MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $E/K, E'/K$ be elliptic curves defined over a number field $K$. Let $\phi: E \to E'$ be a non-constant isogeny defined over $K$. Why must the conductors be equal?

I know that this is an exercise in Chapter IV of Silverman's Advanced Topics in the Arithmetic of Elliptic Curves, but I haven't yet carefully read through that chapter. I'm not looking for a technical answer, but for some intuition as to why this makes sense.

share|cite|improve this question
Dear Sarah, One standard definition of the conductor is that you take $V_{\ell}(E)$ (i.e. the $\mathbb Q_{\ell}$-linearization of the Tate-module) and compute the Artin conductor of this at each $p \neq \ell$. One proves that the Artin conductor is in fact independent of $\ell$ (provided that $\ell \neq p$; in the case that $p$ is a prime of good reduction, this is the Neron--Ogg--Shafarevic criterion, which shows that the condcutor is $1$ at such a $p$ independent of $\ell$). This gives the prime-to-$\ell$ part of conductor of $E$, and varying $\ell$, one obtains the conductor of $E$. ... – Emerton Apr 9 '13 at 2:17
Now if $E \to E'$ is a non-constant isogeny, it induces an isomorphism $V(E) \cong V(E')$, and so with the preceding definition, it is clear that the conductors coincide. As a particular example, by directly applying the criterion of Neron--Ogg--Shafarevic, you can see that if $E$ and $E'$ are isogenous, then $p$ is a prime of good reduction for $E$ if and only if it is so for $E'$. (Something that is not so obvious if you work directly, rather than in terms of Tate modules and their ramification properties.) I would use this good reduction case as a basis for my intuition about the ... – Emerton Apr 9 '13 at 2:20
general case. Regards, Matthew – Emerton Apr 9 '13 at 2:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.