Let $j: \mathbf{C} - \mathbf{R} \rightarrow \mathbf{C}$ denote the classical $j$-function from the theory of elliptic functions. That is, $j(\tau)$ is the $j$-invariant of the elliptic curve $\mathbf{C}/(\mathbf{Z} + \mathbf{Z}\tau)$, so this has the familiar expansion $1/q_{\tau} + 744 + \cdots$ where $q_{\tau} = \exp(2\pi i_{\tau} \cdot \tau)$ with $i_{\tau}$ denoting the square root of $-1$ lying in the same connected component of $\mathbf{C} - \mathbf{R}$ as $\tau$ does. (In particular, $\overline{j(\tau)} = j(\overline{\tau})$ and $j(-\tau) = j(\tau)$.)

Let $\zeta$ denote a primitive cube root of unity in $\mathbf{C}$, and let $n \ge 1$ be an integer. The value $j(n \zeta)$ is an algebraic integer (lying in the ring class field of conductor $n$ over $\mathbf{Q}(\zeta)$, by CM theory). In more conceptual terms, this is the $j$-invariant of the elliptic curve $\mathbf{C}/\mathcal{O}_n$ where $\mathcal{O}_n$ is the unique order of conductor $n$ in the subfield of $\mathbf{C}$ generated by the cube roots of unity.

I was recently asked by someone (based on numerical evidence for $n$ up to 20 or so) to prove that $j(n \zeta) \equiv 0 \bmod 81$ for $n \equiv 1 \bmod 3$ and $j(n \zeta) \equiv -27 \bmod 81$ for $n \equiv -1 \bmod 3$ (congruence modulo 81 in the ring of algebraic integers in $\mathbf{C}$); beware that this is sensitive to the fact that $n > 0$ but is independent of the choice of $\zeta$ in $\mathbf{C}$ (contemplate the behavior of $j$ with negation, as noted above). For $n = 1, 2$ these congruences are easily verified by hand. I eventually came up with an affirmative proof in general using the Grothendieck-Messing crystalline Dieudonne theory for $3$-divisible groups and some concrete calculations.

This leads to the following question. It is natural to use deformation theory (which is what the Grothendieck-Messing theory is part of) to prove congruential properties of $j$-values, but hauling out such high-powered theoretical machinery to prove something as down to earth as a mod 81 congruence on $j(n \zeta)$ for $n > 0$ (not divisible by 3) may seem like killing a fly with a sledgehammer. So...does anyone see a way to determine $j(n \zeta) \bmod 81$ (for all integers $n > 0$ not divisible by 3) by using pre-Grothendieck technology?

[Note that it is equivalent to prove that for $n > 0$ not divisible by 3, $j(n \zeta) \bmod 81$ only depends on $n \bmod 3$, as we can then compute for $n = 1, 2$ to conclude. But it seems not obvious at the outset that this congruence class in the ring of algebraic integers in $\mathbf{C}$ is represented by a rational integer, let alone one that only depends on $n \bmod 3$. It is also relatively easy to prove that $j(n\zeta) \equiv 0 \bmod 27$, so the real difficulty lies in improving things to work modulo 81.]

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    $\begingroup$ The Gross-Zagier paper "On singular moduli" proves that $j(\tau) \equiv 2^6 3^3 \mathrm{mod} 3^6$ when $j(\tau)$ comes from an elliptic curve with CM by $\mathscr{O}_{\sqrt{-d}}$ for $-d$ a fundamental discriminant with $d\equiv 1\; \mathrm{mod}\; 3$. Perhaps their technique generalizes to answer your question? $\endgroup$ Nov 11 '10 at 5:14
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    $\begingroup$ David, let me explain why that kind of approach falls crucially short. For the setup you mention, $3$ is unramified in the relevant ring class field, so the $j$-congruence ultimately stems from the fact that the reduction in char. 3 is supersingular with half size of aut. group equal to 6. The comparable result one would obtain in this way for the question posed is that $j(n \zeta) \equiv 0 \bmod 27$ (since 3 is now quadratically ramified in the relevant ring class field, so $3^3$ is really an order 6 congruence for the relevant uniformizer, akin to $3^6$ in your comment). So too weak, alas. $\endgroup$
    – BCnrd
    Nov 11 '10 at 6:14
  • $\begingroup$ Have you looked at: de Shalit, E.: Kronecker's polynomial, supersingular elliptic curves, andp-adic periods of modular curves. In: Proceedings of the workshop onp-adic monodromy and the Birch-Swinnerton-Dyer conjecture (Boston 1991). Contemporary Mathematics165, AMS (1994), 135-148 ? $\endgroup$ Nov 13 '10 at 4:02
  • $\begingroup$ Dear Felipe: I have not seen that paper, so I just looked at the review on MathSciNet. It addresses a formula for the norm (from $\mathbf{F}_{p^2}$ down to $\mathbf{F}_p$) of a difference of distinct supersingular $j$-values in characteristic $p$ that are moreover distinct from 0 and 1728. If we suppose the method might adapt to handle 0 and 1728 (with more work), for characteristic 3 there is only one $j$-value and anyway it is all about a mod-$p$ congruence. So my gut feeling is that this is unlikely to help in proving a mod-81 congruence for our desired CM $j$-values in char. 0. Oh well! $\endgroup$
    – BCnrd
    Nov 13 '10 at 4:26
  • $\begingroup$ I recall that the paper has some p-adic results as well as some mod p^2 results. It might not answer your question but it does more than what is in MR. $\endgroup$ Nov 13 '10 at 13:43

Since I wasn't yet reading Mathoverflow at the time, I didn't see this question until Brian e-mailed it to me in January. I was eventually able to give a more elementary proof by applying formulas of Vélu to the $n$-isogeny from a curve with $j=0$ to one with $j=j(n\zeta)$, which in fact determined $j(n\zeta) \bmod 3^{9/2}$. With some more work I then proved a congruence $\bmod 3^5$ in the case $n \equiv -1 \bmod 3$, and even obtained some information $\bmod 3^6$ for $n \equiv +1 \bmod 3$. Namely:

@ if $n \equiv -1 \bmod 3$ then $j(n\zeta) \equiv -54 \bmod 3^5$; and

@ if $n \equiv 1 \bmod 3$ then $j(n\zeta)$ has valuation at least $9/2$ at $3$, and every conjugate is congruent to either $0$ or $\pm 324 \sqrt{3} \bmod 3^6$.

Brian soon replied that he can understand these refined congruences using the same "Grothendieck-Messing crystalline Dieudonné theory" that he applied to get the congruences $\bmod 3^4$.

Here's a link to a talk I gave here in mid-February that recounts this story and outlines the proofs and some additional information: http://www.math.harvard.edu/~elkies/j3.pdf


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