A section on automorphisms of schemes over fields. For some information on (infinitesimal) automorphisms of curves, see Algebraic Curves, Section 53.25 and Moduli of Curves, Section 109.7.
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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