Question about characteristic polynomial of the Frobenius endomorphism on elliptic curves.

Question

I have another possibly trivial question about elliptic curves. A lot of papers I've seen state that the characteristic polynomial of the Frobenius endomorphism of an elliptic curve over a finite field of characteristic $q$ is $\varphi^2-t\varphi+q$, where $t$ is the trace of the endomorphism.

Is there some simple derivation of this characteristic polynomial and if so where could I find it?

Thank you very much.

Answer 1

This is shown in Theorem 2.3.1.(b), Chapter V, of Silverman's "The Arithmetic of Elliptic Curves".

Answer 2

This is explained p.117 and p.158 of Silverman's book. From what I understand :

• We have an elliptic curve $E/\mathbb{F}_q$. We show that if $l$ is a prime power coprime with $q$ then $E[l] \simeq C_{l} \times C_{l}$ is a product of two cyclic subgroups of order $l$. Thus $A \in End(E)$ acts on $E[l]$ as a $2\times 2$ modulo $l$ integer matrix $A_l = \begin{pmatrix} a & b \\c & d \end{pmatrix}$ and hence $$A_l^2 - tr(A_l) A_l + \det(A_l) I = 0, \qquad tr(A_l)=a+d, \det(A_l)=ad-bc$$

• Then we show that $\det(A_l) = deg(A) \bmod l$ and we use $tr(A_l)= 1 + \det(A_l)-\det(1-A_l)=1 + \deg(A)-\deg(1-A) \bmod l$.

• Qed. for any $l$ and hence in $End(E)$ : $$A^2 - [1 + \deg(A)-\deg(1-A)] A+[\deg(A)] = 0 $$

• When $A = \varphi$ is the Frobenius endomorphism then $\deg(\varphi) = q$ and $\deg(1-\varphi) = \#E(\mathbb{F}_q) = q+1-t$ therefore $$\varphi^2 - [t] \varphi+ [q] = 0$$

  and the subring of $End(E)$ generated by $[1]$ and $\varphi$ is the commutative ring $$\mathbb{Z}[\varphi] \cong \mathbb{Z}[\alpha], \qquad \alpha= \frac{t \pm \sqrt{t^2-4q}}{2}$$ a subring of $\mathcal{O}_K$ the ring of integers of the imaginary quadratic field $K = \mathbb{Q}(\alpha)$.