4 applies and we win. $\square$

Proof. The property of being an open immersion is stable under base change, see Schemes, Lemma 26.18.2. The property of being an open immersion is Zariski local on the base (this is obvious).

Let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is an open immersion. We claim that $f$ is an open immersion. Then $f'$ is universally open, and universally injective. Hence we conclude that $f$ is universally open by Lemma 35.23.4, and universally injective by Lemma 35.23.8. In particular $f(X) \subset S$ is open. If for every affine open $U \subset f(X)$ we can prove that $f^{-1}(U) \to U$ is an isomorphism, then $f$ is an open immersion and we're done. If $U' \subset S'$ denotes the inverse image of $U$, then $U' \to U$ is a faithfully flat morphism of affines and $(f')^{-1}(U') \to U'$ is an isomorphism (as $f'(X')$ contains $U'$ by our choice of $U$). Thus we reduce to the case discussed in the next paragraph.

Let $S' \to S$ be a flat surjective morphism of affine schemes, let $f : X \to S$ be a morphism, and assume that the base change $f' : X' \to S'$ is an isomorphism. We have to show that $f$ is an isomorphism also. It is clear that $f$ is surjective, universally injective, and universally open (see arguments above for the last two). Hence $f$ is bijective, i.e., $f$ is a homeomorphism. Thus $f$ is affine by Morphisms, Lemma 29.45.4. Since

is an isomorphism and since $\mathcal{O}(S) \to \mathcal{O}(S')$ is faithfully flat this implies that $\mathcal{O}(S) \to \mathcal{O}(X)$ is an isomorphism. Thus $f$ is an isomorphism. This finishes the proof of the claim above. Therefore Lemma 35.22.4 applies and we win. $\square$

Proof. Combine Lemmas 35.23.7 and 35.23.16. $\square$

Proof. A base change of an affine morphism is affine, see Morphisms, Lemma 29.11.8. Being affine is Zariski local on the base, see Morphisms, Lemma 29.11.3. Finally, let $g : S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is affine. In other words, $X'$ is affine, say $X' = \mathop{\mathrm{Spec}}(A')$. Also write $S = \mathop{\mathrm{Spec}}(R)$ and $S' = \mathop{\mathrm{Spec}}(R')$. We have to show that $X$ is affine.

By Lemmas 35.23.1 and 35.23.6 we see that $X \to S$ is separated and quasi-compact. Thus $f_*\mathcal{O}_ X$ is a quasi-coherent sheaf of $\mathcal{O}_ S$-algebras, see Schemes, Lemma 26.24.1. Hence $f_*\mathcal{O}_ X = \widetilde{A}$ for some $R$-algebra $A$. In fact $A = \Gamma (X, \mathcal{O}_ X)$ of course. Also, by flat base change (see for example Cohomology of Schemes, Lemma 30.5.2) we have $g^*f_*\mathcal{O}_ X = f'_*\mathcal{O}_{X'}$. In other words, we have $A' = R' \otimes _ R A$. Consider the canonical morphism

over $S$ from Schemes, Lemma 26.6.4. By the above the base change of this morphism to $S'$ is an isomorphism. Hence it is an isomorphism by Lemma 35.23.17. Therefore Lemma 35.22.4 applies and we win. $\square$

Proof. Let $f : X \to Y$ be a morphism of schemes. Let $\{ Y_ i \to Y\} $ be an fpqc covering. Assume that each $f_ i : Y_ i \times _ Y X \to Y_ i$ is a closed immersion. This implies that each $f_ i$ is affine, see Morphisms, Lemma 29.11.9. By Lemma 35.23.18 we conclude that $f$ is affine. It remains to show that $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ is surjective. For every $y \in Y$ there exists an $i$ and a point $y_ i \in Y_ i$ mapping to $y$. By Cohomology of Schemes, Lemma 30.5.2 the sheaf $f_{i, *}(\mathcal{O}_{Y_ i \times _ Y X})$ is the pullback of $f_*\mathcal{O}_ X$. By assumption it is a quotient of $\mathcal{O}_{Y_ i}$. Hence we see that

is surjective. Since $\mathcal{O}_{Y_ i, y_ i}$ is faithfully flat over $\mathcal{O}_{Y, y}$ this implies the surjectivity of $\mathcal{O}_{Y, y} \longrightarrow (f_*\mathcal{O}_ X)_ y$ as desired. $\square$

Proof. Let $f : X \to Y$ be a morphism of schemes. Let $\{ g_ i : Y_ i \to Y\} $ be an fpqc covering. Assume that each $f_ i : Y_ i \times _ Y X \to Y_ i$ is quasi-affine. This implies that each $f_ i$ is quasi-compact and separated. By Lemmas 35.23.1 and 35.23.6 this implies that $f$ is quasi-compact and separated. Consider the sheaf of $\mathcal{O}_ Y$-algebras $\mathcal{A} = f_*\mathcal{O}_ X$. By Schemes, Lemma 26.24.1 it is a quasi-coherent $\mathcal{O}_ Y$-algebra. Consider the canonical morphism

see Constructions, Lemma 27.4.7. By flat base change (see for example Cohomology of Schemes, Lemma 30.5.2) we have $g_ i^*f_*\mathcal{O}_ X = f_{i, *}\mathcal{O}_{X'}$ where $g_ i : Y_ i \to Y$ are the given flat maps. Hence the base change $j_ i$ of $j$ by $g_ i$ is the canonical morphism of Constructions, Lemma 27.4.7 for the morphism $f_ i$. By assumption and Morphisms, Lemma 29.13.3 all of these morphisms $j_ i$ are quasi-compact open immersions. Hence, by Lemmas 35.23.1 and 35.23.16 we see that $j$ is a quasi-compact open immersion. Hence by Morphisms, Lemma 29.13.3 again we conclude that $f$ is quasi-affine. $\square$

Proof. Let $f : X \to Y$ be a morphism of schemes. Let $\{ Y_ i \to Y\} $ be an fpqc covering. Write $X_ i = Y_ i \times _ Y X$ and $f_ i : X_ i \to Y_ i$ the base change of $f$. Also denote $q_ i : Y_ i \to Y$ the given flat morphisms. Assume each $f_ i$ is a quasi-compact immersion. By Schemes, Lemma 26.23.8 each $f_ i$ is separated. By Lemmas 35.23.1 and 35.23.6 this implies that $f$ is quasi-compact and separated. Let $X \to Z \to Y$ be the factorization of $f$ through its scheme theoretic image. By Morphisms, Lemma 29.6.3 the closed subscheme $Z \subset Y$ is cut out by the quasi-coherent sheaf of ideals $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to f_*\mathcal{O}_ X)$ as $f$ is quasi-compact. By flat base change (see for example Cohomology of Schemes, Lemma 30.5.2; here we use $f$ is separated) we see $f_{i, *}\mathcal{O}_{X_ i}$ is the pullback $q_ i^*f_*\mathcal{O}_ X$. Hence $Y_ i \times _ Y Z$ is cut out by the quasi-coherent sheaf of ideals $q_ i^*\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_{Y_ i} \to f_{i, *}\mathcal{O}_{X_ i})$. By Morphisms, Lemma 29.7.7 the morphisms $X_ i \to Y_ i \times _ Y Z$ are open immersions. Hence by Lemma 35.23.16 we see that $X \to Z$ is an open immersion and hence $f$ is a immersion as desired (we already saw it was quasi-compact). $\square$

Proof. An integral morphism is the same thing as an affine, universally closed morphism. See Morphisms, Lemma 29.44.7. Hence the lemma follows on combining Lemmas 35.23.3 and 35.23.18. $\square$

Proof. An finite morphism is the same thing as an integral morphism which is locally of finite type. See Morphisms, Lemma 29.44.4. Hence the lemma follows on combining Lemmas 35.23.10 and 35.23.22. $\square$

Proof. Let $f : X \to S$ be a morphism of schemes, and let $\{ S_ i \to S\} $ be an fpqc covering such that each base change $f_ i : X_ i \to S_ i$ is locally quasi-finite. We have already seen (Lemma 35.23.10) that “locally of finite type” is fpqc local on the base, and hence we see that $f$ is locally of finite type. Then it follows from Morphisms, Lemma 29.20.13 that $f$ is locally quasi-finite. The quasi-finite case follows as we have already seen that “quasi-compact” is fpqc local on the base (Lemma 35.23.1). $\square$

Proof. This follows immediately from the fact that being locally of finite type is fpqc local on the base and Morphisms, Lemma 29.28.3. $\square$

Proof. A morphism is syntomic if and only if it is locally of finite presentation, flat, and has locally complete intersections as fibres. We have seen already that being flat and locally of finite presentation are fpqc local on the base (Lemmas 35.23.15, and 35.23.11). Hence the result follows for syntomic from Morphisms, Lemma 29.30.12. $\square$

Proof. A morphism is smooth if and only if it is locally of finite presentation, flat, and has smooth fibres. We have seen already that being flat and locally of finite presentation are fpqc local on the base (Lemmas 35.23.15, and 35.23.11). Hence the result follows for smooth from Morphisms, Lemma 29.34.15. $\square$

Proof. A morphism is unramified (resp. G-unramified) if and only if it is locally of finite type (resp. finite presentation) and its diagonal morphism is an open immersion (see Morphisms, Lemma 29.35.13). We have seen already that being locally of finite type (resp. locally of finite presentation) and an open immersion is fpqc local on the base (Lemmas 35.23.11, 35.23.10, and 35.23.16). Hence the result follows formally. $\square$

Proof. A morphism is étale if and only if it flat and G-unramified. See Morphisms, Lemma 29.36.16. We have seen already that being flat and G-unramified are fpqc local on the base (Lemmas 35.23.15, and 35.23.28). Hence the result follows. $\square$

Proof. Being finite locally free is equivalent to being finite, flat and locally of finite presentation (Morphisms, Lemma 29.48.2). Hence this follows from Lemmas 35.23.23, 35.23.15, and 35.23.11. If $f : Z \to U$ is finite locally free, and $\{ U_ i \to U\} $ is a surjective family of morphisms such that each pullback $Z \times _ U U_ i \to U_ i$ has degree $d$, then $Z \to U$ has degree $d$, for example because we can read off the degree in a point $u \in U$ from the fibre $(f_*\mathcal{O}_ Z)_ u \otimes _{\mathcal{O}_{U, u}} \kappa (u)$. $\square$

Proof. Let $f : X \to S$ be a morphism of schemes. Let $\{ S_ i \to S\} $ be an fpqc covering, and assume each of the base changes $f_ i : X_ i \to S_ i$ of $f$ is a monomorphism. Let $a, b : T \to X$ be two morphisms such that $f \circ a = f \circ b$. We have to show that $a = b$. Since $f_ i$ is a monomorphism we see that $a_ i = b_ i$, where $a_ i, b_ i : S_ i \times _ S T \to X_ i$ are the base changes. In particular the compositions $S_ i \times _ S T \to T \to X$ are equal. Since $\coprod S_ i \times _ S T \to T$ is an epimorphism (see e.g. Lemma 35.13.7) we conclude $a = b$. $\square$

Proof. We will use the criterion of Lemma 35.22.4 to prove this. By Divisors, Definition 31.21.1 being a Koszul-regular (resp. $H_1$-regular, quasi-regular) immersion is Zariski local on the base. By Divisors, Lemma 31.21.4 being a Koszul-regular (resp. $H_1$-regular, quasi-regular) immersion is preserved under flat base change. The final hypothesis (3) of Lemma 35.22.4 translates into the following algebra statement: Let $A \to B$ be a faithfully flat ring map. Let $I \subset A$ be an ideal. If $IB$ is locally on $\mathop{\mathrm{Spec}}(B)$ generated by a Koszul-regular (resp. $H_1$-regular, quasi-regular) sequence in $B$, then $I \subset A$ is locally on $\mathop{\mathrm{Spec}}(A)$ generated by a Koszul-regular (resp. $H_1$-regular, quasi-regular) sequence in $A$. This is More on Algebra, Lemma 15.32.4. $\square$