The Hasse invariant at cuspidal elliptic curves in a generically ordinary family

Question

For an elliptic curve of the form $y^2 = f(x)$ where $f(x) \in \mathbb F_q[x]$ is a cubic polynomial with distinct roots, it is known (from Silverman's book, say) that the curve is supersingular precisely when the coefficient of $x^{p-1}$ in $f(x)^\frac{p-1}{2}$ is $0$.

My question is about the behavior of this coefficient in families of curves. Consider the following families of elliptic curves in characteristic $p>3$.

1. $E_1:y^2 = f_1(x,t) = x(x-1)(x-t)$ and
2. $E_2:y^2 = f_2(x,t) = x^3+3tx^2+3x+1$

where we can think of $t$ as an element of $\overline{\mathbb F}_q[t]$. We can plug in any value $\alpha$ from $\overline {\mathbb F}_q$ for $t$ and so long $f_i(x,\alpha)$ doesn't obtain roots with multiplicity $>1$, we still get an elliptic curve. The key difference between the above two families is that the former has node at $t=0,1$ and the latter has a cusp singularity at $t=1$.

Now I want to study the $x^{p-1}$ coefficient of $f_i(x,t)^{\frac{p-1}{2}}$ as an element of $\overline{\mathbb F}_q[t]$. Let us call it $H_i(t)$. Now, by doing some experiments in Sage (you can play around with these equations in the link), I have seen that $H_1(t)$ does not vanish at $t=0,1$ (and can prove it using an old argument of Igusa) but $H_2(t)$ does seem to vanish at $t=1$ with a very large and predictable multiplicity: $\frac{p-1}{6}$ if $p\equiv 1 \pmod 6$ and $\frac{p+1}{6}$ if $p\equiv 5 \pmod 6$.

My question is why does this happen? So I've managed to tweak the equation of $E_2$ a bit so that the curve retains it's cusp but makes bookkeeping easier and have found a combinatorial proof of this high order vanishing. But I am looking for something more geometric that explains it. Somehow the polynomial vanishes on $\mathbb G_a$ but not on $\mathbb G_m$? This feels very much like the mod-$p$ modular form that goes by the name of the Hasse invariant or at least it's pullback under the map to the moduli space of elliptic curves. Is $\mathbb G_a$ "very supersingular" while $\mathbb G_m$ is "ordinary" under a suitable interpretation? Can someone give an explanation or direct me to a book/paper on what is going on and what the relationship of this polynomial is to the different types of singularities?

NB: This phenomenon is clearly not unique to Elliptic Curves: Even for some higher genus cases, I see very similar patterns happening: If there are cusps or higher order singularities for a curve in a family of curves $C_t$ at a point $a$, this large order of vanishing at $a$ is happening for an appropriate analog of the coefficient studied above.

Answer 1

Figuring this out was fun!

I feel like part of my answer should be intuitive, except I’m not sure how to make it rigorous at all. The gist is that $\mathbb{G}_m$ really, really, doesn’t want to be supersingular. I can suggest a few “vibe-based reasons”:

1. $\mathbb{G}_m$ means that your $j$-invariant goes to infinity. On the other hand, being supersingular means that your $j$-invariant lies in a few points of $\mathbb{F}_{p^2}$.

2. whether an elliptic curve over $\overline{\mathbb{F}_p}$ is supersingular depends on the structure of $p$-torsion subscheme, which can have exactly two types: either it is $\mu_p \times \mathbb{Z}/p\mathbb{Z}$, or it is something else entirely which doesn’t contain $\mu_p$. Well, the $p$-torsion of $\mathbb{G}_m$ is exactly $\mu_p$.

3. having multiplicative reduction means that “very locally” (after completing?) your family will look an awful lot like a base changed Tate curve. But the Hasse invariant of the Tate curve is equal to $1$ – this comes back to the same idea that the Tate curve is anything but supersingular.

The case of the second family is interesting because there’s no such heuristic for $\mathbb{A}^1$ at play. That’s because “additive reduction” is a catch-all term that can describe a lot of different kinds of degeneracy – is the curve even potentially multiplicative or does it have potentially good reduction?

TL;DR: after a small extension of the coefficient field, the family acquires good reduction at $t=1$ and the fibre at $t=1$ is a supersingular elliptic curve iff $p \equiv -1\pmod{3}$. We can then compute the vanishing order of the Hasse invariant at $t=1$ by keeping track of the changes of variables that we made.

Anyway, let’s rewrite the family $y^2=x^3+tx^2+3x+1$ as $y_1^2=x_1^3+3(1-t^2)x_1+(1+t+t^2-3t(1+t))(1-t)$, ie $y_1^2=x_1^3+3s(2-s)x_1+s(-3+6s-2s^2)$ with $s=1-t$.

We have just set $x_1=x+t$ and $y_1=y$, this does not change the Hasse invariant.

Obviously, this equation becomes singular cuspidal when $s=0$, so we’re going to try and make the cusp disappear. Write $x_1=s^{1/3}x_2$ and $y_1=s^{1/2}y_2$.

The equation becomes $y_2^2=x_2^3+3s^{1/3}(2-s)x_2+(-3+6s-2s^2)$, and we divided the Hasse invariant by $(s^{1/2}/s^{1/3})^{p-1}=s^{\frac{p-1}{6}}=(1-t)^{\frac{p-1}{6}}$ (because it’s a modular form and we scaled the implied invariant differential by $s^{1/2}/s^{1/3}$).

The good news is that when $s=0$, this equation becomes $y_2^2=x_2^3-3$. This elliptic curve has $j$-invariant zero, so it is supersingular iff $p \equiv -1 \pmod{3}$.

Thus, when $p \equiv 1\pmod{6}$, the vanishing order of the Hasse invariant of the original family at $t=1$ is $\frac{p-1}{6}$. When $p \equiv -1 \pmod{6}$, the argument shows that the Hasse invariant vanishes at $t=1$ with an order greater than $\frac{p-1}{6}$, so it vanishes with order at least $\frac{p+1}{6}$.

To compute the exact vanishing order, we look at the equation modulo $s^{2/3}$, so we get $y_2^2=x_2^3+6s^{1/3}x_2-3$.

Now, in this ring, the $x_2^{p-1}$ coefficient of $(x_2^3+6s^{1/3}x_2-3)^{\frac{p-1}{2}}$ is the same as the $$-th coefficient of $x_2^{p-1}$ in $(x_2^3-3)^{\frac{p-1}{2}} + 3(p-1)s^{1/3}(x_2^3-3)^{\frac{p-3}{2}}x_2$.

Because $p \equiv -1\pmod{3}$, the first term does not contribute to $x_2^{p-1}$. The contribution of the second term to this coefficient is $6s^{1/3}(-3)^{\frac{p-3}{2}-\frac{p-2}{3}}\binom{\frac{p-3}{2}}{\frac{p-2}{3}}$, so it’s a nonzero constant multiple of $s^{1/3}$, whence the exact vanishing order at $t=1$ of $\frac{1}{3}+\frac{p-1}{6}=\frac{p+1}{6}$.