# Hasse's Theorem for Elliptic Curves over Finite Fields + proof clarification

I need a little help understanding Hasse's theorem for elliptic curves over finite fields, as well as the proof of this theorem. (Sorry about my editing)

Hasse’s Theorem: Let $E$ be an elliptic curve defined over $F_q$. Then |#$E(K)-q-1|≤2\sqrt{q}$

So Hasse's theorem is estimating the number of points of the E over $F_q$. It says that there are approximately q points with an error of no more than $2\sqrt{q}$ right?

I also read somewhere that "Hasse’s theorem on elliptic curves, provides a bound for the number of points on an elliptic curve when it is reduced modulo a prime p. It’s also referred to as the Hasse bound, because as a result the value is bounded both above and below." but I don't completely understand this result.

Proof: Consider the Frobenius endomorphism on $E$ in $F_q$ where $p$ is prime. This is the map $ϕ:(x,y)↦(x^p,y^p)$.

I guess I also am struggling with the concept of the Froebnius endomorphism. i can see that (x,y) gets mapped to a power p of itself $x^p,y^p)$, but I don't really see the important of it, although I see Frobenius endomorphism constantly being mentioned in elliptic curve literature.

From Fermat’s little theorem we have $x^p≡x modp$.

I understand this.

So the map fixes $E$ pointwise, that is, $ϕ(P)=P$.

This just means every x maps to an $x^p$ which is congruent to x (modp) right? so actually every x maps to itself...?

Then $ϕ(P)-P=0$, so $(ϕ-1)(P)=0$.

Just alegbra I think.

$P∈ker(ϕ-1)$. Thus $E$ is isomorphic to the kernel of the map $(ϕ-1)$.

I don't get where this result comes from, the concept of kernel has always confused me.

The isomorphism yields $N_p (E)=ker(ϕ-1)=(ϕ-1)$.

is $N_p$ just a different notation for the # of points?

So $|(ϕ-1)-deg⁡(ϕ)-deg91)|≤√(deg⁡(ϕ) deg⁡(1) )$.

Totally confused.

But $deg⁡(ϕ-1)=N_p > (E)$,$deg⁡(ϕ)=p$ and $deg1=1$ so $|N_p (E)-p-1|≤2√p$.

Where did these values for the degrees come from??

Any help would be greatly appreciated. Thanks

Here is the proof I used: Alternatively, here is Silverman's proof, which I understand even less: • What proof are you following? Silverman's? – Álvaro Lozano-Robledo Apr 29 '15 at 18:48
• @ÁlvaroLozano-Robledo I actually can't remember where I found this proof, but I know it's not the one Silverman gives in his "Arithmetic of Elliptic Curves". I totally didn't understand his, this one seemed a bit easier to me – Math Major Apr 29 '15 at 18:51
• @ÁlvaroLozano-Robledo I used this proof: math.washington.edu/~morrow/336_09/papers/Igor.pdf – Math Major Apr 29 '15 at 18:58

## 1 Answer

The idea is, $E$ has lots of points in $\overline{\mathbb F_p}$, not all of which are in $\mathbb F_p$. However, if I have an $\overline{\mathbb F_p}$-point, then it is actually an $\mathbb F_p$-point iff it is fixed by the $p^{th}$ power map. This follows because the solutions to $x^p=x$ in $\overline{\mathbb{F_p}}$ (of which there are $p$) are canonically identified with $\mathbb F_p$.

So computing the number of points of $E$ mod $p$ is the same as counting solutions to $\phi(P)=P$ which is the same as counting solutions to $(\phi-1)P=0$, where $(\phi-1)P = \phi(P)-P$.

Saying that $(\phi-1)P=0$ is exactly saying that $P$ is in the kernel of $(\phi-1)$.

The value of the degrees of these maps is worked out elsewhere in Silverman. Does this help?

• Sorry...I know what $F_p$ is, but what is $\bar{F_p}$? – Math Major Apr 29 '15 at 19:25
• The algebraic closure of $\mathbb{F}_p$. – Álvaro Lozano-Robledo Apr 29 '15 at 20:58