# 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 , ${\mathcal{C}}_{\mathrm{\Delta}}$, by defining a “multiplication^{}” on forms that is based on generalizations^{} of identities^{} such as

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

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

Now, given an algebraic extension^{} $K$ of $\mathbb{Q}$, ideal classes of ${\mathcal{O}}_{K}$ also form an abelian group, called the *http://planetmath.org/node/IdealClassideal class group* of $K$, ${\mathcal{C}}_{K}$. The order of ${\mathcal{C}}_{K}$ 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

$${d}_{K}=\{\begin{array}{cc}4n\hfill & \text{if}n\equiv 2,3\phantom{\rule{veryverythickmathspace}{0ex}}(mod4)\hfill \\ n\hfill & \text{if}n\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(mod4)\hfill \end{array}$$ |

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 ${\mathcal{C}}_{{d}_{K}}\to {\mathcal{C}}_{K}$:

$$a{x}^{2}+bxy+c{y}^{2}\mapsto (a,\frac{b+\sqrt{{d}_{K}}}{2})$$ |

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 $(1)={\mathcal{O}}_{K}$; these forms are the identities in ${\mathcal{C}}_{{d}_{K}}$. 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 $-23\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(4)$, ${\mathcal{O}}_{K}=\mathbb{Z}[\frac{1+\sqrt{-23}}{2}]$ 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:

$(1,1,6)$ | |
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$(2,1,3)$ | |

$(2,-1,3)$ | 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 ${\mathcal{C}}_{K}$ is $3$, so ${\mathcal{C}}_{K}\cong \mathbb{Z}/3\mathbb{Z}$.

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^{}, $A\subset {\mathcal{O}}_{K}$, then $A$ is not a Dedekind domain^{} unless $A={\mathcal{O}}_{K}$. 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 ${\mathcal{O}}_{K}$ 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

$${\mathcal{O}}_{\mathrm{\Delta}}=\{\begin{array}{cc}\mathbb{Z}\left[\frac{m}{2}\sqrt{{\mathrm{\Delta}}^{\prime}}\right],{\mathrm{\Delta}}^{\prime}\equiv 2,3\phantom{\rule{veryverythickmathspace}{0ex}}(mod4)\hfill & \\ \mathbb{Z}\left[m\frac{1+\sqrt{{\mathrm{\Delta}}^{\prime}}}{2}\right],{\mathrm{\Delta}}^{\prime}\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(mod4)\hfill & \end{array}$$ |

is a subring of ${\mathrm{O}}_{K}$, and ${\mathrm{C}}_{\mathrm{\Delta}}\mathrm{\cong}{\mathrm{C}}_{{\mathrm{O}}_{\mathrm{\Delta}}}$. (Note that if ${\mathrm{\Delta}}^{\mathrm{\prime}}\mathrm{\equiv}\mathrm{2}\mathrm{,}\mathrm{3}\phantom{\rule{veryverythickmathspace}{0ex}}\mathrm{(}\mathrm{mod}\mathrm{4}\mathrm{)}$, then $m$ must be even. For otherwise, ${m}^{\mathrm{2}}\mathrm{\equiv}\mathrm{1}\phantom{\rule{veryverythickmathspace}{0ex}}\mathrm{(}\mathrm{mod}\mathrm{4}\mathrm{)}$ and thus $\mathrm{\Delta}\mathrm{\equiv}\mathrm{2}\mathrm{,}\mathrm{3}\phantom{\rule{veryverythickmathspace}{0ex}}\mathrm{(}\mathrm{mod}\mathrm{4}\mathrm{)}$, 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.

Title | equivalence of form class group and class group |
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Canonical name | EquivalenceOfFormClassGroupAndClassGroup |

Date of creation | 2013-03-22 16:56:27 |

Last modified on | 2013-03-22 16:56:27 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 5 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 11E12 |

Classification | msc 11E16 |

Classification | msc 11R29 |