The Stacks project

Proof. Let us use chain complexes in the proof of this lemma. The Koszul complex $K_\bullet (\overline{f}_1, \ldots , \overline{f}_ r)$ is defined in More on Algebra, Definition 15.28.2. By More on Algebra, Lemma 15.75.5 we can represent $K$ by a complex

whose tensor product with $A/\mathfrak mA$ is equal (!) to $K_\bullet (\overline{f}_1, \ldots , \overline{f}_ r)$. Denote $f_1, \ldots , f_ r \in A$ the components of the arrow $A^{\oplus r} \to A$. These $f_ i$ are lifts of the $\overline{f}_ i$. By Algebra, Lemma 10.128.6 $f_1, \ldots , f_ r$ form a regular sequence in $A$ and $A/(f_1, \ldots , f_ r)$ is flat over $R$. Let $J = (f_1, \ldots , f_ r) \subset A$. Consider the diagram

Since $f_1, \ldots , f_ r$ is a regular sequence the south-west arrow is a quasi-isomorphism (see More on Algebra, Lemma 15.30.2). Hence we can find the dotted arrow making the diagram commute for example by Algebra, Lemma 10.71.4. Reducing modulo $\mathfrak m$ we obtain a commutative diagram

by our choice of $K_\bullet $. Thus $\overline{\varphi }$ is an isomorphism in the derived category $D(A/\mathfrak m A)$. It follows that $\overline{\varphi } \otimes _{A/\mathfrak m A}^\mathbf {L} \lambda $ is an isomorphism. Since $\overline{f}_ i \in \mathfrak n / \mathfrak m A$ we see that

Hence $\varphi _ i \bmod \mathfrak n$ is invertible. Since $A$ is local this means that $\varphi _ i$ is an isomorphism and the proof is complete. $\square$

Proof. We can find $g \in S$, $g \not\in \mathfrak q$ with property (1) by the definition of localizations. After replacing $g$ by $gg'$ for some $g' \in S$, $g' \not\in \mathfrak q$ we may assume (2) holds, see Algebra, Lemma 10.68.6. By Algebra, Theorem 10.129.4 we find that $S_ g/(f'_1, \ldots , f'_ r)$ is flat over $R$ in an open neighbourhood of $\mathfrak q$. Hence after once more replacing $g$ by $gg'$ for some $g' \in S$, $g' \not\in \mathfrak q$ we may assume (3) holds as well. Finally, we get (4) for a further replacement by More on Algebra, Lemma 15.74.17. $\square$

Proof. The morphism $p$ is proper by Morphisms, Lemma 29.41.6. By Cohomology of Schemes, Lemma 30.21.2 there is an open $Y_ s \subset V \subset Y$ such that $p|_{p^{-1}(V)} : p^{-1}(V) \to V$ is finite. By More on Morphisms, Theorem 37.16.1 there is an open $X_ s \subset U \subset X$ such that $p|_ U : U \to Y$ is flat. After removing the images of $X \setminus U$ and $Y \setminus V$ (which are closed subsets not containing $s$) we may assume $p$ is flat and finite. Then $p$ is open (Morphisms, Lemma 29.25.10) and $Y_ s \subset p(X) \subset Y$ hence after shrinking $S$ we may assume $p$ is surjective. As $p_ s : X_ s \to Y_ s$ is an isomorphism, the map

of coherent $\mathcal{O}_ Y$-modules ($p$ is finite) becomes an isomorphism after pullback by $i : Y_ s \to Y$ (by Cohomology of Schemes, Lemma 30.5.1 for example). By Nakayama's lemma, this implies that $\mathcal{O}_{Y, y} \to (p_*\mathcal{O}_ X)_ y$ is surjective for all $y \in Y_ s$. Hence there is an open $Y_ s \subset V \subset Y$ such that $p^\sharp |_ V$ is surjective (Modules, Lemma 17.9.4). Hence after shrinking $S$ once more we may assume $p^\sharp $ is surjective which means that $p$ is a closed immersion (as $p$ is already finite). Thus now $p$ is a surjective flat closed immersion of Noetherian schemes and hence an isomorphism, see Morphisms, Section 29.26. $\square$

Proof. Denote $i : Y_ s \times Y_ s = X_ s \times Y_ s \to X \times _ S Y$ the natural closed immersion. (We will write $Y_ s$ and not $X_ s$ for the fibre of $X$ over $s$ from now on.) Let $z \in Y_ s \times Y_ s = (X \times _ S Y)_ s \subset X \times _ S Y$ be a closed point. As indicated we think of $z$ both as a closed point of $Y_ s \times Y_ s$ as well as a closed point of $X \times _ S Y$.

Case I: $z \not\in \Delta _{Y_ s/k}(Y_ s)$. Denote $\mathcal{O}_ z$ the coherent $\mathcal{O}_{Y_ s \times Y_ s}$-module supported at $z$ whose value is $\kappa (z)$. Then $i_*\mathcal{O}_ z$ is the coherent $\mathcal{O}_{X \times _ S Y}$-module supported at $z$ whose value is $\kappa (z)$. Our assumption means that

Hence by Lemma 57.11.3 we find an open neighbourhood $U(z) \subset X \times _ S Y$ of $z$ such that $K|_{U(z)} = 0$. In this case we set $Z(z) = \emptyset $ as closed subscheme of $U(z)$.

Case II: $z \in \Delta _{Y_ s/k}(Y_ s)$. Since $Y_ s$ is smooth over $k$ we know that $\Delta _{Y_ s/k} : Y_ s \to Y_ s \times Y_ s$ is a regular immersion, see More on Morphisms, Lemma 37.62.18. Choose a regular sequence $\overline{f}_1, \ldots , \overline{f}_ r \in \mathcal{O}_{Y_ s \times Y_ s, z}$ cutting out the ideal sheaf of $\Delta _{Y_ s/k}(Y_ s)$. Since a regular sequence is Koszul-regular (More on Algebra, Lemma 15.30.2) our assumption means that

is represented by the Koszul complex on $\overline{f}_1, \ldots , \overline{f}_ r$ over $\mathcal{O}_{Y_ s \times Y_ s, z}$. By Lemma 57.16.1 applied to $\mathcal{O}_{S, s} \to \mathcal{O}_{X \times _ S Y, z}$ we conclude that $K_ z \in D(\mathcal{O}_{X \times _ S Y, z})$ is represented by the Koszul complex on a regular sequence $f_1, \ldots , f_ r \in \mathcal{O}_{X \times _ S Y, z}$ lifting the regular sequence $\overline{f}_1, \ldots , \overline{f}_ r$ such that moreover $\mathcal{O}_{X \times _ S Y}/(f_1, \ldots , f_ r)$ is flat over $\mathcal{O}_{S, s}$. By some limit arguments (Lemma 57.16.2) we conclude that there exists an affine open neighbourhood $U(z) \subset X \times _ S Y$ of $z$ and a closed subscheme $Z(z) \subset U(z)$ such that

1. $Z(z) \to U(z)$ is a regular closed immersion,

2. $K|_{U(z)}$ is quasi-isomorphic to $\mathcal{O}_{Z(z)}$,

3. $Z(z) \to S$ is flat,

4. $Z(z)_ s = \Delta _{Y_ s/k}(Y_ s) \cap U(z)_ s$ as closed subschemes of $U(z)_ s$.

By property (2), for $z, z' \in Y_ s \times Y_ s$, we find that $Z(z) \cap U(z') = Z(z') \cap U(z)$ as closed subschemes. Hence we obtain an open neighbourhood

of $Y_ s \times Y_ s$ in $X \times _ S Y$ and a closed subscheme $Z \subset U$ such that (1) $Z \to U$ is a regular closed immersion, (2) $Z \to S$ is flat, and (3) $Z_ s = \Delta _{Y_ s/k}(Y_ s)$. Since $X \times _ S Y \to S$ is proper, after replacing $S$ by an open neighbourhood of $s$ we may assume $U = X \times _ S Y$. Since the projections $Z_ s \to Y_ s$ and $Z_ s \to X_ s$ are isomorphisms, we conclude that after shrinking $S$ we may assume $Z \to Y$ and $Z \to X$ are isomorphisms, see Lemma 57.16.3. This finishes the proof. $\square$

Proof. Choose a closed point $s \in S$. Since $k$ is algebraically closed this is a $k$-rational point. Set $Y_0 = Y_ s$. The restriction $K_0$ of $K$ to $X \times Y_0$ is the Fourier-Mukai kernel of a relative equivalence from $X$ to $Y_0$ over $\mathop{\mathrm{Spec}}(k)$ by Lemma 57.15.3. Let $K'_0$ in $D_{perf}(\mathcal{O}_{Y_0 \times X})$ be the object assumed to exist in Definition 57.15.1. Then $K'_0$ is the Fourier-Mukai kernel of a relative equivalence from $Y_0$ to $X$ over $\mathop{\mathrm{Spec}}(k)$ by the symmetry inherent in Definition 57.15.1. Hence by Lemma 57.15.3 we see that the pullback

on $(Y_0 \times S) \times _ S (X \times S) = Y_0 \times X \times S$ is the Fourier-Mukai kernel of a relative equivalence from $Y_0 \times S$ to $X \times S$ over $S$. Now consider the kernel

on $(Y_0 \times S) \times _ S Y$. This is the Fourier-Mukai kernel of a relative equivalence from $Y_0 \times S$ to $Y$ over $S$ since it is the composition of two invertible arrows in the category constructed in Section 57.14. Moreover, this composition passes through base change (Lemma 57.14.1). Hence we see that the pullback of $K_{new}$ to $((Y_0 \times S) \times _ S Y)_ s = Y_0 \times Y_0$ is equal to the composition of $K_0$ and $K'_0$ and hence equal to the identity in this category. In other words, we have

Thus by Lemma 57.16.4 we conclude that $Y \to S$ is isomorphic to $Y_0 \times S$ in an open neighbourhood of $s$. This finishes the proof. $\square$