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I want to verify $\ker(1- \phi_q ) = E(F_q) $ by an example where $\phi_q$ is the Frobenius map $(x^q,y^q)$. If we take $E: y^2= x^3 + 3x +2$ over $F_5$, then the points on it are ${(1,1),(1,4),(2,1),(2,3),(4,0),\infty}$. But I do not see how these points are in $ker(1- \phi_q )$? If $\alpha = 1- \phi_5$, can we say $\alpha = (x-x^5,y-y^5)$? I appreciate any help!

Thomas Andrews
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CCCC
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  • The Frobenius map $x\mapsto x^q$ is always the identity on $\mathbb F_q.$ The non-trivial Frobenius maps are $x\mapsto x^{p^k}$ where $p^k<q$ and $p$ is the prime divisor of $q.$ – Thomas Andrews Mar 29 '23 at 19:49
  • Then what does the map $1-\phi_q$ mean? For k=1 does it give the zero map? I try to understand the proposition 3.25 of the notes http://math.uchicago.edu/~may/REU2018/REUPapers/Tang.pdf by this example – CCCC Mar 29 '23 at 19:56
  • My guess, without reading the article, is that the reference applies to points in $\mathbb F_{q^k}^2$ for some $k.$ The the Frobenius isomorphism is not the identity there. But I can't get a look at the article for a little bit. – Thomas Andrews Mar 29 '23 at 21:06
  • Yes, the $1$ here denotes the identity map $(x,y) \mapsto (x,y)$, so $1 - \phi_5$ is given by $(x,y) \mapsto (x-x^5,y-y^5)$ as you say. The rest all basically boils down to the fact that, given $c \in \overline{\mathbb{F}_q}$, then $c^q = c \iff c \in \mathbb{F}_q$. – Viktor Vaughn Mar 30 '23 at 17:26

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