5
$\begingroup$

Let $f$ be a binary quadratic form with integer coefficients and non-zero discriminant $D$. Suppose for simplicity that $D$ is a fundamental discriminant (which in particular implies that $f$ is primitive). We say that $f$ represents a square mod $D$ if for any $a$ representable by $f$ we have

$$a \equiv \square \pmod{p}$$

for each prime $p | D$. Note that this condition only depends on the $\text{GL}_2(\mathbb{Z})$-equivalence class of $f$. We can thus say that the $\text{GL}_2(\mathbb{Z})$-class of $f$, $[f]$, is a square class if for a prime $q$ representable by $f$ and co-prime to $D$ satisfies $q \equiv \square \pmod{p}$ for $p | D$, and if $D \equiv 0 \pmod{4}$ (respectively $8$), then $q \equiv 1 \pmod{4}$ (respectively $8$). By our earlier discussion, the choice of $q$ representable by $f$ does not matter.

How are 'square classes' distributed in the ideal class group of $\mathcal{O} = \mathcal{O}_{\mathbb{Q}(\sqrt{D})}$? Heuristically, for each (odd) prime $p | D$, half of the classes of the ideal class group should be square mod $p$ and the other half should not, so the number of square classes ought to be $h_2(D) 2^{-\omega(D)}$ where $h_2(D)$ is the class number of $\mathcal{O}$ and $\omega(n)$ denotes the number of prime divisors of $n$.

$\endgroup$
  • 1
    $\begingroup$ Are you expecting an answer going beyond genus theory? Or is the principal genus theorem (Gauss, DA 1801) the answer? $\endgroup$ – Franz Lemmermeyer Oct 20 '18 at 9:12
  • $\begingroup$ @FranzLemmermeyer yes, it seems that 'square classes' are precisely the classes in the principal genus. If you want to write that as an answer I will accept it. $\endgroup$ – Stanley Yao Xiao Oct 26 '18 at 0:59
3
$\begingroup$

The standard definition for a form to belong to the principal genus of forms with fundamental discriminant $d$ is that the primes $p$ coprime to $d$ that the form $Q$ represents satisfy $(d_1/p) = \ldots = (d_t/p)$, where $d =d_1 \cdots d_t$ is the factorization of $d$ into prime discriminants. In particular, $Q$ is allowed to represent primes $p \equiv \pm 1 \bmod 8$ if $d_1 = 8$ occurs in the factorization, and primes $p \equiv 1, 3 \bmod 8$ if $d_1 = -8$ occurs (for example, $40 = 5 \cdot 8$ and $-40 = 5 \cdot (-8)$ are factorizations into prime discriminants, $40 = -5 \cdot (-8)$ is not since $-5$ is not a discriminant).

Gauss's principal genus theorem states that a form is in the principal genus if and only if the equivalence classof $Q$ is a square in the class group. The genus of a form is the sign vector $((d_1/p), \ldots, (d_t/p))$, and there are forms for every sign vectors whose entries have product $+1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.