The Stacks project

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34.20 Properties of morphisms local in the fpqc topology on the target

In this section we find a large number of properties of morphisms of schemes which are local on the base in the fpqc topology. By contrast, in Examples, Section 102.57 we will show that the properties “projective” and “quasi-projective” are not local on the base even in the Zariski topology.

Lemma 34.20.1. The property $\mathcal{P}(f) =$“$f$ is quasi-compact” is fpqc local on the base.

Proof. A base change of a quasi-compact morphism is quasi-compact, see Schemes, Lemma 25.19.3. Being quasi-compact is Zariski local on the base, see Schemes, Lemma 25.19.2. 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 34.19.4 applies and we win. $\square$

Lemma 34.20.2. The property $\mathcal{P}(f) =$“$f$ is quasi-separated” is fpqc local on the base.

Proof. Any base change of a quasi-separated morphism is quasi-separated, see Schemes, Lemma 25.21.13. Being quasi-separated is Zariski local on the base (from the definition or by Schemes, Lemma 25.21.7). 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 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 34.20.1 this implies $\Delta $ is quasi-compact, and hence $f$ is quasi-separated. Therefore Lemma 34.19.4 applies and we win. $\square$

Lemma 34.20.3. The property $\mathcal{P}(f) =$“$f$ is universally closed” is fpqc local on the base.

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 28.39.2). 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

\[ \xymatrix{ X' \ar[d] & S' \times _ S T \times _ S X \ar[d] \ar[r] \ar[l] & T \times _ S X \ar[d] \\ S' & S' \times _ S T \ar[r] \ar[l] & T } \]

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 28.24.11. An easy diagram chase shows that the right vertical arrow is closed too, and we conclude $X \to S$ is universally closed. Therefore Lemma 34.19.4 applies and we win. $\square$

Lemma 34.20.4. The property $\mathcal{P}(f) =$“$f$ is universally open” is fpqc local on the base.

Proof. The proof is the same as the proof of Lemma 34.20.3. $\square$

Lemma 34.20.5. The property $\mathcal{P}(f) =$“$f$ is universally submersive” is fpqc local on the base.

Proof. The proof is the same as the proof of Lemma 34.20.3 using that a quasi-compact flat surjective morphism is universally submersive by Morphisms, Lemma 28.24.11. $\square$

Lemma 34.20.6. The property $\mathcal{P}(f) =$“$f$ is separated” is fpqc local on the base.

Proof. A base change of a separated morphism is separated, see Schemes, Lemma 25.21.13. Being separated is Zariski local on the base (from the definition or by Schemes, Lemma 25.21.8). 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 34.20.3 this implies $\Delta $ is universally closed. Since it is an immersion (Schemes, Lemma 25.21.2) we conclude $\Delta $ is a closed immersion. Hence $f$ is separated. Therefore Lemma 34.19.4 applies and we win. $\square$

Lemma 34.20.7. The property $\mathcal{P}(f) =$“$f$ is surjective” is fpqc local on the base.

Proof. This is clear. $\square$

Lemma 34.20.8. The property $\mathcal{P}(f) =$“$f$ is universally injective” is fpqc local on the base.

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 25.13 there exists a field extension $K \subset K'$ and a morphism $\mathop{\mathrm{Spec}}(K') \to S'$ such that the following solid diagram commutes

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[rrd] \ar@{-->}[rd]_{a', b'} \ar[dd] \\ & X' \ar[r] \ar[d] & S' \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar[r]^{a, b} & X \ar[r] & S } \]

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 28.10.2. Clearly this forces $a = b$ (by the discussion in Schemes, Section 25.13). Therefore Lemma 34.19.4 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 28.10.2. The lemma then follows from the fact that the property of being surjective is fpqc local on the base, see Lemma 34.20.7. (Hint: use that the base change of the diagonal is the diagonal of the base change.) $\square$

Lemma 34.20.9. The property $\mathcal{P}(f) =$“$f$ is a universal homeomorphism” is fpqc local on the base.

Proof. This can be proved in exactly the same manner as Lemma 34.20.3. 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 34.20.7, 34.20.8, and 34.20.4. $\square$

Lemma 34.20.10. The property $\mathcal{P}(f) =$“$f$ is locally of finite type” is fpqc local on the base.

Proof. Being locally of finite type is preserved under base change, see Morphisms, Lemma 28.14.4. Being locally of finite type is Zariski local on the base, see Morphisms, Lemma 28.14.2. 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 10.125.1. It follows that $f$ is locally of finite type. Therefore Lemma 34.19.4 applies and we win. $\square$

Lemma 34.20.11. The property $\mathcal{P}(f) =$“$f$ is locally of finite presentation” is fpqc local on the base.

Proof. Being locally of finite presentation is preserved under base change, see Morphisms, Lemma 28.20.4. Being locally of finite type is Zariski local on the base, see Morphisms, Lemma 28.20.2. 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 10.125.2. It follows that $f$ is locally of finite presentation. Therefore Lemma 34.19.4 applies and we win. $\square$

Lemma 34.20.12. The property $\mathcal{P}(f) =$“$f$ is of finite type” is fpqc local on the base.

Lemma 34.20.13. The property $\mathcal{P}(f) =$“$f$ is of finite presentation” is fpqc local on the base.

Lemma 34.20.14. The property $\mathcal{P}(f) =$“$f$ is proper” is fpqc local on the base.

Lemma 34.20.15. The property $\mathcal{P}(f) =$“$f$ is flat” is fpqc local on the base.

Proof. Being flat is preserved under arbitrary base change, see Morphisms, Lemma 28.24.7. 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 10.38.8. Hence $f$ is flat. Therefore Lemma 34.19.4 applies and we win. $\square$

Lemma 34.20.16. The property $\mathcal{P}(f) =$“$f$ is an open immersion” is fpqc local on the base.

Proof. The property of being an open immersion is stable under base change, see Schemes, Lemma 25.18.2. The property of being an open immersion is Zariski local on the base (this is obvious). 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 an open immersion. Then $f'$ is universally open, and universally injective. Hence we conclude that $f$ is universally open by Lemma 34.20.4, and universally injective by Lemma 34.20.8. In particular $f(X) \subset S$ is open, and we may replace $S$ by $f(X)$ and assume that $f$ is surjective. This implies that $f'$ is an isomorphism and we have to show that $f$ is an isomorphism also. Since $f$ is universally injective we see that $f$ is bijective. Hence $f$ is a homeomorphism. Let $x \in X$ and choose $U \subset X$ an affine open neighbourhood of $x$. Since $f(U) \subset S$ is open, and $S$ is affine we may choose a standard open $D(g) \subset f(U)$ containing $f(x)$ where $g \in \Gamma (S, \mathcal{O}_ S)$. It is clear that $U \cap f^{-1}(D(g))$ is still affine and still an open neighbourhood of $x$. Replace $U$ by $U \cap f^{-1}(D(g))$ and write $V = D(g) \subset S$ and $V'$ the inverse image of $V$ in $S'$. Note that $V'$ is a standard open of $S'$ as well and in particular that $V'$ is affine. Since $f'$ is an isomorphism we have $V' \times _ V U \to V'$ is an isomorphism. In terms of rings this means that

\[ \mathcal{O}(V') \longrightarrow \mathcal{O}(V') \otimes _{\mathcal{O}(V)} \mathcal{O}(U) \]

is an isomorphism. Since $\mathcal{O}(V) \to \mathcal{O}(V')$ is faithfully flat this implies that $\mathcal{O}(V) \to \mathcal{O}(U)$ is an isomorphism. Hence $U \cong V$ and we see that $f$ is an isomorphism. Therefore Lemma 34.19.4 applies and we win. $\square$

Lemma 34.20.17. The property $\mathcal{P}(f) =$“$f$ is an isomorphism” is fpqc local on the base.

Lemma 34.20.18. The property $\mathcal{P}(f) =$“$f$ is affine” is fpqc local on the base.

Proof. A base change of an affine morphism is affine, see Morphisms, Lemma 28.11.8. Being affine is Zariski local on the base, see Morphisms, Lemma 28.11.3. 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 34.20.1 and 34.20.6 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 25.24.1. 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 29.5.2) 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

\[ X \longrightarrow \mathop{\mathrm{Spec}}(A) \]

over $S$ from Schemes, Lemma 25.6.4. By the above the base change of this morphism to $S'$ is an isomorphism. Hence it is an isomorphism by Lemma 34.20.17. Therefore Lemma 34.19.4 applies and we win. $\square$

Lemma 34.20.19. The property $\mathcal{P}(f) =$“$f$ is a closed immersion” is fpqc local on the base.

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 28.11.9. By Lemma 34.20.18 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 29.5.2 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

\[ \Big( \mathcal{O}_{Y, y} \longrightarrow (f_*\mathcal{O}_ X)_ y \Big) \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{Y_ i, y_ i} \]

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. $\square$

Lemma 34.20.20. The property $\mathcal{P}(f) =$“$f$ is quasi-affine” is fpqc local on the base.

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 34.20.1 and 34.20.6 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 25.24.1 it is a quasi-coherent $\mathcal{O}_ Y$-algebra. Consider the canonical morphism

\[ j : X \longrightarrow \underline{\mathop{\mathrm{Spec}}}_ Y(\mathcal{A}) \]

see Constructions, Lemma 26.4.7. By flat base change (see for example Cohomology of Schemes, Lemma 29.5.2) 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 26.4.7 for the morphism $f_ i$. By assumption and Morphisms, Lemma 28.12.3 all of these morphisms $j_ i$ are quasi-compact open immersions. Hence, by Lemmas 34.20.1 and 34.20.16 we see that $j$ is a quasi-compact open immersion. Hence by Morphisms, Lemma 28.12.3 again we conclude that $f$ is quasi-affine. $\square$

Lemma 34.20.21. The property $\mathcal{P}(f) =$“$f$ is a quasi-compact immersion” is fpqc local on the base.

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 25.23.8 each $f_ i$ is separated. By Lemmas 34.20.1 and 34.20.6 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 28.6.3 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 29.5.2; 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 28.7.7 the morphisms $X_ i \to Y_ i \times _ Y Z$ are open immersions. Hence by Lemma 34.20.16 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). $\square$

Lemma 34.20.22. The property $\mathcal{P}(f) =$“$f$ is integral” is fpqc local on the base.

Proof. An integral morphism is the same thing as an affine, universally closed morphism. See Morphisms, Lemma 28.42.7. Hence the lemma follows on combining Lemmas 34.20.3 and 34.20.18. $\square$

Lemma 34.20.23. The property $\mathcal{P}(f) =$“$f$ is finite” is fpqc local on the base.

Proof. An finite morphism is the same thing as an integral morphism which is locally of finite type. See Morphisms, Lemma 28.42.4. Hence the lemma follows on combining Lemmas 34.20.10 and 34.20.22. $\square$

Lemma 34.20.24. The properties $\mathcal{P}(f) =$“$f$ is locally quasi-finite” and $\mathcal{P}(f) =$“$f$ is quasi-finite” are fpqc local on the base.

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 34.20.10) 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 28.19.13 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 34.20.1). $\square$

Lemma 34.20.25. The property $\mathcal{P}(f) =$“$f$ is locally of finite type of relative dimension $d$” is fpqc local on the base.

Proof. This follows immediately from the fact that being locally of finite type is fpqc local on the base and Morphisms, Lemma 28.27.3. $\square$

Lemma 34.20.26. The property $\mathcal{P}(f) =$“$f$ is syntomic” is fpqc local on the base.

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 34.20.15, and 34.20.11). Hence the result follows for syntomic from Morphisms, Lemma 28.29.12. $\square$

Lemma 34.20.27. The property $\mathcal{P}(f) =$“$f$ is smooth” is fpqc local on the base.

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 34.20.15, and 34.20.11). Hence the result follows for smooth from Morphisms, Lemma 28.32.15. $\square$

Lemma 34.20.28. The property $\mathcal{P}(f) =$“$f$ is unramified” is fpqc local on the base. The property $\mathcal{P}(f) =$“$f$ is G-unramified” is fpqc local on the base.

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 28.33.13). 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 34.20.11, 34.20.10, and 34.20.16). Hence the result follows formally. $\square$

Lemma 34.20.29. The property $\mathcal{P}(f) =$“$f$ is étale” is fpqc local on the base.

Proof. A morphism is étale if and only if it flat and G-unramified. See Morphisms, Lemma 28.34.16. We have seen already that being flat and G-unramified are fpqc local on the base (Lemmas 34.20.15, and 34.20.28). Hence the result follows. $\square$

Lemma 34.20.30. The property $\mathcal{P}(f) =$“$f$ is finite locally free” is fpqc local on the base. Let $d \geq 0$. The property $\mathcal{P}(f) =$“$f$ is finite locally free of degree $d$” is fpqc local on the base.

Proof. Being finite locally free is equivalent to being finite, flat and locally of finite presentation (Morphisms, Lemma 28.46.2). Hence this follows from Lemmas 34.20.23, 34.20.15, and 34.20.11. 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 $(f_*\mathcal{O}_ Z)_ u \otimes _{\mathcal{O}_{U, u}} \kappa (u)$. $\square$

Lemma 34.20.31. The property $\mathcal{P}(f) =$“$f$ is a monomorphism” is fpqc local on the base.

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 34.10.7) we conclude $a = b$. $\square$

Lemma 34.20.32. The properties

  1. $\mathcal{P}(f) =$“$f$ is a Koszul-regular immersion”,

  2. $\mathcal{P}(f) =$“$f$ is an $H_1$-regular immersion”, and

  3. $\mathcal{P}(f) =$“$f$ is a quasi-regular immersion”

are fpqc local on the base.

Proof. We will use the criterion of Lemma 34.19.4 to prove this. By Divisors, Definition 30.21.1 being a Koszul-regular (resp. $H_1$-regular, quasi-regular) immersion is Zariski local on the base. By Divisors, Lemma 30.21.4 being a Koszul-regular (resp. $H_1$-regular, quasi-regular) immersion is preserved under flat base change. The final hypothesis (3) of Lemma 34.19.4 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 15.31.4. $\square$


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