The Stacks project

Proof. As the problem is local on $X'$ we may assume that $X, X', S, S'$ are affine schemes. Say $S' = \mathop{\mathrm{Spec}}(A')$, $X' = \mathop{\mathrm{Spec}}(B')$, $S = \mathop{\mathrm{Spec}}(A)$, $X = \mathop{\mathrm{Spec}}(B)$ with $A = A'/I$ and $B = B'/J$ for some square zero ideals. Then we obtain the following commutative diagram

with exact rows. The canonical map of the lemma is the map

The assumption that $f$ is flat signifies that $A \to B$ is flat.

Assume (1). Then $A' \to B'$ is flat and $J = IB'$. Flatness implies $\text{Tor}_1^{A'}(B', A) = 0$ (see Algebra, Lemma 10.75.8). This means $I \otimes _{A'} B' \to B'$ is injective (see Algebra, Remark 10.75.9). Hence we see that $I \otimes _ A B \to J$ is an isomorphism.

Assume (2). Then it follows that $J = IB'$, so that $X = S \times _{S'} X'$. Moreover, we get $\text{Tor}_1^{A'}(B', A'/I) = 0$ by reversing the implications in the previous paragraph. Hence $B'$ is flat over $A'$ by Algebra, Lemma 10.99.8. $\square$

Proof. Immediate consequence of Algebra, Lemma 10.101.8. $\square$

Proof. The assumptions on $X$ and $Y$ mean that $f$ is the base change of $f'$ by $X \to X'$. The properties $\mathcal{P}$ listed in (1) – (23) above are all stable under base change, hence if $f'$ has property $\mathcal{P}$, then so does $f$. See Schemes, Lemmas 26.18.2, 26.19.3, 26.21.12, and 26.23.5 and Morphisms, Lemmas 29.9.4, 29.10.4, 29.11.8, 29.15.4, 29.20.13, 29.21.4, 29.29.2, 29.30.4, 29.34.5, 29.35.5, 29.36.4, 29.41.5, 29.44.6, and 29.48.4.

The interesting direction in each case is therefore to assume that $f$ has the property and deduce that $f'$ has it too. By induction on the order of the thickening we may assume that $S \subset S'$ is a first order thickening, see discussion immediately following Definition 37.2.1. We make a couple of general remarks which we will use without further mention in the arguments below. (I) Let $W' \subset S'$ be an affine open and let $U' \subset X'$ and $V' \subset Y'$ be affine opens lying over $W'$ with $f'(U') \subset V'$. Let $W' = \mathop{\mathrm{Spec}}(R')$ and denote $I \subset R'$ be the ideal defining the closed subscheme $W' \cap S$. Say $U' = \mathop{\mathrm{Spec}}(B')$ and $V' = \mathop{\mathrm{Spec}}(A')$. Then we get a commutative diagram

with exact rows. Moreover $IB' \cong I \otimes _ R B$, see proof of Lemma 37.10.1. (II) The morphisms $X \to X'$ and $Y \to Y'$ are universal homeomorphisms. Hence the topology of the maps $f$ and $f'$ (after any base change) is identical. (III) If $f$ is flat, then $f'$ is flat and $Y' \to S'$ is flat at every point in the image of $f'$, see Lemma 37.10.2.

Ad (1). This is general remark (III).

Ad (2). Assume $f$ is an isomorphism. By (III) we see that $Y' \to S'$ is flat. Choose an affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$. Then $V = Y \cap V'$ is affine which implies that $V \cong f^{-1}(V) = U = Y \times _{Y'} U'$ is affine. By Lemma 37.2.3 we see that $U'$ is affine. Thus we have a diagram as in the general remark (I) and moreover $IA \cong I \otimes _ R A$ because $R' \to A'$ is flat. Then $IB' \cong I \otimes _ R B \cong I \otimes _ R A \cong IA'$ and $A \cong B$. By the exactness of the rows in the diagram above we see that $A' \cong B'$, i.e., $U' \cong V'$. Thus $f'$ is an isomorphism.

Ad (3). Assume $f$ is an open immersion. Then $f$ is an isomorphism of $X$ with an open subscheme $V \subset Y$. Let $V' \subset Y'$ be the open subscheme whose underlying topological space is $V$. Then $f'$ is a map from $X'$ to $V'$ which is an isomorphism by (2). Hence $f'$ is an open immersion.

Ad (4). Immediate from remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (5). Immediate from remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (6). Note that $X \times _ Y X = Y \times _{Y'} (X' \times _{Y'} X')$ so that $X' \times _{Y'} X'$ is a thickening of $X \times _ Y X$. Hence the topology of the maps $\Delta _{X/Y}$ and $\Delta _{X'/Y'}$ matches and we win. See also Lemma 37.3.1 for a more general statement.

Ad (7). Assume $f$ is a monomorphism. Consider the diagonal morphism $\Delta _{X'/Y'} : X' \to X' \times _{Y'} X'$. The base change of $\Delta _{X'/Y'}$ by $S \to S'$ is $\Delta _{X/Y}$ which is an isomorphism by assumption. By (2) we conclude that $\Delta _{X'/Y'}$ is an isomorphism.

Ad (8). This is clear. See also Lemma 37.3.1 for a more general statement.

Ad (9). Immediate from remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (10). Assume $f$ is affine. Choose an affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$. Then $V = Y \cap V'$ is affine which implies that $U = Y \times _{Y'} U'$ is affine. By Lemma 37.2.3 we see that $U'$ is affine. Hence $f'$ is affine. See also Lemma 37.3.1 for a more general statement.

Ad (11). Via remark (I) comes down to proving $A' \to B'$ is of finite type if $A \to B$ is of finite type. Suppose that $x_1, \ldots , x_ n \in B'$ are elements whose images in $B$ generate $B$ as an $A$-algebra. Then $A'[x_1, \ldots , x_ n] \to B$ is surjective as both $A'[x_1, \ldots , x_ n] \to B$ is surjective and $I \otimes _ R A[x_1, \ldots , x_ n] \to I \otimes _ R B$ is surjective. See also Lemma 37.3.3 for a more general statement.

Ad (12). Follows from (11) and that quasi-finiteness of a morphism of finite type can be checked on fibres, see Morphisms, Lemma 29.20.6. See also Lemma 37.3.3 for a more general statement.

Ad (13). Via remark (I) comes down to proving $A' \to B'$ is of finite presentation if $A \to B$ is of finite presentation. We may assume that $B' = A'[x_1, \ldots , x_ n]/K'$ for some ideal $K'$ by (11). We get a short exact sequence

As $B'$ is flat over $R'$ we see that $K' \otimes _{R'} R$ is the kernel of the surjection $A[x_1, \ldots , x_ n] \to B$. By assumption on $A \to B$ there exist finitely many $f'_1, \ldots , f'_ m \in K'$ whose images in $A[x_1, \ldots , x_ n]$ generate this kernel. Since $I$ is nilpotent we see that $f'_1, \ldots , f'_ m$ generate $K'$ by Nakayama's lemma, see Algebra, Lemma 10.20.1.

Ad (14). Follows from (11) and general remark (II). See also Lemma 37.3.3 for a more general statement.

Ad (15). Immediate from general remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (16). Assume $f$ is syntomic. By (13) $f'$ is locally of finite presentation, by general remark (III) $f'$ is flat and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is syntomic by Morphisms, Lemma 29.30.11.

Ad (17). Assume $f$ is smooth. By (13) $f'$ is locally of finite presentation, by general remark (III) $f'$ is flat, and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is smooth by Morphisms, Lemma 29.34.3.

Ad (18). Assume $f$ unramified. By (11) $f'$ is locally of finite type and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is unramified by Morphisms, Lemma 29.35.12. See also Lemma 37.3.3 for a more general statement.

Ad (19). Assume $f$ étale. By (13) $f'$ is locally of finite presentation, by general remark (III) $f'$ is flat, and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is étale by Morphisms, Lemma 29.36.8.

Ad (20). This follows from a combination of (6), (11), (4), and (5). See also Lemma 37.3.3 for a more general statement.

Ad (21). Combine (5) and (10) with Morphisms, Lemma 29.44.7. See also Lemma 37.3.1 for a more general statement.

Ad (22). Combine (21), and (11) with Morphisms, Lemma 29.44.4. See also Lemma 37.3.3 for a more general statement.

Ad (23). Assume $f$ finite locally free. By (22) we see that $f'$ is finite, by general remark (III) $f'$ is flat, and by (13) $f'$ is locally of finite presentation. Hence $f'$ is finite locally free by Morphisms, Lemma 29.48.2. $\square$

Proof. Immediate consequence of Algebra, Lemma 10.128.10. $\square$

Proof. The assumptions on $X$ and $Y$ mean that $f$ is the base change of $f'$ by $X \to X'$. The properties $\mathcal{P}$ listed in (1) – (20) above are all stable under base change, hence if $f'$ has property $\mathcal{P}$, then so does $f$. See Schemes, Lemmas 26.18.2, 26.19.3, 26.21.12, and 26.23.5 and Morphisms, Lemmas 29.9.4, 29.10.4, 29.11.8, 29.20.13, 29.29.2, 29.30.4, 29.34.5, 29.35.5, 29.36.4, 29.41.5, 29.44.6, and 29.48.4.

The interesting direction in each case is therefore to assume that $f$ has the property and deduce that $f'$ has it too. We make a couple of general remarks which we will use without further mention in the arguments below. (I) Let $W' \subset S'$ be an affine open and let $U' \subset X'$ and $V' \subset Y'$ be affine opens lying over $W'$ with $f'(U') \subset V'$. Let $W' = \mathop{\mathrm{Spec}}(R')$ and denote $I \subset R'$ be the ideal defining the closed subscheme $W' \cap S$. Say $U' = \mathop{\mathrm{Spec}}(B')$ and $V' = \mathop{\mathrm{Spec}}(A')$. Then we get a commutative diagram

with exact rows. (II) The morphisms $X \to X'$ and $Y \to Y'$ are universal homeomorphisms. Hence the topology of the maps $f$ and $f'$ (after any base change) is identical. (III) If $f$ is flat, then $f'$ is flat and $Y' \to S'$ is flat at every point in the image of $f'$, see Lemma 37.10.2.

Ad (1). This is general remark (III).

Ad (2). Assume $f$ is an isomorphism. Choose an affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$. Then $V = Y \cap V'$ is affine which implies that $V \cong f^{-1}(V) = U = Y \times _{Y'} U'$ is affine. By Lemma 37.2.3 we see that $U'$ is affine. Thus we have a diagram as in the general remark (I). By Algebra, Lemma 10.126.11 we see that $A' \to B'$ is an isomorphism, i.e., $U' \cong V'$. Thus $f'$ is an isomorphism.

Ad (3). Assume $f$ is an open immersion. Then $f$ is an isomorphism of $X$ with an open subscheme $V \subset Y$. Let $V' \subset Y'$ be the open subscheme whose underlying topological space is $V$. Then $f'$ is a map from $X'$ to $V'$ which is an isomorphism by (2). Hence $f'$ is an open immersion.

Ad (4). Immediate from remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (5). Immediate from remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (6). Note that $X \times _ Y X = Y \times _{Y'} (X' \times _{Y'} X')$ so that $X' \times _{Y'} X'$ is a thickening of $X \times _ Y X$. Hence the topology of the maps $\Delta _{X/Y}$ and $\Delta _{X'/Y'}$ matches and we win. See also Lemma 37.3.1 for a more general statement.

Ad (7). Assume $f$ is a monomorphism. Consider the diagonal morphism $\Delta _{X'/Y'} : X' \to X' \times _{Y'} X'$. Observe that $X' \times _{Y'} X' \to S'$ is locally of finite type. The base change of $\Delta _{X'/Y'}$ by $S \to S'$ is $\Delta _{X/Y}$ which is an isomorphism by assumption. By (2) we conclude that $\Delta _{X'/Y'}$ is an isomorphism.

Ad (8). This is clear. See also Lemma 37.3.1 for a more general statement.

Ad (9). Immediate from remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (10). Assume $f$ is affine. Choose an affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$. Then $V = Y \cap V'$ is affine which implies that $U = Y \times _{Y'} U'$ is affine. By Lemma 37.2.3 we see that $U'$ is affine. Hence $f'$ is affine. See also Lemma 37.3.1 for a more general statement.

Ad (11). Follows from the fact that $f'$ is locally of finite type (by Morphisms, Lemma 29.15.8) and that quasi-finiteness of a morphism of finite type can be checked on fibres, see Morphisms, Lemma 29.20.6.

Ad (12). Follows from general remark (II) and the fact that $f'$ is locally of finite type (Morphisms, Lemma 29.15.8).

Ad (13). Immediate from general remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (14). Assume $f$ is syntomic. By Morphisms, Lemma 29.21.11 $f'$ is locally of finite presentation. By general remark (III) $f'$ is flat. The fibres of $f'$ are the fibres of $f$. Hence $f'$ is syntomic by Morphisms, Lemma 29.30.11.

Ad (15). Assume $f$ is smooth. By Morphisms, Lemma 29.21.11 $f'$ is locally of finite presentation. By general remark (III) $f'$ is flat. The fibres of $f'$ are the fibres of $f$. Hence $f'$ is smooth by Morphisms, Lemma 29.34.3.

Ad (16). Assume $f$ unramified. By Morphisms, Lemma 29.15.8 $f'$ is locally of finite type. The fibres of $f'$ are the fibres of $f$. Hence $f'$ is unramified by Morphisms, Lemma 29.35.12.

Ad (17). Assume $f$ étale. By Morphisms, Lemma 29.21.11 $f'$ is locally of finite presentation. By general remark (III) $f'$ is flat. The fibres of $f'$ are the fibres of $f$. Hence $f'$ is étale by Morphisms, Lemma 29.36.8.

Ad (18). This follows from a combination of (6), the fact that $f$ is locally of finite type (Morphisms, Lemma 29.15.8), (4), and (5).

Ad (19). Combine (5), (10), Morphisms, Lemma 29.44.7, the fact that $f$ is locally of finite type (Morphisms, Lemma 29.15.8), and Morphisms, Lemma 29.44.4.

Ad (20). Assume $f$ finite locally free. By (19) we see that $f'$ is finite. By general remark (III) $f'$ is flat. By Morphisms, Lemma 29.21.11 $f'$ is locally of finite presentation. Hence $f'$ is finite locally free by Morphisms, Lemma 29.48.2. $\square$

Proof. We first describe the maps $r'$ and $\theta '$. Observe that $\mathcal{L}'$ is $f'$-ample, see Lemma 37.3.2. There is a canonical map of quasi-coherent graded $\mathcal{O}_{S'}$-algebras $\mathcal{A}' \to \bigoplus _{d \geq 0} (f')_*(\mathcal{L}')^{\otimes d}$ which is an isomorphism in degrees $\geq d_0$. Hence this induces an isomorphism on relative Proj compatible with the Serre twists of the structure sheaf, see Constructions, Lemma 27.18.4. Hence we get the morphism $r'$ by Morphisms, Lemma 29.37.4 (which in turn appeals to the construction given in Constructions, Lemma 27.19.1) and it is an isomorphism by Morphisms, Lemma 29.43.17. We get the map $\theta '$ from Constructions, Lemma 27.19.1. By Properties, Lemma 28.28.2 we find that $\theta '$ is an isomorphism (this also uses that the morphism $r'$ over affine opens of $S'$ is the same as the morphism from Properties, Lemma 28.26.9 as is explained in the proof of Morphisms, Lemma 29.43.17).

Assuming the vanishing and local freeness stated in parts (1) and (2), the functoriality of the construction can be seen as follows. Suppose that $h : T \to S'$ is a morphism of schemes, denote $f_ T : X'_ T \to T$ the base change of $f'$ and $\mathcal{L}_ T$ the pullback of $\mathcal{L}$ to $X'_ T$. By cohomology and base change (as formulated in Derived Categories of Schemes, Lemma 36.22.5 for example) we have the corresponding vanishing over $T$ and moreover $h^*\mathcal{A}'_ d = f_{T, *}\mathcal{L}_ T^{\otimes d}$ (and thus the local freeness of pushforwards as well as the finite generation of the corresponding graded $\mathcal{O}_ T$-algebra $\mathcal{A}_ T$). Hence the morphism $r_ T : X_ T \to \underline{\text{Proj}}_ T(\bigoplus f_{T, *}\mathcal{L}_ T^{\otimes d})$ is simply the base change of $r'$ to $T$ and the pullback of $\theta '$ is the map $\theta _ T$.

Having said all of the above, we see that it suffices to prove (1), (2), and (3). Pick $d_0$ such that $R^ pf_*\mathcal{L}^{\otimes d} = 0$ for all $d \geq d_0$ and $p > 0$, see Cohomology of Schemes, Lemma 30.16.1. We claim that $d_0$ works.

By cohomology and base change (Derived Categories of Schemes, Lemma 36.30.4) we see that $E'_ d = Rf'_*(\mathcal{L}')^{\otimes d}$ is a perfect object of $D(\mathcal{O}_{S'})$ and its formation commutes with arbitrary base change. In particular, $E_ d = Li^*E'_ d = Rf_*\mathcal{L}^{\otimes d}$. By Derived Categories of Schemes, Lemma 36.32.4 we see that for $d \geq d_0$ the complex $E_ d$ is isomorphic to the finite locally free $\mathcal{O}_ S$-module $f_*\mathcal{L}^{\otimes d}$ placed in cohomological degree $0$. Then by Derived Categories of Schemes, Lemma 36.31.3 we conclude that $E'_ d$ is isomorphic to a finite locally free module placed in cohomological degree $0$. Of course this means that $E'_ d = \mathcal{A}'_ d[0]$, that $R^ pf'_*(\mathcal{L}')^{\otimes d} = 0$ for $p > 0$, and that $\mathcal{A}'_ d$ is finite locally free. This proves (1) and (2).

The last thing we have to show is finite presentation of $\mathcal{A}'$ as a sheaf of $\mathcal{O}_{S'}$-algebras (this notion was introduced in Properties, Section 28.22). Let $U' = \mathop{\mathrm{Spec}}(R') \subset S'$ be an affine open. Then $A' = \mathcal{A}'(U')$ is a graded $R'$-algebra whose graded parts are finite projective $R'$-modules. We have to show that $A'$ is a finitely presented $R'$-algebra. We will prove this by reduction to the Noetherian case. Namely, we can find a finite type $\mathbf{Z}$-subalgebra $R'_0 \subset R'$ and a pair¹ $(X'_0, \mathcal{L}'_0)$ over $R'_0$ whose base change is $(X'_{U'}, \mathcal{L}'|_{X'_{U'}})$, see Limits, Lemmas 32.10.2, 32.10.3, 32.13.1, 32.8.7, and 32.4.15. Cohomology of Schemes, Lemma 30.16.1 implies $A'_0 = \bigoplus _{d \geq 0} H^0(X'_0, (\mathcal{L}'_0)^{\otimes d})$ is a finitely generated graded $R'_0$-algebra and implies there exists a $d'_0$ such that $H^ p(X'_0, (\mathcal{L}'_0)^{\otimes d}) = 0$, $p > 0$ for $d \geq d'_0$. By the arguments given above applied to $X'_0 \to \mathop{\mathrm{Spec}}(R'_0)$ and $\mathcal{L}'_0$ we see that $(A'_0)_ d$ is a finite projective $R'_0$-module and that

for $d \geq d'_0$. Now a small twist in the argument is that we don't know that we can choose $d'_0$ equal to $d_0$². To get around this we use the following sequence of arguments to finish the proof:

1. The algebra $B = R'_0 \oplus \bigoplus _{d \geq \max (d_0, d'_0)} (A'_0)_ d$ is an $R'_0$-algebra of finite type: apply the Artin-Tate lemma to $B \subset A'_0$, see Algebra, Lemma 10.51.7.

2. As $R'_0$ is Noetherian we see that $B$ is an $R'_0$-algebra of finite presentation.

3. By right exactness of tensor product we see that $B \otimes _{R'_0} R'$ is an $R'$-algebra of finite presentation.

4. By the displayed equalities this exactly says that $C = R' \oplus \bigoplus _{d \geq \max (d_0, d'_0)} A'_ d$ is an $R'$-algebra of finite presentation.

5. The quotient $A'/C$ is the direct sum of the finite projective $R'$-modules $A'_ d$, $d_0 \leq d \leq \max (d_0, d'_0)$, hence finitely presented as $R'$-module.

6. The quotient $A'/C$ is finitely presented as a $C$-module by Algebra, Lemma 10.6.4.

7. Thus $A'$ is finitely presented as a $C$-module by Algebra, Lemma 10.5.3.

8. By Algebra, Lemma 10.7.4 this implies $A'$ is finitely presented as a $C$-algebra.

9. Finally, by Algebra, Lemma 10.6.2 applied to $R' \to C \to A'$ this implies $A'$ is finitely presented as an $R'$-algebra.

This finishes the proof. $\square$