Given a quadratic form with discriminant $D$, what does the class number of $\mathbb{Q}(\sqrt{D})$ tell us?
(This question is inspired by a comment on the question here)
Given a quadratic form with discriminant $D$, what does the class number of $\mathbb{Q}(\sqrt{D})$ tell us?
(This question is inspired by a comment on the question here)
I guess you mean a binary quadratic form, i.e. of the form $a x^2 + b x y + c y^2$, with $a,b,c$ integers and of discrmimant $D$ (i.e. $b^2 - 4 a c = D$).
If $\mathbb Q(\sqrt{D})$ has class number one, then the conditions for solving $a x^2 + b x y + c y^2 = p$ ($p$ a prime not dividing $D$) depend only on the residue class of $p$ mod $D$, in fact only on the value of the Kronecker symbol of $p$ mod $D$.
If the class group is just a product of cyclic groups of order 2 (note that this can't be detected by the class number alone, which e.g. can't distinguish between $C_2 \times C_2$ and $C_4$) then the condition for solving $a x^2 + b x y + c y^2 = p$ depends only on the residue class of $p$ mod $D$, but one has to consider not just the Kronecker symbol mod $D$, but other Kronecker symbols modulo various divisors of $D$.
If the class group is not a product of cyclic groups of order 2 (e.g. if the class number is not a power of two) then there is no congruence condition on $p$ which guarantees being able to solve $a x^2 + b x y + c y^2$.
A geometric interpretation of the class number of any number field is described at http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/SL2classno.pdf.
I will put a little something here, Lisa wants to know something else but I am not yet sure what that is.
The question linked to (long since deleted) had to do with the fact that the binary quadratic form $$ f(x,y) = x^2 + x y + 3 y^2 $$ is the only form ($SL_2 \mathbb Z$ equivalence class) of its discriminant, $-11.$ This has any number of consequences. For example, if $p, q$ are distinct primes and we have $$ pq = x^2 + xy + 3 y^3, $$ then we know that we can write both $$ p = s^2 + st+ 3 t^2 $$ and $$ q = u^2 + u v + 3 v^2, $$ all in integers.
For comparison, with $g(x,y) = x^2 + x y + 6 y^2,$ we have $$ g(-1,8) = 377 = 13 \cdot 29. $$ But these two primes are represented by $2 x^2 \pm xy + 3 y^2,$ the other two classes of that discriminant. With $h(x,y) = 2 x^2 + xy+ 3 y^2,$ we get $h(2,1) = 13,$ then $h(-2,3) = 29.$ At most one class (along with its opposite) represents a prime.
I also put some stuff at http://math.blogoverflow.com/2014/08/23/binary-quadratic-forms-over-the-rational-integers-and-class-numbers-of-quadratic-%EF%AC%81elds/