Frobenius endomorphism

Suppose that X is a scheme of characteristic p > 0. Choose an open affine subset U = Spec A of X. The ring A is an F_p-algebra, so it admits a Frobenius endomorphism. If V is an open affine subset of U, then by the naturality of Frobenius, the Frobenius morphism on U, when restricted to V, is the Frobenius morphism on V. Consequently, the Frobenius morphism glues to give an endomorphism of X. This endomorphism is called the absolute Frobenius morphism of X, denoted F_X. By definition, it is a homeomorphism of X with itself. The absolute Frobenius morphism is a natural transformation from the identity functor on the category of F_p-schemes to itself.

If X is an S-scheme and the Frobenius morphism of S is the identity, then the absolute Frobenius morphism is a morphism of S-schemes. In general, however, it is not. For example, consider the ring ${\displaystyle A=\mathbf {F} _{p^{2}}}$. Let X and S both equal Spec A with the structure map X → S being the identity. The Frobenius morphism on A sends a to a^p. It is not a morphism of ${\displaystyle \mathbf {F} _{p^{2}}}$-algebras. If it were, then multiplying by an element b in ${\displaystyle \mathbf {F} _{p^{2}}}$ would commute with applying the Frobenius endomorphism. But this is not true because:

${\displaystyle b\cdot a=ba\neq F(b)\cdot a=b^{p}a.}$

The former is the action of b in the ${\displaystyle \mathbf {F} _{p^{2}}}$-algebra structure that A begins with, and the latter is the action of ${\displaystyle \mathbf {F} _{p^{2}}}$ induced by Frobenius. Consequently, the Frobenius morphism on Spec A is not a morphism of ${\displaystyle \mathbf {F} _{p^{2}}}$-schemes.

The absolute Frobenius morphism is a purely inseparable morphism of degree p. Its differential is zero. It preserves products, meaning that for any two schemes X and Y, F_X×Y = F_X × F_Y.

Suppose that φ : X → S is the structure morphism for an S-scheme X. The base scheme S has a Frobenius morphism F_S. Composing φ with F_S results in an S-scheme X_F called the restriction of scalars by Frobenius. The restriction of scalars is actually a functor, because an S-morphism X → Y induces an S-morphism X_F → Y_F.

For example, consider a ring A of characteristic p > 0 and a finitely presented algebra over A:

${\displaystyle R=A[X_{1},\ldots ,X_{n}]/(f_{1},\ldots ,f_{m}).}$

The action of A on R is given by:

${\displaystyle c\cdot \sum a_{\alpha }X^{\alpha }=\sum ca_{\alpha }X^{\alpha },}$

where α is a multi-index. Let X = Spec R. Then X_F is the affine scheme Spec R, but its structure morphism Spec R → Spec A, and hence the action of A on R, is different:

${\displaystyle c\cdot \sum a_{\alpha }X^{\alpha }=\sum F(c)a_{\alpha }X^{\alpha }=\sum c^{p}a_{\alpha }X^{\alpha }.}$

Because restriction of scalars by Frobenius is simply composition, many properties of X are inherited by X_F under appropriate hypotheses on the Frobenius morphism. For example, if X and S_F are both finite type, then so is X_F.

The extension of scalars by Frobenius is defined to be:

${\displaystyle X^{(p)}=X\times _{S}S_{F}.}$

The projection onto the S factor makes X^(p) an S-scheme. If S is not clear from the context, then X^(p) is denoted by X^(p/S). Like restriction of scalars, extension of scalars is a functor: An S-morphism X → Y determines an S-morphism X^(p) → Y^(p).

As before, consider a ring A and a finitely presented algebra R over A, and again let X = Spec R. Then:

${\displaystyle X^{(p)}=\operatorname {Spec} R\otimes _{A}A_{F}.}$

A global section of X^(p) is of the form:

${\displaystyle \sum _{i}\left(\sum _{\alpha }a_{i\alpha }X^{\alpha }\right)\otimes b_{i}=\sum _{i}\sum _{\alpha }X^{\alpha }\otimes a_{i\alpha }^{p}b_{i},}$

where α is a multi-index and every a_iα and b_i is an element of A. The action of an element c of A on this section is:

${\displaystyle c\cdot \sum _{i}\left(\sum _{\alpha }a_{i\alpha }X^{\alpha }\right)\otimes b_{i}=\sum _{i}\left(\sum _{\alpha }a_{i\alpha }X^{\alpha }\right)\otimes b_{i}c.}$

Consequently, X^(p) is isomorphic to:

${\displaystyle \operatorname {Spec} A[X_{1},\ldots ,X_{n}]/\left(f_{1}^{(p)},\ldots ,f_{m}^{(p)}\right),}$

where, if:

${\displaystyle f_{j}=\sum _{\beta }f_{j\beta }X^{\beta },}$

then:

${\displaystyle f_{j}^{(p)}=\sum _{\beta }f_{j\beta }^{p}X^{\beta }.}$

A similar description holds for arbitrary A-algebras R.

Because extension of scalars is base change, it preserves limits and coproducts. This implies in particular that if X has an algebraic structure defined in terms of finite limits (such as being a group scheme), then so does X^(p). Furthermore, being a base change means that extension of scalars preserves properties such as being of finite type, finite presentation, separated, affine, and so on.

Extension of scalars is well-behaved with respect to base change: Given a morphism S′ → S, there is a natural isomorphism:

${\displaystyle X^{(p/S)}\times _{S}S'\cong (X\times _{S}S')^{(p/S')}.}$

Let X be an S-scheme with structure morphism φ. The relative Frobenius morphism of X is the morphism:

${\displaystyle F_{X/S}:X\to X^{(p)}}$

defined by the universal property of the pullback X^(p) (see the diagram above):

${\displaystyle F_{X/S}=(F_{X},\varphi ).}$

Because the absolute Frobenius morphism is natural, the relative Frobenius morphism is a morphism of S-schemes.

Consider, for example, the A-algebra:

${\displaystyle R=A[X_{1},\ldots ,X_{n}]/(f_{1},\ldots ,f_{m}).}$

We have:

${\displaystyle R^{(p)}=A[X_{1},\ldots ,X_{n}]/(f_{1}^{(p)},\ldots ,f_{m}^{(p)}).}$

The relative Frobenius morphism is the homomorphism R^(p) → R defined by:

${\displaystyle \sum _{i}\sum _{\alpha }X^{\alpha }\otimes a_{i\alpha }\mapsto \sum _{i}\sum _{\alpha }a_{i\alpha }X^{p\alpha }.}$

Relative Frobenius is compatible with base change in the sense that, under the natural isomorphism of X^(p/S) ×_S S′ and (X ×_S S′)^(p/S′), we have:

${\displaystyle F_{X/S}\times 1_{S'}=F_{X\times _{S}S'/S'}.}$

Relative Frobenius is a universal homeomorphism. If X → S is an open immersion, then it is the identity. If X → S is a closed immersion determined by an ideal sheaf I of O_S, then X^(p) is determined by the ideal sheaf I^p and relative Frobenius is the augmentation map O_S/I^p → O_S/I.

X is unramified over S if and only if F_X/S is unramified and if and only if F_X/S is a monomorphism. X is étale over S if and only if F_X/S is étale and if and only if F_X/S is an isomorphism.

The arithmetic Frobenius morphism of an S-scheme X is a morphism:

${\displaystyle F_{X/S}^{a}:X^{(p)}\to X\times _{S}S\cong X}$

defined by:

${\displaystyle F_{X/S}^{a}=1_{X}\times F_{S}.}$

That is, it is the base change of F_S by 1_X.

Again, if:

${\displaystyle R=A[X_{1},\ldots ,X_{n}]/(f_{1},\ldots ,f_{m}),}$

${\displaystyle R^{(p)}=A[X_{1},\ldots ,X_{n}]/(f_{1},\ldots ,f_{m})\otimes _{A}A_{F},}$

then the arithmetic Frobenius is the homomorphism:

${\displaystyle \sum _{i}\left(\sum _{\alpha }a_{i\alpha }X^{\alpha }\right)\otimes b_{i}\mapsto \sum _{i}\sum _{\alpha }a_{i\alpha }b_{i}^{p}X^{\alpha }.}$

If we rewrite R^(p) as:

${\displaystyle R^{(p)}=A[X_{1},\ldots ,X_{n}]/\left(f_{1}^{(p)},\ldots ,f_{m}^{(p)}\right),}$

then this homomorphism is:

${\displaystyle \sum a_{\alpha }X^{\alpha }\mapsto \sum a_{\alpha }^{p}X^{\alpha }.}$

Assume that the absolute Frobenius morphism of S is invertible with inverse ${\displaystyle F_{S}^{-1}}$. Let ${\displaystyle S_{F^{-1}}}$ denote the S-scheme ${\displaystyle F_{S}^{-1}:S\to S}$. Then there is an extension of scalars of X by ${\displaystyle F_{S}^{-1}}$:

${\displaystyle X^{(1/p)}=X\times _{S}S_{F^{-1}}.}$

If:

${\displaystyle R=A[X_{1},\ldots ,X_{n}]/(f_{1},\ldots ,f_{m}),}$

then extending scalars by ${\displaystyle F_{S}^{-1}}$ gives:

${\displaystyle R^{(1/p)}=A[X_{1},\ldots ,X_{n}]/(f_{1},\ldots ,f_{m})\otimes _{A}A_{F^{-1}}.}$

If:

${\displaystyle f_{j}=\sum _{\beta }f_{j\beta }X^{\beta },}$

then we write:

${\displaystyle f_{j}^{(1/p)}=\sum _{\beta }f_{j\beta }^{1/p}X^{\beta },}$

and then there is an isomorphism:

${\displaystyle R^{(1/p)}\cong A[X_{1},\ldots ,X_{n}]/(f_{1}^{(1/p)},\ldots ,f_{m}^{(1/p)}).}$

The geometric Frobenius morphism of an S-scheme X is a morphism:

${\displaystyle F_{X/S}^{g}:X^{(1/p)}\to X\times _{S}S\cong X}$

defined by:

${\displaystyle F_{X/S}^{g}=1_{X}\times F_{S}^{-1}.}$

It is the base change of ${\displaystyle F_{S}^{-1}}$ by 1_X.

Continuing our example of A and R above, geometric Frobenius is defined to be:

${\displaystyle \sum _{i}\left(\sum _{\alpha }a_{i\alpha }X^{\alpha }\right)\otimes b_{i}\mapsto \sum _{i}\sum _{\alpha }a_{i\alpha }b_{i}^{1/p}X^{\alpha }.}$

After rewriting R^(1/p) in terms of ${\displaystyle \{f_{j}^{(1/p)}\}}$, geometric Frobenius is:

${\displaystyle \sum a_{\alpha }X^{\alpha }\mapsto \sum a_{\alpha }^{1/p}X^{\alpha }.}$

Suppose that the Frobenius morphism of S is an isomorphism. Then it generates a subgroup of the automorphism group of S. If S = Spec k is the spectrum of a finite field, then its automorphism group is the Galois group of the field over the prime field, and the Frobenius morphism and its inverse are both generators of the automorphism group. In addition, X^(p) and X^(1/p) may be identified with X. The arithmetic and geometric Frobenius morphisms are then endomorphisms of X, and so they lead to an action of the Galois group of k on X.

Consider the set of K-points X(K). This set comes with a Galois action: Each such point x corresponds to a homomorphism O_X → K from the structure sheaf to K, which factors via k(x), the residue field at x, and the action of Frobenius on x is the application of the Frobenius morphism to the residue field. This Galois action agrees with the action of arithmetic Frobenius: The composite morphism

${\displaystyle {\mathcal {O}}_{X}\to k(x){\xrightarrow {\overset {}{F}}}k(x)}$

is the same as the composite morphism:

${\displaystyle {\mathcal {O}}_{X}{\xrightarrow {{\overset {}{F}}_{X/S}^{a}}}{\mathcal {O}}_{X}\to k(x)}$

by the definition of the arithmetic Frobenius. Consequently, arithmetic Frobenius explicitly exhibits the action of the Galois group on points as an endomorphism of X.