Computing p-adic Sigma Functions and Cyclotomic Heights

Barry Mazur, William Stein, John Tate

December 8, 2004

1 Introduction

Fix an elliptic curve E over ℚ, given by a minimal Weierstrass equation

y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}.  (1.1)

Suppose p is a prime of good ordinary reduction for E, so

p ∤ N_{E} · a_{E,p},

where N_{E} is the conductor of E and a_{E,p} = p + 1 − #E(𝔽_{p}).

2 The Canonical p-adic sigma function

Let

σ(t) = t + s_{1}t^{2} + s_{2}t^{3} + s_{3}t^{4} + ⋯ ∈ t(1 + tℤ_{p}[[t]])  (2.1)

be the canonical p-adic σ-function attached to E as in [MT91].

2.1 Computing σ(t)

[[Describe the resultin algorithm here – method to compute σ.]] Theorem 3.1 of [MT91] asserts that if m ∈ ℤ and Q ∈ E^{f}(\overline{R}), then

σ(mQ) = σ(Q)^{m^{2}} · f_{m}(Q).  (2.2)

We use this relation theorem to determine s_{1} and s_{2} in terms of the formal endomorphism [p] and the pth division polynomial. [[Idea – maybe we could just use m = 2 instead of m = p?! It’s probably easier to compute [2] and f_{2} than to compute [p] and f_{p}; if not m = 2, then m = 3?]] Then we use knowledge of s_{1} and s_{2} and the differential equation satisfied by σ to determine the constant c ∈ ℤ_{p} such that x(t) − c is the canonical “eks” (pronounced like the letter “x”) function of [MT91]. Finally, given x(t) − c, we find the unique series σ(t) ∈ t(1 + tℤ_{p}[[t]]) that satisfies the differential equation

x(t) − c = −\frac{d}{ω} (\frac{1}{σ} \frac{dσ}{ω} )  (2.3)

of [MT91, pg. 665].

Let ω = dx/(2y + a_{1}x + a_{3}) be the invariant differential on E. Inverting the power series in equation 15 of [Tat74], we have

\frac{dt}{ω} = 1 − a_{1}t − a_{2}t^{2} − 2a_{3}t^{3} + (−2a_{1}a_{3} − 2a_{4})t^{4} + ⋯ = ∑ b_{n}t^{n},

where the a_{i}’s are as in (1.1). Using the chain rule, etc., and writing σ(t) = ∑_{n≥1} s_{n−1}t^{n} (as in (2.1)), we see that the differential equation (2.3) written in terms of t is

\frac{d}{dt} (\frac{(∑_{n≥1} ns_{n−1}t^{n−1}) · (∑_{n≥0} b_{n}t^{n})}{∑_{n≥1} s_{n−1}t^{n}} ) · ∑_{n≥0} b_{n}t^{n} = c − x(t). (2.4)

Here

x(t) = \frac{1}{t^{2}} − \frac{a_{1}}{t} − a_{2} − a_{3}t − (a_{4} + a_{1}a_{3})t^{2} + ⋯

is the formal expansion of x in terms of t. Expanding the left hand of (2.4) we see that

− \frac{1}{t^{2}} + \frac{a_{1}}{t} + (−a_{1}s_{1} − s_{1}^{2} + 2s_{2})+

(a_{1}^{2}s_{1} + a_{1}a_{2} + 3a_{1}s_{1}^{2} − 6a_{1}s_{2} − 2a_{2}s_{1} − 2a_{3} + 2s_{1}^{3} − 6s_{1}s_{2} + 6s_{3})t+

(−2a_{1}^{2}s_{1}^{2} + 4a_{1}^{2}s_{2} + 3a_{1}a_{2}s_{1} − 5a_{1}s_{1}^{3} + 15a_{1}s_{1}s_{2} − 15a_{1}s_{3} + a_{2}^{2} + 4a_{2}s_{1}^{2}−

8a_{2}s_{2} − 6a_{3}s_{1} − 4a4 − 3s_{1}^{4} + 12s_{1}^{2}s_{2} − 12s_{1}s_{3} − 6s_{2}^{2} + 12s_{4})t^{2} + ⋯

= c − (\frac{1}{t^{2}} − \frac{a_{1}}{t} + −a_{2} − a_{3}t − (a_{4} + a_{1}a_{3})t^{2} + ⋯) .

Equating constant coefficients, we see that

c = a_{2} − a_{1}s_{1} − s_{1}^{2} + 2s_{2}.

We will not use the higher terms yet...

Next we use the relation (2.2) to compute s_{1} and s_{2} in terms of [p](t) and f_{p}(t). Write the formal multiplication by p as

[p](t) = pt + r_{2}t^{2} + r_{3}t^{3} + ⋯ ∈ ℤ_{p}[[t]]

and the formal division “polynomial” f_{p}(t) as

f_{p}(t) = pt^{1−p^{2}} + d_{1}t^{2−p^{2}} + d_{2}t^{3−p^{2}} + ⋯ ∈ (ℤ_{p}[[t]])[1/t].

The identity (2.2) is

∑_{n≥1} s_{n−1} · (pt + r_{2}t^{2} + r_{3}t^{3} + ⋯)^{n} =

(∑_{n≥1} s_{n−1}t^{n})^{p^{2}} · (pt^{1−p^{2}} + d_{1}t^{2−p^{2}} + d_{2}t^{3−p^{2}} + ⋯) .

Multiply both sides by t^{p^{2}−1}, so that both sides are in ℤ_{p}[[t]].

∑_{n≥1} s_{n−1} · (pt + r_{2}t^{2} + r_{3}t^{3} + ⋯)^{n} · t^{p^{2}−1} =

(∑_{n≥1} s_{n−1}t^{n})^{p^{2}} · (p + d_{1}t + d_{2}t^{2} + ⋯) .

[[Since we have this for all m, I’m going to replace p by 2 and see what happens. Also, f_{2}(t) = 2t^{1−2^{2}} + ⋯ and [2](t) = 2t + r_{2}t^{2} + · · ·.]]

t^{3} · ∑_{n≥1} s_{n−1} · (2t + r_{2}t^{2} + r_{3}t^{3} + ⋯)^{n} =

(∑_{n≥1} s_{n−1}t^{n})^{4} · (2 + d_{1}t + d_{2}t^{2} + ⋯) .

Expanding both sides we get

2t^{4} + (4s_{1} + r_{2})t^{5} + (4s_{1}r_{2} + 8s_{2} + r_{3})t^{6} + ⋯ =

2t^{4} + (8s_{1} + d_{1})t^{5} + (12s_{1}^{2} + 4s_{1}d_{1} + 8s_{2} + d_{2})t^{6} + · · · .

Equating coefficients of t^{5} we see that

4s_{1} + r_{2} = 8s_{1} + d_{1}

so

s_{1} = \frac{r_{2} − d_{1}}{4} .

Equating coefficients of t^{6} shows that

4s_{1}r_{2} + 8s_{2} + r_{3} = 12s_{1}^{2} + 4s_{1}d_{1} + 8s_{2} + d_{2},

so we get NO INFORMATION ABOUT s_{2} from this.

[[Observation: Actually one could compute σ directly from equating coefficients!!! See the commented MAGMA session below, where it appears to me that one can simply read off the other coefficients of σ from (??) with m = 2.]]

However, we do get information from the coefficient of t^{7}. Taking differences, we have that

0 = (-4*s1 - d1 + r2)*t^5 + (-12*s1^2 - 4*s1*d1 + 4*s1*r2 - d2 +
    r3)*t^6 + (-8*s1^3 - 6*s1^2*d1 - 24*s1*s2 - 4*s1*d2 + s1*r2^2
    + 4*s1*r3 - 4*s2*d1 + 12*s2*r2 + 8*s3 - d3 + r4)*t^7 +
    (-2*s1^4 - 4*s1^3*d1 - 24*s1^2*s2 - 6*s1^2*d2 - 12*s1*s2*d1 -
    24*s1*s3 - 4*s1*d3 + 2*s1*r2*r3 + 4*s1*r4 - 12*s2^2 - 4*s2*d2
    + 6*s2*r2^2 + 12*s2*r3 - 4*s3*d1 + 32*s3*r2 + 24*s4 - d4 +
    r5)*t^8 + O(t^9)

Thus

−8s_{1}^{3}−6s_{1}^{2}d_{1}−24s_{1}s_{2}−4s_{1}d_{2}+s_{1}r_{2}^{2}+4s_{1}r_{3}−4s_{2}d_{1}+12s_{2}r_{2}+8s_{3}−d_{3}+r_{4} = 0.

Unfortunately we have both s_{1} and s_{3} together in this formula, so this doesn’t determine s_{2}. Using m = −1 we get

0 = (2*s1 - d1 + r2)*t^2 + (-s1*d1 - 2*s1*r2 - d2 + r3)*t^3 + (-s1*d2
    + s1*r2^2 - 2*s1*r3 - s2*d1 + 3*s2*r2 + 2*s3 - d3 + r4)*t^4 +
    (-s1*d3 + 2*s1*r2*r3 - 2*s1*r4 - s2*d2 - 3*s2*r2^2 + 3*s2*r3
    - s3*d1 - 4*s3*r2 - d4 + r5)*t^5 + O(t^6).

The coefficient of t^{4} gives a different linear equation for s_{2} and s_{3}, and probably taking two of these together we’ll be able to solve for both.

For m = 3 it’s:

(-18*s1 - d1 + r2)*t^10 + (-108*s1^2 - 9*s1*d1 + 6*s1*r2 - d2 +
    r3)*t^11 + (-252*s1^3 - 36*s1^2*d1 - 216*s1*s2 - 9*s1*d2 +
    s1*r2^2 + 6*s1*r3 - 9*s2*d1 + 27*s2*r2 + 54*s3 - d3 +
    r4)*t^12 + (-378*s1^4 - 84*s1^3*d1 - 756*s1^2*s2 - 36*s1^2*d2
    - 72*s1*s2*d1 - 216*s1*s3 - 9*s1*d3 + 2*s1*r2*r3 + 6*s1*r4 -
    108*s2^2 - 9*s2*d2 + 9*s2*r2^2 + 27*s2*r3 - 9*s3*d1 +
    108*s3*r2 + 216*s4 - d4 + r5)*t^13 + O(t^14)

2.2 Computing formal multiplication by p

2.3 Computing division polynomials

3 The p-adic cyclotomic height

Let E and p be as in the introduction.

Suppose Q ∈ E(ℚ) is a point that reduces to the point at infinity modulo p, and lies in the identity component modulo every bad prime. For example, if M is the least common multiple of #E(𝔽_{p}) and the Tamagawa numbers c_{ℓ} of E at bad primes, then every element of ME(ℚ) satisfies our hypothesis on Q. Since ...??, we can write

Q = (x, y) = (\frac{a}{e^{2}} , \frac{b}{e^{3}} ) ,

with a, b, e ∈ ℤ and gcd(a, e) = 1. Let t = −x/y. Then the p-adic cyclotomic height h_{p}(Q) is

h_{p}(Q) = log_{p} (\frac{σ(t)}{e}) .  (3.1)

[[I think exactly this formula appears on the upper right page of a preprint Tate was holding yesterday, which perhaps I don’t have a copy.]] Here

log_{p} : ℚ_{p}^{∗} → ℚ_{p}

is the the unique extension of the usual p-adic logarithm on 1 + pℤ_{p} to a homomorphism from the multiplicative group of ℚ_{p}^{∗} to the additive group of ℚ_{p}. Thus if au ∈ ℚ_{p}^{∗} with u ∈ 1 + pℤ_{p}, then log_{p}(au) = log_{p}(u), and one computes log_{p}(u) using the formula

log_{p}(1 − x) = − (x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + ⋯)

with x = 1 − u. [[Again, there’s a discussion of this in a paper Tate was holding, that I don’t seem to have.]]

The cyclotomic height h_{p}(Q) extends uniquely to a ℤ_{p}-valued function on E(ℚ) that satisfies h_{p}(nQ) = n^{2}Q for all integers n. If M is as above, then we can compute h_{p}(P) for any P ∈ E(ℚ) by computing

h_{p}(P) = \frac{h_{p}([M]P)}{M^{2}} = \frac{1}{M^{2}} · log_{p} (\frac{σ(t)}{e}) ,

where t = −x/y and e are for Q = [M]P = (x, y) = (a/e^{2}, b/e^{3}).

References

[MT91] B. Mazur and J. Tate, The p-adic sigma function, Duke Math. J. 62 (1991), no. 3, 663–688. MR 93d:11059

[Tat74] J. Tate, The arithmetic of elliptic curves, Invent. Math. 23 (1974), 179–206. MR 54 #7380