1

Let $E$ be an elliptic curve over the rational number field $\mathbb{Q}$, and I have seen $\text{End}_{\mathbb{Q}}(E)=\mathbb{Z}$ in many places ,but no proof is given. So I want to ask for a proof or a refernce!

Thanks

  • 2
    Related:https://math.stackexchange.com/questions/2973865/endomorphism-ring-of-elliptic-curves-over-bbb-q-vs-over-bbb-c – Watson Sep 29 '19 at 14:21

1 Answers1

1

The point is that $E$ may have complex multiplication, so it has endomorphisms over $\Bbb C$ which are outside $\Bbb Z$. But these are not defined over $\Bbb Q$.

To see this note that the endomorphism ring acts on the vector space of invariant differentials on $E$ (spanned by $\omega=dx/(2y)$ for $y^2=x^3+ax+b$). An endomorphism defined over $\Bbb Q$ takes $\omega$ to a rational multiple of $\omega$, and so we get a ring map $\text{End}_{\Bbb Q}(E)\to \Bbb Q$. No element of $\text{End}_{\Bbb C}(E)$ other that those in $\Bbb Z$ can map to and element of $\Bbb Q$ as in the CM case the endomorphism ring in an order in an imaginary quadratic field.

Angina Seng
  • 158,341
  • Thanks for your answer! Since $\text{char} (\mathbb{Q})=0$, $\text{End}{\mathbb{Q}}(E)$ is isomorphism to $\mathbb{Z}$ or is an order in an imaginary quadratic field $F$. We also have a map $\text{End}{\mathbb{Q}}(E)\rightarrow \mathbb{Q}$, so this map extends to the fraction field of $\text{End}{\mathbb{Q}}(E)$, so we have a ring homomorphism $F\rightarrow \mathbb{Q}$, which is obviously impossible, so $\text{End}{\mathbb{Q}}(E) \cong \mathbb{Z}$, am I right? –  Sep 29 '19 at 12:07
  • @Eiang Yes, you are. – Angina Seng Sep 29 '19 at 12:09
  • Sorry, one more time to disturb you, I want to know if there exists some nonzero element in $\text{End}_{\mathbb{Q}}(E)$ such that it takes $omega$ to $0$(because if there is no such an element, then the map can be extended to the fraction field). –  Sep 29 '19 at 12:20
  • If $R$ is an order in an imaginary quadratic field, then any non-trivial quotient ring of $R$ is finite. – Angina Seng Sep 29 '19 at 16:46
  • Yes, I see. And I also have another proof for this, assume there is a complex number $t$ in the order, then $t$ satisfies a quadratic equation with integer coefficients of discriminant less than 0, assume the map which you give takes $t$ to $s\in \mathbb{Q}$, than $s$ must also satisfy this equation by the additive group homomorphism, which is obviously impossible. –  Sep 29 '19 at 17:44