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

$(s^*\dashv s_*):S.Mod\stackrel{s_*}{\to} R.Mod$

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

$s^*:A\mapsto A\otimes_s S$

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

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

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

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 $F_X:X\to X^{(p)}$is a monomorphism (resp. an isomorphism).

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 $F_{X\otimes_k k^\prime}=F_X\otimes_k k^\prime$.

Last revised on July 18, 2012 at 14:55:21. See the history of this page for a list of all contributions to it.