nLab
Frobenius morphism

Frobenius morphism of fields

Suppose $k$ is a field of positive characteristic $p$. The Frobenius morphism is an endomorphism of the field $F:k\to k$ defined by $F(a)=a^p$.

Notice that this is indeed a homomorphism of fields: the identity $(ab{)}^p=a^p{b}^p$ evidently holds for all $a,b\in k$ and the characteristic of the field is used to show $(a+b{)}^p=a^p+b^p$.

Properties

• Frobenius is always injective. Note that the Frobenius morphism of schemes (see below) is not always a monomorphism.

• The image of Frobenius is the set of elements of $k$ with a $p$-th root and is sometimes denoted $k^{1/p}$.

• Frobenius is surjective if and only if $k$ is perfect.

Frobenius morphism of schemes

In terms of schemes as locally ringed spaces

Suppose $(X,{𝒪}_X)$ is an $S$-scheme where $S$ is a scheme over $k$. The absolute Frobenius is the map $F^{\mathrm{ab}}:(X,{𝒪}_X)\to (X,{𝒪}_X)$ which is the identity on the topological space $X$ and on the structure sheaves $F_{*}:{𝒪}_X\to {𝒪}_X$ is the $p$-th power map. This is not a map of $S$-schemes in general since it doesn’t respect the structure of $X$ as an $S$-scheme, i.e. the diagram:

${\displaystyle \begin{array}{ccc}X& \stackrel{F^{\mathrm{ab}}}{\to }& X\\ \downarrow & & \downarrow \\ S& \stackrel{F^{\mathrm{ab}}}{\to }& S\end{array}}$,

so in order for the map to be an $S$-scheme morphism, $F^{\mathrm{ab}}$ must be the identity on $S$, i.e. $S=\mathrm{Spec}({𝔽}_p)$.

Now we can form the fiber product using this square: $X^{(p)}:=X{\times }_{S}S$. By the universal property of pullbacks there is a map $F^{\mathrm{rel}}:X\to X^{(p)}$ so that the composition $X\to X^{(p)}\to X$ is $F^{\mathrm{ab}}$. This is called the relative Frobenius. By construction the relative Frobenius is a map of $S$-schemes.

Properties

For the purposes below $k$ will be a perfect field of characteristic $p$>$0$.

• $X$ is smooth over $k$ if and only if $F$ is a vector bundle, i.e. $F_{*}{𝒪}_X$ is a free ${𝒪}_X$-module of rank $p$. One can study singularities of $X$ by studying properties of $F_{*}{𝒪}_X$.

• If $X$ is smooth and proper over $k$, the sequence $0\to {𝒪}_X\stackrel{F^{\mathrm{ab}}}{\to }{F}_{*}{𝒪}_X\to d{𝒪}_X\to 0$ is exact and if it splits then $X$ has a lifting to $W_2(k)$.

In terms of schemes as sheaves on $C{\mathrm{Ring}}^{\mathrm{op}}$

Let $p$ be a prime number, let $k$ be a field of characteristic $p$. For a $k$-ring $A$ we define

$$f_A:\{\begin{array}{l}A\to A\\ x\mapsto x^p\end{array}$$f_A: \begin{cases} A\to A \\ x\mapsto x^p \end{cases}

The $k$-ring obtained from $A$ by scalar restriction along $f_k:k\to k$ is denoted by $A_f$.

The $k$-ring obtained from $A$ by scalar extension along $f_k:k\to k$ is denoted by $A^{(p)}:=A{\otimes }_{k,f}k$.

There are $k$-ring morphisms $f_A:A\to A_f$ and $F_A:\{\begin{array}{l}{A}^{(p)}\to A\\ x\otimes \lambda \mapsto x^p\lambda \end{array}$.

For a $k$-functor $X$ we define $X^{(p)}:X{\otimes }_{k,f_k}k$ which satisfies $X^{(p)}(R)=X(R_f)$. The Frobenius morphism for $X$ is the transformation of $k$-functors defined by

$$F_X:\{\begin{array}{l}X\to X^{(p)}\\ X(f_R):X(R)\to X(R_f)\end{array}$$F_X: \begin{cases} X\to X^{(p)} \\ X(f_R):X(R)\to X(R_f) \end{cases}

If $X$ is a $k$-scheme $X^{(p)}$ is a $k$-scheme, too.

Since the completion functor ${}^{\hat{}}:{\mathrm{Sch}}_k\to {\mathrm{fSch}}_k$ commutes with the above constructions the Frobenius morphism can be defined for formal k-schemes, too.

In terms of symmetric products

We give here another characterization of the Frobenius morphism in terms of symmetric products.

Let $p$ be a prime number, let $k$ be a field of characteristic $p$, let $V$ be a $k$-vector space, let ${\otimes }^{p}V$ denote the $p$-fold tensor power of $V$, let ${\mathrm{TS}}^{p}V$ denote the subspace of symmetric tensors. Then we have the symmetrization operator

$$s_V:\{\begin{array}{l}{\otimes }^{p}V\to {\mathrm{TS}}^{p}V\\ a_1\otimes \cdots \otimes a_n\mapsto {\Sigma }_{\sigma \in S_p}{a}_{\sigma (1)}\otimes \cdots \otimes a_{\sigma (n)}\end{array}$$s_V: \begin{cases} \otimes^p V\to TS^p V \\ a_1\otimes\cdots\otimes a_n\mapsto \Sigma_{\sigma\in S_p}a_{\sigma(1)}\otimes\cdots\otimes a_{\sigma(n)} \end{cases}

end the linear map

$${\alpha }_V:\{\begin{array}{l}{\mathrm{TS}}^{p}V\to {\otimes }^{p}V\\ a\otimes \lambda \mapsto \lambda (a\otimes \cdots \otimes a)\end{array}$$\alpha_V: \begin{cases} TS^p V\to\otimes^p V \\ a\otimes \lambda\mapsto\lambda(a\otimes\cdots\otimes a) \end{cases}

then the map $V^{(p)}\stackrel{{\alpha }_V}{\to }{\mathrm{TS}}^{p}V\to {\mathrm{TS}}^{p}V/s({\otimes }^{p}V)$ is bijective and we define ${\lambda }_V:{\mathrm{TS}}^{p}V\to V^{(p)}$ by

$${\lambda }_V\circ s=0$$\lambda_V\circ s=0

and

$${\lambda }_V\circ {\alpha }_V=\mathrm{id}$$\lambda_V \circ \alpha_V= id

If $A$ is a $k$-ring we have that ${\mathrm{TS}}^{p}A$ is a $k$-ring and ${\lambda }_A$ is a $k$-ring morphism.

If $X={\mathrm{Sp}}_{k}A$ is a ring spectrum we abbreviate $S^{p}X=S_k^{p}X:={\mathrm{Sp}}_k({\mathrm{TS}}^{p}A)$ and the following diagram is commutative.

$$\begin{array}{ccc}X& \stackrel{F_X}{\to }& X^{(p)}\\ \downarrow & & \downarrow \\ X^p& \stackrel{\mathrm{can}}{\to }& S^{p}X\end{array}$$\array{ X &\stackrel{F_X}{\to}& X^{(p)} \\ \downarrow&&\downarrow \\ X^p &\stackrel{can}{\to}& S^p X }

Properties

Proposition

Let $X$ be a $k$-formal scheme (resp. a locally algebraic scheme) then $X$ is étale iff the Frobenius morphism $F_X:X\to X^{(p)}$is a monomorphism (resp. an isomorphism).

The Frobenius as a morphism (natural transformation) of (affine) group schemes is one operation among other (related) operations of interest:

Remark

For any commutative affine group scheme $G$ the Frobenius- and the Verschiebung morphism? correspond by ”completed Cartier duality”; i.e. we have

$$\hat{D}(V_G)=F_{\hat{D}(G)}$$\hat D(V_G)=F_{\hat D(G)}

For a more detailed account of the relationship of Frobenius-, Verschiebung-? and homothety morphism? see Hazewinkel

Frobenius morphism of $\lambda$-rings

Examples

If $X={\mathrm{Sp}}_{k}A$ is a $k$-ring spectrum we have $X^{(p)}={\mathrm{Sp}}_k{A}^{(p)}$ and $F_X={\mathrm{Sp}}_k{F}_A$.

If $k=𝔽$ is a finite field we have $X^{(p)}=X$ however $F_X$ will not equal ${\mathrm{id}}_X$ in general.

If $k\hookrightarrow k^{\prime }$ is a field extension we have $F_{X{\otimes }_k{k}^{\prime }}=F_X{\otimes }_k{k}^{\prime }$.

References

• Michel Demazure, lectures on p-divisible groups web

• Michiel Hazewinkel, witt vectors. part 1, arXiv:0804.3888v1

• Karen Smith, Brief Guide to Some of the Literature on F-singularities, American Institute of Mathematics