The Stacks project

Proof. Observe that $S \to S'$ and $X \to X'$ are universal homeomorphisms (see for example Morphisms, Lemma 29.45.6). This immediately implies parts (2), (3), (4), (7), and (8). Part (1) follows from Lemma 37.2.3 which tells us that there is a 1-to-1 correspondence between affine opens of $S$ and $S'$ and between affine opens of $X$ and $X'$. Part (9) follows from Limits, Lemma 32.11.5 and the remark just made about affine opens of $S$ and $S'$. Part (5) follows from (1) and (4) by Morphisms, Lemma 29.44.7. Finally, note that

is a thickening (the two arrows are thickenings by Lemma 37.2.4). Hence applying (3) and (4) to the morphism $(S \subset S') \to (S \times _ X S \to S' \times _{X'} S')$ we obtain (6). $\square$

Proof. Recall that being relatively ample is a condition for each affine open in the base, see Morphisms, Definition 29.37.1. By Lemma 37.2.3 there is a 1-to-1 correspondence between affine opens of $S$ and $S'$. Thus we may assume $S$ and $S'$ are affine and we reduce to proving that $\mathcal{L}'$ is ample if and only if $\mathcal{L}$ is ample. This is Limits, Lemma 32.11.4. $\square$

Proof. The properties $\mathcal{P}$ listed in the lemma are all stable under base change, hence if $f'$ has property $\mathcal{P}$, then so does $f$. See Schemes, Lemmas 26.18.2 and 26.23.5 and Morphisms, Lemmas 29.15.4, 29.20.13, 29.29.2, 29.32.10, 29.35.5, 29.41.5, and 29.44.6.

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.

Most of the proofs will use a reduction to the affine case. Let $U' \subset S'$ be an affine open and let $V' \subset X'$ be an affine open lying over $U'$. Let $U' = \mathop{\mathrm{Spec}}(A')$ and denote $I \subset A'$ be the ideal defining the closed subscheme $U' \cap S$. Say $V' = \mathop{\mathrm{Spec}}(B')$. Then $V' \cap X = \mathop{\mathrm{Spec}}(B'/IB')$. Setting $A = A'/I$ and $B = B'/IB'$ we get a commutative diagram

with exact rows and $I^2 = 0$.

The translation of (1) into algebra: If $A \to B$ is surjective, then $A' \to B'$ is surjective. This follows from Nakayama's lemma (Algebra, Lemma 10.20.1).

The translation of (2) into algebra: If $A \to B$ is a finite type ring map, then $A' \to B'$ is a finite type ring map. This follows from Nakayama's lemma (Algebra, Lemma 10.20.1) applied to a map $A'[x_1, \ldots , x_ n] \to B'$ such that $A[x_1, \ldots , x_ n] \to B$ is surjective.

Proof of (3). Follows from (2) and that quasi-finiteness of a morphism which is locally of finite type can be checked on fibres, see Morphisms, Lemma 29.20.6.

Proof of (4). Follows from (2) and that the additional property of “being of relative dimension $d$” can be checked on fibres (by definition, see Morphisms, Definition 29.29.1.

The translation of (5) into algebra: If $\Omega _{B/A} = 0$, then $\Omega _{B'/A'} = 0$. By Algebra, Lemma 10.131.12 we have $0 = \Omega _{B/A} = \Omega _{B'/A'}/I\Omega _{B'/A'}$. Hence $\Omega _{B'/A'} = 0$ by Nakayama's lemma (Algebra, Lemma 10.20.1).

The translation of (6) into algebra: If $A \to B$ is unramified map, then $A' \to B'$ is unramified. Since $A \to B$ is of finite type we see that $A' \to B'$ is of finite type by (2) above. Since $A \to B$ is unramified we have $\Omega _{B/A} = 0$. By part (5) we have $\Omega _{B'/A'} = 0$. Thus $A' \to B'$ is unramified.

Proof of (7). Follows by combining (2) with results of Lemma 37.3.1 and the fact that proper equals quasi-compact $+$ separated $+$ locally of finite type $+$ universally closed.

Proof of (8). Follows by combining (2) with results of Lemma 37.3.1 and using the fact that finite equals integral $+$ locally of finite type (Morphisms, Lemma 29.44.4).

Proof of (9). As $f$ is a monomorphism we have $X = X \times _ S X$. We may apply the results proved so far to the morphism of thickenings $(X \subset X') \to (X \times _ S X \subset X' \times _{S'} X')$. We conclude $X' \to X' \times _{S'} X'$ is a closed immersion by (1). In fact, it is a first order thickening as the ideal defining the closed immersion $X' \to X' \times _{S'} X'$ is contained in the pullback of the ideal $\mathcal{I} \subset \mathcal{O}_{S'}$ cutting out $S$ in $S'$. Indeed, $X = X \times _ S X = (X' \times _{S'} X') \times _{S'} S$ is contained in $X'$. Hence by Morphisms, Lemma 29.32.7 it suffices to show that $\Omega _{X'/S'} = 0$ which follows from (5) and the corresponding statement for $X/S$.

Proof of (10). If $f : X \to S$ is an immersion, then it factors as $X \to U \to S$ where $U \to S$ is an open immersion and $X \to U$ is a closed immersion. Let $U' \subset S'$ be the open subscheme whose underlying topological space is the same as $U$. Then $X' \to S'$ factors through $U'$ and we conclude that $X' \to U'$ is a closed immersion by part (1). This finishes the proof. $\square$

Proof. The properties $\mathcal{P}$ listed in the lemma are all stable under base change, hence if $f'$ has property $\mathcal{P}$, then so does $f$. See Schemes, Lemmas 26.18.2 and 26.23.5 and Morphisms, Lemmas 29.20.13, 29.29.2, 29.32.10, 29.35.5, 29.41.5, and 29.44.6. Hence in each case we need only to prove that if $f$ has the desired property, so does $f'$.

A morphism is locally quasi-finite if and only if it is locally of finite type and the scheme theoretic fibres are discrete spaces, see Morphisms, Lemma 29.20.8. Since the underlying topological space is unchanged by passing to a thickening, we see that $f'$ is locally quasi-finite if (and only if) $f$ is. This proves (1).

Case (2) follows from case (5) of Lemma 37.3.1 and the fact that the finite morphisms are precisely the integral morphisms that are locally of finite type (Morphisms, Lemma 29.44.4).

Case (3). This follows immediately from Morphisms, Lemma 29.45.7.

Case (4) follows from the following algebra statement: Let $A$ be a ring and let $I \subset A$ be a locally nilpotent ideal. Let $B$ be a finite type $A$-algebra. If $\Omega _{(B/IB)/(A/I)} = 0$, then $\Omega _{B/A} = 0$. Namely, the assumption means that $I\Omega _{B/A} = 0$, see Algebra, Lemma 10.131.12. On the other hand $\Omega _{B/A}$ is a finite $B$-module, see Algebra, Lemma 10.131.16. Hence the vanishing of $\Omega _{B/A}$ follows from Nakayama's lemma (Algebra, Lemma 10.20.1) and the fact that $IB$ is contained in the Jacobson radical of $B$.

Case (5) follows immediately from (4) and Morphisms, Lemma 29.35.2.

Proof of (6). As $f$ is a monomorphism we have $X = X \times _ Y X$. We may apply the results proved so far to the morphism of thickenings $(X \subset X') \to (X \times _ Y X \subset X' \times _{Y'} X')$. We conclude $\Delta _{X'/Y'} : X' \to X' \times _{Y'} X'$ is a closed immersion by (3). In fact $\Delta _{X'/Y'}$ is a bijection on underlying sets, hence $\Delta _{X'/Y'}$ is a thickening. On the other hand $\Delta _{X'/Y'}$ is locally of finite presentation by Morphisms, Lemma 29.21.12. In other words, $\Delta _{X'/Y'}(X')$ is cut out by a quasi-coherent sheaf of ideals $\mathcal{J} \subset \mathcal{O}_{X' \times _{Y'} X'}$ of finite type. Since $\Omega _{X'/Y'} = 0$ by (5) we see that the conormal sheaf of $X' \to X' \times _{Y'} X'$ is zero by Morphisms, Lemma 29.32.7. In other words, $\mathcal{J}/\mathcal{J}^2 = 0$. This implies $\Delta _{X'/Y'}$ is an isomorphism, for example by Algebra, Lemma 10.21.5.

Proof of (7). If $f : X \to Y$ is an immersion, then it factors as $X \to V \to Y$ where $V \to Y$ is an open immersion and $X \to V$ is a closed immersion. Let $V' \subset Y'$ be the open subscheme whose underlying topological space is the same as $V$. Then $X' \to V'$ factors through $V'$ and we conclude that $X' \to V'$ is a closed immersion by part (3).

Case (8) follows from Lemma 37.3.1 and the definition of proper morphisms as being the quasi-compact, universally closed, and separated morphisms that are locally of finite type. $\square$