Is $\sharp E((\mathbb{F}_{p^{2}})/E(\mathbb{F}_{p}))=1$ for almost all primes $p$? [closed]

Question

Let $E$ be an elliptic curve over $\mathbb{Q}.$

Is $\sharp E((\mathbb{F}_{p^{2}})/E(\mathbb{F}_{p}))=1$ for almost all primes $p$?

Answer 1

No, this is almost never true. $|\# E(\mathbb{F}_q) -q - 1| \leq 2 \sqrt{q}$. So $\#E(\mathbb{F}_{p^2}) \geq p^2 - 2p +1=(p-1)^2$ and $\#E(\mathbb{F}_p) \leq p + 2 \sqrt{p} +1= (\sqrt{p}+1)^2$. So $\#E(\mathbb{F}_{p^2})/\#E(\mathbb{F}_p) \geq (\sqrt{p}-1)^2$. If $p \geq 5$, this is $>1$.

Let's look a little closer at $p=2$ or $p=3$. Let the characteristic polynomial of Frobenius be $T^2-aT+1=(T-\alpha)(T-\beta)$. We have $\#E(\mathbb{F}_p) = (\alpha-1)(\beta-1)$ and $\#E(\mathbb{F}_{p^2}) = (\alpha^2-1)(\beta^2-1)$. So $\#E(\mathbb{F}_{p^2})/\#E(\mathbb{F}_p)=(\alpha+1)(\beta+1) = p+a+1$. So we will succeed if and only if $a=-p$.

Such curves do exist for $p=2$, $3$. For $p=2$, consider $y^2+y=x^3+x$. This has $4$ points in the $(x,y)$ plane (namely, $(0,0)$, $(0,1)$, $(1,0)$ and $(1,1)$) plus the point at infinity, so $\#E(\mathbb{F}_2) = 5$ and $a=-2$. For $p=3$, consider $y^2 = x^3-x+1$, with $6$ points in the $(x,y)$ plane (namely, $(0,\pm 1)$, $(1, \pm 1)$ and $(2, \pm 1)$.) So $\#E(\mathbb{F}_3) = 7$ and $a = -3$.