Definition 33.36.1. Let $p$ be a prime number. Let $X$ be a scheme in characteristic $p$. The absolute frobenius of $X$ is the morphism $F_ X : X \to X$ given by the identity on the underlying topological space and with $F_ X^\sharp : \mathcal{O}_ X \to \mathcal{O}_ X$ given by $g \mapsto g^ p$.
33.36 Frobenii
Let $p$ be a prime number. If $X$ is a scheme, then we say “$X$ has characteristic $p$”, or “$X$ is of characteristic $p$”, or “$X$ is in characteristic $p$” if $p$ is zero in $\mathcal{O}_ X$.
This makes sense because for any ring $A$ of characteristic $p$ the map $F_ A : A \to A$, $a \mapsto a^ p$ is a ring endomorphism which induces the identity on $\mathop{\mathrm{Spec}}(A)$. Moreover, if $A$ is local, then $F_ A$ is a local homomorphism. In this way we see that the absolute frobenius of $X$ is an endomorphism of $X$ in the category of schemes. It turns out that the absolute frobenius defines a self map of the identity functor on the category of schemes in characteristic $p$.
Lemma 33.36.2. Let $p > 0$ be a prime number. Let $f : X \to Y$ be a morphism of schemes in characteristic $p$. Then the diagram
\[ \xymatrix{ X \ar[d]_ f \ar[r]_{F_ X} & X \ar[d]^ f \\ Y \ar[r]^{F_ Y} & Y } \]
commutes.
Proof. This follows from the following trivial algebraic fact: if $\varphi : A \to B$ is a homomorphism of rings of characteristic $p$, then $\varphi (a^ p) = \varphi (a)^ p$. $\square$
Lemma 33.36.3. Let $p > 0$ be a prime number. Let $X$ be a scheme in characteristic $p$. Then the absolute frobenius $F_ X : X \to X$ is a universal homeomorphism, is integral, and induces purely inseparable residue field extensions.
Proof. This follows from the corresponding results for the frobenius endomorphism $F_ A : A \to A$ of a ring $A$ of characteristic $p > 0$. See the discussion in Algebra, Section 10.46, for example Lemma 10.46.7. $\square$
If we are working with schemes over a fixed base, then there is a relative version of the frobenius morphism.
Definition 33.36.4. Let $p > 0$ be a prime number. Let $S$ be a scheme in characteristic $p$. Let $X$ be a scheme over $S$. We define
\[ X^{(p)} = X^{(p/S)} = X \times _{S, F_ S} S \]
viewed as a scheme over $S$. Applying Lemma 33.36.2 we see there is a unique morphism $F_{X/S} : X \longrightarrow X^{(p)}$ over $S$ fitting into the commutative diagram
\[ \xymatrix{ X \ar[rr]_{F_{X/S}} \ar[rrd] \ar@/^1em/[rrrr]^{F_ X} & & X^{(p)} \ar[rr] \ar[d] & & X \ar[d] \\ & & S \ar[rr]^{F_ S} & & S } \]
where the right square is cartesian. The morphism $F_{X/S}$ is called the relative Frobenius morphism of $X/S$.
Observe that $X \mapsto X^{(p)}$ is a functor; it is the base change functor for the absolute frobenius morphism $F_ S : S \to S$. We have the same lemmas as before regarding the relative Frobenius morphism.
Lemma 33.36.5. Let $p > 0$ be a prime number. Let $S$ be a scheme in characteristic $p$. Let $f : X \to Y$ be a morphism of schemes over $S$ . Then the diagram
\[ \xymatrix{ X \ar[d]_ f \ar[r]_{F_{X/S}} & X^{(p)} \ar[d]^{f^{(p)}} \\ Y \ar[r]^{F_{Y/S}} & Y^{(p)} } \]
commutes.
Proof. This follows from Lemma 33.36.2 and the definitions. $\square$
Lemma 33.36.6. Let $p > 0$ be a prime number. Let $S$ be a scheme in characteristic $p$. Let $X$ be a scheme over $S$. Then the relative frobenius $F_{X/S} : X \to X^{(p)}$ is a universal homeomorphism, is integral, and induces purely inseparable residue field extensions.
Proof. By Lemma 33.36.3 the morphisms $F_ X : X \to X$ and the base change $h : X^{(p)} \to X$ of $F_ S$ are universal homeomorphisms. Since $h \circ F_{X/S} = F_ X$ we conclude that $F_{X/S}$ is a universal homeomorphism (Morphisms, Lemma 29.45.8). By Morphisms, Lemmas 29.45.5 and 29.10.2 we conclude that $F_{X/S}$ has the other properties as well. $\square$
Lemma 33.36.7. Let $p > 0$ be a prime number. Let $S$ be a scheme in characteristic $p$. Let $X$ be a scheme over $S$. Then $\Omega _{X/S} = \Omega _{X/X^{(p)}}$.
Proof. This translates into the following algebra fact. Let $A \to B$ be a homomorphism of rings of characteristic $p$. Set $B' = B \otimes _{A, F_ A} A$ and consider the ring map $F_{B/A} : B' \to B$, $b \otimes a \mapsto b^ pa$. Then our assertion is that $\Omega _{B/A} = \Omega _{B/B'}$. This is true because $\text{d}(b^ pa) = 0$ if $\text{d} : B \to \Omega _{B/A}$ is the universal derivation and hence $\text{d}$ is a $B'$-derivation. $\square$
Lemma 33.36.8. Let $p > 0$ be a prime number. Let $S$ be a scheme in characteristic $p$. Let $X$ be a scheme over $S$. If $X \to S$ is locally of finite type, then $F_{X/S}$ is finite.
Proof. This translates into the following algebra fact. Let $A \to B$ be a finite type homomorphism of rings of characteristic $p$. Set $B' = B \otimes _{A, F_ A} A$ and consider the ring map $F_{B/A} : B' \to B$, $b \otimes a \mapsto b^ pa$. Then our assertion is that $F_{B/A}$ is finite. Namely, if $x_1, \ldots , x_ n \in B$ are generators over $A$, then $x_ i$ is integral over $B'$ because $x_ i^ p = F_{B/A}(x_ i \otimes 1)$. Hence $F_{B/A} : B' \to B$ is finite by Algebra, Lemma 10.36.5. $\square$
Lemma 33.36.9. Let $k$ be a field of characteristic $p > 0$. Let $X$ be a scheme over $k$. Then $X$ is geometrically reduced if and only if $X^{(p)}$ is reduced.
Proof. Consider the absolute frobenius $F_ k : k \to k$. Then $F_ k(k) = k^ p$ in other words, $F_ k : k \to k$ is isomorphic to the embedding of $k$ into $k^{1/p}$. Thus the lemma follows from Lemma 33.6.4. $\square$
Lemma 33.36.10. Let $k$ be a field of characteristic $p > 0$. Let $X$ be a variety over $k$. The following are equivalent
$X^{(p)}$ is reduced,
$X$ is geometrically reduced,
there is a nonempty open $U \subset X$ smooth over $k$.
In this case $X^{(p)}$ is a variety over $k$ and $F_{X/k} : X \to X^{(p)}$ is a finite dominant morphism of degree $p^{\dim (X)}$.
Proof. We have seen the equivalence of (1) and (2) in Lemma 33.36.9. We have seen that (2) implies (3) in Lemma 33.25.7. If (3) holds, then $U$ is geometrically reduced (see for example Lemma 33.12.6) and hence $X$ is geometrically reduced by Lemma 33.6.8. In this way we see that (1), (2), and (3) are equivalent.
Assume (1), (2), and (3) hold. Since $F_{X/k}$ is a homeomorphism (Lemma 33.36.6) we see that $X^{(p)}$ is a variety. Then $F_{X/k}$ is finite by Lemma 33.36.8. It is dominant as it is surjective. To compute the degree (Morphisms, Definition 29.51.8) it suffices to compute the degree of $F_{U/k} : U \to U^{(p)}$ (as $F_{U/k} = F_{X/k}|_ U$ by Lemma 33.36.5). After shrinking $U$ a bit we may assume there exists an étale morphism $h : U \to \mathbf{A}^ n_ k$, see Morphisms, Lemma 29.36.20. Of course $n = \dim (U)$ because $\mathbf{A}^ n_ k \to \mathop{\mathrm{Spec}}(k)$ is smooth of relative dimension $n$, the étale morphism $h$ is smooth of relative dimension $0$, and $U \to \mathop{\mathrm{Spec}}(k)$ is smooth of relative dimension $\dim (U)$ and relative dimensions add up correctly (Morphisms, Lemma 29.29.3). Observe that $h$ is a generically finite dominant morphism of varieties, and hence $\deg (h)$ is defined. By Lemma 33.36.5 we have a commutative diagram
\[ \xymatrix{ X \ar[rr]_{F_{X/k}} \ar[d]_ h & & X^{(p)} \ar[d]^{h^{(p)}} \\ \mathbf{A}^ n_ k \ar[rr]^{F_{\mathbf{A}^ n_ k/k}} & & (\mathbf{A}^ n_ k)^{(p)} } \]
Since $h^{(p)}$ is a base change of $h$ it is étale as well and it follows that $h^{(p)}$ is a generically finite dominant morphism of varieties as well. The degree of $h^{(p)}$ is the degree of the extension $k(X^{(p)})/k((\mathbf{A}^ n_ k)^{(p)})$ which is the same as the degree of the extension $k(X)/k(\mathbf{A}^ n_ k)$ because $h^{(p)}$ is the base change of $h$ (small detail omitted). By multiplicativity of degrees (Morphisms, Lemma 29.51.9) it suffices to show that the degree of $F_{\mathbf{A}^ n_ k/k}$ is $p^ n$. To see this observe that $(\mathbf{A}^ n_ k)^{(p)} = \mathbf{A}^ n_ k$ and that $F_{\mathbf{A}^ n_ k/k}$ is given by the map sending the coordinates to their $p$th powers. $\square$
Remark 33.36.11. Let $p > 0$ be a prime number. Let $S$ be a scheme in characteristic $p$. Let $X$ be a scheme over $S$. For $n \geq 1$
\[ X^{(p^ n)} = X^{(p^ n/S)} = X \times _{S, F_ S^ n} S \]
viewed as a scheme over $S$. Observe that $X \mapsto X^{(p^ n)}$ is a functor. Applying Lemma 33.36.2 we see $F_{X/S, n} = (F_ X^ n, \text{id}_ S) : X \longrightarrow X^{(p^ n)}$ is a morphism over $S$ fitting into the commutative diagram
\[ \xymatrix{ X \ar[rr]_{F_{X/S, n}} \ar[rrd] \ar@/^1em/[rrrr]^{F_ X^ n} & & X^{(p^ n)} \ar[rr] \ar[d] & & X \ar[d] \\ & & S \ar[rr]^{F_ S^ n} & & S } \]
where the right square is cartesian. The morphism $F_{X/S, n}$ is sometimes called the $n$-fold relative Frobenius morphism of $X/S$. This makes sense because we have the formula
\[ F_{X/S, n} = F_{X^{(p^{n - 1})}/S} \circ \ldots \circ F_{X^{(p)}/S} \circ F_{X/S} \]
which shows that $F_{X/S, n}$ is the composition of $n$ relative Frobenii. Since we have
\[ F_{X^{(p^ m)}/S} = F_{X^{(p^{m - 1})}/S}^{(p)} = \ldots = F_{X/S}^{(p^ m)} \]
(details omitted) we get also that
\[ F_{X/S, n} = F_{X/S}^{(p^{n - 1})} \circ \ldots \circ F_{X/S}^{(p)} \circ F_{X/S} \]
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