Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)

share|cite|improve this question
Lisa, what else do you want to know? – Will Jagy Sep 23 '14 at 21:29
Maybe she should have gone ahead and posted her question even though the system was dead set on finding a "duplicate." – Robert Soupe Sep 24 '14 at 5:24
@WillJagy I know that $\mathcal{O}_{\mathbb{Q}(\sqrt{D})}$ has unique factorization if its class number is 1, and doesn't if the class number is 2 or higher. But what else does the class number us? Let's say $\mathcal{O}_{\mathbb{Q}(\sqrt{m})}$ has class number 2 and $\mathcal{O}_{\mathbb{Q}(\sqrt{n})}$ has class number 3. Neither is a UFD. Obviously $m \neq n$. Without knowing the actual values of $m$ and $n$, can we contrast $\mathcal{O}_{\mathbb{Q}(\sqrt{m})}$ and $\mathcal{O}_{\mathbb{Q}(\sqrt{n})}$ in any meaningful way? – Lisa Sep 24 '14 at 21:07
@Will The ring with $m$ has numbers with multiple factorizations but in each case the total number of factors is the same, but the one with $n$ there can be multiple factorizations of different lengths. e.g., $\mathbb{Z}[\sqrt{-14}]$ has class number 3, and for 81 we have $3^4 = (5 - 2 \sqrt{-14})(5 + 2 \sqrt{-14})$. So, as Finch says, "the class number measures how far $\mathcal{O}_D$ is from being a UFD." – Robert Soupe Sep 25 '14 at 3:58
@WillJagy Oops, I meant to cite Harold Whistler's dissertation "Ethical Considerations of High Frequency Econometrics." ;-) – Robert Soupe Sep 25 '14 at 12:28
up vote 8 down vote accepted

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 $a x^2 + b x y + c y^2$.

share|cite|improve this answer

A geometric interpretation of the class number of any number field is described at

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.