Hasse theorem proof

Question

I don't understand the following points 1) why is $E$ isomorphic to the $\ker(\phi - 1)$? 2) Why is $\#\ker(\phi - 1) = \deg(\phi - 1)$? The proof is taken from here page 11 https://www.math.washington.edu/~morrow/336_09/papers/Igor.pdf

Answer 1

By Fermat's little theorem, $x^p \equiv x \mod{p}$ so in particular $\mathbb{F}_p$ is the fixed field of $\overline{\mathbb{F}_p}$ fixedby the Frobenius automorphism, so a point $(x,y)$ of $E$ is defined over $\mathbb{F}_p$ if and only if both of its coordinates are.

This is the same as saying that their image under the Frobenius map $\phi$ is fixed and therefore $\phi(P)=P$, which is equivalent to $(\phi -1)(P)=0$. This shows we have a bijection of finite sets, and since the kernel is naturally a subgroup, it must be the whole group and hence this is the isomorphism.

The second part is due to the fact that for separable endomorphisms, the size of the kernel equals the degree of the map (for inseparable endomorphisms, we have to take the separable degree). In general, $m\phi+n$ is separable if and only if $p$ doesn't divide $n$.

In your case, $m=n=1$ so this map is separable, and therefore its degree equals the size of its kernel.

Edit: Definition of degree

Let $E$ be given by the equation $y^2=x^3+Ax+B$. Then an endomorphism $\phi:E \rightarrow E$ can be written in the unique form $\phi(x,y)=(f_1(x),f_2(x)y)$ where $f_1,f_2 \in \mathbb{F}_p(x)$.

This means we can write $f_1$ as the quotient of two polynomials, i.e. $f_1(x)=r(x)/s(x)$ where $r,s \in \mathbb{F}_p[x]$ are polynomials. We can then just define the degree of $\phi$ to be $\max \{\deg(r),\deg(s) \}$, unless $\phi$ is just the zero map in which case we define the degree to be zero.

This is the easiest definition, although it seems rather random. The proper way to define it is as the degree of the function field extension $\mathbb{F}_p(E)/\phi^*(\mathbb{F}_p(E))$, which is more useful for theory but slightly more abstract.