what was/is motivation and short history/story behind "class number"?

Question

We know that there are some formula to calculate class number of a quadratic field $Q(\sqrt{d})$, where $d\in Z-\{0,1\}$, in terms of Dirichlet L-functions.

Please correct me if I am wrong: for example one motivation to build $Q(\sqrt{d})$ is to define a field (extension of $Q$) in which the equation $x^2-d=0$ is solvable.

I looked at some books and I did not see a simple explanation of what is a class number (without using the Ideals language) in a simple way. I saw some explanation in "An introduction to number theory Book by Harold Stark", but it was very short.

I would like to know a little bit more about what was/is the motivation to define class number of a quadratic field and what are its applications? what is the use of knowing that a class number of a field is 1,2,...?

giving some references would be appreciated as well.

Answer 1

Let $K/\mathbb{Q}$ be a finite extension. The basic idea for defining the class number is to determine "whether unique factorisation fails in $\mathcal{O}_K$". In the sense that, when the class number is $1$, $\mathcal{O}_K$ has unique factorisation. This, of course, is quite an interesting question in itself.

One application to a concrete question is the following.

Suppose that $n$ is a squarefree positive integer. When can we may write a prime $p$ as $$ p = a^2 + nb^2 $$ for some $a, b \in \mathbb{Z}$?

Suppose that $n \not\equiv 3 \mod 4$ (this is basically to avoid the case that $\mathbb{Z}[\sqrt{-n}]$ is not the ring of integers of $\mathbb{Q}(\sqrt{-n})$). Suppose that the class number of $K = \mathbb{Q}(\sqrt{-n})$ is $1$, and let $p$ be a prime not dividing $4n$. Then we may write $p$ as $a^2 + nb^2$ if and only if $-n$ is a square mod $p$.

One consequence of this is a result of Fermat (proved by Euler) that $p = a^2 + b^2$ if and only if $p \equiv 1 \mod 4$.

As requested I tried to avoid using the language of ideals, however the above paragraph seems much more natural if I say "let $p$ be unramified in $K/\mathbb{Q}$, then $p = a^2 + nb^2$ if and only if $p$ splits in $K$". Moreover I don't think one can really avoid such language if we want to deal with class numbers greater than 1 (although I would love to be proven wrong).

Answer 2

The class group is a measure of the extent to which unique factorization, and hence PID, fails in the ring of integers of a number field $K$. In particular, class number equal to $1$ is equivalent to $\mathcal{O}_K$ being a PID. Already the question which rings of integers have class number $1$ is very interesting. See for example here.

Here is a fascinating example of the class numbers of $\Bbb Q(\zeta_p)$ for $p$ prime:

\begin{array}{c|c} p & h_{\Bbb Q(\zeta_p)} \\ \hline 2 & 1 \\ 3 & 1 \\ 5 & 1 \\ 7 & 1 \\ 11 & 1 \\ 13 & 1 \\ 17 & 1 \\ 19 & 1 \\ 23 & 3 \\ 29 & 8 \\ 31 & 9 \\ 37 & 37 \\ 41 & 121 \\ 43 & 211 \\ 47 & 695 \\ 53 & 4889 \\ 59 & 41241 \\ 61 & 76301 \\ 67 & 853513 \\ 71 & 3882809 \\ 73 & 11957417 \\ 79 & 100146415 \\ 83 & 838216959 \\ 89 & 13379363737 \\ 97 & 411322824001 \\ 101 & 3547404378125 \\ 103 & 9069094643165 \\ 107 & 63434933542623 \\ 109 & 161784800122409 \\ 113 & 1612072001362952 \\ 127 & 2604529186263992195 \\ 131 & 28496379729272136525 \\ 137 & 646901570175200968153 \\ 139 & 1753848916484925681747 \\ 149 & 687887859687174720123201 \end{array}

As an application, one can give an easy proof of FLT $$ X^p+Y^p=Z^p $$ for primes $p$, where the class number is $1$ - so exactly for all primes $p\le 19$.