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N.B. Some sections here are written up formally; others are just notes.  Note also I need to learn how to do bibliographies.

\section{Introducing \Sha}

For an elliptic curve $E$ defined over a number field $K$, let $\Sha(E/K)= \hbox{ker} (H^1(K, E) \rightarrow \prod_v H^1(K_v, E))$ be the Shafarevich-Tate group of $E/K$.  Recall that $H^1(K, E)= \varinjlim H^1(L/K, E)$, where the direct limit is taken over all finite Galois extensions $L/K$; since each one of these groups is torsion, killed by $[L:K]$, so is $H^1(K, E)$.  Therefore, $\Sha(E/K)$ is a torsion group, a fact we will later to use to conclude that $\Sha(E/K)=0$ for curves having $\Sha(E/K)[p]=0$ for all primes $p$.

\section{Known Results for CM Elliptic Curves}
NEED MORE SPECIFICALLY WHAT RUBIN GIVES-- 
We will now summarize and give references for the main theoretical results that give information about $\Sha$ and the rank of elliptic curves over $\Q$.  Progress was first made in the case of elliptic curves with complex multiplication: J. Coates and A. Wiles showed in [J. Coates, A. Wiles, ``On the Conjecture of Birch and Swinnerton-Dyer'', Invent. Math., 39 (1977), 223-251.] that for such an elliptic curve $E/\Q$, $L(E/\Q, 1)\neq 0 \Rightarrow E(\Q)$ is finite.  Improving on their techniques, K. Rubin proved two major results in his paper [K. Rubin, ``Tate-Shafarevich groups and L-functions of elliptic curves with complex multiplication'', Invent. Math., 89 (1987), 527-560.] about elliptic curves over $\Q$ with complex multiplication.  Rubin's Theorem $A$ implies that whenever $L(E/\Q, 1) \neq 0$, $\Sha(E/\Q)$ is finite.  Theorem $A$ also includes a local result, providing evidence for the full Birch and Swinnerton-Dyer conjecture, describing for which primes $p$ $\Sha(E/\Q)[p]$ can be non-trivial.  Note that before Rubin's result, not a single Shafarevich-Tate group was known to be finite; his local result makes $\Sha(E/\Q)$ effectively computable in some cases.  Under the same hypotheses, Theorem $B$ of Rubin's paper states that $\hbox{rk}_\Z(E/\Q) \geq 2 \Rightarrow \hbox{ord}_{s=1}L(E/\Q, s)\geq 2$.  The previous year B. Gross and D. Zagier proved ([B. H. Gross, D. B. Zagier, ``Heegner points and derivatives of L-series'', Invent. Math., 84 (1986), 225-320.])their limit formula \[L^{\prime}(E/K,1)= \frac{\iint_{E(\C)} \omega \wedge \overline{i \omega}}{\sqrt{D}} \hat{h}(y_K),\] where $y_K$ denotes the usual Heegner point.  Combining Rubin's result with the work of Coates-Wiles and the theorem, of Gross-Zagier, one obtains the 
\begin{theorem}
Let $E/\Q$ be an elliptic curve with complex multiplication by an order in a quadratic imaginary field.  Then \[\hbox{ord}_{s=1}L(E/\Q, s) \leq 1 \Rightarrow \hbox{rk}_\Z(E/\Q)= \hbox{ord}_{s=1}L(E/\Q, s).\]
\end{theorem}

\section{Kolyvagin and Consequences}
Kolyvagin shows that if $y_K$ has infinite order (i.e., $L^{\prime}(E/K, 1)\neq 0$), then $E(K)$ has rank $1$.  Moreover, he proves $\Sha(E/K)$ is finite, with the following bounds.  Writing $C_D= [E(K): \Z y_K]$, and letting $c_3$ and $c_4$ be constants I don't understand yet that are mired in Kolyvagin's paper, $c_3C_D$ kills $\Sha(E/K)$, and $\# \Sha(E/K)| c_4C_D^2$.

Under the same hypothesis, he obtains the corresponding result for $E/\Q$ and $E_D/\Q$: $2c_3C_D$ kills $\Sha(E/\Q)$ and $\Sha(E_D/\Q)$, and both groups have order dividing $c_42^aC_D^2$, where $a$ is the 2-rank of both groups.  Moreover, he gives a condition for determining whether $E/\Q$ or its twist $E_D/\Q$ contains the point of infinite order.  If $E$ has modular parametrization $\gamma : X_0(N) \rightarrow E$, then $\gamma \circ w_N = \varepsilon \gamma + \gamma(0)$, where $w_N$ is the principal (Fricke) involution on the modular curve (alternatively, $\varepsilon$ is the eigenvalue on the normalized newform associated to E of the involution $w_N$).  Kolyvagin deduces that $\varepsilon=1 \Rightarrow$ $E(\Q)$ contains the point of infinite order, and $\varepsilon=-1 \Rightarrow$ $E_D(\Q)$ contains the point of infinite order. 
(Note that our calculations for $E/\Q$ will give us some info free of charge about $E_D/\Q$)
$\varepsilon=1$ makes our calculations easier, because while we'll be using low-level $E/\Q$, we don't know how high the conductor of $E_D/\Q$ might be.

In [] Kolyvagin shows that $23$ elliptic curves over $\Q$ have trivial Shafarevich-Tate group, and he verifies the full Birch and Swinnerton-Dyer conjecture for $5$ of them, reducing the verification in other cases to a computation we will check.  Note, however, that all of these curves have rank $0$: they are quadratic twists of the rank $1$ curve $E: y^2= 4x^3-4x+1$ having $\varepsilon=1$.  In particular, he shows for the $23$ curves $E_D: -Dy^2= 4x^3-4x+1$ with $D\in \{7,11,47,71,83,84,127,159,164,219,231,263,271,\\
287,292,303,308,359,371,404,443,447,471\}$, $\Sha(E_D/\Q)=0$.  These are the twists with $D<500$ (the extent of the tables of computations Kolyvagin had available), $D$ prime to the conductor of $E$ ($37$), and as usual ruling out $D=3,4$, for which $C_D=1$.  This suffices to conclude that $\Sha(E_D/\Q)=0$ because in Theorem $B$ of his paper, Kolyvagin shows that for whichever of $E/\Q$ and $E_D/\Q$ that does not contain the point of infinite order, $\Sha$ is killed by $C_D$ (as opposed to $c_3C_D$, which works for both curves).  

By explicitly working with the curve $E$, Kolyvagin reduces the full BSD conjecture in this case to showing $r_2(D)=0$, where $r_2(D)= \#\{q\, \hbox{an odd prime}: q|D, (\frac{q}{37})=1, \hbox{and}\, a_q\, \hbox{is even}\}$.  $(\frac{q}{37})$ denotes the Legendre symbol, and $a_q$ is the Hecke eigenvalue.  
NOTE Cremona's tables won't be good for handling these, because conductors may get very high.  His outputs of $a_q$ also only go up to $100$, and some of our $D$'s have prime divisors $>100$.    

\section{Weakening the Hypotheses for Triviality of $\Sha(E/K)[p]$}


S. Donnelly has explained a way to weaken the hypotheses of Gross's Proposition $2.1$, replacing $G(\Q(E[p])/\Q)= GL_2(/Z/\/Z)$ with the statement that $E$ admits no $p$-isogenies.  Gross only uses the hypothesis on $G(\Q(E[p])/\Q)$ at two points in the paper: section $4$, where he constructs Kolyvagin's cohomology classes and shows that restriction gives an isomorphism $H^1(K, E[p]) \cong H^1(K_n, E[p])^{\mathcal{G}_n}$, where $K_n$ denotes the ring class field of conductor $n$ over $K$ and $\mathcal{G}_n= G(K_n/K)$; and section $9$, where he requires $H^n(K(E[p])/K, E[p])=0$ for all $n \geq 0$, for then by the Hochshild-Serre spectral sequence he obtains the isomorphism $H^1(K, E[p]) \cong H^1(K(E[p]), E[p])^{G(K(E[p])/K)}$.

The next proposition extends the result of section $4$ of Gross's paper.  Note that if $E$ has no $p$-isogeny (defined over $K$), then certainly it has no $p$-torsion over $K$.

\begin{proposition}
Let $E$ be an elliptic curve with $E(K)[p]=0$, where $p>3$ or, if $p=3$, $K \neq \Q(\mu_3)$.  Then $H^m(K_n/K, E(K_n)[p]) =0$ for all $m \geq 1$. 
\end{proposition}
\proof
We may write the abelian group $G(K_n/K)$ as a direct sum $P \oplus P^\prime$, where $P$ is its Sylow $p$-subgroup and $(p, \#{P^\prime})=1$.  We claim that the subgroup of $E(K_n)[p]$ invariant under $P^\prime$ is trivial.  Let $G = G(K_n/K)/H$, where $H$ is the subgroup of $G(K_n/K)$ that acts trivially on $E(K_n[p])$.  If $(\#G, p)=1$, $P\subseteq H$, so $P^\prime$ surjects onto $G$.  As there is no subgroup of $E(K_n)[p]$ invariant under all of $G(K_n/K)$, the same then holds for $P^\prime$.

If $p|\#G$, we cannot have $\#E(K_n)[p]= p$, for then a subgroup of $G$ of order $p$ would act trivially on $E(K_n)[p]$, contradicting the definition of $G$.  Thus, $E(K_n)[p]$ is the full $p$-torsion subgroup of $E$, and we can identify $G$ with a subgroup of $GL_2(\Z/p\Z)$ acting on $E(K_n)[p]= (\Z/p\Z)^2$.

We can choose a basis of $(\Z/p\Z)^2$ so that $G$ contains the subgroup $\left( \begin{array}{cc}
1 & x \\
0 & 1 \end{array} \right)$, where $x \in \Z/p\Z$.  Being abelian, $G$ must be contained in the normalizer of this subgroup, so $G \subseteq \{ \left( \begin{array}{cc} 
a & b \\
0 & a \end{array} \right) | a,b \in \Z/p\Z \}$, and we claim that $G$ contains an element with $a \neq 1$.  Since $E(K_n)$ contains all $E$'s full $p$-torsion subgroup, $\mu_p \subset K_n$ (this follows from basic properties of the Weil pairing).  Then $G(K_n/K)$ acts on $\mu_p$, inducing a well-defined action of $G$ on $\mu_p$, for the kernel of $G(K_n/K)\onto G$ consists of automorphisms acting trivially on $E(K_n)[p]$, and therefore acting trivially on $\mu_p$ as well.  Choosing a primitive $p^{th}$ root of unity $\zeta$, we claim that the action of $G$ on $\mu_p$ is given by \[g \cdot \zeta= \zeta^{det(g)}\] for all $g \in G$.  That is, $\hbox{det}(g)$ is the cyclotomic character.  NOW WHAT?

Having shown that $P^\prime$ leaves no subgroup of $E(K_n)[p]$ invariant, the result will follow from an application of the inflation-restriction exact sequence: \[0\rightarrow  H^1(P, E(K_n)[p]^{P^\prime} \rightarrow  H^1(K_n/K, E(K_n)[p]) \rightarrow H^1(P^{\prime}, E(K_n)[p])\]

The first group vanishes since $E(K_n)[p]^{P^{\prime}}=0$, and the third group vanishes since the order of $P^{\prime}$ is prime to $p$, and thus to the order of $E(K_n)[p]$.  We conclude that the middle group is trivial, as desired.  We can therefore extend the sequence to the second cohomology groups, deduce the same triviality result, and by induction conclude that $H^m(K_n/K, E(K_n)[p])=0$ for all $m \geq 1$.    
\endproof

The next proposition shows the weakened hypothesis is sufficient in section $9$ of Gross's paper.
\begin{proposition}
Suppose $E$ has no $p$-isogeny.  Then $H^1(K(E[p])/K, E[p])=0$.
\end{proposition} 

\proof
\endproof


\section{The Decomposition of $E(K)$ under Complex Conjugation}

To compute the index $I_K= \#{E(K)/{\Z y_K}}$ needed to verify the full Birch and Swinnerton-Dyer conjecture, we compare the canonical height of the Heegner point $y_K$ with the height of a generator of $E(K)$, which we know to have algebraic rank 1 by Kolyvagin's theorem.  To assist in finding the (infinite order) generator of $E(K)$, we use the 
\begin{proposition}
For a quadratic imaginary field $K=\Q(\sqrt{D})$, $E(K)$ decomposes, up to $2$-torsion, as a direct sum $E(\Q) \oplus E_D(\Q)$, where $E_D$ is the quadratic twist of $E$.  (Recall that from a Weierstrass equation of $E/\Q$, we obtain a Weierstrass equation of $E_D/\Q$ as follows: \[\begin{array}{c} 
E: y^2= x^3+ax+b \\
E_D: y^2= x^3+D^2ax+D^3b \end{array})\]
\end{proposition} 
\proof
Denote complex conjugation by $\tau$.  We will decompose $E(K)$ into its eigenspaces under the action of $\tau$.  First, we kill $E(K)[2]$ (by tensoring with $\Z[\frac{1}{2}]$, for instance), so that, using the fact that $\tau^2=1$, we can define the projections from $E(K)$ to its $+1$ and $-1$ eigenspaces under $\tau$: for $P\in E(K)$, \[P= \frac{1+\tau}{2}P + \frac{1-\tau}{2}P.\]  But if $P=(x,y)\in E(K)$ satisfies $\tau P=P$, then $P\in E(\Q)$; if $\tau P=-P= (x, -y)$, then $x\in \Q$ and $y\in \sqrt{D}\Q$.  In particular, $(Dx, D\sqrt{D}y)\in E_D(\Q)$, and conversely we can obtain any such point in the $-1$ eigenspace of $E(K)$ from a point of $E_D(\Q)$.  Thus, having eliminated the $2$-torsion, the decomposition of $E(K)$ into its $\tau$-eigenspaces just reads \[E(K)= E(\Q) \oplus E_D(\Q).\] 
\endproof

\section{Application of Gross-Zagier to our calculations}

In light of Donnelly's observation, we can apply Kolyvagin's result
that $\Sha(E/K)[p]=0$ for all odd primes $p$ not dividing the Heegner
point $y_K$ (i.e., $p$ does not divide $C_D$) and such that $E/K$ has
no $p$-isogenies.  By the inflation-restriction sequence,
$\Sha(E/\Q)[p]=$ for all of these primes as well, since
$\Sha(E/\Q)[p]=$ maps to $\Sha(E/K)[p]$ with kernel contained in
$H^1(K/\Q, E(K)$, a finite $2$-group.  Thus, it will be essential to
compute the constant $C_D= [E(K): \Z y_K]$.  First we will have to
find a generator $P$ of the free part of $E(K)$ and compute its
canonical height $\hat{h}(P)$.  We can use another form of the
Gross-Zagier formula to compute $\hat{h}(y_K)$: 
\[\hat{h}(y_K)=
\frac{u^2 \sqrt{D}}{\|\omega_f\|} L^{\prime}(E_D/\Q, 1)L(E/\Q, 1),\]
where $u$ is half the number of units in $K= \Q(\sqrt{-D})$, and
$\|\omega_f\|$ is the Peterson norm of the newform $f$ corresponding
to $E$ (note that for $D\neq 3, 4$, $u=1$).

With these preliminaries, we can compute the index $C_D=$
 
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