Conductor CM abelian variety

This is probably well known but I am not an expert in the subject.

Given an abelian variety $A$ of dimension $g$ with CM by $O_K$ where $K$ is a CM field of degree $2g$, let $N_A$ be the norm of the conductor of $A$ as defined here

http://en.wikipedia.org/wiki/Conductor_of_an_abelian_variety

My question is : can one compare $N_A$ to $d_K$, the absolute discriminant of $K$?

More precisely, can one expect $\log(N_A) < c \log (d_K)$ where $c$ depends on $g$ only.

Many thanks in advance!

• Is this known for elliptic curves? – abx Feb 2 '14 at 10:00

The short answer is no, since if we take the quadratic twist of $A$ by $\sqrt{D}$, then the conductor of $A$ will more-or-less acquire divisibility by the primes dividing $D$. So we can make the conductor arbitrarily large without changing the CM field $K$. For example, if $D$ is a square-free odd integer, then the conductor of $E_D:y^2=x^3+Dx$ is divisible by $D^2$, but the CM field $\mathbb{Q}(i)$ is independent of $D$.
On the other hand, you might find it useful to know that for any prime $\mathfrak{p}$, if $A$ has potential good reduction at $\mathfrak{p}$ (which is the case for CM abelian varieties), then the power of $\mathfrak{p}$ dividing $N_A$ is bounded by a constant that depends only on $g$ and $[K:\mathbb{Q}]$ (independent of $A$ and $\mathfrak{p}$). For example, for an elliptic curve $E/\mathbb{Q}$ with potential good reduction at $p$, we have $\operatorname{ord_p(N_E)}\le 2$ for $p\ge5$, $\operatorname{ord_p(N_E)}\le 5$ for $p=3$, and $\operatorname{ord_p(N_E)}\le 8$ for $p=2$.
• But note that the OP asked for complex multiplication by $O_K$, which I assume is the ring of integers of $K$. Then your example $E_D$ does not apply. – abx Feb 2 '14 at 12:49
• @abx I guess it depends whether $E/F$ having CM by $O_K$ means that $\mbox{End}(E/F)=O_K$ or $\mbox{End}(E/FK)=O_K$. But if you prefer, then consider $E_D$ to be defined over $\mathbb{Q}(i)$, and twist by a square-free $D\in\mathbb{Z}[i]$ whose norm is odd. Then the same conclusion applies. – Joe Silverman Feb 2 '14 at 13:41