\input amssym.def
\input amssym.tex
%\def\Bbb{\bf}
\nopagenumbers
\magnification=\magstep1
%\hoffset=1truecm
%\voffset=2truecm
\baselineskip = 5.2 true mm
\font\frkkk=eufm10
\font\twelverm=cmr12
\font\tenrm=cmr10
\font\ninerm=cmr9
\font\ninebf=cmbx9
\font\eightrm=cmr8
\font\sevrm=cmr7
\font\sixrm=cmr6
\font\scrpp=eusm10
\font\frkk=eufm10
\font\deffont=cmssi10
\font\chaptitle=cmbx10 at 14 pt
\tolerance=10000
\def\sqr{\ifmmode\square\else{$\square$}\fi}
\def\square{\vcenter{
\hrule height.1mm
\hbox{\vrule width.1mm height2.2mm\kern2.18mm\vrule width.1mm}
\hrule height.1mm}}                  % This is a slimmer sqr.
\null
\def\le{\leqslant}
\def\ge{\geqslant}
\def\etq{{\cal E}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})}
\def\etqp{{\cal E}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q}_p)}
\def\cotq{{\cal C}_{\lower 1pt\hbox{\eightrm oddtors}}({\Bbb Q})}
\def\dotq{{\cal D}_{\lower 1pt\hbox{\eightrm oddtors}}({\Bbb Q})}
\def\c{{\cal C}}
\def\d{{\cal D}}
\def\e{{\cal E}}
\def\pk{\phi _\kappa}
\def\im{{\hbox{\sl im}}}
\def\hs{H_{\varsigma}}
\def\hpk{\hat \phi _\kappa}
\font\sc=cmssqi8
\def\scc#1{\hbox{\sc #1}}
\def\sf{{\scc F}}
\def\pnbq{{\Bbb P}^n(\overline {\Bbb Q} )}
\def\hk{{\hat \kappa}}
\def\bq{{\overline {\Bbb Q}}}
\def\hq{{\hat q}}
\def\pv{\prod\limits_v }
\def\pnk{{\Bbb P}^n(K)}
\def\mnkvw{{\Bbb M}^n(K[{\bf v}^2,{\bf w}^2])}
\def\pnkv{{\Bbb P}^n(K[{\bf v}^2])}
\def\kj{\kappa (J)}
\def \qmods {{\Bbb Q}^*/({\Bbb Q}^*)^2}
\def \qmodss { {\Bbb Q}^*/({\Bbb Q}^*)^2 \times
{\Bbb Q}^*/({\Bbb Q}^*)^2 }
\def \qs{{\Bbb Q}^*}
\def \qss{({\Bbb Q}^*)^2}
\def\bbQ{\Bbb Q}
\def\bbF{\Bbb F}
\def\bbZ{\Bbb Z}
\def\bbR{\Bbb R}
\def\bbC{\Bbb C}
\def\notdiv{{\not\hskip-.5pt |\ }}
\def\Q{{\Bbb Q}}
\def\F{{\Bbb F}}
\def\Z{{\Bbb Z}}
\def\R{{\Bbb R}}
\def\C{{\Bbb C}}
%
\chaptitle
\noindent
\centerline{Elliptic Curves. Sheet 6. To be handed in during 7th Week.}
\rm
\bigskip
\medskip\noindent {\bf 1.} Find the torsion group over~$\Q$
for each of:
\par\noindent {\bf (a).} $Y^2 = X^3 + 1$.
\par\noindent {\bf (b).} $Y^2 = X(X-1)(X-2)$.
\par\noindent {\bf (c).} $Y^2 = X^3 + 1/3^6$.
\par\noindent {\bf (d) [optional].} $Y^2 = X^3 - 219X + 1654$.
%\par\noindent {\bf (a).} $Y^2 = X^3 + 1$.
%\par\noindent {\bf (b).} $Y^2 = X^3 - 219X + 1654$.
%%\par\noindent {\bf (c).} $Y^2 = X(X+1)(X+4)$.
%\par\noindent {\bf (c).} $Y^2 = X(X+81)(X+256)$.
%\par\noindent {\bf (d).} $Y^2 = X(X-1)(X-2)$.
%\par\noindent {\bf (e).} $Y^2 = X^3 + 1/3^6$.
%\par\noindent {\bf (f).} $Y^2 + Y = X^3 - X + 13$.
%%\par\noindent {\bf (g).} $Y^2 = X^3 - X^2 + 1/4$.
\par\noindent Note: (d) requires significant computation
(you should do an initial search for points with $x$-coordinate
in the range $|x| \leqslant 20$, with $x\in \Z$)
and is of course much more time consuming than anything which
would be asked in a timed exam. It is mainly intended to demonstrate
that interesting torsion groups can occur.
\medskip\noindent {\bf 2.} Let, as usual, $\c : Y^2 = X(X^2 + aX + b)$
and $\d : Y^2 = X(X^2 + a_1X + b_1)$, where $a,b\in\Z$, $a_1 = -2a,
b_1 = a^2 - 4b$ and $b(a^2-4b)\not= 0$. 
Let $\cotq$ denote the set of torsion elements of $\c (\Q)$ which have
odd order, and let $\dotq$ denote the set of torsion elements of $\d (\Q)$
which have odd order. Show that $\cotq$ and $\dotq$ are isomorphic.
\medskip\noindent {\bf 3.} Let $\c$ and $\d$ be as in question~2.
Let the homomorphism~$\phi, \hat\phi$ be defined as usual by 
$$\phi : \c (\Q ) \rightarrow \d (\Q) : (x,y)
\mapsto \Bigl( \bigl( {y\over x}\bigr)^2 , 
y - {by\over x^2} \Bigr),$$
$$
\hat\phi : \d (\Q) \rightarrow \c (\Q) : (u,v)
\mapsto \Bigl( {1\over 4} \bigl( {v\over u} \bigr)^2,
{1\over 8} \bigl( v - {b_1 v\over u^2}\bigr) \Bigr).
$$
What are the preimages of $(0,0)$ under $\hat\phi$?
Show that $(0,0) \in 2\c (\Q)$ if and only if there
exist~$m,n\in \Z$ such that $b = m^2$ and $a+2m = n^2$. 
\bigskip
\hrule
\medskip
{\it The following question is compulsory for students taking
the MSc in MFoCS (Mathematics and the Foundations of Computer
Science). For everyone else, it is optional.}
\medskip\medskip\noindent
{\bf 4.} Show that any elliptic curve over $\Q$ with
a rational point of order~$4$ is birationally equivalent to:
$$ Y^2 + XY + v Y = X^3 + v X^2,$$
for some $v\in \Q$.
\vfil \eject \end