**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

\[ \xymatrix{ 0 \ar[r] & IB' \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & IA' \ar[r] \ar[u] & A' \ar[r] \ar[u] & A \ar[r] \ar[u] & 0 } \]

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.
$\square$

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