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

This entry is about a section of the text

• Michel Demazure, lectures on p-divisible groups, web

Reminder

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

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

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

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

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

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 $X={\mathrm{Sp}}_{k}A$ 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 $F_{X{\otimes }_k{k}^{\prime }}=F_X{\otimes }_k{k}^{\prime }$.