In this section we prove some results on functors of a special type that we will use later in this chapter.

**Proof.**
By Lemma 56.14.2 we conclude that for every coherent $\mathcal{O}_ X$-module $\mathcal{F}$ whose support is a closed point there are isomorphisms

\[ H^0(X, M \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{F}) = H^0(X, F(M) \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{F}) \]

functorial in $M$.

Let $x \in X$ be a closed point and apply the above with $\mathcal{F} = \mathcal{O}_ x$ the skyscraper sheaf with value $\kappa (x)$ at $x$. We find

\[ \dim _{\kappa (x)} \text{Tor}^{\mathcal{O}_{X, x}}_ p(M_ x, \kappa (x)) = \dim _{\kappa (x)} \text{Tor}^{\mathcal{O}_{X, x}}_ p(F(M)_ x, \kappa (x)) \]

for all $p \in \mathbf{Z}$. In particular, if $H^ i(M) = 0$ for $i > 0$, then $H^ i(F(M)) = 0$ for $i > 0$ by Lemma 56.14.3.

If $\mathcal{E}$ is locally free of rank $r$, then $F(\mathcal{E})$ is locally free of rank $r$. This is true because a perfect complex $K$ over $\mathcal{O}_{X, x}$ with

\[ \dim _{\kappa (x)} \text{Tor}^{\mathcal{O}_{X, x}}_ i(K, \kappa (x)) = \left\{ \begin{matrix} r
& \text{if}
& i = 0
\\ 0
& \text{if}
& i \not= 0
\end{matrix} \right. \]

is equal to a free module of rank $r$ placed in degree $0$. See for example More on Algebra, Lemma 15.74.6.

If $M$ is supported on a closed subscheme $Z \subset X$, then $F(M)$ is also supported on $Z$. This is clear because we will have $M \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ x = 0$ for $x \not\in Z$ and hence the same will be true for $F(M)$ and hence we get the conclusion from Lemma 56.14.3.

In particular $F(\mathcal{O}_ x)$ is supported at $\{ x\} $. Let $i \in \mathbf{Z}$ be the minimal integer such that $H^ i(\mathcal{O}_ x) \not= 0$. We know that $i \leq 0$. If $i < 0$, then there is a morphism $\mathcal{O}_ x[-i] \to F(\mathcal{O}_ x)$ which contradicts the fact that all morphisms $\mathcal{O}_ x[-i] \to \mathcal{O}_ x$ are zero. Thus $F(\mathcal{O}_ x) = \mathcal{H}[0]$ where $\mathcal{H}$ is a skyscraper sheaf at $x$.

Let $\mathcal{G}$ be a coherent $\mathcal{O}_ X$-module with $\dim (\text{Supp}(\mathcal{G})) = 0$. Then there exists a filtration

\[ 0 = \mathcal{G}_0 \subset \mathcal{G}_1 \subset \ldots \subset \mathcal{G}_ n = \mathcal{G} \]

such that for $n \geq i \geq 1$ the quotient $\mathcal{G}_ i/\mathcal{G}_{i - 1}$ is isomorphic to $\mathcal{O}_{x_ i}$ for some closed point $x_ i \in X$. Then we get distinguished triangles

\[ F(\mathcal{G}_{i - 1}) \to F(\mathcal{G}_ i) \to F(\mathcal{O}_{x_ i}) \]

and using induction we find that $F(\mathcal{G}_ i)$ is a coherent sheaf placed in degree $0$.

Let $\mathcal{G}$ be a coherent $\mathcal{O}_ X$-module. We know that $H^ i(F(\mathcal{G})) = 0$ for $i > 0$. To get a contradiction assume that $H^ i(F(\mathcal{G}))$ is nonzero for some $i < 0$. We choose $i$ minimal with this property so that we have a morphism $H^ i(F(\mathcal{G}))[-i] \to F(\mathcal{G})$ in $D_{perf}(\mathcal{O}_ X)$. Choose a closed point $x \in X$ in the support of $H^ i(F(\mathcal{G}))$. By More on Algebra, Lemma 15.98.2 there exists an $n > 0$ such that

\[ H^ i(F(\mathcal{G}))_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{X, x}/\mathfrak m_ x^ n \longrightarrow \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(F(\mathcal{G})_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \]

is nonzero. Next, we take $m \geq 1$ and we consider the short exact sequence

\[ 0 \to \mathfrak m_ x^ m \mathcal{G} \to \mathcal{G} \to \mathcal{G}/\mathfrak m_ x^ m\mathcal{G} \to 0 \]

By the above we know that $F(\mathcal{G}/\mathfrak m_ x^ m\mathcal{G})$ is a sheaf placed in degree $0$. Hence $H^ i(F(\mathfrak m_ x^ m \mathcal{G})) \to H^ i(F(\mathcal{G}))$ is an isomorphism. Consider the commutative diagram

\[ \xymatrix{ H^ i(F(\mathfrak m_ x^ m\mathcal{G}))_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{X, x}/\mathfrak m_ x^ n \ar[r] \ar[d] & \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(F(\mathfrak m_ x^ m\mathcal{G})_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \ar[d] \\ H^ i(F(\mathcal{G}))_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{X, x}/\mathfrak m_ x^ n \ar[r] & \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(F(\mathcal{G})_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) } \]

Since the left vertical arrow is an isomorphism and the bottom arrow is nonzero, we conclude that the right vertical arrow is nonzero for all $m \geq 1$. On the other hand, by the first paragraph of the proof, we know this arrow is isomorphic to the arrow

\[ \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(\mathfrak m_ x^ m\mathcal{G}_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \longrightarrow \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(\mathcal{G}_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \]

However, this arrow is zero for $m \gg n$ by More on Algebra, Lemma 15.100.2 which is the contradiction we're looking for.

Thus we know that $F$ preserves coherent modules. By Lemma 56.15.2 we find $F$ is a sibling to the Fourier-Mukai functor $F'$ given by a coherent $\mathcal{O}_{X \times X}$-module $\mathcal{K}$ flat over $X$ via $\text{pr}_1$ and finite over $X$ via $\text{pr}_2$. Since $F(\mathcal{O}_ X)$ is an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ placed in degree $0$ we see that

\[ \mathcal{L} \cong F(\mathcal{O}_ X) \cong F'(\mathcal{O}_ X) \cong \text{pr}_{2, *}\mathcal{K} \]

Thus by Lemma 56.12.4 there is a morphism $s : X \to X \times X$ with $\text{pr}_2 \circ s = \text{id}_ X$ such that $\mathcal{K} = s_*\mathcal{L}$. Set $f = \text{pr}_1 \circ s$. Then we have

\begin{align*} F'(M) & = R\text{pr}_{2, *}(L\text{pr}_1^*K \otimes \mathcal{K}) \\ & = R\text{pr}_{2, *}(L\text{pr}_1^*M \otimes s_*\mathcal{L}) \\ & = R\text{pr}_{2, *}(Rs_*(Lf^*M \otimes \mathcal{L})) \\ & = Lf^*M \otimes \mathcal{L} \end{align*}

where we have used Derived Categories of Schemes, Lemma 36.22.1 in the third step. Since for all closed points $x \in X$ the module $F(\mathcal{O}_ x)$ is supported at $x$, we see that $f$ induces the identity on the underlying topological space of $X$. We still have to show that $f$ is an isomorphism which we will do in the next paragraph.

Let $x \in X$ be a closed point. For $n \geq 1$ denote $\mathcal{O}_{x, n}$ the skyscaper sheaf at $x$ with value $\mathcal{O}_{X, x}/\mathfrak m_ x^ n$. We have

\[ \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, \mathcal{O}_{x, n}) \cong \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, F(\mathcal{O}_{x, n})) \cong \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, f^*\mathcal{O}_{x, n} \otimes \mathcal{L}) \]

functorially with respect to $\mathcal{O}_ X$-module homomorphisms between the $\mathcal{O}_{x, n}$. (The first isomorphism exists by assumption and the second isomorphism because $F$ and $F'$ are siblings.) For $m \geq n$ we have $\mathcal{O}_{X, x}/\mathfrak m^ n = \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, \mathcal{O}_{x, n})$ via the action on $\mathcal{O}_{x, n}$ we conclude that $f^\sharp : \mathcal{O}_{X, x}/\mathfrak m_ x^ n \to \mathcal{O}_{X, x}/\mathfrak m_ x^ n$ is bijective for all $n$. Thus $f$ induces isomorphisms on complete local rings at closed points and hence is étale (Étale Morphisms, Lemma 41.11.3). Looking at closed points we see that $\Delta _ f : X \to X \times _{f, X, f} X$ (which is an open immersion as $f$ is étale) is bijective hence an isomorphism. Hence $f$ is a monomorphism. Finally, we conclude $f$ is an isomorphism as Descent, Lemma 35.22.1 tells us it is an open immersion.
$\square$

**Proof.**
Part (1) follows from part (2) and the fact that the connected components of $X$ of dimension $0$ are spectra of fields.

Let $Z \subset X$ be an irreducible component viewed as an integral closed subscheme. Clearly $f(Z) \subset Z$ and $f|_ Z : Z \to Z$ is an automorphism over $k$ which induces the identity map on the underlying topological space of $Z$. Since $X$ is reduced, it suffices to show that the arrows $f|_ Z : Z \to Z$ are the identity. This reduces us to the case discussed in the next paragraph.

Assume $X$ is irreducible of dimension $> 0$. Choose a nonempty affine open $U \subset X$. Since $f(U) \subset U$ and since $U \subset X$ is scheme theoretically dense it suffices to prove that $f|_ U : U \to U$ is the identity.

Assume $X = \mathop{\mathrm{Spec}}(A)$ is affine, irreducible, of dimension $> 0$ and $k$ is an infinite field. Let $g \in A$ be nonconstant. The set

\[ S = \bigcup \nolimits _{\lambda \in k} V(g - \lambda ) \]

is dense in $X$ because it is the inverse image of the dense subset $\mathbf{A}^1_ k(k)$ by the nonconstant morphism $g : X \to \mathbf{A}^1_ k$. If $x \in S$, then the image $g(x)$ of $g$ in $\kappa (x)$ is in the image of $k \to \kappa (x)$. Hence $f^\sharp : \kappa (x) \to \kappa (x)$ fixes $g(x)$. Thus the image of $f^\sharp (g)$ in $\kappa (x)$ is equal to $g(x)$. We conclude that

\[ S \subset V(g - f^\sharp (g)) \]

and since $X$ is reduced and $S$ is dense we conclude $g=f^\sharp (g)$. This proves $f^\sharp = \text{id}_ A$ as $A$ is generated as a $k$-algebra by elements $g$ as above (details omitted; hint: the set of constant functions is a finite dimensional $k$-subvector space of $A$). We conclude that $f = \text{id}_ X$.

Assume $X = \mathop{\mathrm{Spec}}(A)$ is affine, irreducible, of dimension $> 0$ and $k$ is a finite field. If for every $1$-dimensional integral closed subscheme $C \subset X$ the restriction $f|_ C : C \to C$ is the identity, then $f$ is the identity. This reduces us to the case where $X$ is a curve. A curve over a finite field has a finite automorphism group (details omitted). Hence $f$ has finite order, say $n$. Then we pick $g : X \to \mathbf{A}^1_ k$ nonconstant as above and we consider

\[ S = \{ x \in X\text{ closed such that }[\kappa (g(x)) : k] \text{ is prime to }n\} \]

Arguing as before we find that $S$ is dense in $X$. Since for $x \in X$ closed the map $f^\sharp : \kappa (x) \to \kappa (x)$ is an automorphism of order dividing $n$ we see that for $x \in S$ this automorphism acts trivially on the subfield generated by the image of $g$ in $\kappa (x)$. Thus we conclude that $S \subset V(g - f^\sharp (g))$ and we win as before.
$\square$

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