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!

  • 1
    $\begingroup$ Getting a grasp on the classgroup/class number is tricky from either the algebraic or the analytic approaches. Goldfeld used the Deuring-Heilbronn phenomenon for the Gauss Class number one problem : ams.org/notices/200608/fea-stopple.pdf .. In the algebraic side, the relevant keywords are Hilbert Classfield and Golod-Shafarevich theorem. $\endgroup$
    – user96815
    Commented Nov 1, 2013 at 10:48

2 Answers 2


The class number of a number field is a highly subtle and nontrivial invariant. To illustrate the extent of our ignorance: it is unknown if there are infinitely many number fields of class number $1$. However, a plausible conjecture of Gauss says that even among real quadratic fields, there are already infinitely many of them!

As you know, the size of the class group $\text{Pic}(\mathcal O_K)$ is related to special values of $\zeta_K(s)$, by the class number formula. The class number formula is most elegantly expressed by saying that $\zeta_K(s)$ has a zero of rank $r_1+r_2-1 = \text{rank}_\mathbf{Z}\: \mathcal O_K^\times$ at $s=0$, and that its leading coefficient $\zeta_K^{(r_1+r_2-1)}(s)/(r_1+r_2-1)!$ equals $hR/\omega$, where $h = |\text{Pic}(\mathcal O_K)|$, $R$ is the regulator of $K$ and $\omega = |(\mathcal O_K^{\times})_{\text{tors}}|$. In practice, however, one cannot hope to rely on the class number formula, say, to bound the class number in a family of number fields. As the case of real quadratic fields shows, the behavior in families is highly irregular.

Choose an embedding of $K$ into $\mathbf C$. Let us write $G$ for the absolute Galois group $\text{Gal}(\overline{K}/K)$, where $\overline{K}$ is the algebraic closure of $K$ in $\mathbf C$.

I will try to show how the problem of bounding the class group is a problem in Galois cohomology.

A word of warning: In what follows, we need to ensure that the Kummer sequence

$$1 \to \mu_n \to \mathbf G_m \xrightarrow{\cdot^n} \mathbf G_m \to 1$$

of étale sheaves on $X=\text{Spec }\mathcal O_K$ is exact, and this is true if and only if $n$ is a unit in $\mathcal O_K$. Therefore, everything that follows is only true for $n=1$, where it is vacuous. This is not a problem we have with curves; thanks to the existence of a base field, we have a large supply of $n$'s which are units. In order to apply this method to a number ring, we should invert a finite set of primes in $\mathcal O_K$ to make some room for ramification. Take what follows as wishful thinking of the fruitful kind.

One deduces from the Kummer sequence (with $n=1$!) the "Kummer-Mordell-Weil" exact sequence

$$1 \to \mathcal{O}_K^\times/(\mathcal{O}_K^\times)^n \to H^1_{ét}(\text{Spec }\mathcal O_K, \mu_n) \to \text{Pic}(\mathcal O_K)[n] \to 1,$$

where $\text{Pic}(\mathcal O_K)[n]$ is the $n$-torsion of the class group.

Base-change to $K$ determines a map $H^1_{ét}(\text{Spec }\mathcal O_K, \mu_n) \to H^1_{ét}(K, \mu_n) = H^1(G, \mu_n)$. According to Kummer theory, $H^1(G, \mu_n) \simeq K^\times/(K^\times)^n$, and in this, $H^1_{ét}(\text{Spec }\mathcal O_K, \mu_n)$ becomes identified with $$(K^\times/(K^\times)^n)(n) = \{ x \in K^\times/(K^\times)^n : \nu(x) \equiv 0 \mod n \quad \forall \nu\},$$ a finite group. The sequence becomes

$$1 \to \mathcal{O}_K^\times/(\mathcal{O}_K^\times)^n \to (K^\times/(K^\times)^n)(n) \to \text{Pic}(\mathcal O_K)[n] \to 1.$$

(If we identify $\text{Pic}(\mathcal O_K)[n]$ via global class field theory with the $n$-torsion of the Galois group of the Hilbert class field of $K$, then the map $(K^\times/(K^\times)^n)(n) \to \text{Pic}(\mathcal O_K)[n]$ identifies with the reciprocity map of global class field theory. The group $(K^\times/(K^\times)^n)(n)$ is the analogue of the $n$-Selmer group of an elliptic curve.)

Thus, the size of $\text{Pic}(\mathcal O_K)[n]$ is controlled by the cokernel of the map $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^n \to (K^\times/(K^\times)^n)(n)$. This resembles very much the construction of the regulator. In a way, $|\text{Pic}(\mathcal O_K)[n]|$ contributes to the "finite part" of the regulator ("finite part" in the sense that $\widehat{\mathbf Z} \otimes \mathbf Q$ is the finite part of $\mathbf A_\mathbf{Q}$). This suggests that the quantity $hR$ (= class number times regulator) is a more natural invariant of the field $K$, and suggests why it is so difficult to isolate the two factors.

The group $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^n$ is easy to write down explicitly using Dirichlet's unit theorem. The real difficulty lies in controlling the size of $(K^\times/(K^\times)^n)(n)$. For instance, in order to bound $\text{Pic}(\mathcal O_K)[n]$ below, one would need to construct a sufficiently large collection of nontrivial elements of $(K^\times/(K^\times)^n)(n)$. This is a particular case of "the art of constructing cohomology classes", which is an active area of research.

A general principle says that cohomology classes on an object $X$ can be constructed from objects "lying over $X$" (think of sheaf cohomology). In this context, this point of view is the object of Iwasawa theory and the theory of Euler systems. The construction of cohomology classes is usually an "artisanal" thing. One has to make use of whatever special properties the object considered may have. For instance, the "Euler system of cyclotomic units" allows us to bound the size of class groups of cyclotomic fields. The construction of this Euler system depends crucially on the special arithmetic properties of the tower of cyclotomic fields. For the analogous questions for an elliptic curve over $\mathbf Q$, namely the question of bounding the Selmer group or the Tate-Shafarevich group, one has the theory of homogeneous spaces, Kolyvagin's Euler system...

Finally, one can relate this point of view with the theory of $L$-values via $p$-adic $L$-functions, but that is a whole other story.

  • 4
    $\begingroup$ Nice answer Bruno! $\endgroup$ Commented Nov 1, 2013 at 6:50
  • $\begingroup$ @AlexYoucis Thank you! I hope there are no mistakes. $\endgroup$ Commented Nov 1, 2013 at 6:53
  • $\begingroup$ Bruno - from this point of view of Galois cohomology, can you see any heuristic for why the class numbers of totally real fields tend to be very small? $\endgroup$
    – John M
    Commented Dec 20, 2013 at 6:41

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 $\chi: \text{Gal}(K/\mathbb{Q}) \rightarrow \mathbb{C}?$

  1. Let $\chi$ be a character of $K$ and $\ell$ be a prime in $\mathbb{Z}$.
  2. Then $\ker(\chi) \subseteq \text{Gal}(K/\mathbb{Q}).$ Let $K^{\chi}$ be the fixed field of this subgroup.
  3. If $\ell$ ramifies in $K^{\chi}$, then $\chi(\ell) = 0$.
  4. Otherwise, let $\ell = \mathfrak{p}_1\dots \mathfrak{p}_r$ in $\mathcal{O}_{K^{\chi}}$.
  5. 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.

  • $\begingroup$ a detail : usually the term Dirichlet characters is for characters of $\mathbb{Z}/(n)^\times$ (extended to completely multiplicative functions on $\mathbb{Z}$), sometimes also for characters of $O_K/\mathfrak{a}^\times$, and the composition of the Artin map and the one-dimensional representations of $Gal(L/\mathbb{Q})$ are Dirichlet characters by CFT or Kronecker-Weber $\endgroup$
    – reuns
    Commented Oct 14, 2018 at 15:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .