Frobenius morphism

Let $K$ be a field of characteristic $p>0$ and let $q=p^r$. Let$C$ be a curve defined over $K$ contained in ${\mathbb{P}}^N$, theprojective space of dimension $N$. Define the homogeneous ideal of$C$ to be (the ideal generated by):

$$I(C)=\{f\in K[X_0,\mathrm{\dots },X_N]\mid \forall P\in C,f(P)=0,f\text{ is homogeneous}\}$$

For $f\in K[X_0,\mathrm{\dots },X_N]$, of the form $f={\sum }_i{a}_i{X}_0^{i_0}\mathrm{\dots }{X}_N^{i_N}$ we define $f^{(q)}={\sum }_i{a}_i^q{X}_0^{i_0}\mathrm{\dots }{X}_N^{i_N}$. We define a new curve $C^{(q)}$ asthe zero set of the ideal (generated by):

$$I(C^{(q)})=\{f^{(q)}\mid f\in I(C)\}$$

Definition 1.

The $q^{t\mathit{}h}$-power Frobenius morphism is defined to be:

$$\varphi :C\to C^{(q)}$$

$$\varphi ([x_0,\mathrm{\dots },x_N])=[x_0^q,\mathrm{\dots }{x}_N^q]$$

In order to check that the Frobenius morphism is well defined weneed to prove that

$$P=[x_0,\mathrm{\dots },x_N]\in C\Rightarrow \varphi (P)=[x_0^q,\mathrm{\dots }{x}_N^q]\in C^{(q)}$$

This is equivalent toproving that for any $g\in I(C^{(q)})$ we have $g(\varphi (P))=0$.Without loss of generality we can assume that $g$ is a generatorof $I(C^{(q)})$, i.e. $g$ is of the form $g=f^{(q)}$ for some$f\in I(C)$. Then:

$g(\varphi (P))=f^{(q)}(\varphi (P))$$=$$f^{(q)}([x_0^q,\mathrm{\dots },x_N^q])$

$=$${(f([x_0,\mathrm{\dots },x_N]))}^q,[a^q+b^q={(a+b)}^q\text{in characteristic }p]$

$=$${(f(P))}^q$

$=$$0,[P\in C,f\in I(C)]$

as desired.

Example: Suppose $E$ is an elliptic curve defined over$K={𝔽}_q$, the field of $p^r$ elements. In this case theFrobenius map is an automorphism of $K$, therefore

$$E=E^{(q)}$$

Hence the Frobenius morphism is an endomorphism (orisogeny) of the elliptic curve.