# Contents

## 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

$F(a) \coloneqq a^p \,.$

Notice that this is indeed a homomorphism of fields: the identity $(a b)^p=a^p b^p$ 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 $(X,\mathcal{O}_X)$ is an $S$-scheme where $S$ is a scheme over $k$. The absolute Frobenius is the map $F^{ab}:(X,\mathcal{O}_X)\to (X,\mathcal{O}_X)$ which is the identity on the topological space $X$ and on the structure sheaves $F_*:\mathcal{O}_X\to \mathcal{O}_X$ 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:

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

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

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 $F^{rel}:X\to X^{(p)}$ so that the composition $X\to X^{(p)}\to X$ is $F^{ab}$. 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. $F_*\mathcal{O}_X$ is a free $\mathcal{O}_X$-module of rank $p$. One can study singularities of $X$ by studying properties of $F_*\mathcal{O}_X$.

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

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

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

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

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 $F_A:\begin{cases} A^{(p)}\to A \\ x\otimes \lambda\mapsto x^p \lambda \end{cases}$.

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

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

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

Since the completion functor ${}^\hat\;:Sch_k\to fSch_k$ 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 $TS^p V$ denote the subspace of symmetric tensors. Then we have the symmetrization operator

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

end the linear map

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

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

$\lambda_V\circ s=0$

and

$\lambda_V \circ \alpha_V= id$

If $A$ is a $k$-ring we have that $TS^p A$ is a $k$-ring and $\lambda_A$ is a $k$-ring morphism.

If $X=Sp_k A$ is a ring spectrum we abbreviate $S^p X=S^p_k X:=Sp_k (TS^p A)$ and the following diagram is commutative.

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

### 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

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

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

## Frobenius morphism of $\lambda$-rings

### Examples

If $X=Sp_k A$ is a $k$-ring spectrum we have $X^{(p)}=Sp_k A^{(p)}$ and $F_X=Sp_k F_A$.

If $k=\mathbb{F}$ is a finite field we have $X^{(p)}=X$ however $F_X$ will not equal $id_X$ 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

Revised on November 24, 2013 08:46:30 by Urs Schreiber (89.204.137.79)