# nLab Demazure, lectures on p-divisible groups, I.9, the Frobenius morphism

This entry is about a section of the text

## Reminder

Let $s:R\to S$ be a morphism of rings. Then we have an adjunction

$\left({s}^{*}⊣{s}_{*}\right):S.\mathrm{Mod}\stackrel{{s}_{*}}{\to }R.\mathrm{Mod}$(s^*\dashv s_*):S.Mod\stackrel{s_*}{\to} R.Mod

from the category of $S$-modules to that of $R$-modules where

${s}^{*}:A↦A{\otimes }_{s}S$s^*:A\mapsto A\otimes_s S

is called scalar extension and ${s}_{*}$ is called scalar restriction.

## Idea

(Frobenius recognizes p-torsion)

## Definition

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

${f}_{A}:\left\{\begin{array}{l}A\to A\\ x↦{x}^{p}\end{array}$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 ${A}^{\left(p\right)}:=A{\otimes }_{k,f}k$.

There are $k$-ring morphisms ${f}_{A}:A\to {A}_{f}$ and ${F}_{A}:\left\{\begin{array}{l}{A}^{\left(p\right)}\to A\\ x\otimes \lambda ↦{x}^{p}\lambda \end{array}$.

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

${F}_{X}:\left\{\begin{array}{l}X\to {X}^{\left(p\right)}\\ X\left({f}_{R}\right):X\left(R\right)\to X\left({R}_{f}\right)\end{array}$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 ${X}^{\left(p\right)}$ is a $k$-scheme, too.

Since the completion functor ${}^{\stackrel{^}{\phantom{\rule{thickmathspace}{0ex}}}}:{\mathrm{Sch}}_{k}\to {\mathrm{fSch}}_{k}$ 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

${s}_{V}:\left\{\begin{array}{l}{\otimes }^{p}V\to {\mathrm{TS}}^{p}V\\ {a}_{1}\otimes \cdots \otimes {a}_{n}↦{\Sigma }_{\sigma \in {S}_{p}}{a}_{\sigma \left(1\right)}\otimes \cdots \otimes {a}_{\sigma \left(n\right)}\end{array}$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

${\alpha }_{V}:\left\{\begin{array}{l}{\mathrm{TS}}^{p}V\to {\otimes }^{p}V\\ a\otimes \lambda ↦\lambda \left(a\otimes \cdots \otimes a\right)\end{array}$\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 ${V}^{\left(p\right)}\stackrel{{\alpha }_{V}}{\to }{\mathrm{TS}}^{p}V\to {\mathrm{TS}}^{p}V/s\left({\otimes }^{p}V\right)$ is bijective and we define ${\lambda }_{V}:{\mathrm{TS}}^{p}V\to {V}^{\left(p\right)}$ 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 ${S}^{p}X={S}_{k}^{p}X:={\mathrm{Sp}}_{k}\left({\mathrm{TS}}^{p}A\right)$ and the following diagram is commutative.

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

## Properties

Note that the Frobenius ${F}_{p}$ is an endomorphism of a field $R$ only if the characteristic of $R$ is $p$. In this case it is automatically a monomorphism, since field homomorphisms always are.

However if we pass from rings to schemes, in general it is not true that Frobenius is a monomorphism. The following proposition gives necessary and sufficient conditions for the Frobenius to be a monomorphism in case of formal schemes.

###### Proposition

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

## Examples

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

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

If $k↪{k}^{\prime }$ is a field extension we have ${F}_{X{\otimes }_{k}{k}^{\prime }}={F}_{X}{\otimes }_{k}{k}^{\prime }$.

Revised on July 18, 2012 14:55:21 by Stephan Alexander Spahn (79.227.139.9)