\documentclass[11pt]{article} \hoffset=-0.05\textwidth \textwidth=1.1\textwidth \newcommand{\bndone}{87} \newcommand{\bndzero}{19} \include{macros} \begin{document} Here we do some sample verifications of full BSD. \section{$14A1$} \begin{enumerate} \item $t=6$ (Cremona) \item We need $2,7$ to split in $K$, with $D$ prime to $14$. $D=-1, -2$ no good. $D=-3$ works. [[Actually $D=-3$ doesn't satsify the hypothesis, since $2$ doesn't split.]] \item For e=ellinit([1,0,1,4,-6]), ellrootno(e)= 1, so we use $E^D$. \item reducing $E^D$--[126, [1, 0, 1/2, 1/2], 36] ellanalyticrank(e2) INPUT CURVE NOT MINIMAL MINIMISED TO [1, -1, 1, 40, 155] This also has analytic rank 0, so $E/K$ has rank 0, and our choice of $D$ was bad, so we go back to step $2$. $D=-19$ gives the next $K$ satisfying the Heegner hypothesis. Compute $E^{-19}$: %? e2=ellinit([0,-19/4,0,19^2*9/2,23*19^3/4]); %? ellglobalred(e2) %24 = [5054, [1, 2, 1/2, 1/2], 36] %? ellanalyticrank(e2) %INPUT CURVE NOT MINIMAL %MINIMISED TO [1, 1, 1, 1617, 42677] %Summing 142 a_n terms %Rank is even %L^(0)=5.473594665263139297916944347 %25 = [0, 5.473594665263139297916944347, 1.000000154638710257400827784] SO STILL RANK ZERO. We have to try higher $D$'s. try $-31$. new twist is $[0,-31/4,0, 31^2*9/2,23*31^3/4]$. This finally has rank $1$, so $-31$ is okay, but it has conductor $13454$. But now this isn't listed in Cremona's table (he lists a different rep of the isogeny class, apparently), so I can't find a M-W generator. Magma gives free generator $(2635 : -203732 : 1)$ of $E^-31$. Height is $3.65369257693788807017...$ \item $\alpha=\sqrt{31}/\omega_E$. $L(E/\Q, 1)= 0.3302239655664549236801327451$, $L^{\prime}(E^D/\Q,1)= 5.218903558405753668020265818$. $\omega_E \sim 2.626251405582016186497854965$. Then $$\hat{h}(y_K)\sim \sqrt{31}* 0.3302239655664549236801327451* 5.218903558405753668020265818/ 2.626251405582016186497854965 \sim 3.653695993510223589610801579$$. \item $I_K \sim 3.653695993510223589610801579/3.65369257693788807017\sim 1.000000935101206129067749992$, which is good news. (didn't sqrt) \item $\Sha(E/\Q)[p]=0$ for all $p\neq 2,3$. \item Perform $2$ and $3$-descent to show triviality of $\Sha$ there as well. \end{enumerate} \section{15A1} \begin{enumerate} \item $t=2$. \item $D=-11$ is first to satisfy Heegner hypothesis. \item root number is $1$, so we need the twist $E^-11$ $e2=ellinit([0,-55/4,0,-19*11^2/2,39*11^3/4]); ellanalyticrank(e2) INPUT CURVE NOT MINIMAL MINIMISED TO [1, 1, 0, -1212, 7011] Summing 85 a_n terms Rank is odd L^(1)=3.207426598649529335635931247 36 = [1, 3.207426598649529335635931247, 1.666073986584487822978044210]$ \item Has conductor 1815, and a M-W generator is $(1527 : 38880 : 1)$ according to Magma. Height is $1.66607395687729953337$. \item $L^1(E^D/\Q, 1) \sim 3.207426598649529335635931247$, $L(E/\Q, 1) \sim 0.3501506898191986108956287229$. $\omega_E \sim 2.235701712617529101766742778$, so $$\hat{h}(y_K) \sim 1.666073649878152404321456746$$ \item $I_K \sim .9999998157349823146788753820$ (didn't bother w/ sqrt) \item $\Sha[p]=0$ for $p\neq 2$. \item Run $2$-descent. \end{enumerate} So twice the problem of twists with high conductor has arisen. We'll now try an example where $E/\Q$ has rank $1$ to avoid this problem. \section{37A1} \begin{enumerate} \item $t=2$ \item $D=-4$ is ok (remember, have to use discriminant of $K$.) \item root number is $-1$, so we can compute with $E$. \item a M-W generator is $(x,y)= (0,0)$. \item height of Heegner point. $0.3059997890560586325057899275= L^{\prime}(E/\Q, 1)$ the twist by $-1$ is isomorphic to $[0,0,0,-1,-1/4]$ $2.451389679618052965620183312= L(E^{-1}/\Q, 1)$. $u=2, \sqrt{|D|}=2, \omega_E= 7.338132740789576739070720999$. then $\alpha=0.5450977982131191944086112918$, and finally $$\hat{h}(y_K)=.4088913359049488054582505759*2$$ \item $\hat{h}(0,0)=0.05111140823996884023588609975$ (twice the pari value), so $$I_K=\sqrt{2* 8.000001369267654590724456920}$$ \item Conclude that $\Sha(E/\Q)[p]=0$ for $p \neq 2$. \item Perform $2$-descent. \end{enumerate} \section{17A1} \begin{enumerate} \item $t=2$ \item $D=-4$ works. \item $E= ([1,-1,1,-1,-14])$ root number is $1$, so the twist has rank $1$. A W.E. for the twist is $y^2 = x^3 - 14256*x + 41523840$, and this has conductor $272$, so we can look it up in Cremona's tables. Using MAGMA to find a minimal model, the twist is isomorphic to 272B4. Cremona only gives gen's for strong Weil curves, so we use MAGMA again to find $z= (-36 : 6480 : 1)$. For future reference we find the height to be $2.036297434690738317186$. \item $L(E/\Q,1)=0.3867705802219959913534634548, L^{\prime}(E^D/\Q,1)=1.397785337254319520109796667, \omega_E= 2.123938699064983080555718069$. Then $$\hat{h}(y_K)\sim 2.036300750689832872531797350$$ \item $I_K \sim 1.000000814222344949138440895$. \item $\Sha(E/\Q)[p]=0$ for all $p \neq 2$. \item perform $2$-descent. \end{enumerate} Now we will try a couple of high-conductor examples, one for each possible root number. \section{462F1} \begin{enumerate} \item $t=2$ \item $462=2*3*7*11$. $D=-17$ works. \item $E=([1,0,0,-97,1337])$. root number is $1$, so we need to find generator of $E^D$. $E^D$ has a W.E. $y^2 = x^3 - 36338571*x - 308321515206$, and its conductor is $1068144$. Magma's not going to find a generator here, so we're stuck for the moment. Perhaps trying higher values of $|D|$ we might luck out and get a lower conductor twist? Try $D=-227$ just for kicks. $23806398$ is the new conductor, even worse. There probably just aren't enough curves of low conductor... My experimentation in the next section allows me to return here with a twist of relatively low conductor, letting $D=-841$ (conductor 3696). But unfortunately this has rank $0$, so we're still out of luck: $E/\Q(\sqrt{-841}$ has rank $0$. \end{enumerate} \section{462E1} \begin{enumerate} \item $t=2$ \item We can take $D=-17$ as above. \item $E= ([1,1,1,-405,4731])$. root number is $-1$, so we need the free generator of $E$. \item a M-W generator is $(-17,92)$ from Cremona's tables. its height is $\sim .039411059881925059045$. \item $u=1$, $L(E^D/\Q,1) \sim$ $L^{\prime}(E/\Q,1)\sim 2.036386747116844897014687695, \omega_E \sim 0.3161575749688195692943622398$. Both MAGMA and PARI say that $E^D$ also has rank $1$, so the original $E/K$ must have had rank $2$. Now again we'll go back and try $D=-227$. Again I find that the twist also has rank $1$, so $E$ has rank $2$ over $\Q(\sqrt{-227})$ and $\Q(\sqrt{-17})$. Yet another value of $D$ must be tried. $D=-229$ works as well (N.B. I've gotten all my values of $D$ by applying Q.R. and solving a system of congruences). Again, the twist has rank $1$. Contrast this with the situation with $14A1$, where we needed to try a few values of $D$ before getting $E/K$ to have rank $>0$. Now we're having trouble getting this rank $<2$. Try $-17-462=-479$. I'm getting that the twist $E^{-479}$ has rank $2$ over $\Q$!. Now I try $D=-841$, and get rank zero, with $L(E^{-841}/\Q,1) \sim 1.027947342001393896663733734$. A minimal model, incidentally, is $[0,1,0,-6480,-315756]$. We can find the height of a Heegner point here: $$\hat{h}(y_K) \sim 192.0107464816242787400060455$$ \item $I_K \sim 69.79972486951426762605949488$. This is not convincingly close to an integer... how precise are all of the above calculations? Or did I make an error along the way? \item ? \item ? \end{enumerate} http://magma.maths.usyd.edu.au/magma/ps/hb7.ps is a handy reference for everything about ell. curves and magma \end{document}