Let $K$ be a field of characteristic $p>0$ and let $q=p^r$ Let $C$ be a curve defined over $K$ contained in $\mathbb{P}^N$ the projective space of dimension $N$ Define the homogeneous ideal of $C$ to be (the ideal generated by): $$I(C)=\{f\in K[X_0,...,X_N] \mid \forall P \in C,\quad f(P)=0,\quad f\text{ is homogeneous}\}$$ For $f\in K[X_0,...,X_N]$ of the form $f=\sum_i a_iX_0^{i_0}...X_N^{i_N}$ we define $f^{(q)}=\sum_i a_i^qX_0^{i_0}...X_N^{i_N}$ We define a new curve $C^{(q)}$ as the zero set of the ideal (generated by): $$I(C^{(q)})=\{f^{(q)}\mid f\in I(C)\}$$

In order to check that the Frobenius morphism is well defined we need to prove that $$P=[x_0,...,x_N]\in C \Rightarrow \phi(P)=[x_0^q,...x_N^q]\in C^{(q)}$$ This is equivalent to proving that for any $g \in I(C^{(q)})$ we have $g(\phi(P))=0$ Without loss of generality we can assume that $g$ is a generator of $I(C^{(q)})$ i.e. $g$ is of the form $g=f^{(q)}$ for some $f\in I(C)$ Then: \begin{eqnarray*} g(\phi(P))=f^{(q)}(\phi(P)) &=& f^{(q)}([x_0^q,...,x_N^q])\\ &=& (f([x_0,...,x_N]))^q,\quad [a^q+b^q=(a+b)^q \text{in characteristic $p$}] \\ &=& (f(P))^q\\ &=& 0,\quad [P\in C, f\in I(C)] \end{eqnarray*}as desired.

Example: Suppose $E$ is an elliptic curve defined over $K=\mathbb{F}_q$ the field of $p^r$ elements. In this case the Frobenius map is an automorphism of $K$ therefore $$E=E^{(q)}$$ Hence the Frobenius morphism is an endomorphism (or isogeny) of the elliptic curve.

Bibliography

1

Joseph H. Silverman, The Arithmetic of Elliptic Curves. Springer-Verlag, New York, 1986.