Elliptic curves over $\mathbf{F}_q$ with $q = p^{2m}$

Question

I am reading Washingtons book about elliptic curves and struggling with an exercise there (4.9), which is the following:

Let $E$ be an elliptic curve over $\mathbf{F}_q$ with $q = p^{2m}$. Suppose that $\#E(\mathbf{F}_q) = q + 1 − 2 \sqrt{q}$.

In part (b) it is to show that $\phi_q - p^m = 0$, $\phi_q$ being the Frobenius endomorphism. The hint given is using Theorem 2.22 stating any endomorphism $\alpha \neq 0$ defined on $E(K)$ is surjective over the algebraic closure of the underlying field (so $\alpha: E(\overline{K})\rightarrow E(\overline{K})$ is surjective). By the hint I would assume to proof this as follows:

• Assume $\phi_q - p^m \neq 0$
• Show that the endomorphism is not surjective on $E(\overline{K})$ which is a contradiction

However I have no clue how to show this not being surjective, any hints?

Thx

Answer 1

Well, use part a of the question, where you have proved that $(\phi_q-p^m)^2=0$. Suppose that $\phi_q-p^m\neq0$. Then $\phi_q-p^m$ is surjective on $E(\bar{\mathbf F}_q)$ by Theorem 2.22. Then $(\phi_q-p^m)^2=(\phi_q-p^m)\circ(\phi_q-p^m)$ is surjective too. But this endomorphism was shown to be $0$ on $E(\bar{\mathbf F}_q)$. Hence $E(\bar{\mathbf F}_q)=\{0\}$. Contradiction.

Answer 2

As we are in the case of algebraically closed fields, statements about $E(\bar K)$ can be proved using (algebraic) geometry. The image of $\alpha$ (when it is non-constant) would be an irreducible subvariety of the 1-dimensional $E(\bar K)$. The theorem is: In an $n$-dimensional irreducible variety any proper subvariety is strictly lower dimensional. (This does not happen in differential geometry).