TITLE: Intuition for Class Numbers
QUESTION [28 upvotes]: So I've been thinking about the analytic class number formula lately, and class numbers in general and I'm trying to develop a good intuition for them. My basic question, which may be too general/difficult is: $\textbf{What forces the class number of a number field to be large?}$
If you like, an answer for a class of fields such as quadratic fields would be helpful. 
Here's my intuition: I know that every ideal class has to contain a split prime, so if I wanted to construct a quadratic extension of Q with a large class number, I would try to find one where lots of small primes split, because once the primes get too large, they have to compete with things such as the Minkowski bound, and no longer help the class number grow. A prime $q$ splits in $\mathbb Q(\sqrt{p})$ iff $\left ( \frac{p}{q} \right) =1$. Now you can apply quadratic reciprocity to translate this to a congruence condition mod $p$. To me, this says that primes splitting should be related to the class number.
More concretely, by the analytic class number formula, for real quadratic fields  we have that $h=\frac{\sqrt{D}}{log(\epsilon)} L(1, \chi)$, where $\epsilon$ is the unique fundamental unit greater than $1$. For imaginary quadratic fields (excluding $\mathbb Q(i)$ and $\mathbb Q(\sqrt{-3})$, we have $h=\frac{\sqrt{|D|}}{\pi} L(1, \chi) $. So in general it seems like to make $h$ large, we need to make $L(1,\chi)$ large, which (I believe) is trying to find lots of small quadratic residues mod $m$, where your field is $\mathbb Q(\sqrt{m})$. This seems to work okay for imaginary quadratic fields. For example, $\mathbb Q(\sqrt{-163})$ is able to have a small class number because $\left (\frac{p}{163} \right )= -1$  for $p<41$ when $p$ is prime.
Real quadratic fields seem trickier, because it isn't enough to control $L(1, \chi)$, you also have to worry about the fundamental unit, and I'm afraid I don't have much intuition for that. Feel free to correct me if I've said anything above that was incorrect. Thanks!

REPLY [2 votes]: This doesn't exactly answer your question of how we can force a class number to be large (and in fact, that's a question I'm interested in as well), but I thought the following might be of interest to you. We can show fairly explicitly (at least in some specific cases) how primes splitting in number fields relates to the class number!
(1) This is based off things I've learned from research written up here. If others see anything that is incorrect, please let me know!
(2) This will essentially be copied from my talk, which is in slide format if you like that better.
Long story short: If $K$ is an abelian number field, then we can rewrite the residue of its Dedekind zeta function $\zeta_K$ at $s = 1$ as a product over the nontrivial characters of $K$ of Dirichlet $L$-functions. These are Dirichlet characters, so $\chi(\ell)$ depends highly on the factorization shape of $\ell$ in the ring of integers of $K$. Since the class number is some constant multiple of this residue, the class number is very closely related to the factorization of primes in $\mathcal{O}_K$.
Although the above resources are all specifically about totally imaginary abelian extensions of certain degrees, I'm pretty sure this interpretation works for any abelian number field $K$.
The analytic class number formula tells us the residue of the Dedekind zeta function $\zeta_K$ at $s = 1$ contains a lot of information about arithmetic invariants of $K$. 
$$\lim_{s \rightarrow 1} (s - 1)\zeta_K(s) = \frac{2^{r_1}(2\pi)^{r_2}h_K R_K}{\omega_K \sqrt{|D_K|}}.$$
Since our extension is abelian, we can rewrite the left-hand side as the product of Dirichlet $L$-functions over the nontrivial characters of $K$.
$$\prod_{\chi \in X_K\backslash \chi_0}L(1,\chi) = \frac{2^{r_1}(2\pi)^{r_2}h_K R_K}{\omega_K \sqrt{|D_K|}}.$$
If we're just interested in the class number of $K$, then after rearranging, we see that
$$h_K = \gamma \prod_{\chi \in X_K\backslash \chi_0}L(1, \chi),$$
for some known $\gamma$.
We can expand each $L$-function as an Euler product to get
$$h_K = \gamma \prod_{\chi \in X_K\backslash \chi_0}\prod_{\ell \text{prime}}\left(1 - \frac{\chi(\ell)}{\ell}\right)^{-1}.$$
After worrying about convergence questions, we can swap the order in which we do this product to get
$$\begin{align*}
h_K &= \gamma\prod_{\ell \text{prime}}\left(\prod_{\chi \in X_K\backslash \chi_0}\left(1 - \frac{\chi(\ell)}{\ell}\right)^{-1}\right) \\
&= \gamma\prod_{\ell \text{prime}}\nu_K(\ell)
\end{align*},$$
where we define $\nu_K(\ell)$ to be the term inside the parenthesis - a "local factor" at $\ell$.
So up to a constant, the class number is made up of these local factors that boil down to evaluating a character at a prime $\ell$. How do we do this, since characters are usually maps $\chi: \text{Gal}(K/\mathbb{Q}) \rightarrow \mathbb{C}?$

Let $\chi$ be a character of $K$ and $\ell$ be a prime in $\mathbb{Z}$.
Then $\ker(\chi) \subseteq \text{Gal}(K/\mathbb{Q}).$ Let $K^{\chi}$ be the fixed field of this subgroup.
If $\ell$ ramifies in $K^{\chi}$, then $\chi(\ell) = 0$.
Otherwise, let $\ell = \mathfrak{p}_1\dots \mathfrak{p}_r$ in $\mathcal{O}_{K^{\chi}}$.
Since our extension is Galois, the decomposition group $D_{K^{\chi}}$ of any prime $\mathfrak{p}_i$ lying over $\ell$ is isomorphic. Let $\text{Frob}_{K^{\chi}}$ be the generator of $D_{K^{\chi}}$. Then we define
$$\chi(\ell) = \chi(\text{Frob}_{K^{\chi}}).$$

So the class number is very closely related to the way to primes factor in your extension! I have some more specific examples, but I just noticed this was asked almost 5 years ago! So if anyone is still interested, let me know.
Final Fun Fact: That constant term $\gamma$ can actually be shown to come from the Archimedean place/the Sato-Tate distribution. So we can really think about the class number as telling us about ALL places, not just finite ones.