Skip to main content
MathOverflow

Return to Answer

error corrected
Source Link
paul Monsky
paul Monsky
  • 5.4k
  • 2
  • 26
  • 45

The way Gauss did things was in terms of SL(2,Z) equivalence classes of (primitive) binary quadratic forms over Z. So lets consider such definite forms, axx+bxy+cyy with b^2-4ac=-N. For simplicity suppose N is squarefree and odd. Gauss defines a composition on the classes, making the set of classes into a finite group. For each prime dividing N he defines a genus character from this group to Z/2. The product of these characters is the trivial map, and the joint kernel of them all consists of the squares, a subgroup that he calls the principal genus. In the case N=pq, the character attached to p maps a form to (M/p) where M is any integer prime to p represented by the form. So the squares form, in your case, a subgroup of index 2, and there is a unique non-trivial class of order 2 in the group. Gauss calls the classes of order 2 the "ambiguous classes". They're easily written down in general and are represented by "reduced" forms axx+bxy+cyy with b=0 or with a=b (or c). So in your case the non-trivial ambiguous class is represented by pxx+qyy. The genus character attached to p maps this class to (q/p). So the class is a square if and only if (q/p)=1, giving the result you want. The theory also works for even N not necessarily square-free, though there are 2 genus characters attached to the prime 2 when 32 divides N, and it works for indefinite forms as well.

Oops--I should have said that the non-trivial ambiguous class is represented by qxx+qxy+(1/4)(p+q)yy.

The way Gauss did things was in terms of SL(2,Z) equivalence classes of (primitive) binary quadratic forms over Z. So lets consider such definite forms, axx+bxy+cyy with b^2-4ac=-N. For simplicity suppose N is squarefree and odd. Gauss defines a composition on the classes, making the set of classes into a finite group. For each prime dividing N he defines a genus character from this group to Z/2. The product of these characters is the trivial map, and the joint kernel of them all consists of the squares, a subgroup that he calls the principal genus. In the case N=pq, the character attached to p maps a form to (M/p) where M is any integer prime to p represented by the form. So the squares form, in your case, a subgroup of index 2, and there is a unique non-trivial class of order 2 in the group. Gauss calls the classes of order 2 the "ambiguous classes". They're easily written down in general and are represented by "reduced" forms axx+bxy+cyy with b=0 or with a=b. So in your case the non-trivial ambiguous class is represented by pxx+qyy. The genus character attached to p maps this class to (q/p). So the class is a square if and only if (q/p)=1, giving the result you want. The theory also works for even N not necessarily square-free, though there are 2 genus characters attached to the prime 2 when 32 divides N, and it works for indefinite forms as well.

The way Gauss did things was in terms of SL(2,Z) equivalence classes of (primitive) binary quadratic forms over Z. So lets consider such definite forms, axx+bxy+cyy with b^2-4ac=-N. For simplicity suppose N is squarefree and odd. Gauss defines a composition on the classes, making the set of classes into a finite group. For each prime dividing N he defines a genus character from this group to Z/2. The product of these characters is the trivial map, and the joint kernel of them all consists of the squares, a subgroup that he calls the principal genus. In the case N=pq, the character attached to p maps a form to (M/p) where M is any integer prime to p represented by the form. So the squares form, in your case, a subgroup of index 2, and there is a unique non-trivial class of order 2 in the group. Gauss calls the classes of order 2 the "ambiguous classes". They're easily written down in general and are represented by "reduced" forms axx+bxy+cyy with b=0 or with a=b (or c). So in your case the non-trivial ambiguous class is represented by pxx+qyy. The genus character attached to p maps this class to (q/p). So the class is a square if and only if (q/p)=1, giving the result you want. The theory also works for even N not necessarily square-free, though there are 2 genus characters attached to the prime 2 when 32 divides N, and it works for indefinite forms as well.

Oops--I should have said that the non-trivial ambiguous class is represented by qxx+qxy+(1/4)(p+q)yy.

Source Link
paul Monsky
paul Monsky
  • 5.4k
  • 2
  • 26
  • 45

The way Gauss did things was in terms of SL(2,Z) equivalence classes of (primitive) binary quadratic forms over Z. So lets consider such definite forms, axx+bxy+cyy with b^2-4ac=-N. For simplicity suppose N is squarefree and odd. Gauss defines a composition on the classes, making the set of classes into a finite group. For each prime dividing N he defines a genus character from this group to Z/2. The product of these characters is the trivial map, and the joint kernel of them all consists of the squares, a subgroup that he calls the principal genus. In the case N=pq, the character attached to p maps a form to (M/p) where M is any integer prime to p represented by the form. So the squares form, in your case, a subgroup of index 2, and there is a unique non-trivial class of order 2 in the group. Gauss calls the classes of order 2 the "ambiguous classes". They're easily written down in general and are represented by "reduced" forms axx+bxy+cyy with b=0 or with a=b. So in your case the non-trivial ambiguous class is represented by pxx+qyy. The genus character attached to p maps this class to (q/p). So the class is a square if and only if (q/p)=1, giving the result you want. The theory also works for even N not necessarily square-free, though there are 2 genus characters attached to the prime 2 when 32 divides N, and it works for indefinite forms as well.