## 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 $\mathcal{O}_ X$.

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 $g \mapsto g^ p$.

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

1. $X^{(p)}$ is reduced,

2. $X$ is geometrically reduced,

3. 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

$F_{X/S, n} = F_{X/S}^{(p^{n - 1})} \circ \ldots \circ F_{X/S}^{(p)} \circ F_{X/S}$

Comment #3007 by Fei Hu on

In (33.36.4), the relative Frobenius morphism is defined as $F_{X/S} = (F_X, \text{id}_S)$. It looks like that $F_{X/S}$ follows the universal property of the fibre product. So I guess the second $\text{id}_S$ should be the morphism $X \to S$?

Comment #3129 by on

Indeed, you are correct. When you comment on something, please click on it and then comment on the page of the thing you are commenting on. Fixed here.

Comment #5893 by Zhenhua Wu on

I suggest adding one tag about group scheme: Let $p>0$ be a prime number. Let $S$ be a scheme in characteristic $p$. Let $X$ be a group scheme over $S$. Then $F_{X/S}$ is a morphism of group schemes.

This tag should be added into somewhere in the chapter of ''Groupoid Schemes'', e.g. tag 047F. And then refer to that tag here. Otherwise people may not found when they search about Frobenius morphism.

Comment #6096 by on

@#5893. This is a good suggestion. We could do this. Anybody?

Frobenii are divided into "frobenius" (e.g., Definition 03SM) and "Frobenius" (e.g., in Definition 0CC9 and Remark 0CCG). I guess these are just misprints and "frobenius" should always be "Frobenius".

Comment #8228 by on

Yes, but perhaps not essential to change this now. I will take a pull request for this.

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