Equivalent definition of supersingular elliptic curves

Question

Using the definitions that an elliptic curve $E$ over a finite field $K$ of characteristics $p$ is ordinary if $E[p]\cong \mathbb{Z}/p\mathbb{Z}$ and supersingular if $E[p]\cong 0$, how can I show that $E$ is supersingular if and only if $\#E(K)\equiv 1 \mod p$?

Answer 1

This is explained in the reference textbook Silverman AEC

From that $\phi_q(x,y)=(x^q,y^q)\in End(E)$ has a dual endomorphism $\phi_q^*$ such that $\phi_q^*\phi_q=q\in End(E)$ and $\phi_q^*+\phi_q=t\in \Bbb{Z}\subset End(E)$. Then $\phi_q$ is a root of $X^2-Xt+q$

With $a=\frac{t+\sqrt{t^2-4q}}{2},b=\frac{t-\sqrt{t^2-4q}}{2}$

and $$\#E(\Bbb{F}_{q^n})=\deg(\phi_q^n-1)= (a^n-1)(b^n-1)$$

• If $X^2-Xt+q$ is irreducible then $\Bbb{Z}[\phi_q]$ is isomorphic to $\Bbb{Z}[a]\subset \Bbb{Q}(a)$, if one of $a,b$ is a unit of $\Bbb{Z}[a]/(p)$ then for some $n$ we'll have that $(a^n -1)(b^n-1)\in (p)$ so $p | \#E(\Bbb{F}_{q^n})$.

  Thus $E$ is supersingular iff $a,b$ are not units which implies that $X^2-Xt+q = (X-0)(X-0) \bmod (p)$ ie. $\#E(\Bbb{F}_q)=q+1-t = 1\bmod p$.

• If $X^2-Xt+q$ is reducible then $a,b$ are integers and $ab=q$, since $\deg(\phi)=\deg(\phi^*)=q$ then $a=b$, $q$ is a square and $X^2-Xt+q=X^2\pm 2q^{1/2} X+q$ so $\#E(\Bbb{F}_q)=q+1\pm q^{1/2} = 1\bmod p$.