The Stacks project

Proof. Omitted. $\square$

Proof. It suffices to prove the statement for $D(\mathcal{A})$ as the others are full subcategories of this one. Since every object in $D(\mathcal{A})$ is a complex of objects of $\mathcal{A}$ it is immediate that the set of isomorphism classes of objects of $D^ b(\mathcal{A})$ is countable. Moreover, for bounded complexes $A^\bullet$ and $B^\bullet$ of $\mathcal{A}$ it is clear that $\mathop{\mathrm{Hom}}\nolimits _{K^ b(\mathcal{A})}(A^\bullet , B^\bullet )$ is countable. We have

by Derived Categories, Lemma 13.11.6. Thus this is a countable set as a countable colimit of $\square$

Proof. Observe that $X$ is Noetherian by Morphisms, Lemma 29.15.6. Hence $D_{perf}(\mathcal{O}_ X)$ is a full subcategory of $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.11.6. Thus it suffices to prove the result for $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$. Recall that $D^ b_{\textit{Coh}}(\mathcal{O}_ X) = D^ b(\textit{Coh}(\mathcal{O}_ X))$ by Derived Categories of Schemes, Proposition 36.11.2. Hence by Lemma 57.17.2 it suffices to prove that $\textit{Coh}(\mathcal{O}_ X)$ is countable. This we omit. $\square$

Proof. The field $K$ is the filtered union of its countable subfields. Dually, $\mathop{\mathrm{Spec}}(K)$ is the cofiltered limit of the spectra of the countable subfields of $K$. Hence Limits, Lemma 32.10.1 guarantees that we can find a countable subfield $k$ and morphisms $X_0 \to S_0$ and $Y_0 \to S_0$ of schemes of finite type over $k$ such that $X \to S$ and $Y \to S$ are the base changes of these.

By Lemma 57.17.1 there is a countable set $I$ and pairs $(S_{0, i} \to S_0, h_{0, i})$ such that

1. $S_{0, i} \to S_0$ is a morphism of finite type, set $X_{0, i} = X_0 \times _{S_0} S_{0, i}$ and $Y_{0, i} = Y_0 \times _{S_0} S_{0, i}$,

2. $h_{0, i} : X_{0, i} \to Y_{0, i}$ is an isomorphism over $S_{0, i}$.

such that every pair $(T \to S_0, h_ T)$ with $T \to S_0$ of finite type and $h_ T : X_0 \times _{S_0} T \to Y_0 \times _{S_0} T$ an isomorphism is isomorphic to one of these. Denote $(S_ i \to S, h_ i)$ the base change of $(S_{0, i} \to S_0, h_{0, i})$ by $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(k)$. We claim this works.

Let $s \in S(K)$ and let $h_ s : X_ s \to Y_ s$ be an isomorphism over $K = \kappa (s)$. We can write $K$ as the filtered union of its finitely generated $k$-subalgebras. Hence by Limits, Proposition 32.6.1 and Lemma 32.10.1 we can find such a finitely generated $k$-subalgebra $K \supset A \supset k$ such that

1. there is a commutative diagram

   $$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[d]_ s \ar[r] & \mathop{\mathrm{Spec}}(A) \ar[d]^{s'} \\ S \ar[r] & S_0}$$

   for some morphism $s' : \mathop{\mathrm{Spec}}(A) \to S_0$ over $k$,

2. $h_ s$ is the base change of an isomorphism $h_{s'} : X_0 \times _{S_0, s'} \mathop{\mathrm{Spec}}(A) \to X_0 \times _{S_0, s'} \mathop{\mathrm{Spec}}(A)$ over $A$.

Of course, then $(s' : \mathop{\mathrm{Spec}}(A) \to S_0, h_{s'})$ is isomorphic to the pair $(S_{0, i} \to S_0, h_{0, i})$ for some $i \in I$. This concludes the proof because the commutative diagram in (1) shows that $s$ is in the image of the base change of $s'$ to $\mathop{\mathrm{Spec}}(K)$. $\square$

Proof. Choose a countable subfield $k \subset K$ for example the prime field. By Lemmas 57.17.1 and 57.17.3 there exists a countable set of isomorphism classes of systems over $k$ satisfying parts (1), (2), (3) of the lemma. Thus we can choose a countable set $I$ and for each $i \in I$ such a system

over $k$ such that each isomorphism class occurs at least once. Denote $(S_ i/K, X_ i \to S_ i, Y_ i \to S_ i, M_ i)$ the base change of the displayed system to $K$. This system has properties (1), (2), (3), see Lemma 57.15.3. Let us prove property (4).

Consider smooth proper schemes $X$ and $Y$ over $K$ such that there is a $K$-linear exact equivalence $F : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ Y)$. By Proposition 57.13.4 we may assume that there exists an object $M \in D_{perf}(\mathcal{O}_{X \times Y})$ such that $F = \Phi _ M$ is the corresponding Fourier-Mukai functor. By Lemma 57.8.9 there is an $M'$ in $D_{perf}(\mathcal{O}_{Y \times X})$ such that $\Phi _{M'}$ is the right adjoint to $\Phi _ M$. Since $\Phi _ M$ is an equivalence, this means that $\Phi _{M'}$ is the quasi-inverse to $\Phi _ M$. By Lemma 57.8.9 we see that the Fourier-Mukai functors defined by the objects

in $D_{perf}(\mathcal{O}_{X \times X})$ and

in $D_{perf}(\mathcal{O}_{Y \times Y})$ are isomorphic to $\text{id} : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ X)$ and $\text{id} : D_{perf}(\mathcal{O}_ Y) \to D_{perf}(\mathcal{O}_ Y)$ Hence $A \cong \Delta _{X/K, *}\mathcal{O}_ X$ and $B \cong \Delta _{Y/K, *}\mathcal{O}_ Y$ by Lemma 57.13.5. Hence we see that $M$ is the Fourier-Mukai kernel of a relative equivalence from $X$ to $Y$ over $K$ by definition.

We can write $K$ as the filtered colimit of its finite type $k$-subalgebras $A \subset K$. By Limits, Lemma 32.10.1 we can find $X_0, Y_0$ of finite type over $A$ whose base changes to $K$ produces $X$ and $Y$. By Limits, Lemmas 32.13.1 and 32.8.9 after enlarging $A$ we may assume $X_0$ and $Y_0$ are smooth and proper over $A$. By Lemma 57.15.4 after enlarging $A$ we may assume $M$ is the pullback of some $M_0 \in D_{perf}(\mathcal{O}_{X_0 \times _{\mathop{\mathrm{Spec}}(A)} Y_0})$ which is the Fourier-Mukai kernel of a relative equivalence from $X_0$ to $Y_0$ over $\mathop{\mathrm{Spec}}(A)$. Thus we see that $(S_0/k, X_0 \to S_0, Y_0 \to S_0, M_0)$ is isomorphic to $(S_{0, i}/k, X_{0, i} \to S_{0, i}, Y_{0, i} \to S_{0, i}, M_{0, i})$ for some $i \in I$. Since $S_ i = S_{0, i} \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(K)$ we conclude that (4) is true with $s : \mathop{\mathrm{Spec}}(K) \to S_ i$ induced by the morphism $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(A) \cong S_{0, i}$ we get from $A \subset K$. $\square$