Lemma 33.32.1. Let $X$ be a reduced scheme of finite type over a field $k$. Let $f : X \to X$ be an automorphism over $k$ which induces the identity map on the underlying topological space of $X$. Then

1. $f^*\mathcal{F} \cong \mathcal{F}$ for every coherent $\mathcal{O}_ X$-module, and

2. if $\dim (Z) > 0$ for every irreducible component $Z \subset X$, then $f$ is the identity.

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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