The Stacks project

Slight improvement of [AT]

Theorem 57.18.2. Let $K$ be an algebraically closed field. Let $\mathbf{X}$ be a smooth proper scheme over $K$. There are at most countably many isomorphism classes of smooth proper schemes $\mathbf{Y}$ over $K$ which are derived equivalent to $\mathbf{X}$.

Proof. Choose a countable set $I$ and for $i \in I$ systems $(S_ i/K, X_ i \to S_ i, Y_ i \to S_ i, M_ i)$ satisfying properties (1), (2), (3), and (4) of Lemma 57.17.5. Pick $i \in I$ and set $S = S_ i$, $X = X_ i$, $Y = Y_ i$, and $M = M_ i$. Clearly it suffice to show that the set of isomorphism classes of fibres $Y_ s$ for $s \in S(K)$ such that $X_ s \cong \mathbf{X}$ is countable. This we prove in the next paragraph.

Let $S$ be a finite type scheme over $K$, let $X \to S$ and $Y \to S$ be proper smooth morphisms, and let $M \in D_{perf}(\mathcal{O}_{X \times _ S Y})$ be the Fourier-Mukai kernel of a relative equivalence from $X$ to $Y$ over $S$. We will show the set of isomorphism classes of fibres $Y_ s$ for $s \in S(K)$ such that $X_ s \cong \mathbf{X}$ is countable. By Lemma 57.17.4 applied to the families $\mathbf{X} \times S \to S$ and $X \to S$ there exists a countable set $I$ and for $i \in I$ a pair $(S_ i \to S, h_ i)$ with the following properties

  1. $S_ i \to S$ is a morphism of finite type, set $X_ i = X \times _ S S_ i$,

  2. $h_ i : \mathbf{X} \times S_ i \to X_ i$ is an isomorphism over $S_ i$, and

  3. for any closed point $s \in S(K)$ if $\mathbf{X} \cong X_ s$ over $K = \kappa (s)$ then $s$ is in the image of $S_ i \to S$ for some $i$.

Set $Y_ i = Y \times _ S S_ i$. Denote $M_ i \in D_{perf}(\mathcal{O}_{X_ i \times _{S_ i} Y_ i})$ the pullback of $M$. By Lemma 57.15.3 $M_ i$ is the Fourier-Mukai kernel of a relative equivalence from $X_ i$ to $Y_ i$ over $S_ i$. Since $I$ is countable, by property (3) it suffices to prove that the set of isomorphism classes of fibres $Y_{i, s}$ for $s \in S_ i(K)$ is countable. In fact, this number is finite by Lemma 57.16.5 and the proof is complete. $\square$


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