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If we take E to be an elliptic curve over $\mathbb{F}_q$ given by $$y^2=x^3+Ax+B$$ and $F_q$ to be the Frobenius endomorphism given by: $$F_q (x,y) = (x^q,y^q)$$ I am told that: $$\#E(\mathbb{F}_q) = deg(F_q-1)$$

In the very first line of the proof of why $\#(E(\mathbb{F}_q)) = \deg(F_q -1)$, it states that:

$$a \in E(\mathbb{F}_q) \Leftrightarrow F_q(a) = a$$

but I can't understand the backwards implication. Surely for any element $x \in \mathbb{F}_q$ we have that $x^q = x$, so it seems like $\forall a \in \mathbb{F}_q$, we should have that $F_q(a) = a$, not just for elements of $E(\mathbb{F}_q)$. What am I misunderstanding?

user75617
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  • $a$ is probably assumed to be an element of $E$. – Wojowu Feb 26 '20 at 13:47
  • I'm sure the equation of the curve should read $$y^2=x^3+Ax+B.$$ In other words, the right hand side is a (depressed) cubic. – Jyrki Lahtonen Feb 26 '20 at 13:52
  • Anyway, I'm also morally certain that $F_q-1$ here refers to the isogeny mapping the point $P\in E(k)$, $k$ an algebraic closure of $\Bbb{F}_q$, to the point $F_q(P)-P$. We see that the kernel of this isogeny consists exactly of the $\Bbb{F}_q$-rational points of $E$. And the cardinality of the kernel of an isogeny is equal to the degree. Take a look at Silverman's book, chapters III and V in particular. – Jyrki Lahtonen Feb 26 '20 at 14:06
  • The relevant geometric properties of varieties and morphisms between them. elliptic curves and their isogenies in particular, need to properly be studied in a setting that allows points to have coordinates in extension fields also (a simple way is to go all the way to an algebraic closure). The Frobenius automorphism is the identity mapping exactly when restricted to $\Bbb{F}_q$. That is sort of the point here, and makes the degree formula work. – Jyrki Lahtonen Feb 26 '20 at 14:10
  • (cont'd) If we don't include points with coordinates in (infinitely many) extension fields, we don't really have a curve. Instead we only have a finite straggle of isolated points. – Jyrki Lahtonen Feb 26 '20 at 14:13
  • For example, if $q=7$ and $E:y^2=x^3+x$, then we see that there is no square root of $-1$ in $\Bbb{F}_7$. Yet, denoting a square root of $-1$ by $i$, the point $P=(i,0)$ is on the curve as $x^3+x=x(x^2+1)$ vanishes when $x=i$. The point $P$ is not a fixed point of the Frobenius for $F_7(i,0)=(i^7,0^7)=(-i,0)\neq P$, which is the way it should be. Therefore $P$ is not in the kernel of $id-F$ either, and won't contribute to the degree. – Jyrki Lahtonen Feb 26 '20 at 14:19
  • So, if I correctly understood your question from the last sentence. We are not applying Frobenius to points outside the curve $E$. But we are applying Frobenius to those points on the curve $E$ whose coordinates lie outside $\Bbb{F}_q$. – Jyrki Lahtonen Feb 26 '20 at 14:23
  • Apologies, I made a typo with the elliptic curve equation. Was meant to be cubic, it's fixed now – user75617 Feb 26 '20 at 18:17
  • Just so I make sure I'm understanding correctly, it's likely that a is just assumed to be a point on the elliptic curve? – user75617 Feb 26 '20 at 18:23
  • At least that's my interpretation: $a$ is on the curve, and we are testing whether its coordinates are in $\Bbb{F}_q$. – Jyrki Lahtonen Feb 29 '20 at 20:39

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Isn't it obvious? For an element $x\in \bar{F_p}$, then we have $x^q=a$ if and only if $x\in F_q$. So $a\in E(\mathbb{F}_q)$ if and only if it is fixed by the action of Frobenius.