A FORMULA FOR THE SUPERSINGULAR POLYNOMIAL: ADDENDUM

LUÍS R. A. FINOTTI

Abstract. In these notes we prove give a (different) elementary proof that the the equation given for the supersingular polynomial in [Fin08] has simple roots.

Last revised: September 10, 2008.

1. Introduction

In [Fin08], it was proved that the supersingular polynomial

ss_{p}(X) =^{def} ∏_{j supersig.}(X − j)  (1.1)

can be explicitly written as

ss_{p}(X) = (−2)^{r} ∑_{i=r_{1}}^{r_{2}} \binom{r}{i}\binom{i}{3i − r} (−\frac{27}{4})^{i} X^{i−r′_{1}}(X − 1728)^{r′_{2}−i},  (1.2)

where, if p is the characteristic of the field of definition of the curve, then r =^{def} (p − 1)/2,

r_{1} =^{def} ⌈r/3⌉, r_{2} =^{def} ⌊r/2⌋, r′_{1} =^{def} ⌊r/3⌋, and r′_{2} =^{def} ⌈r/2⌉.

This formula was deduced by using the fact the an elliptic curve is supersingular if, and only if, its Hasse invariant is zero. This was enough to obtain an expression quite close to the one above, where only a factor of X or (X − 1728) would be missing. On the other hand, to show that this polynomial only has simple roots, we quoted the well-know result that there are exactly (r′_{2} − r′_{1}) supersingular invariants j-invariants.

At the end of the same reference, an alternative proof, which is completely elementary. We first observe that ss_{p}(X) has simple roots if, and only if,

G(X) =^{def} ∑_{i=r_{1}}^{r_{2}} \binom{r}{i}\binom{i}{3i − r} (−\frac{27}{4})^{i} X^{i−r_{1}}(X − 1728)^{r_{2}−i}  (1.3)

has simple roots, as it differs from ss_{p}(X) only by a constant multiple or possible factors of X or (X − 1728). This second proof is then given by means of the differential equation

X(X − 1728)G′′ + ((−2r_{2} + 2r_{1} + 1)X − 1728(2r_{1} + 1))G′ + (r_{2} − r_{1})^{2}G = 0, (1.4)

1991 Mathematics Subject Classification. Primary 11G20; Secondary 11T71.

Key words and phrases. arithmetic geometry, elliptic curves, supersingular polynomial.

2 LUÍS R. A. FINOTTI

which is deduced in that same paper.

Although this latter proof is completely elementary, the deduction of Eq. (1.4) is not completely natural. It was motivated by similar proofs, and its deduction involved educated guesses and variation of parameters. Although this is quite fine, and give a nice and short proof of the statement without having to rely on any previous knowledge, as with the first proof given, we here would like to give a more direct proof of it, without using the differential equation. This proof is longer, but hopefully can be of some interest.

Our goal is then to prove the following:

Proposition 1.1. The polynomial G(X) (as in (1.3)) has only simple roots.

2. The Proof

As a trivial consequence of the Lemma 2.2 from [Fin08], we have:

Proposition 2.1. If E is given by

E/k : y^{2} = f(x) =^{def} x^{3} + ax + b  (2.1)

and a, b ≠ 0, then the Hasse invariant of E is

(\frac{b}{a})^{r} F̃(\frac{a^{3}}{b^{2}} ) ,

where

F̃(X) =^{def} ∑_{i=r_{1}}^{r_{2}} \binom{r}{i}\binom{i}{3i − r}X^{i}.  (2.2)

So, if a, b ≠ 0, an E is supersingular if, and only if, F̃(a^{3}/b^{2}) = 0.

Proof. Just remember that an elliptic curve is supersingular if, and only if, the Hasse invariant is zero, and that the Hasse invariant a of curve as in (2.1) is the coefficient of x^{p−1} in f(x)^{r}. (Remember, r =^{def} (p − 1)/2.) □

Observe that if indeed a, b ≠ 0, then a^{3}/b^{2} is an invariant (under isomorphisms) of E. In fact, as done in [Fin08], we have that if

F (X) =^{def} \frac{F̃(X)}{X^{r_{1}}} = ∑_{i=r_{1}}^{r_{2}} \binom{r}{i}\binom{i}{3i − r}X^{i−r_{1}},  (2.3)

then

G(X) = (−\frac{27}{4})^{r_{1}} (X − 1728)^{r_{2}−r_{1}}F (−\frac{27}{4} · \frac{X}{X − 1728}) .  (2.4)

So, by construction, we have that X = 1728 is not a root of G(X). Also, if T (X) = −27X/4(X − 1728), then

G′(X) = (−\frac{27}{4})^{r_{1}} (X − 1728)^{r_{2}−r_{1}−2} [(r_{2} − r_{1})(X − 1728)F (T (X)) + 1164F′(T (X))] .

A FORM. FOR THE SUPERSING. POL.: ADDENDUM 3

Thus, if X = x_{0} is a root of G(X), and so x_{0} ≠ 1728, then T (x_{0}) is a root of F (X), and if x_{0} is a double root of G(X), then T (x_{0}) is a also a double root of F (X). Therefore, if F (X) has no double roots, then neither does G(X). Moreover, observe that T (x_{0}) ≠ −27/4, so it suffices that F (X) has no double roots different from X = −27/4. (In fact, X = −27/4 is not a root of F (X), as seen in [Fin08].)

To make our computations a bit more straight forward, we deal with F̃(X) instead of F (X) itself. So, our goal now is to prove the following proposition, which, from our previous remarks, is enough to prove Proposition 1.1.

Proposition 2.2. If λ is a double (or higher order) root of F̃(X), then λ is either 0 or −27/4.

The rest of these notes is devoted to the proof of the proposition above. We proceed by contradiction. Assume then that we have a double root. If this root is non-zero and different from −27/4, we can assume that it has the form a^{3}/b^{2} with a, b ≠ 0, with a and b defining an elliptic curve as in Eq. (2.1). So, assume that F̃(a^{3}/b^{2}) = 0 and F̃′(a^{3}/b^{2}) = 0.

Let a_{i} and b_{i} be such that

f(x)^{r} = ∑_{i=0}^{3r} a_{i}x^{i}  and  f(x)^{r−1} = ∑_{i=0}^{3r−3} b_{i}x^{i}.  (2.5)

By Lemma 2.2 from [Fin08],

b_{p−4} = (\frac{b}{a})^{r} ∑_{i=r_{1}}^{r_{2}} \binom{r − 1}{i}\binom{i}{3i − r} (\frac{a^{3}}{b^{2}} )^{i}

= (\frac{b}{a})^{r} ∑_{i=r_{1}}^{r_{2}} (1 − \frac{i}{r} ) \binom{r}{i}\binom{i}{3i − r} (\frac{a^{3}}{b^{2}} )^{i}

= (\frac{b}{a})^{r} (F̃(\frac{a^{3}}{b^{2}} ) − \frac{1}{r} \frac{a^{3}}{b^{2}} F̃′(\frac{a^{3}}{b^{2}} )) .

So, if F̃(a^{3}/b^{2}) = F̃′(a^{3}/b^{2}) = 0, then a_{p−1} = b_{p−4} = 0.

Let

f^{r} = f_{1}x^{p} + f_{2}, with deg f_{1} = r − 1, deg f_{2} ≤ p − 2;
f^{r−1} = g_{1}x^{p} + g_{2}, with deg g_{1} = r − 4, deg g_{2} ≤ p − 1.

(Note that a_{p−1} = 0.)

The proof of the proposition will be broken in smaller steps:

Step 1. deg g_{2} ≤ p − 5.

4 LUÍS R. A. FINOTTI

Proof. Observing that

\frac{d}{dx}(f^{r}) = ∑_{i=0}^{3r−1}(i + 1)a_{i+1}x^{i},

but also

\frac{d}{dx}(f^{r}) = −\frac{1}{2}(3x^{2} + a) ∑_{i=0}^{3r−3} b_{i}x^{i},

comparing the terms x^{p−2} in these equations, we obtain b_{p−2} = 0 (since b_{p−4} = 0), and comparing the terms x^{p−2}, we obtain b_{p−3} = −a b_{p−1}/3. Also, since

∑_{i=0}^{3r} a_{i}x^{i} = (x^{3} + ax + b) ∑_{i=0}^{3r−3} b_{i}x^{i},

comparing the terms in x^{p−1} gives that b_{p−1} = 0, and hence also b_{p−3} = 0.

□

Step 2.

f g′_{2} = −\frac{3}{2}f′g_{2}.  (2.6)

Proof. We have,

\frac{d}{dx}(f^{r}) = −\frac{1}{2}f^{r−1}f′ = −\frac{1}{2}(f′g_{1}x^{p} + f′g_{2}).

On the other hand, also

\frac{d}{dx}(f^{r}) = f′_{1}x^{p} + f′_{2}.

So, by Step 1, we have

f′_{2} = (−1/2)f′g_{2}.  (2.7)

Again by Step 1, f_{i} = f g_{i}, for i = 1, 2, and so,

f′_{2} = \frac{d}{dx}(f g_{2}) = f′g_{2} + fg′_{2},

which, together with equation (2.7), gives

f g′_{2} = −\frac{3}{2}f′g_{2}.

□

Step 3. deg g_{2} = (p − 9)/2.

Proof. By Step 1, we have

g_{2} = ∑_{i=0}^{k} b_{i}x^{i},

for some k ≤ (p − 5). Comparing coefficients of x^{k+2} in equation (2.6), we have k b_{k} = (9/2)b_{k}. Hence, if p does not divide (2k + 9), then deg g_{2} ≤ (k − 1).

A FORM. FOR THE SUPERSING. POL.: ADDENDUM 5

Since k ≤ (p − 5), we have that 2k + 9 ≤ 2p − 1, and therefore, if p divides 2k + 9, then p = 2k + 9.

If b_{(p−9)/2} = 0, then we can proceed as above for all k ≥ 0, thus obtaining that g_{2} = 0. Otherwise, deg g_{2} = (p − 9)/2.

If g_{2} = 0, then f_{2} = 0, but f_{2}(0) = b^{r} ≠ 0. Therefore, deg g_{2} = (p − 9)/2. □

Step 4.

f_{1} = \frac{1}{a_{r−1}} f_{2}.

Proof. By the previous step, and since f_{2} = f g_{2}, we have that deg f_{2} = (p − 3)/2 = (r − 1). Now,

(x^{3p} + a^{p}x^{p} + b^{p}) = f^{p} = f(f^{r})^{2} = f (f_{1}^{2}x^{2p} + 2f_{1}f_{2}x^{p} + f_{2}^{2}).

Thus, since deg f_{1}^{2} = deg f_{2}^{2} = deg f_{1} f_{2} = (p − 3) (observe that f_{1} is monic),

f (\frac{1}{a_{r−1}} f_{2})^{2} = x^{p} + C_{1},
f f_{1}^{2} = x^{p} + C_{2},

with C_{1}, C_{2} ∈ k. It follows, by the uniqueness of the quotient of the division of x^{p} by f, that f_{1}^{2} = (1/a_{r−1} f_{2})^{2}. Hence, since f_{1} and (1/a_{r−1})f_{2} are monic, f_{1} = (1/a_{r−1})f_{2}.

□

This last step gives us:

(x^{3} + ax + b)(f_{1}^{2}x^{2p} + 2a_{r−1}f_{1}^{2}x^{p} + a_{r−1}^{2}f_{1}^{2}) = (x^{3p} + a^{p}x^{p} + b^{p}).  (2.8)

Comparing the terms in x^{2p}, x^{p} and constant term, we have

bf_{1}(0)^{2} + 2a_{r−1} = 0,
b2a_{r−1}f_{1}(0)^{2} + a_{r−1}^{2} = a^{p},
ba_{r−1}^{2}f_{1}(0)^{2} = b^{p}.

It follows that

bf_{1}(0)^{2} = −2a_{r−1},
b^{p} = −2a_{r−1}^{3},
a^{p} = −3a_{r−1}^{2}.

Hence, (a^{3}/b^{2})^{p} = −27/4, and so a^{3}/b^{2} = −27/4, which contradicts our assumptions and concludes the proof of Proposition 2.2.

References

[Fin08] L. R. A. Finotti. A new formula for the supersingular polynomial. Available at

    http://www.math.utk.edu/~finotti/, 2008.

Department of Mathematics, University of Tennessee, Knoxville, TN – 37996

E-mail address: finotti@math.utk.edu