Given the j-function,


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,

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

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

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


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

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

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

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 $h(-d) = 2^n$.

  • $\begingroup$ 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

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


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