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


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*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:


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*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.


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*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 $d|G \implies d = G$ 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 $7^2$.
 A: 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}}.
