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\title{Computing a Heegner Point $p$-adic Height,\\(motivated
  by a conjecture of Mazur-Rubin)}

\author{Jennifer Balakrishnan, Mirela \c{C}iperiani, and William Stein}


\begin{document}
\maketitle

\begin{abstract} 
\end{abstract}

\section{Introduction}\label{sec:intro}

\section{Example (first layer)}

Let $p=5$, let $E$ be the elliptic curve 57a1,
and let $K=\Q(\sqrt{-2})$.
Let $K[c]$ denote the ring class field extension of
$K$ associated to the conductor $c$.
Let $x_{25} \in X_0(57)(K[{25}])$ be the Heegner
point for $c=25$ and via the modular parametrization $\pi: X_0(57) \ra E$, let 
$y_{25} = \pi(x_{25}) \in E(K[{25}])$ be its
image.  

We have $K[1] = K$ (i.e., class number 1), and 
$$
 \Gal(K[{25}]/K) \isom (\cO_K/25\cO_K)^* / (\Z/25\Z)^* \ncisom \P^1(\Z/25\Z),
$$
which is cyclic of order $5\cdot 6$.   Thus we have the tower 

$$
  K \subset K_1 \subset K[{25}]
$$

fitting into the following diagram

\begin{displaymath}
    \xymatrix{ 
     & K[{25}] \ar@{-}[dl]_5 \ar@{-}[dr]^6 & \\
     K[{5}] \ar@{-}[dr]_5 & & K_{1} \ar@{-}[dl]^5\\
     & K \ar@{-}[d]_2 & \\
     & \Q & \\
     }
    %A \ar[r] & B \ar[d] \\
       %        D \ar[u] & C \ar[l] }
\end{displaymath}



where $K_1/K$ is cyclic of degree $5$ and $K[{25}]/K_1$ is cyclic of
degree $6$.

Let $\sigma$ be a generator of $\Gal(K[{25}]/K_1)$, so $\sigma$ is of order $6$.
The points we're interested in are
$$
 z = \Tr_{K[{25}]/K_1}(y_{25}) = \sum_{i=0}^5 \sigma^i(y_{25}),
$$
and its Galois conjugates.    

\begin{question}[Mazur]
What are the 5-adic heights (i.e., mod $5^n$, with
$n$ not too large being perfectly OK for us, and here we mean  standard $p$-adic
heights---related to the $p$-cyclotomic extension) of the conjugates of $z$?
\end{question}


\subsection{Computing $z$}

How this is done:
\begin{enumerate}

\item Using the fact that conjugates of $x_{c}$ correspond to binary quadratic forms of discriminant
$\disc(O_c)$, we begin by enumerating the 30 primitive binary quadratic forms up to equivalence
  of discriminant $\disc(\O_c) = D \cdot{} c^2 = -8\cdot 5^4$
that correspond to Heegner points of this conductor, i.e., {\em for which
$N\mid A$ and $\gcd(A/N,B,CN)=1$ (see prop 2.2 of \cite{watkins:heegner}). }
\textbf{Actually, this needs William's generalization of Prop 2.2 of Watkins.}
\item For each of these 30 primitive binary quadratic forms $Q=Ax^2+Bxy+Cy^2$, we compute
the ideal $I=(A, (-B+c\sqrt{D})/2)\O_K$ and find a generator $\alpha_Q$
of this principal ideal.  Then we compute the multiplicative order
of the class of $\alpha_Q$ in $(\cO_K/25\cO_K)^* / (\Z/25\Z)^*$.  We
find an element with multiplicative order $6$. 
\item Now we have $\alpha = \alpha_Q$ whose class has multiplicative order $6$, so
it corresponds to the $\sigma$ of order $6$ chosen above.  
To compute $\sigma^i(y_{25})$ for $i=0,\ldots, 5$ we proceed as follows:
\begin{enumerate}
\item Binary quadratic forms correspond to conjugates of $x_{25}$ as follows.
The form $f=Ax^2 + Bxy + Cy^2$ corresponding to $[\tau_f] \in X_0(N)$, where
$f(\tau_f,1)=0$ and $\tau_f\in\h$; explicitly, $\tau_f = (-B + c\sqrt{D})/(2A)$.

\item Choose any $[\alpha_0] \in (\cO_K/25\cO_K)^* / (\Z/25\Z)^*$ corresponding
to some $\sigma_0 \in G$, and let $y_{25}' =\sigma(y_{25})$.

\item Then $\sigma^i(y_{25}')$ is the point corresponding to $[\alpha^i \alpha_0]$.
Since our group is abelian, we can thus compute a conjugate of
$z$ by computing 
$$
  \sum \sigma^i(y_{25}') = \sum \sigma^i \sigma_0(y_{25}) = \sigma_0(z),
$$
and that is good enough for our application. 

\end{enumerate}

\item Recognize the $x$-coordinate of $z \in E(\C)$ using algdep, and
  get a $y$-coordinate.  Make sure that $z$ is defined over
  a Galois dihedral field of degree $10$ that is ramified exactly at
  $2$ and $5$, by checking that discriminant of defining poly of $x$
  coordinate is divisible by $2$ and $5$ and the coprime to $10$ part
  is a perfect square.

%\item This didn't work (with 1000 bits).  But we can instead
%directly numerically compute $z$ and all of its conjugates, say
%$z_0, \ldots, z_4$, then consider the polynomial
%$
% F(X) = \prod_{i=0}^4 (X - x(z_i)) \in K[X]
%$$
%and also
%$$
% G(X) = \prod_{i=0}^4 (X - x(z_i))\cdot \prod_{i=0}^4(X-\overline{x(z_i)}) \in \Q[X].
%$$
%Let's see what we get.  We can also compute
%$$
 % \gamma = \sum_{i=0}^4 z_i \in E(K) = E(\Q) \ncisom \Z,
%$$
%which should be exciting.   In fact, we must (by the ``distribution relation'') get
%$$
%  \gamma = \Tr_{K_{25}/K_1}(y_{25}) = a_{25} y_K = 4 y_K = 8 g ,
%$$
%where $y_K$ is the base Heegner point and $g$ is a generator for $E(\Q)$... probably.
%So probably $x$-coordinate $2$ or $1$ or $0$ or $13/9$ or $-1976/7569$...?
\end{enumerate}


%After a lot of work, we found that 

The $x$-coordinate
of $z$ satisfies
$$
18034072681 x^{5} - 126430131580 x^{4} + 352783410220 x^{3} -
489834319200 x^{2} + 338504989540 x - 93144838864.
$$
The point $z$ is defined over $$L:= \Q(b_3),$$ where $b_3$ is a root of
$$
x^{10} - 10 x^{8} - 20 x^{7} + 165 x^{6} - 12 x^{5} - 760 x^{3} + 2220
x^{2} + 5280 x + 7744.
$$
(This is not a {\bf theorem} because there are no proven error
bounds on our computations; but it's inconceivable that this is
wrong.)

Explicitly, we will compute with $z$ with $x$-coordinate
\begin{align*}
x(z) &= \frac{96698852571685}{2145672615243325696} b_{3}^{9} + \frac{2472249905907}{195061146840302336} b_{3}^{8} + \frac{916693155514421}{2145672615243325696} b_{3}^{7} + \frac{1348520950997779}{2145672615243325696} b_{3}^{6}\\
& \quad  - \frac{82344497086595}{12191321677518896} b_{3}^{5} + \frac{2627122040194919}{536418153810831424} b_{3}^{4}  - \frac{452199105143745}{48765286710075584} b_{3}^{3} \\
& \quad+ \frac{4317002771457621}{536418153810831424} b_{3}^{2} + \frac{2050725777454935}{67052269226353928} b_{3} + \frac{3711967683469209}{3047830419379724} .\end{align*}


\subsection{Computing $p$-adic Heights}

By the work of Mazur-Stein-Tate \cite{mazur-stein-tate:padic}, we have that the cyclotomic $p$-adic height of $P$ can be computed as
$$
  h_p(P) = {\frac{1}{p}} \cdot \left(\sum_{v\mid p} 
     \log_p(N_{K_v/\Q_p}(\sigma_v(P))) - 
      \sum_{w\nmid p}\ord_w(d_w(P)) \cdot \log_p(\#k_w)\right).$$

We will return to this formula later\footnote{TODO: It'd be another good numerical consistency check to compute the height directly this way...}, but for now, we use the following:

If
we assume that $P$ lies in a sufficiently small (finite index)
subgroup of $E(K)$ (see \cite[Prop.~2]{wuthrich:heightfamily}), 
then there will be a global choice of denominator
$d(P)$, and the formula simplifies to
\begin{equation}\label{height-short}
  h_p(P) = \frac{1}{p} \cdot \log_p\left( \prod_{v\mid p} N_{K_v/\Q_p}\left(\frac{\sigma_v(P)}{d(P)}
\right) \right).\end{equation}

More precisely, since our point $z$ does not necessarily lie in this sufficiently small subgroup of $E(K)$, we compute an appropriate $n$ such that $nz$ does lie in the subgroup, compute the height of $nz$ by \eqref{height-short}, then use the fact that the height pairing is a quadratic form to recover the height of our original point.


We begin by computing the denominator $d(nz)$ of $nz$, which in turn requires that we compute the denominator $d:=d(z)$ of $z$. Note that this requires that we work over a PID, since we wish to write each coordinate of $z$ as the ratio of two integral elements.  The goal then is to write $z = (x,y)$ as $(\frac{a}{d^2}, \frac{b}{d^3}),$ such that $(a,d) = (b,d) = 1$ and $a,b,d \in \cO_L$.
\subsubsection{Computing $d(z)$ }


\begin{enumerate}\item We start with $z = (x,y)$ as a point on the elliptic curve, where $x,y \in L$. We begin by writing the ideal $I$ generated by $(x,y)$, and compute its GCD by writing it as a principal ideal $I = (g)$ and reading off the generator $g$. In our case, 
\begin{align*}g &=
\frac{123984834273901806131}{45238689824261707813521152} b_3^{9} +
\frac{18086333051389614235}{4112608165841973437592832} b_{3}^{8} \\
& \quad -
\frac{2034338252029802812739}{45238689824261707813521152} b_{3}^{7}  -
\frac{1210137739879154041477}{45238689824261707813521152} b_{3}^{6} \\
& \quad +
\frac{45162732628910301745}{257038010365123339849552} b_{3}^{5} +
\frac{20860129147970487107559}{11309672456065426953380288} b_{3}^{4} \\
& \quad -
\frac{5232862309605254732233}{1028152041460493359398208} b_{3}^{3} +
\frac{73696187951449076337469}{11309672456065426953380288} b_{3}^{2} \\
& \quad -
\frac{11829570856332514083089}{1413709057008178369172536} b_{3} +
\frac{2848093589238986978981}{64259502591280834962388}.\end{align*}

\item Read off the integer denominator $d(g)$ of $g$, and re-express the coordinates of $z$ in terms of a numerator (ideal) over this integer denominator (ideal). In our example, $$d(g) = 45238689824261707813521152.$$
\item Simplify the coordinates of $z$ by canceling common prime ideals in the numerator and denominator ideals. 
\item What is left in the factored denominator ideal is a perfect cube of prime ideals in the ring of integers of $L$, and a cube root of this is the desired denominator $d(z)$. In our example, we find that \begin{align*} d(z) &= -\frac{170066107}{18679674112} b_{3}^{9} + \frac{46616573}{1698152192}
b_{3}^{8} + \frac{3760482603}{18679674112} b_{3}^{7} -
\frac{11188479427}{18679674112} b_{3}^{6} - \frac{263947335}{106134512}
b_{3}^{5} \\
& \quad  + \frac{40187214425}{4669918528} b_{3}^{4} +
\frac{1074830385}{424538048} b_{3}^{3} - \frac{67626028101}{4669918528}
b_{3}^{2}  - \frac{15616668599}{583739816} b_{3} +
\frac{738093651}{26533628}.\end{align*}\end{enumerate}

%for our example, we computed everything for $3P, 9P...$

\subsubsection{Computing $d(nz)$}

One could repeat the above process for $nz$, but this would be slightly tedious due to the numerical explosion in the coordinates of $nz$. Instead, we make use of $n$-division polynomials to write $d(nz)$ in terms of $d(z)$.

\begin{lemma} Let $f_n$ be the  $n$th division polynomial of $E$.
Then $$d(nz) = f_n(x(z)) d(z)^{n^2}.$$ \end{lemma}

%Note that the exponent of $d(z)$ is determined by the degree of $f_n$ and the fact the the exponent of $x$ in the numerator of $x(nz)$ equals one more than the exponent of $x$ in the denominator.

\subsubsection{An algorithm for the height of $z$}

\begin{enumerate}
\item Compute $z$.
\item Compute $n$ and $nz$. This $n$ is a positive integer such that $nz$ reduces\footnote{depending on which is relevant; since $p$ is inert, this amounts to how complex conjugation $\tau \in \Gal(\F_{p^2}/\F_p)$ acts.  In our case, since $E$ has rank 1, $\eps = -1$, and $(nz)^{\tau} = -\eps nz  = nz$. So $nz \in E(\F_p)$.} to $\cO \in E(\F_p)$ (resp. $E(\F_{p^2})$) and to the connected component of $\mathcal{E}_{\F_{\ell}}$ at all bad primes $\ell$. If the torsion subgroup of $E$ is trivial, then as $P$ is a Heegner point, by a theorem of Gross-Zagier, this second condition is automatic, so $n$ can be taken to be $\#E(\F_p)$ (resp. $\#E(\F_{p^2})$). (Note that in practice, one wants $n$ to be as small as possible, to reduce numerical explosion in the coordinates.)
\item Compute $d(z)$ and $d(nz).$ 
%\textbf{We should say more here about computing $n$. See remark below.}
\item Compute $\sigma_v(t)$ the $v$-adic $\sigma$ function above $p$ as in \cite{mazur-stein-tate:padic}. Note that $p$ is \emph{a priori} totally ramified in $L$, so there is just one $v$ above $p$.
\item Let $t_n = -\frac{x(nz)}{y(nz)}$. Evaluate $\sigma_v(t_n)$ and compute  $$h_p(z) = \frac{1}{n^2 p} \log_p\left(N_{K_v/\Q_p}\left(\frac{\sigma_v(t_n)}{d(nz)}\right)\right).$$
\end{enumerate}

%\begin{remark}A note on finding ``appropriate'' $n$: what I  (Jen) did was to look at $x_n := x(nz) \in L$ for small $n$, and select those $x_n$ whose ``constant'' term (when considered as a polynomial in $b_3$)  had positive $5$-adic valuation. This seemed to work in practice, since this ensured that $\sigma(t_n)$ converged. However, I'm not sure if this is equivalent to the finite index criterion in Wuthrich's paper. We should check this. \end{remark}

For our example, we may take $n = 3$. Indeed, the point $z$ reduces to $(4,1) \in E(\F_5)$, and we see that $3z = \cO \in E(\F_5)$.
 
Moreover, we have \begin{align*}\sigma_5(t) &= \left(1 + O(5^{10})\right)t + O(5^{9})t^{2} + \left(1 + 2
\cdot 5 + 2 \cdot 5^{3} + 4 \cdot 5^{5} + 4 \cdot 5^{6} +
O(5^{8})\right)t^{3} \\
&\quad + \left(3 + 2 \cdot 5 + 2 \cdot 5^{2} + 2 \cdot 5^{3} + 2 \cdot 5^{4} + 2 \cdot 5^{5} + 2 \cdot 5^{6} +
O(5^{7})\right)t^{4} \\
&\quad + \left(1 + 4 \cdot 5 + 3 \cdot 5^{3} + 4 \cdot
5^{4} + O(5^{6})\right)t^{5} + \left(1 + 3 \cdot 5 + 3 \cdot 5^{3} +
O(5^{5})\right)t^{6} \\
&\quad + \left(5^{2} + 5^{3} + O(5^{4})\right)t^{7} +
\left(3 + 4 \cdot 5 + 3 \cdot 5^{2} + O(5^{3})\right)t^{8} \\
& \quad + \left(2 + 2 \cdot 5 + O(5^{2})\right)t^{9} + \left(1 + O(5)\right)t^{10} + O(t^{11}).\end{align*}



%\begin{align*}t_3 &=        	
%-\frac{10519334895626295182081280398573415348690061530280150146193927410\
%841491396912151575517574133173060624004209572132952146984742206159786176\
%4800496957}{200252342634655778416630819200467164870533140916732894868509\
%689008533280176549546063274139462951606616836040400540029170007579963229\
%057094037246864896} b_{3}^{9}\\
%&\quad +
%\frac{337130444092975382194991665226437910999932571527943119566686142783\
%279551688616461475676886732361670623977008938000082335201156995796368031\
%171245}{2600679774476049070345854794811261881435495336580946686604021935\
%175756885409734364458105707311059826192675849357662716493604934587390351\
%870613595648} b_{3}^{8} \\
%&\quad +
%\frac{115702301484323578450742738154959274574242945701330104866215101531\
%736013472722015938029867849388563643409293670101929506160114254586623516\
%0723562141}{200252342634655778416630819200467164870533140916732894868509\
%689008533280176549546063274139462951606616836040400540029170007579963229\
%057094037246864896} b_{3}^{7} \\
%&\quad+
%\frac{194167403319272189875830428189295256549605491292749128740206633909\
%650217734250480984083386574402301721856997907945342224856742148683075138\
%189842029}{2860747751923653977380440274292388069579044870239041355264424\
%128693332573950707800903916278042165808811943434293428988142965428046129\
%3870576749552128} b_{3}^{6} \\
%&\quad -
%\frac{253393049633462725703811373650337277939970879208998571374433271642\
%051345578781185973690726064802929290573453984953597074253895910394975265\
%09586563}{28444935033331786706907786818248176828200730243854104384731489\
%915984840934168969611260531173714716848982392102349435961648803972049581\
%9735848362024} b_{3}^{5} \\
%&\quad +
%\frac{245064680754611936765249426415583959526680025514414288269657010847\
%169063975929612577717606678043093938665140642369784398489795709357827383\
%6216874095}{500630856586639446041577048001167912176332852291832237171274\
%222521333200441373865158185348657379016542090101001350072925018949908072\
%64273509311716224} b_{3}^{4} \\
%&\quad +
%\frac{482582062329423326594612516059644227722543127283559831691886930303\
%688315679965494888950316086371282234756683000721565367465848696113857612\
%088354743}{4551189605333085873105245890919708292512116839016656701557038\
%386557574549467035137801684987794354695837182736375909753863808635527933\
%115773573792384} b_{3}^{3}\\
%&\quad  -
%\frac{508475213343352584876671291004662118907545500225303555780404434487\
%079275171643050440778750015788938788415287444829140388222700700814077250\
%5908176179}{500630856586639446041577048001167912176332852291832237171274\
%222521333200441373865158185348657379016542090101001350072925018949908072\
%64273509311716224} b_{3}^{2}\\
%&\quad -
%\frac{665682350830381456544113115695728530120538903379710358080220551717\
%479097571794107631654338078647743451103491422614927145117492488848385251\
%193376981}{8939836724761418679313875857163712717434515219497004235201325\
%402166664293595961877824738368881768152537323232166965587946766962644154\
%33455523423504} b_{3}\\
%&\quad -
%\frac{577597748406133949841855661490540409013473678192766389872775387492\
%569418590936994705653339950969026216602096967861339521338861966493860666\
%565574595}{2844493503333178670690778681824817682820073024385410438473148\
%991598484093416896961126053117371471684898239210234943596164880397204958\
%19735848362024},\end{align*}\end{table} 
Then substituting the appropriate parameters into \eqref{height-short}, we find that

$$h_5(z) = 2 + 4 \cdot 5 + 2 \cdot 5^{2} + 2 \cdot 5^{3} + 5^{4} + 3 \cdot 5^{5} + O(5^6).$$

Repeating the computation with $n = 9$ serves as a numerical consistency check.
% so now we look at appropriate n and compute $\sigma(nP)$ and $d(nP)$ to recover the desired height.
%we compute $\sigma_v(t)$ the $p$-adic sigma function as in M_1ST (note that while the sigma function itself is a power series with coefficients in $Q_p$, it is evaluated at an element %in an extension of $Q_p$, so the computation has to be done in this extension of $Q_p$, which M_1agma doesn't know  it can handle...what one does is create $Q_p[x]/$(polynomial %defining the extension) and then work in this ring ... also, division is sketchy in this ring, so instead of computing norm $(\sigma/d)$, compute $norm(\sigma)/norm(d)$

\section{Example: twisting}

We take the previous example and take a quadratic twist by $\sqrt{41}$. This gives us the elliptic curve $$E_{41}  :  y^2 + y = x^3 + x^2 - 3922x + 102712.$$ We have that $E_{41}$ has rank 1 over $\Q$ as well as over $K$. A generator $P$ of $E_{41}$ has height divisible by $p=5$ as well. So we repeat the computation as above. 

We find that a traced Heegner point $z$  in $K_5$ has $x$-coordinate

\begin{align*}
...
\end{align*}

with denominator $d(z) = $.

As $z$ reduces to $ $ in $E(\F_5)$, we have that $n =  $, and we use this to compute

$$h(z) = .$$

\section{Example (second layer)}

We could try to compute with the next layer of the anti-cyclotomic tower. 

For convenience, here are the relevant fields in the tower along with their relative degrees:

\begin{displaymath}
    \xymatrix{ 
     & K[{125}] \ar@{-}[dl]_5 \ar@{-}[dr]^6 & \\
     K[{25}] \ar@{-}[d]_5 \ar@{-}[drr] _6& & K_{2} \ar@{-}[d]^5\\
     K[5] \ar@{-}[dr]_6 & & K_1 \ar@{-}[dl]^5 \\
     & K \ar@{-}[d]_2 & \\
     & \Q & \\
     }
    %A \ar[r] & B \ar[d] \\
       %        D \ar[u] & C \ar[l] }
\end{displaymath}

The idea is very similar to before: we use a generator of $\Gal(K[{125}]/K_{2})$ (again, of order 6) to compute the trace of $y_{125}$ from $K[125]$ to $K_{2}$, then use algebraic dependency to recognize the extension $K_{2}/K$, which we then write as a degree 50 extension over $\Q$.


\begin{remark} We have gotten as far as finding the $x$-coordinate of the Heegner point as an element of a field that has the right properties (degree 50 over $\Q$ and with discriminant divisible by $(\disc(K_1/\Q))^5$)\end{remark}

\section{Example: fixing $E,p$, varying $K$}
Let $E$ be the elliptic curve 89a1, and fix $p = 5$. We have the fields with discriminants $D = -8,-11,-67$ satisfying the necessary hypotheses. 

We have that the height of a Mordell-Weil generator $P$ is

$$h_p(P) = 2 \cdot 5 + 5^{2} + 3 \cdot 5^{3} + 2 \cdot 5^{4} + 3 \cdot 5^{5} + 3 \cdot 5^{6} + 4 \cdot 5^{8} + O(5^{10}).$$
\subsection{$D = -8$}
$$K_1 =  \Q[b_{3}]/(b_{3}^{10} - 10 b_{3}^{8} - 20 b_{3}^{7} + 165 b_{3}^{6} - 12 b_{3}^{5} - 760 b_{3}^{3} + 2220 b_{3}^{2} + 5280 b_{3} + 7744)$$

\begin{align*}d(z) &= -\frac{184960682133}{583739816} b_{3}^{9} + \frac{201372405891}{106134512} b_{3}^{8} - \frac{4427090691257}{1167479632} b_{3}^{7} + \frac{2576122895483}{1167479632} b_{3}^{6} - \frac{1047171187695}{106134512} b_{3}^{5} \\
&\quad + \frac{3808621305257}{145934954} b_{3}^{4} - \frac{402890182961}{26533628} b_{3}^{3} - \frac{13515253550229}{291869908} b_{3}^{2} - \frac{50649651812995}{291869908} b_{3} - \frac{123445723611}{6633407}\end{align*}

$$z = \left(\frac{1802706589}{409822531712} b_{3}^{9} - \frac{64304775}{37256593792} b_{3}^{8} - \frac{20879156681}{409822531712} b_{3}^{7} - \frac{20852393943}{409822531712} b_{3}^{6} + \frac{7470954657}{9314148448} b_{3}^{5} - \frac{38479444339}{102455632928} b_{3}^{4} - \frac{11381532251}{9314148448} b_{3}^{3} + \frac{27684625039}{102455632928} b_{3}^{2} + \frac{154372580447}{25613908232} b_{3} + \frac{16052722259}{582134278} : \frac{120010271549}{6184594569472} b_{3}^{9} - \frac{40411852593}{6184594569472} b_{3}^{8} - \frac{1382035611513}{6184594569472} b_{3}^{7} - \frac{1480634228779}{6184594569472} b_{3}^{6} + \frac{492283188531}{140558967488} b_{3}^{5} - \frac{2271125318647}{1546148642368} b_{3}^{4} - \frac{7872707189561}{1546148642368} b_{3}^{3} + \frac{766000006683}{1546148642368} b_{3}^{2} + \frac{9448457886457}{386537160592} b_{3} + \frac{282711261904}{2196233867} : 1\right)$$

which reduces to $(0,0)$ in $E(\F_5)$, and this point has order $7$. We use $n=7,14$ to compute the $p$-adic height.

$$h_p(z) = 4 + 3 \cdot 5 + 4 \cdot 5^{2} + 5^{3} + 4 \cdot 5^{4} + 5^{5} + 5^{7} + 4 \cdot 5^{8} + O(5^9).$$


%Let $E$ be the elliptic curve 83a1, and fix $p = 5$. We have the fields with discriminants $D = -8,-19,-43,-67$ satisfying the necessary hypotheses.
%
%\subsection{$D = -8$}
%We begin with $D = -8$
%
%$$K_1 =  \Q[b_{3}]/(b_{3}^{10} - 10 b_{3}^{8} - 20 b_{3}^{7} + 165 b_{3}^{6} - 12 b_{3}^{5} - 760 b_{3}^{3} + 2220 b_{3}^{2} + 5280 b_{3} + 7744)$$
%
%\begin{align*}d(z) &= \frac{33566043}{118978816} b_{3}^{9} - \frac{1613965}{10816256} b_{3}^{8} - \frac{399252123}{118978816} b_{3}^{7} - \frac{332992397}{118978816} b_{3}^{6} + \frac{18099753}{338008} b_{3}^{5}\\
%&\quad  - \frac{813842121}{29744704} b_{3}^{4}- \frac{250659857}{2704064} b_{3}^{3} + \frac{558702965}{29744704} b_{3}^{2} + \frac{1591596313}{3718088} b_{3} + \frac{348422849}{169004}.\end{align*}
%
%\begin{align*}x(z) &=  \frac{4386450218107993}{210911893968651392} b_{3}^{9} - \frac{136813195828847}{19173808542604672} b_{3}^{8} - \frac{50610077277194873}{210911893968651392} b_{3}^{7}\\
%&\quad - \frac{53727126409126639}{210911893968651392} b_{3}^{6} + \frac{2254543811196519}{599181516956396} b_{3}^{5} - \frac{83728439142665323}{52727973492162848} b_{3}^{4} \\
%&\quad - \frac{26627832386947707}{4793452135651168} b_{3}^{3} + \frac{32156371185457271}{52727973492162848} b_{3}^{2} + \frac{175644837822162471}{6590996686520356} b_{3} + \frac{41120251788665117}{299590758478198} ,\end{align*}
%\begin{align*}y(z) &= \frac{5805379138118797982725}{25116653805150819667712} b_{3}^{9} - \frac{170995738038296944211}{2283332164104619969792} b_{3}^{8} - \frac{67057155215545541401221}{25116653805150819667712} b_{3}^{7}\\
%&\quad  - \frac{72446741059925791814867}{25116653805150819667712} b_{3}^{6} + \frac{2981233818896545830713}{71354130128269374056} b_{3}^{5} - \frac{105634918359579856036279}{6279163451287704916928} b_{3}^{4}\\
%&\quad - \frac{35796027123815537655695}{570833041026154992448} b_{3}^{3} + \frac{37341446633524392065195}{6279163451287704916928} b_{3}^{2} + \frac{234429069751103991398815}{784895431410963114616} b_{3}\\
%&\quad + \frac{54634123366300964263703}{35677065064134687028}\end{align*}
%
%which reduces to $(1,1)$ in $E(\F_5)$, and this point has order 4. Thus we use $n  = 4$ to compute the $p$-adic height.
%
%$$h_p(z) =  2 + 3 \cdot 5 + 3 \cdot 5^{2} + 3 \cdot 5^{3} + 4 \cdot 5^{6} + 4 \cdot 5^{8} + O(5^{9}).$$
%
%Checking with $n=8,16$ confirms the computation.
%
%%\subsection{$D = -19$}
%\subsection{$D = -43$}
%
%$$K_1 = \Q[b_{3}]/(b_{3}^{10} + 10 b_{3}^{8} - 210 b_{3}^{7} + 1040 b_{3}^{6} - 834 b_{3}^{5} + 2125 b_{3}^{4} + 2220 b_{3}^{3} - 4580 b_{3}^{2} + 14160 b_{3} + 17856)$$
%
%
%\begin{align*}d(z) &= -\frac{1065039690681671}{8307611887304592} b_{3}^{9} - \frac{14340939904765}{173075247652179} b_{3}^{8} - \frac{3526258931335411}{4153805943652296} b_{3}^{7}\\
%&\quad + \frac{36901109765952349}{1384601981217432} b_{3}^{6} - \frac{114027245100098255}{1038451485913074} b_{3}^{5} - \frac{149581719588876751}{1384601981217432} b_{3}^{4}\\
%&\quad + \frac{59167836993688927}{193200276448944} b_{3}^{3} - \frac{97867023627100873}{692300990608716} b_{3}^{2} - \frac{1926937375873063127}{2076902971826148} b_{3} - \frac{2832479048170657}{5583072504909}\end{align*}
%
%
%\begin{align*}x(z) &=-\frac{71053408662216827675}{107260171976230298219592} b_{3}^{9} - \frac{28123134726555234509}{35753390658743432739864} b_{3}^{8} - \frac{78863248339692153610}{13407521497028787277449} b_{3}^{7}\\
%&\quad + \frac{403243299731508861115}{2979449221561952728322} b_{3}^{6} - \frac{27195045619104491218055}{53630085988115149109796} b_{3}^{5} - \frac{2198927741855979337233}{5958898443123905456644} b_{3}^{4}\\
%&\quad - \frac{79809519772443874536329}{107260171976230298219592} b_{3}^{3} - \frac{16262886060300678179787}{11917796886247810913288} b_{3}^{2} \\
%&\quad + \frac{353576748971457221649797}{53630085988115149109796} b_{3} + \frac{1045208797934202700807}{144166897817513841693} \end{align*}
%
%\begin{align*} y(z) &= \frac{442880517915505604682128525}{98668381927199640128173501752} b_{3}^{9} + \frac{137530443818227901669830007}{32889460642399880042724500584} b_{3}^{8} \\
%&\quad + \frac{1016345232866400296232162809}{24667095481799910032043375438} b_{3}^{7} - \frac{3808030200635496323292987333}{4111182580299985005340562573} b_{3}^{6}\\
%&\quad + \frac{182369718805264091222311436729}{49334190963599820064086750876} b_{3}^{5}  + \frac{17381369307168451931388315659}{16444730321199940021362250292} b_{3}^{4} \\
%&\quad + \frac{692503294108282906461717396755}{98668381927199640128173501752} b_{3}^{3}  + \frac{226894599199530661796356092603}{32889460642399880042724500584} b_{3}^{2} \\
%&\quad - \frac{2054146320007162634080610515511}{49334190963599820064086750876} b_{3} - \frac{5240865210417667232206638482}{132618792912902742107760083}\end{align*}
%
%
%
%which reduces to $(2,1) \in E(\F_5)$, and this point has order 2.
%
%$$h_p(z) = 2 + 2 \cdot 5 + 3 \cdot 5^{2} + 5^{3} + 4 \cdot 5^{4} + 2 \cdot 5^{5} + 4 \cdot 5^{6} + 4 \cdot 5^{7} + 4 \cdot 5^{8} + O(5^{9}).$$
%
%Checking with $n =4$ confirms.
%
%\subsection{$D = -67$}
%
%$$K_1 =  \Q[b_{3}]/(b_{3}^{10} - 40 b_{3}^{8} + 480 b_{3}^{6} + 75 b_{3}^{4} - 43960 b_{3}^{2} + 309808)$$
%
%$$d(z) = $$
%
%
%\begin{align*}x(z) &= -\frac{1076354443271673792679874461184580759721}{9952035736636323014955631505740623888936096} b_{3}^{8}\\
%&\quad + \frac{3137161687112805655402903220078661925671}{1658672622772720502492605250956770648156016} b_{3}^{6}\\
%&\quad + \frac{1612141194280853173998289607684268938887}{236953231824674357498943607279538664022288} b_{3}^{4}\\
%&\quad - \frac{737367512900831310725424833095271174842167}{3317345245545441004985210501913541296312032} b_{3}^{2} \\
%&\quad + \frac{6002533435550561917797466058185046124061}{8609027453837649666916636250640678104616} \end{align*}
%\begin{align*}
%y(z) &= \frac{30852840117749097095703705040268386092270088106331240401415}{145301658250476595692926152272725357273274818168755698561361088} b_{3}^{8}\\
%&\quad  - \frac{785728515020580523262885856195752687970158623403533814875111}{145301658250476595692926152272725357273274818168755698561361088} b_{3}^{6} \\
%&\quad + \frac{489372774521266458546450702120234854924543519029623347025947}{20757379750068085098989450324675051039039259738393671223051584} b_{3}^{4}\\
%&\quad + \frac{24687572523436666522201724935553837409838067523653832590452987}{72650829125238297846463076136362678636637409084377849280680544} b_{3}^{2}\\
%&\quad - \frac{216593694151661662396912343886014618598105523965830407843909}{62846737997610984296248335758099202972869731041849350588824}\end{align*}
%
%which reduces to $(2,1) \in E(\F_5)$ as well.
%
%\section{Example: fixing $E,p$, varying $K$}
%Let $E$ be the elliptic curve 83a1, and fix $p = 7$. We have the fields with discriminants $D = -8,-19,-43,-67$ satisfying the necessary hypotheses.
%
%\subsection{$D = -8$}
%
%$$K_1 = \Q[b_{6}]/(b_{6}^{14} - 56 b_{6}^{12} + 966 b_{6}^{10} - 1792 b_{6}^{8} - 95991 b_{6}^{6} + 1237992 b_{6}^{4} - 6135808 b_{6}^{2} + 10913792)$$
%
%\begin{align*}d(z) &= -\frac{19977011733880094554884410099}{2565980134400} b_{6}^{13} + \frac{24883365981437653747891024759}{641495033600} b_{6}^{12} \\
%&\quad + \frac{78977177359007471251535773207}{320747516800} b_{6}^{11} - \frac{101158090202677363895293387757}{80186879200} b_{6}^{10} \\
%&\quad - \frac{1509353125720892807730367189081}{1282990067200} b_{6}^{9} + \frac{1970517345416509502876466669181}{320747516800} b_{6}^{8}\\
%&\quad - \frac{1403864289349544821688550923023}{80186879200} b_{6}^{7} + \frac{3552743329387098562943055081821}{40093439600} b_{6}^{6} \\
%&\quad + \frac{787656734344271785046176645258677}{2565980134400} b_{6}^{5} - \frac{1010344437279339524081541313475777}{641495033600} b_{6}^{4} \\
%&\quad - \frac{547354874266048173730793382172573}{320747516800} b_{6}^{3} + \frac{705654074642337535648273947664173}{80186879200} b_{6}^{2} \\
%&\quad + \frac{64896161904432083032324856059803}{20046719800} b_{6} - \frac{1149494382335867875560405776871}{68653150}\end{align*}
%
%\begin{align*}x(z) &=\frac{4603807418998512776461543}{6319758384021122409752272793600} b_{6}^{13} - \frac{43628887388042069833993}{24686556187582509413094815600} b_{6}^{12} \\
%&\quad - \frac{32196013994731232572469439}{789969798002640301219034099200} b_{6}^{11} + \frac{1990435032891273429632787}{24686556187582509413094815600} b_{6}^{10} \\
%&\quad + \frac{2163706213316988731176074837}{3159879192010561204876136396800} b_{6}^{9} - \frac{20138482460026943729910949}{24686556187582509413094815600} b_{6}^{8} \\
%&\quad  - \frac{69932542065194400495261629}{197492449500660075304758524800} b_{6}^{7} - \frac{97504548692488604361357361}{12343278093791254706547407800} b_{6}^{6} \\
%&\quad  - \frac{528466121701406665885627372129}{6319758384021122409752272793600} b_{6}^{5} + \frac{1443767383631373738467827927}{12343278093791254706547407800} b_{6}^{4}\\
%&\quad  + \frac{668085056439976573656832795821}{789969798002640301219034099200} b_{6}^{3} - \frac{22984652089346618658380749243}{24686556187582509413094815600} b_{6}^{2} \\
%&\quad - \frac{115761492944011966880914705541}{49373112375165018826189631200} b_{6} + \frac{114137538141517734095073997}{42271500321202927077217150}\end{align*}
%\begin{align*}y(z) &= \frac{16908060832040445156298651692600171}{39671996642010363801276365376748264652800} b_{6}^{13} - \frac{4520999786716075437898319079696639}{9917999160502590950319091344187066163200} b_{6}^{12} \\
%&\quad - \frac{57735469376455916579277854706584103}{4958999580251295475159545672093533081600} b_{6}^{11} + \frac{67205695948122375911671184965151247}{1239749895062823868789886418023383270400} b_{6}^{10}\\
%&\quad - \frac{3018474308007219460649081835536945951}{19835998321005181900638182688374132326400} b_{6}^{9} - \frac{8936928617970496629003643662890893901}{4958999580251295475159545672093533081600} b_{6}^{8} \\
%&\quad + \frac{5949505794339331420332729588208433517}{1239749895062823868789886418023383270400} b_{6}^{7} + \frac{2476654997096247608741382183184930921}{154968736882852983598735802252922908800} b_{6}^{6}\\
%&\quad + \frac{956680609422513240875663655635597465667}{39671996642010363801276365376748264652800} b_{6}^{5} + \frac{1549492806892879917226734678657718365417}{9917999160502590950319091344187066163200} b_{6}^{4}\\
%&\quad - \frac{1586675241175468374241028378953063753083}{4958999580251295475159545672093533081600} b_{6}^{3} - \frac{2642653455372871528913675684741527466433}{1239749895062823868789886418023383270400} b_{6}^{2} \\
%&\quad + \frac{205465035568534643521366597040721751013}{309937473765705967197471604505845817600} b_{6} + \frac{5989274573768330710380230369791649791}{1061429704677075230128327412691252800}\end{align*}
%
%which reduces to $(4,  2) \in E(\F_7)$, which has order $11$.

\subsection{$D=-67$}
%this requires new pari
$$K_1 = \Q[b_3]/(b_3^{10} - 40b_3^8 + 480b_3^6 + 75b_3^4 - 43960b_3^2 + 309808)$$

\begin{align*}d(z)  &= \frac{71875903151330714536048309590595}{7117152} b_{3}^{9} + \frac{6119579507704859976213805383739}{145248} b_{3}^{8}\\
&\quad  - \frac{135897523348126877771180458895081}{593096} b_{3}^{7} - \frac{11663885727697022972855494183141}{12104} b_{3}^{6}\\
&\quad  + \frac{8661175411226096481365225859910}{10591} b_{3}^{5} + \frac{10481186280925635052626481325999}{3026}  b_{3}^{4}\\
&\quad  + \frac{36391171963438963519069902532957435}{2372384} b_{3}^{3} + \frac{3108265948391742685428761709774419}{48416} b_{3}^{2} \\
&\quad - \frac{314460166167253356319310621824691521}{1779288} b_{3} - \frac{1592994568241374601136501734183561}{2136}\end{align*}

$$z = \left(-\frac{2856067519095106012560371454223181}{16830149471465739403360770257217915024} b_{3}^{8} + \frac{79604975003336290995007108094484043}{16830149471465739403360770257217915024} b_{3}^{6} - \frac{438744543341115137469720587224680241}{16830149471465739403360770257217915024} b_{3}^{4} - \frac{1106919686358222710858747268381898523}{4207537367866434850840192564304478756} b_{3}^{2} + \frac{3624228811500041853690467790130454983}{1051884341966608712710048141076119689} ,\frac{1766794679497628443673527334628418471373993857338323}{7318750738406812139516059492171638085059640297155952704} b_{3}^{8} - \frac{43131006543371602479173018347181160105174122280751993}{7318750738406812139516059492171638085059640297155952704} b_{3}^{6} + \frac{201972981163145388457877386877436891972210891659369515}{7318750738406812139516059492171638085059640297155952704} b_{3}^{4} + \frac{123325663276752706970187278408833578276412670955769745}{457421921150425758719753718260727380316227518572247044} b_{3}^{2} - \frac{2312173024694283148577628746668337046387693085182534679}{457421921150425758719753718260727380316227518572247044}\right)$$
which reduces to $(2,0)$ in $E(\F_5)$, and we use $n = 7,14$ to compute the $p$-adic height:

$$h_p(z) = 3 + 2 \cdot 5 + 5^{2} + 5^{3} + 2 \cdot 5^{4} + 2 \cdot 5^{6} + 4 \cdot 5^{7} + O(5^{8}).$$
\section{Going up the anti-cyclotomic tower}
This has non-linear polynomial:

$E = 158b1, D = -7, p = 5$

Height of a MW generator $P$:

$$h_p(P) = 2 \cdot 5 + 5^{2} + 2 \cdot 5^{5} + 3 \cdot 5^{6} + 4 \cdot 5^{7} + 5^{8} + 3 \cdot 5^{9} + O(5^{10})$$

We have $$K_1 = \Q[b_{4}]/(b_{4}^{10} - 20 b_{4}^{8} - 25 b_{4}^{7} + 260 b_{4}^{6} + 49 b_{4}^{5} + 525 b_{4}^{4} - 200 b_{4}^{3} + 705 b_{4}^{2} - 85 b_{4} + 617).$$

We have \begin{align*}d(z) &= -\frac{6492339886}{3631485879} b_{4}^{9} + \frac{3731642890}{3631485879} b_{4}^{8} + \frac{136176671707}{3631485879} b_{4}^{7} + \frac{1211499568}{57642633} b_{4}^{6} - \frac{65643271261}{125223651} b_{4}^{5}\\
&\quad + \frac{2822926228}{14021181} b_{4}^{4} - \frac{199895608687}{518783697} b_{4}^{3} + \frac{104410975006}{1210495293} b_{4}^{2} - \frac{444144852182}{1210495293} b_{4} - \frac{903689351762}{3631485879}\end{align*}

and 

$$z = \left(\frac{2591369386796327995028}{380006066301795301136013} b_{4}^{9} + \frac{14888842387016594709925}{380006066301795301136013} b_{4}^{8} - \frac{60132612773564347609430}{380006066301795301136013} b_{4}^{7} - \frac{42044016343105427822755}{42222896255755033459557} b_{4}^{6} + \frac{16443654628828499069507}{13103657458682596590897} b_{4}^{5} + \frac{122114863458982979778265}{10270434224372845976649} b_{4}^{4} + \frac{109285805357277975893444}{380006066301795301136013} b_{4}^{3} + \frac{898984319969367978522121}{126668688767265100378671} b_{4}^{2} - \frac{782872679842638907592195}{126668688767265100378671} b_{4} + \frac{8906081956030183015509046}{380006066301795301136013}, \frac{241793584242385995303453511820}{5815444242657926058736938940899} b_{4}^{9} + \frac{1134912224499894111510846668762}{5815444242657926058736938940899} b_{4}^{8} - \frac{5522319739507395368665547944825}{5815444242657926058736938940899} b_{4}^{7} - \frac{29921352498897455528134016066444}{5815444242657926058736938940899} b_{4}^{6} + \frac{562160808127169912273773327028}{66844186697217540905022286677} b_{4}^{5} + \frac{9395648691023334684246881655259}{157174168720484488073971322727} b_{4}^{4} + \frac{16918458906849853301941931115106}{5815444242657926058736938940899} b_{4}^{3} + \frac{196994012062915894180654189356080}{5815444242657926058736938940899} b_{4}^{2} - \frac{158362370600385860588837297899838}{5815444242657926058736938940899} b_{4} + \frac{223652232107205363700149561009977}{1938481414219308686245646313633}\right)$$

This reduces to $(1,0)$ in $E(\F_5)$, which has order $7$. We use $n=7,14$ to compute the height:

$$h_p(z) = 3 \cdot 5 + 4 \cdot 5^{2} + 2 \cdot 5^{3} + 4 \cdot 5^{4} + 4 \cdot 5^{5} + O(5^{6}).$$

So we'd like to go the next level up...


\section{Higher valuation}

$E = 214b1, D = -7, p = 5$


Height of MW gen:

$$h_p(P) = 4 \cdot 5^{2} + 3 \cdot 5^{3} + 3 \cdot 5^{4} + 5^{8} + 2 \cdot 5^{12} + 3 \cdot 5^{13} + 2 \cdot 5^{14} + 3 \cdot 5^{15} + 4 \cdot 5^{17} + 2 \cdot 5^{18} + O(5^{19})$$

We have $$K_1 = \Q[b_{4}]/(b_{4}^{10} - 20 b_{4}^{8} - 25 b_{4}^{7} + 260 b_{4}^{6} + 49 b_{4}^{5} + 525 b_{4}^{4} - 200 b_{4}^{3} + 705 b_{4}^{2} - 85 b_{4} + 617).$$

\begin{align*}d(z) &= -\frac{55207226496320}{2824489017} b_{4}^{9} - \frac{21659558622586}{941496339} b_{4}^{8} + \frac{425994429247118}{941496339} b_{4}^{7} + \frac{2546474905061947}{2824489017} b_{4}^{6}\\
&\quad - \frac{561982398059725}{97396173} b_{4}^{5} - \frac{566184447353632}{76337541} b_{4}^{4} + \frac{1987085301419457}{313832113} b_{4}^{3} - \frac{53653231928520598}{2824489017} b_{4}^{2} \\
&\quad + \frac{33500836567973212}{2824489017} b_{4} - \frac{55170079769161475}{2824489017}\end{align*}

$$z = \left(-\frac{2078748072945839}{3434191491764983131} b_{4}^{9} + \frac{16748042702008406}{3434191491764983131} b_{4}^{8} + \frac{5136977149233923}{490598784537854733} b_{4}^{7} - \frac{103545523100105017}{1144730497254994377} b_{4}^{6} - \frac{28797757987817540}{118420396267758039} b_{4}^{5} + \frac{133405370751498302}{92815986263918463} b_{4}^{4} - \frac{1310865157686021320}{3434191491764983131} b_{4}^{3} + \frac{2921442846747143}{6056775117751293} b_{4}^{2} - \frac{1710299362485145022}{1144730497254994377} b_{4} - \frac{2660213075206547956}{3434191491764983131} : \frac{6290457363957322827554}{1254019062558425945166567} b_{4}^{9} + \frac{4678270047353411882182}{1254019062558425945166567} b_{4}^{8} - \frac{129408413980581522853685}{1254019062558425945166567} b_{4}^{7} - \frac{11201630479622762696120}{59715193455163140246027} b_{4}^{6} + \frac{54818066426718044386589}{43242036639945722247123} b_{4}^{5} + \frac{34548421198028294525692}{33892407096173674193691} b_{4}^{4} + \frac{2249644736088251027834513}{1254019062558425945166567} b_{4}^{3} + \frac{23093004743419152096979}{5160572273902987428669} b_{4}^{2} + \frac{105153038236884456831158}{59715193455163140246027} b_{4} + \frac{427339209333707206716829}{179145580365489420738081} \right)$$

which reduces to $(0,4)$ in $E(\F_5)$, which has order 7.

We use $n=7,14$ to compute the heights:

$$h_p(z) = 2 + 2 \cdot 5 + 3 \cdot 5^{2} + 2 \cdot 5^{3} + 2 \cdot 5^{4} + 2 \cdot 5^{5} + 3 \cdot 5^{6} + O(5^{7})$$


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