0

I have a questiion about a statement about Galoistheory in Silvermans book on elliptic curves. In particular i want to know why enter image description here

this statement holds. of course, one direction is clear, but i cant seem to understand why P is already in $E(\mathbb{F}_q)$ just because $\Phi(P)=P$. An elementary explanation would be nice too, because my course in Galois theory was too long ago for me to actually remember how everything fitted together. Thanks!

Adronic
  • 327
  • 1
  • 11
  • 2
    If $P=(x,y)$ then $\Phi(P)=(x^q,y^q)$. This is equal to $P$ if and only if $x^q=x$ and $y^q=y$ if and only if $P\in E(\Bbb{F}_q)$. – Jyrki Lahtonen Nov 20 '23 at 18:01
  • Yeah, but why? i dont see iwhy thats the case if (x,y) are potentially elements of the algebraic closure. – Adronic Nov 20 '23 at 18:55
  • @Adronic Does this https://math.stackexchange.com/questions/1873807/what-is-the-algebraic-closure-of-mathbb-f-q help? If not, I'd suggest thinking about the Galois theory of $\overline{\mathbb{F}q}/\mathbb{F}{q^n}$ for various $n$ carefully. –  Nov 20 '23 at 19:11
  • 1
    An element $x\in\overline{\Bbb{F}_q}$ satisfies the equation $x^q=x$ if and only if $x\in\Bbb{F}_q$. Either by Galois theory or by the properties of $\Bbb{F}_q$. – Jyrki Lahtonen Nov 20 '23 at 20:16
  • I am asking for precisely that argument – Adronic Nov 22 '23 at 11:28

0 Answers0