equivalence of form class group and class group

There are only a finite number of reduced primitive positive integral binary quadratic forms of a given negative http://planetmath.org/node/IntegralBinaryQuadraticFormsdiscriminant $\mathrm{\Delta }$. Given $\mathrm{\Delta }$, call this number $h_{\mathrm{\Delta }}$, the form of $\mathrm{\Delta }$.

Thus, for example, since there is only one reduced form of discriminant $-163$, we have that $h_{-163}=1$.

It turns out that the set of reduced forms of a given negative discriminant can be turned into an abelian group, called the , ${𝒞}_{\mathrm{\Delta }}$, by defining a “multiplication” on forms that is based on generalizations of identities such as

$$(2x^2+2xy+3y^2)(2z^2+2zw+3w^2)={(2xz+xw+yz+3yw)}^2+5{(xw+yz)}^2$$

where all of these forms have discriminant $-20$.

Now, given an algebraic extension $K$ of $\mathbb{Q}$, ideal classes of ${𝒪}_K$ also form an abelian group, called the http://planetmath.org/node/IdealClassideal class group of $K$, ${𝒞}_K$. The order of ${𝒞}_K$ is called the class number of $K$ and is denoted $h_K$. See the ideal class entry for more detail.

For an algebraic extension $K/\mathbb{Q}$, one also defines the http://planetmath.org/node/DiscriminantOfANumberFielddiscriminant of the extension, $d_K$. For quadratic extensions $K=\mathbb{Q}[\sqrt{n}]$, where $n$ is assumed squarefree, the discriminant can be explicitly computed to be

For imaginary quadratic extensions, the form class group and the class group turn out to be the same!

Theorem 1.

Let squarefree, be a quadratic extension. Then ${\mathrm{C}}_K$, the class group of $K$, is isomorphic to the group of reduced forms of discriminant $d_K$, ${\mathrm{C}}_{d_K}$.

One can in fact exhibit an explicit correspondence ${𝒞}_{d_K}\to {𝒞}_K$:

$$a{x}^2+bxy+c{y}^2\mapsto (a,\frac{b+\sqrt{d_K}}{2})$$

Note in particular that the simplest, or principal, form of discriminant $d_K$ ($x^2-d_K{y}^2$ or $x^2+xy+\frac{1-d_K}{4}{y}^2$) maps to the ideal $(1)={𝒪}_K$; these forms are the identities in ${𝒞}_{d_K}$. Showing that the map is 1-1 and onto is not difficult; showing that it is a group isomorphism is more difficult but nevertheless essentially amounts to a computation.

This theorem allows us to simply compute at least the size of the class group for quadratic extensions $K$ by computing the number of reduced forms of discriminant $d_K$. For example, suppose $K=\mathbb{Q}(\sqrt{-23})$. Since $-23\equiv 1(4)$, ${𝒪}_K=\mathbb{Z}[\frac{1+\sqrt{-23}}{2}]$ and $d_K=-23$.

What are the forms of discriminant $-23$? , and $b$ is odd, so $b=\pm 1$. $4ac-b^2=23$, so $ac=6$. We thus get three reduced forms:

$(1,1,6)$
$(2,1,3)$
$(2,-1,3)$	reduced since $|b|\ne a,a\ne c$

Note that $(1,-1,6)$ is not reduced, since but $|b|=a$.

So we know that the order of the class group ${𝒞}_K$ is $3$, so ${𝒞}_K\cong \mathbb{Z}/3\mathbb{Z}$.

We can use the explicit correspondence above to find representatives of the three elements of the class group using the map from forms to ideals.

$(1,1,6)$$\to (1,{\displaystyle \frac{1+\sqrt{-23}}{2}})=(1)$

$(2,1,3)$$\to (2,{\displaystyle \frac{1+\sqrt{-23}}{2}})$

$(2,-1,3)$$\to (2,{\displaystyle \frac{-1+\sqrt{-23}}{2}})$

In fact, a more general form of Theorem 1 is true. If $K$ is an algebraic number field, $A\subset {𝒪}_K$, then $A$ is not a Dedekind domain unless $A={𝒪}_K$. But even in this case, if one considers only those ideals that are invertible in $A$, one can define a group structure in a similar way; this is once again called the class group of $A$. In the case that $K$ is a quadratic extension, these subrings of ${𝒪}_K$ are called orders of $K$.

It is the case that each discriminant corresponds to a unique order in a quadratic extension of $\mathbb{Q}$. Specifically,

Theorem 2.

Let . Write $\mathrm{\Delta }\mathrm{=}{m}^{\mathrm{2}}\mathit{}{\mathrm{\Delta }}^{\mathrm{\prime }}$ where ${\mathrm{\Delta }}^{\mathrm{\prime }}$ is squarefree. Let $K\mathrm{=}\mathrm{Q}\mathit{}\mathrm{(}\sqrt{{\mathrm{\Delta }}^{\mathrm{\prime }}}\mathrm{)}$. Then

$${𝒪}_{\mathrm{\Delta }}=\{\begin{array}{cc}\mathbb{Z}\left[\frac{m}{2}\sqrt{{\mathrm{\Delta }}^{\prime }}\right],{\mathrm{\Delta }}^{\prime }\equiv 2,3(mod4)\hfill & \\ \mathbb{Z}\left[m\frac{1+\sqrt{{\mathrm{\Delta }}^{\prime }}}{2}\right],{\mathrm{\Delta }}^{\prime }\equiv 1(mod4)\hfill & \end{array}$$

is a subring of ${\mathrm{O}}_K$, and ${\mathrm{C}}_{\mathrm{\Delta }}\mathrm{\cong }{\mathrm{C}}_{{\mathrm{O}}_{\mathrm{\Delta }}}$.(Note that if ${\mathrm{\Delta }}^{\mathrm{\prime }}\mathrm{\equiv }\mathrm{2}\mathrm{,}\mathrm{3}\mathrm{(}\mathrm{mod}\mathrm{4}\mathrm{)}$, then $m$ must be even. For otherwise, $m^{\mathrm{2}}\mathrm{\equiv }\mathrm{1}\mathrm{(}\mathrm{mod}\mathrm{4}\mathrm{)}$ and thus $\mathrm{\Delta }\mathrm{\equiv }\mathrm{2}\mathrm{,}\mathrm{3}\mathrm{(}\mathrm{mod}\mathrm{4}\mathrm{)}$, which is impossible. Thus $m\mathrm{/}\mathrm{2}$ is an integer in this case)

This reduces to the first theorem in the event that $\mathrm{\Delta }=d_K$.

Thus there is a $1-1$ correspondence between discriminants and orders of quadratic fields; in particular, the ring of algebraic integers of any quadratic field corresponds to the forms of discriminant equal to the discriminant of the field.