Let $E: Y^2=X^3+Ax+B$ be an elliptic curve, defined over $\mathbb{F}_p$ where $p$ is a prime. Define: $$\phi: E(\bar{\mathbb{F}}) \rightarrow E(\bar{\mathbb{F}})$$ by $$\phi(P) = \left\{ \begin{array}{ll} \mathcal{O} & \mbox{if } P = \mathcal O\\ (x^p,y^p) & \mbox{if } P=(x,y) \end{array} \right.$$ Show that $\phi$ is a well-defined map and that it is a group homorphism.

Outline of my answer:

To show that $\phi$ is a map, is it enough to show that $(x^p,y^p)=(x,y)\in E(\bar{\mathbb{F}})$ As $p$ is the group order of the multiplicative $\mathbb{F}_p$?

And to show that it it is a homorphism is it enough to use: $$\phi(a+b)= \phi(c)=c=a+b=\phi(a)+\phi(b)$$ and $$\phi(ab)= \phi(d)=d=ab=\phi(a)\phi(b)$$ $a,b,c,d \in E(\overline{\mathbb{F}})$ (Over-looking trivial cases)

With this however, the function becomes the identity.. Any ideas of how I can interpret the function?

  • 1
    $\begingroup$ Over $\overline{\Bbb F}$, the algebraic closure of $\Bbb F=\Bbb F_p$, $(x^p,y^p)\ne(x,y)$ in general. You need to show that if $(x,y)$ lies on $E$, then so does $(x^p,y^p)$. $\endgroup$ – Lord Shark the Unknown Nov 6 '18 at 20:13
  • $\begingroup$ $\phi(c) \ne c$ for $c\in E(\overline{\mathbb{F}_p}),c \not \in E({\mathbb{F}_p})$. $\phi$ is an homomorphism because the addition is a rational function with coefficients in $\mathbb{F}_p(A,B)$, so any $\sigma \in Aut(\overline{\mathbb{F}_p}/\mathbb{F}_p(A,B))$ commutes with that rational function. $\endgroup$ – reuns Nov 6 '18 at 20:14
  • $\begingroup$ There is "addition" of points on $E$, but not multiplication. $\endgroup$ – Lord Shark the Unknown Nov 6 '18 at 20:14
  • $\begingroup$ So should i insert $x^p $ into the equation of the curve to show that I get $y^{p^3}$? $\endgroup$ – User123456789 Nov 6 '18 at 20:15
  • $\begingroup$ Let $f(x,y) = y^2-(x^3+Ax+B)$. Then $f(x,y)^p = ?$ $\endgroup$ – reuns Nov 6 '18 at 21:32

From Galois theory we know that in $\bar{\mathbb{F}}_p$ the fixed field of $x\mapsto x^p$ is precisely $\mathbb{F}_p$.

To see that this is well defined, let $$E:y^2=x^3+ax+b$$ and note that if $(x,y)\in E$, then $\phi(x,y)=(x^p,y^p)$ and \begin{align*} (y^p)^2-(x^p)^3-a(x^p)-b&=(y^p)^2-(x^p)^3-a(x^p)-b\\ &=(y^p)^2-(x^p)^3-a^p(x^p)-b^p\\ &=(y^2)^p-(x^3)^p-a^p(x)^p-b^p\\ &=(y^2-x^3-ax-b)^p\\ &=0^p=0 \end{align*} so certainly $\phi(x,y)\in E$.

To see it is a homomorphism, you can note that it is an isogeny (morphism which sends the identity $O_E$ to the identity $O_E$) and so is a group homomorphism. If this is outside of your current knowledge, Isogenies as well as the general theory of elliptic curves are well worth reading on (see, for instance, Silverman 'Arithmetic of Elliptic Curves').

An alternative way to see the homomorphism, is noting that the natural isomorphism on the group structure $E \to Pic^0(E)$ given by $P\mapsto (P)-(O_E)$ commutes with the frobenius map (which is well defined on $Pic^0(E)$). Thus the question of this being a homomorphism can be answered by showing it is such on $Pic^0(E)$, which is immediate since it just distributes over the summands in the divisor representatives of $Pic^0(E)$.


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