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Let $p$ be a prime number with $p \equiv 3 \mod 4$, and consider the elliptic curve $$ E/\mathbb{F}_{p^2}: \quad y^2 = x^3 - x. $$ I suspect that $E(\mathbb{F}_{p^2})$ has order $(p+1)^2$ as I observed it with MAGMA (the code I used can be found below).

I found a similar-looking argument in Jyrki Lahtonen's answer here (where he dealt with the same curve over $\mathbb{F}_p$ instead of $\mathbb{F}_{p^2}$). It seems that the crucial argument there was that $-1$ is not a quadratic residue modulo $p \equiv 3 \mod 4$. However, I think $-1$ will become a square in $\mathbb{F}_{p^2}$, so the argument fails I think.

Question How can I show my claim instead?

Appendix: My MAGMA code (Output gives only "true").

L := [ x : x in [3 .. 100] | IsPrime(x) and x mod 4 eq 3];

for p in L do
    K := GF(p^2);
    R<x> := PolynomialRing(K);
    f := x^3 - x;
    E := EllipticCurve(f);
    #E eq (p+1)^2;
end for;
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    It gets much easier when learning the norm, trace and minimal polynomial of Frobenius stuffs. Follow Silverman AEC. – reuns Apr 29 '21 at 03:53
  • The theory of zeta functions of curves over finite fields tells you the following. If $E$ is an elliptic curve defined over $\Bbb{F}p$, and $N=N_1=# E(\Bbb{F}_p)$, then we can find complex numbers $\omega_1$ and $\omega_2=\overline{\omega_1}$ such that $$N=p+1-\omega_1-\omega_2,$$ $$|\omega_1|=|\omega_2|=\sqrt p.$$ Furthermore (Hasse-Davenport), the number of points $N_m$ on $E$ that are rational over $\Bbb{F}{p^m}$ is then given by $$N_m=p^m+1-\omega_1^m-\omega_2^m.$$ With this curve $E$ the information about $N_1$ allows us to deduce that $\omega_{1,2}=\pm i\sqrt{p}$. – Jyrki Lahtonen Apr 29 '21 at 04:56
  • (cont'd) Therefore the number of points of $y^2=x^3-x$ over $\Bbb{F}_{p^m}$ is $$N_m=p^m+1-(i\sqrt p)^m-(-i\sqrt p)^m.$$ Plugging in $m=2$ gives $$N_2=p^2+1+p+p=(p+1)^2.$$ – Jyrki Lahtonen Apr 29 '21 at 04:58
  • What this means is that for any elliptic curve defined over the prime field knowing $N_1$ gives you $N_m$ for all $m$. The same piece of information can be coded into a recurrence relation satisfied by $N_i$, $i=1,2,\ldots$, but I am used to going via the route of finding the $\omega$s. Similar technique is available for smooth higher genus $g>1$ curves also. The number of $\omega$s is equal to $2g$. You then need to find $N_1,N_2,\ldots, N_g$ to determine the values of the $\omega$s, and then get the rest of the $N_i$s. – Jyrki Lahtonen Apr 29 '21 at 05:03
  • (cont'd) The $\omega$s come in complex conjugate pairs. And the (proven) Riemann hypothesis for function fields states that $|\omega_i|=p^{1/2}$ for all $i$. – Jyrki Lahtonen Apr 29 '21 at 05:04
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    Related – Viktor Vaughn Apr 29 '21 at 07:26
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    Great recollection @ViktorVaughn. I can leave this to your able hands (and brain) :-) – Jyrki Lahtonen Apr 29 '21 at 21:20

1 Answers1

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For $\mathbb{F}_p$, the argument is that $$\#E(\mathbb{F}_p) = 1 + \sum_{x\in \mathbb{F}_p} \left(1+{{\genfrac{(}{)}{}{}{x^3-x}{p}}}\right)$$

Since $x^3-x$ is an odd function, and $p \equiv 3 \pmod{4}$, it follows that ${{\genfrac{(}{)}{}{}{x^3-x}{p}}}$ is also an odd function, so $$\sum_{x\in \mathbb{F}_p} {{\genfrac{(}{)}{}{}{x^3-x}{p}}} = 0$$ which implies $\#E(\mathbb{F}_p) = p+1.$

For $\mathbb{F}_{p^2}$, the easiest way to see that $\#E(\mathbb{F}_{p^2}) = (p+1)^2$ is to take the complex multiplication endomorphism $\phi\colon E \to E$ sending $(x,y)$ to $(-x,y\cdot \sqrt{-1})$. From standard theory, $$\#E(\mathbb{F}_p) = \deg(1-\pi)$$

$$\#E(\mathbb{F}_{p^2}) = \deg(1-\pi^2)$$ where $\pi$ denotes the $p$-th power Frobenius map. But $1-\pi^2 = (1-\pi)(1+\pi)$, and $\deg(1+\pi) = \#E(\mathbb{F}_p)$ because $1+\pi$ is separable and the kernel of $1+\pi$ is equal to the image of $E(\mathbb{F}_p)$ under $\phi$, so $$\deg(1-\pi^2) = \deg(1-\pi)\deg(1+\pi) = (p+1)^2.$$

For a general elliptic curve $E: y^2 = x^3 + ax + b$ and $p \equiv 3 \pmod{4}$, a version of this argument still works. The map $\phi$ as defined above maps from $E$ to $E' : y^2 = x^3 + ax - b$ and we have $\#E'(\mathbb{F}_p) = p+1+t$ where $t$ is the trace of $E/\mathbb{F}_p$, so we obtain $$\#E(\mathbb{F}_{p^2}) = (p+1-t)(p+1+t).$$ Alternatively, the proof of rationality of the zeta function for $E$ immediately gives you the above formula in full generality, as well as formulas for $\#E(\mathbb{F}_{p^k})$ for all $k$.

djao
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