TITLE: Growth of Class Numbers
QUESTION [6 upvotes]: There is a classical formula stating:
Let $K$ be a number field with ring of integers $\mathcal{O}_K\subseteq K$ and 
let $\mathcal{O}\subseteq \mathcal{O}_K$ be any non-maximal order with conductor $\mathfrak{n}$. Then
$$\frac{\#(\mathcal{O}_K/\mathfrak{n}\mathcal{O}_K)^\times}{\#(\mathcal{O}/\mathfrak{n}\mathcal{O})^\times} = [\mathcal{O}_K^\times:\mathcal{O}^\times]\frac{\# \text{Pic}(\mathcal{O})}{\#\text{Pic}(\mathcal{O}_K)}.$$
My question is: do we know anything about the growth on the RHS, especially in the case K is a CM field? In particular, can we bound from below the LHS by the conductor $\mathfrak{n}$? I am in particular interested when the CM field corresponds to an isotypic CM abelian variety of dimension greater than $1$.

REPLY [7 votes]: Not a complete answer, but for $K/\mathbb Q$ imaginary quadratic, this is answered by Exercise 4.12 in Shimura's Arithmetic Theory of Automorphic Forms. The exercise reads (in your notation)
$$
\#\text{Pic}(\mathcal O)=\#\text{Pic}(\mathcal O_K)\cdot n\cdot [\mathcal{O}_K^\times:\mathcal{O}^\times]^{-1}\cdot \prod_{p\mid n}
\left[1-\left(\frac{K}{p}\right)p^{-1}\right],
$$
where the conductor of the order is $\mathfrak n=n\mathbb Z$ with $n$ a positive integer, and with $\left(\frac{K}{p}\right)$ is $1$, $-1$, $0$ according to whether $p$ splits, is inert, or is ramified in $K$. So the quantity that you have written is
$$
n\cdot \prod_{p\mid\mathfrak{n}}
\left[1-\left(\frac{K}{p}\right)p^{-1}\right].
$$
This gives the desired lower bound of $\gg n$ unless $n$ is composed of a very large number of small primes that split in $K$. In any case, one can use the standard estimate
$$
\prod_{p\mid n} [1-p^{-1}] \gg \frac{1}{\log\log  n}
$$
to get a lower bound that's just a bit worse than $\gg n$.
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