# A question about endomorphism rings of elliptic curves

This is probably a very trivial question, but I haven't been able to find a rigorous explanation anywhere so far or at least haven't understood it.

Assume we have an elliptic curve $E$ over $\mathbb{F}_p$ where $p$ is a prime. Then we often talk about $\text{End}(E)$, the ring of endomorphisms of the curve $E$. The thing I'm confused about is the following. The claim generally is that $\text{End}(E)\cong \mathbb{Z}+\delta\mathbb{Z}$ where $\delta$ is a complex root of the discriminant of a complex quadratic order. So far so good.

Now the thing that confuses me is what is the $\text{End}(E)$ the endomorphism ring of? It can't be the endomorphism ring of the curve $E(\mathbb{F}_p)$ since that is a finite group and it's endomorphism ring must thus also be finite (by a combinatorial argument). My guess is it's the endomorphism ring of $E(\bar{\mathbb{F}_p})$ i.e. $E$ interpreted over the algebraic closure. But then there are places which specifically deal with $\text{End}_{\bar{\mathbb{F}_p}}(E)$ distinguishing it from $\text{End}(E)$.

My other guess was that it might be the endomorphisms of $E(\bar{\mathbb{{F}_p}})$ which are defined over $\mathbb{F}_p$ as morphisms (i.e. (slightly restricted) rational functions if I understand it correctly). The second interpretation is the one I'm leaning towards the most but it would be nice to get confirmation or explanation what I'm missing. Thank you.=

Edit After the discussion with Ferra I've determined that at least part of my issue is in not correctly understanding the abuse of notation used when we say an elliptic curve $E$ (defined over $\mathbb{F}_p$). Without a given base field I'm not sure what this is supposed to mean. In other words, when we write $\text{End}(E)$ which $E$ is meant? Are we in some way thinking of the homogenous polynomial defining $E$? Or are we thinking of the projective variety over the field $E(\overline{\mathbb{F}_p})$? Or are we thinking of the group $(E(\mathbb{F}_p),+,0)$ or $(E(\overline{\mathbb{F}_p}),+,0)$? Or are we thinking of both the group structure and the projective variety structure? I would think the last though I'm still not sure if we have the closure field or the base field. I'm also not sure whether there are group endomorphisms which are not algebraic curve morphisms.

Your are making some confusion between morphisms of groups and morphisms of curves. A morphism of groups $A,B$ is a map $A\to B$ which respects the operations, a morphism of (algebraic) curves $X,Y$ is a map $X\to Y$ whose components can be written as polynomials.

An isogeny of elliptic curves $E,E'$ over a field $k$ is a morphism of algebraic curves $E\to E'$ which is also a morphism of groups $E(\overline{k})\to E'(\overline{k})$.

The set $\text{End}(E)$ is defined as the set of isogenies $E\to E$. Note that if $E$ is defined over $\mathbb F_p$, we are not requiring all elements of $\text{End}(E)$ to be defined over $\mathbb F_p$, but only over $\overline{\mathbb F_p}$. The set of isogenies $E\to E$ defined over $\mathbb F_p$ is (usually, but not always) denoted by $\text{End}_{\mathbb F_p}(E)$. Whenever $\text{End}(E)\simeq \mathbb Z+\delta\mathbb Z$, the two sets coincide because the Frobenius map, which corresponds to $\delta$ in the latter isomorphism, is defined over $\mathbb F_p$. Note that the Frobenius map, which is the one sending $(x,y)\mapsto (x^p,y^p)$ is both a morphism of curves, because its components are polynomials and the coefficients of $E,E'$ lie in $\mathbb F_p$, and a morphism of groups $E(\overline{\mathbb F_p}) \to E(\overline{\mathbb F_p})$ because the addition map $E\times E\to E$ is defined over $\mathbb F_p$.

Note that there is a map $\text{End(E)}\to \text{End}(E(\overline{\mathbb F_p}))$ which sends every isogeny to the induced map on the group of $\overline{\mathbb F_p}$-rational points of $E$. This map is injective, but far from being surjective, since on one hand isogenies have finite kernel while on the other side $E(\overline{\mathbb F_p})$ is an infinite abelian group, so it has many endomorphisms with infinite kernel.

Analogously, if $E$ is defined over $\mathbb F_p$ then there is a map $\text{End(E)}\to \text{End}(E(\mathbb F_p))$ which is not injective since the LHS is infinite while the RHS is not.

• First of all thank you for the answer. I understand the difference between the morphism of groups and the morphism of curves, but as far as I can tell the difference is surprisingly small given that any morphism of curves which preserves the point at infinity (the 0 of the group) is an isogeny and thus turns out to be (to my bafflement) a homomorphism of groups. I also understand the fact that $\text{End}(E)$ contains isogenies not definable over $\mathbb{F}_p$ but even then if $\text{End}(E)$ is the ring of isogenies of $E(\mathbb{F}_p)$ it must be finite.
– DRF
May 5, 2015 at 10:06
• continued: Does this then mean that $\text{End}(E)$ is always meant to mean $\text{End}((E(\overline{\mathbb{F}_p}),+,0))$ when written in group notation?
– DRF
May 5, 2015 at 10:08
• Saying that "$\text{End}(E)$ is the ring of isogenies of $E(\mathbb F_p)$" makes no sense, because the first is the ring of endomorphisms of an object in the category of elliptic curves with isogenies, while the second is a finite abelian group, and (if nothing else is specified) it lives in the category of abelian groups with homomorphisms of groups. You are right in saying that $E(\mathbb F_p)$ has finitely many endomorphisms, since it is finite, but note that $E(\overline{\mathbb F_p})$ clearly has infinitely many, since multiplication by $m$ is a homomorphism for every integer $m$. May 5, 2015 at 11:17
• It is extremely important to understand that when we speak about elliptic curves, we are dealing with object which have a richer structure than algebraic curves: they are algebraic curves together with a group structure; hence their morphisms are maps which preserve both structure at the same time! May 5, 2015 at 11:20
• I'm sorry if I'm thick, but I still don't think I understand this fully. You say that saying "$\text{End}(E)$ is the ...." doesn't make sense. To me $E(\mathbb{F}_p)$ is the zero set of a homogenous polynomial over a projective space. I see $\text{End}(E)$ used to refer to the endomorphisms of $E$ (my question is which $E$) which respect the group structure. This (and correct me if I'm wrong) is for any given $E(K)$ the same as the set of all morphisms of algebraic curves which send the point of infinity to itself.
– DRF
May 5, 2015 at 12:29