CoCalc Public Fileswww / talks / padic_heights / e2heights / cycloht.texOpen with one click!
Author: William A. Stein
1
\documentclass[11pt]{article}
2
\bibliographystyle{amsalpha}
3
\include{macros}
4
\title{Computing $p$-adic Sigma Functions and Cyclotomic Heights}
5
\author{Barry Mazur, William Stein, John Tate}
6
\begin{document}
7
\maketitle
8
9
\section{Introduction}
10
Fix an elliptic curve $E$ over $\Q$, given by a minimal
11
Weierstrass equation
12
\begin{equation}\label{eqn:weq}
13
y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6.
14
\end{equation}
15
Suppose $p$ is a prime
16
of good ordinary reduction for $E$, so $$p \nmid N_E \cdot a_{E,p},$$
17
where $N_E$ is the conductor of $E$ and $a_{E,p}=p+1-\#E(\F_p)$.
18
19
\section{The Canonical $p$-adic sigma function}
20
Let
21
\begin{equation}\label{eqn:sigma}
22
\sigma(t) = t + s_1t^2 + s_2 t^3 + s_3 t^4 + \cdots
23
\in t (1+t\Z_p[[t]])
24
\end{equation}
25
be the canonical $p$-adic $\sigma$-function attached to $E$
26
as in \cite{mazur-tate:sigma}.
27
28
\subsection{Computing $\sigma(t)$}
29
30
[[Describe the resultin algorithm here -- method to compute $\sigma$.]]
31
32
33
Theorem 3.1 of \cite{mazur-tate:sigma} asserts that if $m\in \Z$ and
34
$Q\in E^f(\overline{R})$, then
35
\begin{equation}\label{eqn:sigma_mq}
36
\sigma(m Q) = \sigma(Q)^{m^2} \cdot f_m(Q).
37
\end{equation}
38
We use this relation
39
theorem to determine $s_1$ and $s_2$ in terms of the formal
40
endomorphism $[p]$ and the $p$th division polynomial. [[Idea -- maybe
41
we could just use $m=2$ instead of $m=p$?! It's probably easier to
42
compute $[2]$ and $f_2$ than to compute $[p]$ and $f_p$; if not $m=2$,
43
then $m=3$?]] Then we use knowledge of $s_1$ and $s_2$ and the
44
differential equation satisfied by $\sigma$ to determine the constant
45
$c\in\Z_p$ such that $x(t) - c$ is the canonical ``eks'' (pronounced
46
like the letter ``x'') function of
47
\cite{mazur-tate:sigma}. Finally, given $x(t) - c$, we find the
48
unique series $\sigma(t) \in t(1+t\Z_p[[t]])$
49
that satisfies the differential equation
50
\begin{equation}\label{eqn:de1}
51
x(t) - c = -\frac{d}{\omega}\left(\frac{1}{\sigma}
52
\frac{d\sigma}{\omega}\right)
53
\end{equation}
54
of \cite[pg.~665]{mazur-tate:sigma}.
55
56
Let $\omega = dx/(2y+a_1x + a_3)$ be the invariant differential
57
on $E$. Inverting the power series in equation 15 of
58
\cite{tate:arithmetic}, we have
59
$$
60
\frac{dt}{\omega} =
61
1 - a_1 t - a_2 t^2 - 2a_3 t^3 + (-2 a_1 a_3 - 2 a_4)t^4 + \cdots
62
= \sum b_n t^n,
63
$$
64
where the $a_i$'s are as in (\ref{eqn:weq}).
65
\comment{
66
> K<a1,a2,a3,a4> := FieldOfFractions(PolynomialRing(Q,4));
67
> R<t> := PowerSeriesRing(K);
68
> f := 1 + a1*t + (a1^2+a2)*t^2 + (a1^3+2*a1*a2+2*a3)*t^3 + (a1^4+3*a1^2*a2+6*a1*a3+a2^2+2*a4)*t^4 + O(t^5);
69
> 1/f;
70
1 - a1*t - a2*t^2 - 2*a3*t^3 + (-2*a1*a3 - 2*a4)*t^4 + O(t^5)
71
}
72
Using the chain rule, etc., and writing
73
$\sigma(t) = \sum_{n\geq 1} s_{n-1} t^n$ (as in (\ref{eqn:sigma})),
74
we see that
75
the differential equation (\ref{eqn:de1}) written in terms of~$t$
76
is
77
\begin{equation}\label{eqn:de_big}
78
\frac{d}{dt}\left(
79
\frac{\left(\sum_{n\geq 1} n s_{n-1} t^{n-1}\right) \cdot
80
\left(\sum_{n\geq 0} b_n t^n\right)}{\sum_{n\geq 1} s_{n-1} t^n}
81
\right)\cdot \sum_{n \geq 0} b_n t^n = c - x(t).
82
\end{equation}
83
Here
84
$$
85
x(t) = \frac{1}{t^2} - \frac{a_1}{t} - a_2 - a_3 t - (a_4+a_1 a_3)t^2 + \cdots
86
$$
87
is the formal expansion of $x$ in terms of $t$.
88
\comment{
89
> K<a1,a2,a3,a4,s1,s2,s3,s4,s5> := FieldOfFractions(PolynomialRing(Q,9));
90
> R<t> := LaurentSeriesRing(K);
91
> w := 1 - a1*t - a2*t^2 - 2*a3*t^3 + (-2*a1*a3 - 2*a4)*t^4 + O(t^5);
92
> sig := t + s1*t^2 + s2*t^3 + s3*t^4 + s4*t^5 + s5*t^6 + O(t^7);
93
> sigprime := Derivative(sig);
94
> Derivative(((sigprime*w)/sig))*w;
95
-t^-2 + a1*t^-1 + -a1*s1 - s1^2 + 2*s2 + (a1^2*s1 + a1*a2 +
96
3*a1*s1^2 - 6*a1*s2 - 2*a2*s1 - 2*a3 + 2*s1^3 - 6*s1*s2 +
97
6*s3)*t + (-2*a1^2*s1^2 + 4*a1^2*s2 + 3*a1*a2*s1 - 5*a1*s1^3
98
+ 15*a1*s1*s2 - 15*a1*s3 + a2^2 + 4*a2*s1^2 - 8*a2*s2 -
99
6*a3*s1 - 4*a4 - 3*s1^4 + 12*s1^2*s2 - 12*s1*s3 - 6*s2^2 +
100
12*s4)*t^2 + O(t^3)
101
}
102
Expanding the left hand of (\ref{eqn:de_big})
103
we see that
104
\begin{align*}
105
&-\frac{1}{t^2} + \frac{a_1}{t} + (-a_1 s_1 - s_1^2 + 2s_2)
106
+\\
107
&\quad (a_1^2 s_1 + a_1 a_2 +
108
3 a_1 s_1^2 - 6 a_1 s_2 - 2 a_2 s_1 - 2 a_3 + 2 s_1^3 - 6 s_1 s_2 +
109
6 s_3) t +\\
110
&\quad (-2 a_1^2 s_1^2 + 4 a_1^2 s_2 + 3 a_1 a_2 s_1 - 5 a_1 s_1^3
111
+ 15 a_1 s_1 s_2 - 15 a_1 s_3 + a_2^2 + 4 a_2 s_1^2 - \\
112
&\qquad8 a_2 s_2 -
113
6 a_3 s_1 - 4 a4 - 3 s_1^4 + 12 s_1^2 s_2 - 12 s_1 s_3 - 6 s_2^2 +
114
12 s_4) t^2 + \cdots
115
\\
116
&=
117
c - \left(\frac{1}{t^2} - \frac{a_1}{t} + - a_2 - a_3 t - (a_4+a_1 a_3)t^2 + \cdots\right).
118
\end{align*}
119
Equating constant coefficients, we see that
120
$$
121
c = a_2 - a_1 s_1 - s_1^2 + 2s_2.
122
$$
123
We will not use the higher terms yet...
124
125
Next we use the relation (\ref{eqn:sigma_mq}) to compute
126
$s_1$ and $s_2$ in terms of $[p](t)$ and $f_p(t)$.
127
Write the formal multiplication by $p$ as
128
$$
129
[p](t) = pt + r_2 t^2 + r_3 t^3 + \cdots \in \Z_p[[t]]
130
$$
131
and the formal division ``polynomial'' $f_p(t)$ as
132
$$
133
f_p(t) = pt^{1-p^2} + d_1 t^{2-p^2} + d_2 t^{3-p^2} + \cdots
134
\in (\Z_p[[t]])[1/t]. %\left[\frac{1}{t}\right].
135
$$
136
The identity (\ref{eqn:sigma_mq})
137
is
138
$$
139
\sum_{n\geq 1} s_{n-1}\cdot
140
\left(pt + r_2 t^2 + r_3 t^3 + \cdots \right)^n
141
=\qquad\qquad\qquad\qquad$$
142
$$\qquad\qquad\qquad\left(\sum_{n \geq 1} s_{n-1} t^n\right)^{p^2} \cdot
143
\left(pt^{1-p^2} + d_1 t^{2-p^2} + d_2t^{3-p^2} + \cdots\right).
144
$$
145
Multiply both sides by $t^{p^2-1}$, so that both sides are
146
in $\Z_p[[t]]$.
147
$$
148
\sum_{n\geq 1} s_{n-1}\cdot
149
\left(pt + r_2 t^2 + r_3 t^3 + \cdots \right)^n\cdot t^{p^2-1}
150
=\qquad\qquad\qquad\qquad$$
151
$$\qquad\qquad\qquad\left(\sum_{n \geq 1} s_{n-1} t^n\right)^{p^2} \cdot
152
\left(p + d_1 t + d_2 t^2 + \cdots\right).
153
$$
154
155
156
[[Since we have this for all $m$, I'm going to replace $p$ by $2$ and
157
see what happens. Also, $f_2(t) = 2t^{1-2^2} + \cdots$ and
158
$[2](t)=2t+r_2t^2+\cdots$.]]
159
$$
160
t^3 \cdot \sum_{n\geq 1} s_{n-1}\cdot
161
\left(2t + r_2 t^2 + r_3 t^3 + \cdots \right)^n
162
=\qquad\qquad\qquad\qquad$$
163
$$\qquad\qquad\qquad\left(\sum_{n \geq 1} s_{n-1} t^n\right)^{4} \cdot
164
\left(2 + d_1 t + d_2 t^2 + \cdots\right).
165
$$
166
Expanding both sides we get
167
$$
168
2 t^4 + (4 s_1 + r_2) t^5 + (4 s_1 r_2 + 8 s_2 + r_3) t^6 + \cdots =
169
\qquad\qquad\qquad
170
$$
171
$$
172
\qquad\qquad\qquad
173
2 t^4 + (8 s_1 + d_1) t^5 + (12 s_1^2 + 4 s_1 d_1 + 8 s_2 + d_2) t^6 + \cdots.
174
$$
175
Equating coefficients of $t^5$ we see that
176
$$
177
4 s_1 + r_2 = 8 s_1 + d_1
178
$$
179
so
180
$$
181
s_1 = \frac{r_2 - d_1}{4}.
182
$$
183
Equating coefficients of $t^6$ shows that
184
$$
185
4 s_1 r_2 + 8 s_2 + r_3 =
186
12 s_1^2 + 4 s_1 d_1 + 8 s_2 + d_2,
187
$$
188
so we get NO INFORMATION ABOUT $s_2$ from this.
189
190
[[Observation: Actually one could compute $\sigma$ directly from
191
equating coefficients!!! See the commented MAGMA session below, where
192
it appears to me that one can simply read off the other coefficients
193
of $\sigma$ from (\ref{eqn:sigma_ma}) with $m=2$.]]
194
195
However, we do get information from the coefficient of $t^7$.
196
Taking differences, we have that
197
\begin{verbatim}
198
0 = (-4*s1 - d1 + r2)*t^5 + (-12*s1^2 - 4*s1*d1 + 4*s1*r2 - d2 +
199
r3)*t^6 + (-8*s1^3 - 6*s1^2*d1 - 24*s1*s2 - 4*s1*d2 + s1*r2^2
200
+ 4*s1*r3 - 4*s2*d1 + 12*s2*r2 + 8*s3 - d3 + r4)*t^7 +
201
(-2*s1^4 - 4*s1^3*d1 - 24*s1^2*s2 - 6*s1^2*d2 - 12*s1*s2*d1 -
202
24*s1*s3 - 4*s1*d3 + 2*s1*r2*r3 + 4*s1*r4 - 12*s2^2 - 4*s2*d2
203
+ 6*s2*r2^2 + 12*s2*r3 - 4*s3*d1 + 32*s3*r2 + 24*s4 - d4 +
204
r5)*t^8 + O(t^9)
205
\end{verbatim}
206
Thus
207
$$
208
-8 s_1^3 - 6 s_1^2 d_1 - 24 s_1 s_2 - 4 s_1 d_2 + s_1 r_2^2
209
+ 4 s_1 r_3 - 4 s_2 d_1 + 12 s_2 r_2 + 8 s_3 - d_3 + r_4 = 0.
210
$$
211
Unfortunately we have both $s_1$ and $s_3$ together
212
in this formula, so this doesn't determine $s_2$.
213
Using $m=-1$ we get
214
\begin{verbatim}
215
0 = (2*s1 - d1 + r2)*t^2 + (-s1*d1 - 2*s1*r2 - d2 + r3)*t^3 + (-s1*d2
216
+ s1*r2^2 - 2*s1*r3 - s2*d1 + 3*s2*r2 + 2*s3 - d3 + r4)*t^4 +
217
(-s1*d3 + 2*s1*r2*r3 - 2*s1*r4 - s2*d2 - 3*s2*r2^2 + 3*s2*r3
218
- s3*d1 - 4*s3*r2 - d4 + r5)*t^5 + O(t^6).
219
\end{verbatim}
220
The coefficient of $t^4$ gives a different linear equation for $s_2$
221
and $s_3$, and probably taking two of these together we'll be able to
222
solve for both.
223
224
For $m=3$ it's:
225
\begin{verbatim}
226
(-18*s1 - d1 + r2)*t^10 + (-108*s1^2 - 9*s1*d1 + 6*s1*r2 - d2 +
227
r3)*t^11 + (-252*s1^3 - 36*s1^2*d1 - 216*s1*s2 - 9*s1*d2 +
228
s1*r2^2 + 6*s1*r3 - 9*s2*d1 + 27*s2*r2 + 54*s3 - d3 +
229
r4)*t^12 + (-378*s1^4 - 84*s1^3*d1 - 756*s1^2*s2 - 36*s1^2*d2
230
- 72*s1*s2*d1 - 216*s1*s3 - 9*s1*d3 + 2*s1*r2*r3 + 6*s1*r4 -
231
108*s2^2 - 9*s2*d2 + 9*s2*r2^2 + 27*s2*r3 - 9*s3*d1 +
232
108*s3*r2 + 216*s4 - d4 + r5)*t^13 + O(t^14)
233
\end{verbatim}
234
235
\subsection{Computing formal multiplication by $p$}
236
237
\subsection{Computing division polynomials}
238
239
\section{The $p$-adic cyclotomic height}
240
Let $E$ and $p$ be as in the introduction.
241
242
Suppose $Q \in E(\Q)$ is a point that reduces to the point at infinity
243
modulo $p$, and lies in the identity component modulo every bad prime.
244
For example, if $M$ is the least common multiple of $\#E(\F_p)$ and
245
the Tamagawa numbers $c_\ell$ of $E$ at bad primes, then every element
246
of $M E(\Q)$ satisfies our hypothesis on $Q$.
247
Since ...??, we can write
248
$$
249
Q=(x,y)=\left(\frac{a}{e^2},\frac{b}{e^3}\right),
250
$$
251
with $a,b,e\in\Z$ and $\gcd(a,e)=1$. Let $t=-x/y$. Then the $p$-adic
252
cyclotomic height $h_p(Q)$ is
253
\begin{equation}\label{eqn:hpq}
254
h_p(Q) = \log_p\left(\frac{\sigma(t)}{e}\right).
255
\end{equation}
256
[[I think exactly this formula appears on the upper right page of a preprint
257
Tate was holding yesterday, which perhaps I don't have a copy.]]
258
Here
259
$$
260
\log_p : \Q_p^* \to \Q_p
261
$$
262
is the the unique extension of the usual $p$-adic logarithm
263
on $1+p\Z_p$ to a homomorphism from the multiplicative group
264
of $\Q_p^*$ to the additive group of $\Q_p$. Thus if $au \in \Q_p^*$
265
with $u\in 1 + p\Z_p$, then $\log_p(au) = \log_p(u)$, and
266
one computes $\log_p(u)$ using the formula
267
$$
268
\log_p(1-x) = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots\right)
269
$$
270
with $x=1-u$.
271
[[Again, there's a discussion of this in a paper Tate was holding,
272
that I don't seem to have.]]
273
274
The cyclotomic height $h_p(Q)$ extends uniquely to a $\Z_p$-valued
275
function on $E(\Q)$ that satisfies $h_p(nQ) = n^2 Q$ for all
276
integers $n$. If $M$ is as above, then
277
we can compute $h_p(P)$ for any $P\in E(\Q)$ by computing
278
$$
279
h_p(P) = \frac{h_p([M]P)}{M^2} = \frac{1}{M^2}\cdot
280
\log_p\left(\frac{\sigma(t)}{e}\right),
281
$$
282
where $t=-x/y$ and $e$ are for $Q=[M]P=(x,y)=(a/e^2,b/e^3)$.
283
284
285
286
\bibliography{biblio}
287
\end{document}
288
289
290
Using $m=2$ still,
291
taking the difference of the two sides and setting the
292
$s_{2n}=0$, since $\sigma$ is odd, we see that
293
\begin{verbatim}
294
0 = (-d1 + r2)*t^5 + (-d2 + r3)*t^6 + (-4*s2*d1 + 12*s2*r2 - d3 +
295
r4)*t^7 + (-12*s2^2 - 4*s2*d2 + 6*s2*r2^2 + 12*s2*r3 + 24*s4
296
- d4 + r5)*t^8 + O(t^9)
297
\end{verbatim}
298
\comment{
299
> K<s1,s2,s3,s4,s5,d1,d2,d3,d4,d5,r2,r3,r4,r5> := FieldOfFractions(PolynomialRing(Q,14));
300
> R<t> := LaurentSeriesRing(K);
301
> sig := t + 0*t^2 + s2*t^3 + 0*t^4 + s4*t^5 + 0*t^6 + O(t^7);
302
> m:=2; mult := m*t + r2*t^2+r3*t^3+r4*t^4+r5*t^5 + O(t^6); divpol := m+d1*t+d2*t^2+d3*t^3+d4*t^4+d5*t^5+O(t^6);t^(m^2-1)*Evaluate(sig,mult) - sig^(m^2)*divpol;
303
(-d1 + r2)*t^5 + (-d2 + r3)*t^6 + (-4*s2*d1 + 12*s2*r2 - d3 +
304
r4)*t^7 + (-12*s2^2 - 4*s2*d2 + 6*s2*r2^2 + 12*s2*r3 + 24*s4
305
- d4 + r5)*t^8 + O(t^9)
306
}
307
The coefficient of $t^7$ determines $s_2$, so long as
308
$r_2 - 3d_1\neq 0$. Since $r_2 = d_1$, we're OK, at least as
309
long as $d_1\neq 0$.
310
So we get
311
$$
312
s_2 = \frac{d_3-r_4}{8d_1}.
313
$$
314
Now that we know $s_2$, the coefficient of $t^8$ determines
315
$s_4$. Etc., and it seems likely we can easily determine $\sigma(t)$
316
from the formal $[2]$ and $f_2(t)$, without having to mess with
317
any differential equations. Surprising if right.
318
319
Here is a relevant MAGMA session, which suggests that the
320
information coming from (\ref{eqn:sigma_ma}) for different $m$ is
321
equivalent:
322
323
\begin{verbatim}
324
> K<s1,s2,s3,s4,s5,d1,d2,d3,d4,d5,r2,r3,r4,r5> :=
325
FieldOfFractions(PolynomialRing(Q,14));
326
> R<t> := LaurentSeriesRing(K);
327
> sig := t + 0*t^2 + s2*t^3 + 0*t^4 + s4*t^5 + 0*t^6 + O(t^7);
328
> m:=2; mult := m*t + r2*t^2+r3*t^3+r4*t^4+r5*t^5 + O(t^6);
329
divpol := m+d1*t+d2*t^2+d3*t^3+d4*t^4+d5*t^5+O(t^6);
330
t^(m^2-1)*Evaluate(sig,mult) - sig^(m^2)*divpol;
331
(-d1 + r2)*t^5 + (-d2 + r3)*t^6 + (-4*s2*d1 + 12*s2*r2 - d3 +
332
r4)*t^7 + (-12*s2^2 - 4*s2*d2 + 6*s2*r2^2 + 12*s2*r3 + 24*s4
333
- d4 + r5)*t^8 + O(t^9)
334
> m:=3; mult := m*t + r2*t^2+r3*t^3+r4*t^4+r5*t^5 + O(t^6);
335
divpol := m+d1*t+d2*t^2+d3*t^3+d4*t^4+d5*t^5+O(t^6);
336
t^(m^2-1)*Evaluate(sig,mult) - sig^(m^2)*divpol;
337
(-d1 + r2)*t^10 + (-d2 + r3)*t^11 + (-9*s2*d1 + 27*s2*r2 - d3 +
338
r4)*t^12 + (-108*s2^2 - 9*s2*d2 + 9*s2*r2^2 + 27*s2*r3 +
339
216*s4 - d4 + r5)*t^13 + O(t^14)
340
> m:=5; mult := m*t + r2*t^2+r3*t^3+r4*t^4+r5*t^5 + O(t^6);
341
divpol := m+d1*t+d2*t^2+d3*t^3+d4*t^4+d5*t^5+O(t^6);
342
t^(m^2-1)*Evaluate(sig,mult) - sig^(m^2)*divpol;
343
(-d1 + r2)*t^26 + (-d2 + r3)*t^27 + (-25*s2*d1 + 75*s2*r2 - d3 +
344
r4)*t^28 + (-1500*s2^2 - 25*s2*d2 + 15*s2*r2^2 + 75*s2*r3 +
345
3000*s4 - d4 + r5)*t^29 + O(t^30)
346
> m:=1; mult := m*t + r2*t^2+r3*t^3+r4*t^4+r5*t^5 + O(t^6);
347
divpol := m+d1*t+d2*t^2+d3*t^3+d4*t^4+d5*t^5+O(t^6);
348
t^(m^2-1)*Evaluate(sig,mult) - sig^(m^2)*divpol;
349
(-d1 + r2)*t^2 + (-d2 + r3)*t^3 + (-s2*d1 + 3*s2*r2 - d3 +
350
r4)*t^4 + (-s2*d2 + 3*s2*r2^2 + 3*s2*r3 - d4 + r5)*t^5 +
351
O(t^6)
352
\end{verbatim}
353
Thankfully, we can't use $m=1$ since $d_1=0$ there.
354