9
$\begingroup$

Given the j-function,

$$j(\tau)=\frac{1}{q}+744+196884q+21493760q^2+\dots$$

it is well-known that for $\tau=\tfrac{1+\sqrt{-d}}{2}$, positive integer $d$, then $j(\tau)$ is an algebraic integer of degree = class number $h(-d)$. Thus,

$$j(\tfrac{1+\sqrt{-163}}{2}) = -12^3(231^2-1)^3$$

$$j(\tfrac{1+\sqrt{-427}}{2}) = -12^3\big((7215+924\sqrt{61})^2-1\big)^3$$

where the squares are due to a certain Eisenstein series. However, not all d yield a cube.

Conjecture: Is it true that given fundamental discriminant $d = 3m$, $m \neq$ square, $h(-d) = 2^n$, then $j(\tau) = (U_m)^k\,x^3$ for some integer k, where $U_m$ is a fundamental unit, and $x$ is an algebraic integer of degree $h(-d)$?

For example, given $d = 51, 483, 651$ which have $h(-d) = 2,4,8$, respectively, and fundamental units,

$$U_{17} = 4+\sqrt{17}$$

$$U_{161} = 11775 + 928\sqrt{161}$$

$$U_{217} = 3844063+260952\sqrt{217}$$

then,

$$j(\tfrac{1+\sqrt{-51}}{2}) = -48^3 (U_{17})^2 (5+\sqrt{17})^3$$

$$j(\tfrac{1+\sqrt{-483}}{2}) = -120^3 (U_{161}) {x_1}^3$$

$$j(\tfrac{1+\sqrt{-651}}{2}) = -96^3 (U_{217})^2 {x_2}^3$$

where $x_1,x_2$ are algebraic integers (rather tedious to write down) of degree 4,8, respectively.

Is the conjecture true?

P.S. I tested it with $h(-d) = 6$ and it does not work, so I think it is only for $h(-d) = 2^n$.

$\endgroup$
  • $\begingroup$ Perhaps it works for $h$ prime to $3$? $\endgroup$ – Will Sawin Nov 6 '13 at 5:39
  • 1
    $\begingroup$ Borel / Chowla / Herz / Iwasawa / Serre Seminar on Complex Multiplication $\endgroup$ – Franz Lemmermeyer Nov 6 '13 at 5:40
10
$\begingroup$

So you're really asking when the ideal generated by $j(\tau)$ is the cube of an ideal. There's a famous paper of Gross and Zagier that describes the prime factorization of the ideal generated by the difference $j(\tau_1)-j(\tau_2)$ of two CM $j$-invariants. In your case $j(\tau_2)=0$ is a CM value. This description is likely to be useful in trying to prove (or disprove) your conjecture. The reference is

Gross, Benedict H.; Zagier, Don B. On singular moduli. J. Reine Angew. Math. 355 (1985), 191–220. MR0772491

$\endgroup$

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.