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 . Given , call this number , the form of .
Thus, for example, since there is only one reduced form of discriminant , we have that .
It turns out that the set of reduced forms of a given negative discriminant can be turned into an abelian group, called the , , by defining a “multiplication” on forms that is based on generalizations of identities such as
where all of these forms have discriminant .
Now, given an algebraic extension of , ideal classes of also form an abelian group, called the http://planetmath.org/node/IdealClassideal class group of , . The order of is called the class number of and is denoted . See the ideal class entry for more detail.
For an algebraic extension , one also defines the http://planetmath.org/node/DiscriminantOfANumberFielddiscriminant of the extension, . For quadratic extensions , where is assumed squarefree, the discriminant can be explicitly computed to be
Let squarefree, be a quadratic extension. Then , the class group of , is isomorphic to the group of reduced forms of discriminant , .
One can in fact exhibit an explicit correspondence :
Note in particular that the simplest, or principal, form of discriminant ( or ) maps to the ideal ; these forms are the identities in . 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 by computing the number of reduced forms of discriminant . For example, suppose . Since , and .
What are the forms of discriminant ? , and is odd, so . , so . We thus get three reduced forms:
Note that is not reduced, since but .
So we know that the order of the class group is , so .
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.
In fact, a more general form of Theorem 1 is true. If is an algebraic number field, , then is not a Dedekind domain unless . But even in this case, if one considers only those ideals that are invertible in , one can define a group structure in a similar way; this is once again called the class group of . In the case that is a quadratic extension, these subrings of are called orders of .
It is the case that each discriminant corresponds to a unique order in a quadratic extension of . Specifically,
Let . Write where is squarefree. Let . Then
is a subring of , and . (Note that if , then must be even. For otherwise, and thus , which is impossible. Thus is an integer in this case)
This reduces to the first theorem in the event that .
Thus there is a 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|
|Date of creation||2013-03-22 16:56:27|
|Last modified on||2013-03-22 16:56:27|
|Last modified by||rm50 (10146)|