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The quadratic forms with discriminant -23 up to change of variables are:

  • A(x,y): $x^2 + xy + 6 y^2$
  • B(x,y): $2 x^2 - xy + 3 y^2$
  • C(x,y): $2 x^2 + xy + 3 y^2$

Viewed as number fields it's relatively easy then compute:

  • A(x,y)A(a,b): A(xa - 6yb,ya + (x - y)b).
  • BB: B(xa - (3/2)yb,ya + (x+(1/2)y)b)
  • CC: C(xa - (3/2)yb,ya + (x-(1/2)y)b)

I have not found any number of the form C which is not of the form A or B, maybe I just didn't look for enough though.

  • Can we also compute A(x,y)B(u,v), AC and BC?

  • Is there any way to think about these forms as ideals?

  • Since 23 = A, 3 = B but 23*3 = B = C so it doesn't seem like there is a simple group structure here, but B and C are 'conjugate' in some sense so perhaps there is a group structure on {{A},{B,C}} or maybe the structure is different than a group?

  • What about the converse problem? If d|A then d = A,B or C? update 7 is not of the form A, B or C but 7^2 = A.

So in this case the converse problem is not solvable, but I wonder if there are examples of multiple forms where the converse problem does hold?

For an example of a single form where the converse problem works is G(x,y)=$x^2 + y^2$ I have the answer, it's just d|G => d = G since every factor of a sum of two squares is a sum of two squares or a square (= x^2 + 0^2).


Maybe it would have been better to write this question for discriminant -36 since it has the forms {x^2 + 9y^2, 2x^2 + 2xy + 5y^2, 3x^2 + 3y^2} none of which are conjugate... This set seems a bit simpler but it still has the strange non-multiplicative phenomenon with 7^2.

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Please clean up your typos in the equations where you multiply: on the left you write u and v but on the right there is a and b instead. –  KCd Mar 30 '11 at 9:25
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Every value of C is a value of B since C(x,y) = B(-x,y). –  KCd Mar 30 '11 at 9:49
    
Notice that the example $x^2+y^2$ works exactly because there is only one reduced form in that class. Maybe you already know this, but I think it could be better to say it explicitly. Regards. –  awllower Mar 29 '13 at 18:54

1 Answer 1

up vote 4 down vote accepted

You have noticed that there seems to be a group lurking around when you consider (primitive) quadratic forms with a fixed discriminant, but also there seem to be problems. This is the reason why trying to make a group out of quadratic forms is subtle! Historically, Lagrange first defined equivalent quadratic forms to be those linked to one another by any invertible integral change of variables. Such a change of variables has determinant 1 or -1. Later Gauss found that by using a finer equivalence relation, where only invertible integral changes of variable with determinant 1 are permitted, there is an associated group law on his equivalence classes of quadratic forms.

Lagrange's equivalence classes are unions of Gauss's equivalence classes in the following algebraic sense: start with a finite abelian group G and identify each g in G with its inverse g^(-1). If g^2 is trivial then g = g^(-1) and otherwise g is not g^(-1). Let G* be the equivalence classes of sets {g,g^(-1)} in G, which have size 1 or 2. So G* is a kind of collapsing of G, but it is very awkward to try to make G* a group. You just can't do it in general. For example, if G = Z/3 = {0,1,2 mod 3} then G is an additive group and G* = {{0},{1,2}}, which doesn't make any sense as a group using some natural addition operation on representatives of the sets making up G*. Lagrange was basically dealing with something like G*, albeit in the language of quadratic forms, which is why he never saw a group law.

The connection with your example is that the change of variables (x,y) --> (-x,y) turns your B into C but this change has determinant -1 so it wouldn't be "legal" for Gauss's equivalence relation. I think you can see how G* is like your {{A},{B,C}}.

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