[Theorem 2.2, Orlov-K3]; this is shown in [Noah] without the assumption that $X$ be projective

Theorem 57.14.3 (Orlov). Let $k$ be a field. Let $X$ and $Y$ be smooth proper schemes over $k$ with $X$ projective over $k$. Any $k$-linear fully faithful exact functor $F : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ Y)$ is a Fourier-Mukai functor for some kernel in $D_{perf}(\mathcal{O}_{X \times Y})$.

Proof. Let $F'$ be the Fourier-Mukai functor which is a sibling of $F$ as in Lemma 57.14.2. By Proposition 57.11.6 we have $F \cong F'$ provided we can show that $\textit{Coh}(\mathcal{O}_ X)$ has enough negative objects. However, if $X = \mathop{\mathrm{Spec}}(k)$ for example, then this isn't true. Thus we first decompose $X = \coprod X_ i$ into its connected (and irreducible) components and we argue that it suffices to prove the result for each of the (fully faithful) composition functors

$F_ i : D_{perf}(\mathcal{O}_{X_ i}) \to D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ Y)$

Details omitted. Thus we may assume $X$ is irreducible.

The case $\dim (X) = 0$. Here $X$ is the spectrum of a finite (separable) extension $k'/k$ and hence $D_{perf}(\mathcal{O}_ X)$ is equivalent to the category of graded $k'$-vector spaces such that $\mathcal{O}_ X$ corresponds to the trivial $1$-dimensional vector space in degree $0$. It is straightforward to see that any two siblings $F, F' : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ Y)$ are isomorphic. Namely, we are given an isomorphism $F(\mathcal{O}_ X) \cong F'(\mathcal{O}_ X)$ compatible the action of the $k$-algebra $k' = \text{End}_{D_{perf}(\mathcal{O}_ X)}(\mathcal{O}_ X)$ which extends canonically to an isomorphism on any graded $k'$-vector space.

The case $\dim (X) > 0$. Here $X$ is a projective smooth variety of dimension $> 1$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. We have to show there exists a coherent module $\mathcal{N}$ such that

1. there is a surjection $\mathcal{N} \to \mathcal{F}$ and

2. $\mathop{\mathrm{Hom}}\nolimits (\mathcal{F}, \mathcal{N}) = 0$.

Choose an ample invertible $\mathcal{O}_ X$-module $\mathcal{L}$. We claim that $\mathcal{N} = (\mathcal{L}^{\otimes n})^{\oplus r}$ will work for $n \ll 0$ and $r$ large enough. Condition (1) follows from Properties, Proposition 28.26.13. Finally, we have

$\mathop{\mathrm{Hom}}\nolimits (\mathcal{F}, \mathcal{L}^{\otimes n}) = H^0(X, \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{F}, \mathcal{L}^{\otimes n})) = H^0(X, \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{F}, \mathcal{O}_ X) \otimes \mathcal{L}^{\otimes n})$

Since the dual $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{F}, \mathcal{O}_ X)$ is torsion free, this vanishes for $n \ll 0$ by Varieties, Lemma 33.48.1. This finishes the proof. $\square$

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