3

Please apologize, if that's a stupid question. Why is the Frobenius Endomorphism of an elliptic curve over a finite field not regarded as an automorphism?

  • Since it is an Isogeny, it is surjective
  • Since it has trivial kernel, it is injective

Why is separability required?

Jyrki Lahtonen
  • 133,153
weissnix
  • 31
  • It is not an isomorphism in the sense of algebraic geometry, i.e. its inverse is not an algebraic map (given by polynomials). – Feanor Mar 01 '22 at 10:15

1 Answers1

3

Because it doesn't induce an automorphism of the function field of the curve. Automorphisms of an elliptic curve $E$ are endomorphisms $\varphi \colon E\to E$ which have degree $1$. Recall that the degree is defined as $[k(E)\colon \varphi^*(k(E)) ]$, where $k(E)$ is the function field of the curve and $\varphi^*$ is the induced map on $k(E)$. The Frobenius has degree $p$, hence it is not an automorphism. This somehow reflects the fact that map of curves are NOT maps of the underlying points, but maps of the underlying points AND of function fields.

Ferra
  • 5,437