Question about Theorem 4.2 from Rational Points on Elliptic Curves
问题内容
I have a question about a certain part of the proof of Theorem 4.2 from Rational Points on Elliptic Curves. Let $p$ be a prime so that $p=1\;\mathrm{mod}\;3$. Let $R=\{x^{3}|x\in\mathbb{F}_{p},x\neq 0\}$. Notation: $[XYZ]$ is the number of triples $(x,y,z)$ so that $x+y+z=0,x\in X,y\in Y,z\in Z$, where $X,Y,Z\subset\mathbb{F}_{p}$. The book claims that $[RR\mathbb{F}_{p}]=(\frac{p-1}{3})^2$. Somehow this does not seem to be obvious to me. Is there an elementary argument for this?
回答 (1)
Just to get it off the unanswered list since OP has already realized how to solve it.
The third coordinate is actually fixed once $x,y\in\mathbb F_p$ are chosen because $x+y=-z$ and $x+y\in\mathbb F_p$. So the only thing to do is count the number of elements in $R$. Since the map $\mathbb F_p^*\to\mathbb F_p^*$ sending $x$ to $x^3$ has kernel as cube roots of unity and since $3\mid p-1$, there are exactly $3$ such roots. Since $R$ is the size of the image, we have $|R|=(p-1)/3$. Thus $[RR\mathbb F_p]=((p-1)/3)^2$.