\input amssym.def \input amssym.tex %\def\Bbb{\bf} %\nopagenumbers \magnification=\magstep1 %\hoffset=1truecm %\voffset=2truecm \baselineskip = 5.2 true mm \def\notdiv{{\not\hskip-.5pt |\ }} \font\frkkk=eufm10 \font\twelverm=cmr12 \font\tenrm=cmr10 \font\ninerm=cmr9 \font\ninebf=cmbx9 \font\eightrm=cmr8 \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. %\def\sqr{$\vcenter{\hrule height .3mm %\hbox {\vrule width .3mm height 2mm \kern 1.4mm %\vrule width .3mm} \hrule height .3mm}$} % \null % %\vsize=19.5 true cm %\hsize=11.5 true cm %\vskip 5 true cm %\def\leqslant{\le} \def\c{{\cal C}} \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} % \chaptitle \noindent \centerline{Elliptic Curves. Solutions to Sheet 1.} \rm \par \bigskip \noindent {\bf 1.} \noindent {\bf (a).} The only points on $Y^2 = X^3 + 2 X$ over $\bbF_5$ are: ${\bf o}, (0,0)$. The group table is the same as that of $C_2$ (``cyclic~2'' group; i.e. the integers modulo~2 under addition), with $0,1$ replaced by ${\bf o}, (0,0)$, respectively. \par \noindent {\bf (b).} The only points on $Y^2 = X^3 + 1$ over $\bbF_5$ are: ${\bf o}, (0,1), (0,4), (4,0), (2,2), (2,3)$. The point~$(2,2)$ has order~6, and the group table is the same as that of $C_6$ (``cyclic~6'' group; i.e. the integers modulo~6 under addition), with $0,1,2,3,4,5$ replaced by ${\bf o}, (2,2), (0,4), (4,0), (0,1), (2,3)$, respectively. Note that it's also correct to say that the group is $C_2 \times C_3$, since $C_2\times C_3$ is isomorphic to $C_6$. \medskip \noindent {\bf 2.} We have: $2YY' = 3X^2 + 4$, and so the slope at the point~$(2,4)$ is: $(3\cdot 2^2 + 4)/(2\cdot 4) = 2$. The line tangent to the curve at~$(2,4)$ is: $Y=2X$. The $x$-coordinate of the third point of intersection is therefore: $2^2 - 2 - 2 = 0$, with $y$-coordinate $y=2\cdot 0 = 0$. Hence, $(2,4) + (2,4) + (0,0) = {\bf o}$, and so $(2,4) + (2,4) = (0,-0) = (0,0)$. Now, note that we always have $-(x,y) = (x,-y)$ and so $-(0,0) = (0,0)$, giving: $2(0,0) = {\bf o}$. Hence, $4(2,4) = 2(0,0) = {\bf o}$. Also, check that $3(2,4) = 4(2,4) - (2,4) = -(2,4) = (2,-4)$. Summarising: the first~4 multiples of~$(2,4)$ are: $1(2,4) = (2,4),\, 2(2,4) = (0,0),\, 3(2,4) = (2,-4),\, 4(2,4) = {\bf o}$, and so~4 is the smallest positive multiple of $(2,4)$ equal to~{\bf o}. \medskip \noindent {\bf 3.} \par\noindent {\bf (a).} Case 1. $m$ is odd. The birational transformation $(X,Y) \mapsto (X/f_m , Y/f_m^{(m+1)/2})$ [with inverse transformation: $(X,Y) \mapsto (Xf_m , Yf_m^{(m+1)/2})$] maps the given curve to: $f_m^{(m+1)} Y^2 = f_m^{(m+1)} X^m + \ldots + f_0$, which is the same as: $Y^2 = X^m + c_{m-1} X^{m-1} + \ldots + c_0$, where $c_i = f_i/f_m^{m+1-i} \in \bbQ$ [N.B. our birational transformation is legitimate, since $(m+1)/2$ is an integer]. So, our new right hand side is monic and is still defined over~$\bbQ$. Now, write each $c_i = n_i/d_i$, where each $n_i,d_i\in \bbZ$, with each $n_i,d_i$ coprime. Let $d$ be the least common multiple of $d_0,\ldots d_{(m-1)}$; that is to say, $d$ is the smallest integer divisible by all of the denominators of the coefficients~$c_i$. Now, the birational transformation $(X,Y) \mapsto (Xd^2 , Yd^m)$ [with inverse transformation: $(X,Y) \mapsto (X/d^2 , Y/d^m)$] maps to the curve: $Y^2/d^{2m} = X^m/d^{2m} + c_{m-1} X^{m-1}/d^{2m-2} \ldots + c_0$, which is the same as: $Y^2 = X^m + g_{m-1} X^{m-1} + \ldots + g_0$, where $g_i = c_i d^{2(m-i)}$. But now, each $d_i | d$ and so each $c_i d \in \bbZ$, giving that each $g_i \in \bbZ$, as required. \par Case 2. $f_m$ is a square. Let $w$ be such that $f_m = w^2$. The birational transformation $(X,Y) \mapsto (X , Y/w)$ [with inverse transformation: $(X,Y) \mapsto (X, Yw)$] maps the given curve to: $f_m Y^2 = f_m X^m + \ldots + f_0$, which is the same as: $Y^2 = X^m + c_{m-1} X^{m-1} + \ldots + c_0$, where $c_i = f_i/f_m \in \bbQ$. We are again in the situation where our right hand side is monic and is still defined over~$\bbQ$. So now apply the same second birational transformation as in Case~1. \par\noindent {\bf (b).} The birational transformation $(X,Y) \mapsto (5X , 5^2Y )$ [with inverse $(X,Y) \mapsto ((1/5)X , (1/5^2)Y )$] maps the given curve to the curve: $Y^2 = X^3 + 75 X^2 + 625$. The birational transformation $(X,Y) \mapsto (X + 25 , Y )$ [with inverse $(X,Y) \mapsto (X - 25 , Y )$] maps this to the curve: $Y^2 = X^3 - 1875 X + 31875$, as required. \medskip \noindent {\bf 4.} \par\noindent {\bf (a).} Since $p \equiv 2$~(mod~$3$), we have $\hbox{gcd}(3,p-1) = 1$, and so there exist $\lambda, \mu \in \bbZ$ such that $3\lambda + (p-1)\mu = 1$. For any $x,y\in\bbF_p^*$, we have $x^{p-1} = y^{p-1} = 1$ [by Fermat's Little Theorem], and so: $$ x^3 = y^3 \Rightarrow \bigl( x^3 \bigr)^\lambda \cdot 1^\mu = \bigl( y^3 \bigr)^\lambda \cdot 1^\mu \Rightarrow \bigl( x^3 \bigr)^\lambda \cdot \bigl(x^{p-1}\bigr)^\mu = \bigl( y^3 \bigr)^\lambda \cdot \bigl(y^{p-1}\bigr)^\mu $$ $$ \ \ \ \ \ \Rightarrow x^{ 3\lambda + (p-1)\mu } = y^{ 3\lambda + (p-1)\mu } \Rightarrow x = y. $$ Thus the map $x \mapsto x^3$ is injective on $\bbF_p^*$ and so is a bijection. Since $0^3 = 0$, the map is also a bijection on $\bbF_p$. Now, rearrange the equation of the curve as: $X^3 = Y^2 - A$. We see that, for each $Y = 0,\ldots, p-1$ in $\bbF_p$, there is precisely one $X \in \bbF_p$ such that $X^3 = Y^2 - A$, giving rise to a total of precisely~$p$ affine points, which becomes $p+1$ points when we include the point at infinity. \smallskip \par\noindent {\bf (b).} Since $p\equiv 3$~(mod~$4$), we have that $\bigl( {-1 \over p} \bigr) = -1$ [i.e.\ $-1$ is not congruent to a square mod~$p$]. First note that $X=0$ gives rise to precisely one point~$(0,0)$. Now, pair each $X\in\bbF_p^*$ with its negative $-X$ mod~$p$. If $X(X^2 + B) = 0$ then the pair $X,-X$ gives rise to precisely two points: $(X,0),(-X,0)$. If $X(X^2 + B) \not= 0$ then one of $X(X^2 + B)$ and $-X(X^2 + B)$ will be a nonzero quadratic residue mod~$p$ and the other will be a quadratic nonresidue mod~$p$ [using the fact that $-1$ is a quadratic nonresidue mod~$p$ and the standard property of Legendre symbols that: $\bigl( {mn \over p} \bigr) \bigl( {m \over p} \bigr) \bigl( {n \over p} \bigr)$]. Whichever one is a nonresidue will contribute no points; which one is a residue will contribute two points: $(X,Y),(X,-Y)$. In all cases, each pair $X,-X$ will contribute two points. This gives rise to a total of precisely~$p$ affine points, which becomes $p+1$ points when we include the point at infinity. \medskip \noindent {\bf 5.} \par\noindent {\bf (a).} We first solve for $g_1,g_0$ and $h_1,h_0$ such that $Q(X) = (X^2 + g_1X + g_0)^2 + (h_1X + h_0)$. Equating coefficients of $X^3$ forces $g_1 = f_3/2$. Successiviely equating coefficients of $X^2,X^1,X^0$ now solves for $g_0,h_1,h_0$, giving: $$ g_1 = f_3/2,\ g_0 = (4f_2-f_3^2)/8,\ h_1 = f_1 - f_3(4f_2-f_3^2)/8,\ h_0 = f_0 - (4f_2-f_3^2)^2/64. $$ \noindent Now define $G(X) = X^2 + g_1 X + g_0$ and $H(X) = h_1 X + h_0$, where $g_1,g_0,h_1,h_0$ are as in the above equation, and we see that indeed $Q(X) = G(X)^2 + H(X)$. Our curve can be written as: $(Y + G(X))(Y-G(X)) = H(X)$. The birational transformation $(X,Y) \mapsto (Y+G(X), X(Y+G(X))$ [that is: $(X,Y) \mapsto (T,S)$, given in the hint] has inverse $(X,Y) \mapsto (Y/X, X - G(Y/X) )$ and transforms the given curve to: $ X ( X - 2 G(Y/X) ) = H(Y/X) $. After multiplying both sides by~$X$ and expanding, this can be rewritten as: $$ 2Y^2 + 2g_1XY + h_1Y = X^3 - 2g_0 X^2 - h_0 X.$$ [You could write this as: $(Y + G(X))(Y-G(X)) = H(X)$ $\leftrightarrow$ $T( T - 2 G(S/T) ) = H(S/T)$ [where $T = Y + G(X),\ S = X(Y + G(X)),\ X = S/T,\ Y = T - G(S/T)]$, which in turn is: $2S^2 + 2g_1 TS + h_1 S = T^3 - 2g_0 T^2 - h_0T$, if you prefer]. One can now easily proceed to remove the term $2g_1XY$ by $(X,Y) \mapsto (X,Y+g_1X/2)$ and the term $h_1Y$ by $(X,Y) \mapsto (X,Y+h_1/4)$, giving the curve: $Y^2 = (1/2) X^3 + \ldots$. Finally, remove the $X^2$ term and then make the right hand side monic \& defined over~$\bbZ$, exactly as in Question~1~(a). \par\noindent {\bf (b).} First apply to $\c_1$ the birational transformation $(X,Y)\mapsto (1/X , Y/X^2)$ [self-inverse] which maps $\c_1$ to: $Y^2 = 9 X^4 + 2$. The map $(X,Y) \mapsto (X,Y/3)$ [with inverse $(X,Y) \mapsto (X,3Y)$] maps this to $Y^2 = X^4 + 2/9$. In the notation of part~(a), we can take $G(X) = X^2$ and $H(X) = 2/9$, and so $g_1=g_0=h_1=0$ and $h_0 = 2/9$. The curve is therefore birationally equivalent to: $ 2Y^2 = X^3 - (2/9)X$; that is, $Y^2 = (1/2)X^3 - (1/9)X$. We finally apply $(X,Y) \mapsto (18X , 108Y)$ [with inverse $(X,Y) \mapsto (X/18 , Y/108)$], which maps this curve to: $(Y/108)^2 = (1/2)(X/18)^3 - (1/9)(X/18)$, which is the same as: $Y^2 = X^3 - 72X$, as required. \par For $\c_2$, the important thing to recognise is that it has the $\bbQ$-rational point $(1,3)$, which allows us to make the constant term a square after a linear change in~$X$. Namely, $(X,Y) \mapsto (X-1,Y)$ [with inverse $(X,Y) \mapsto (X+1,Y)$] maps $\c_2$ to the curve: $Y^2 = 2(X+1)^4 + 7$; that is, $Y^2 = 2X^4 + 8X^3 + 12X^2 + 8X + 9$. Now, $(X,Y) \mapsto (1/X, Y/X^2)$ [self-inverse] maps this to $Y^2 = 9X^4 + 8X^3 + 12X^2 + 8X + 2$. Since the coefficient of~$X^4$ is now a square, we can clearly proceed as for $\c_2$ to get the curve in the required form. \par Although the question does not require it, here in fact (for those who are curious) is a series of birational transformations which transform $Y^2 = 9X^4 + 8X^3 + 12X^2 + 8X + 2$ to the form $Y^2 = X^3 + AX +B$. This gives you an idea of how large the numbers can become (although the coefficients of the final answer aren't too bad). In the notation of part~(c), we have: $g_1 = 4/9, g_0 = 46/81, h_1 = 280/729, h_0 = -658/6561$, and so our curve is birationally equivalent to: $Y^2 +(4/9)XY +(140/729)Y = (1/2)X^3 - (46/81)X^2 + 329/6561$. One now removes the $XY$ and $Y$ terms by the map $(X,Y) \mapsto (X, Y + (2/9)X + 70/729)$ [with inverse: $(X,Y) \mapsto (X, Y - (2/9)X - 70/729)$], which birationally transforms this curve to the curve: $Y^2 = (1/2) X^3 - (14/27)X^2 + (203/2187)X + 4900/531441$. Now apply: $(X,Y)\mapsto (X-28/81,Y)$ [with inverse $(X,Y)\mapsto (X+28/81,Y)$], which maps this curve to the curve: $Y^2 = (1/2)X^3 - (7/81)X$. Finally, the map: $(X,Y) \mapsto (18X, 108Y)$ [with inverse $(X,Y) \mapsto (X/18, Y/108)$] maps this to the curve: $Y^2 = X^3 - 56X$, which is in the required form. In summary: $\c_2 : Y^2 = 2X^4 + 7$ is birationally equivalent to $Y^2 = X^3 - 56X$. \par\noindent {\bf (c).} First, the map $(X,Y) \mapsto (X,Y/i)$ [with inverse: $(X,Y) \mapsto (X,iY)$] maps the given curve to the curve: $Y^2 = X^4 + 1$. In the notation of part~(a), we can take $G(X) = X^2$ and $H(X) = 1$, so that our curve can be birationally transformed over~$\bbQ$ to $2Y^2 = X^3 - X$; that is: $Y^2 = (1/2)X^3 - (1/2)X$. Finally, applying $(X,Y) \mapsto (2X, 4Y)$ [with inverse: $(X,Y) \mapsto (X/2, Y/4)$] birationally transforms this curve to: $Y^2 = X^3 - 4X$, as required. \par It is not possible to birationally transform over $\bbR$ the curve $Y^2 = -X^4 - 1$ to the form $Y^2 = X^3 + AX + B$, since any such birational transformation would give a map between the $\bbR$-rational points of the two curves; but the second curve has infinitely many $\bbR$-rational points (take all $X$ sufficiently large, e.g. all $X > 2\hbox{max}(1,|A|,|B|)$), whereas the first curve has no $\bbR$-rational points. [Note that this also implies that there is no such birational transformation over~$\bbQ$, either.] The curves $Y^2 = -X^4 - 1$ and $Y^2 = X^3 - 4X$ are called {\it twists} of each other, since they are both defined over~$\bbQ$, are birationally equivalent over~$\bbC$, but are not birationally equivalent over~$\bbQ$]. \par Over $\bbQ (\sqrt{-2})$, the curve $Y^2 = -X^4 - 1$ has the $\bbQ (\sqrt{-2})$-rational point $(1,\sqrt{-2})$. Applying the transformation $(X,Y) \mapsto (X-1, Y)$ [with inverse $(X,Y) \mapsto (X+1, Y)$] we get: $Y^2 = -X^4 - 4X^3 - 6X^2 - 4X - 2$. Now, applying $(X,Y) \mapsto (1/X , Y/X^2)$ [self-inverse], we get: $Y^2 = -2 X^4 - 4X^3 - 6X^2 - 4X - 1$. Now, we apply $(X,Y) \mapsto (X, Y/\sqrt{-2})$ [with inverse $(X,Y) \mapsto (X, Y\sqrt{-2})$] to get: $Y^2 = X^4 + 2X^3 +3X^2 + 2X + 1/2$. From part~(a), we can now say that this is birationally equivalent (over~$\bbQ$ and so over $\bbQ (\sqrt{-2})$) to a curve of the form $Y^2 = X^3 + AX + B$, with $A,B\in \bbZ$. \medskip \vfil \eject %\end \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\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. %\def\sqr{$\vcenter{\hrule height .3mm %\hbox {\vrule width .3mm height 2mm \kern 1.4mm %\vrule width .3mm} \hrule height .3mm}$} % \null % %\vsize=19.5 true cm %\hsize=11.5 true cm %\vskip 5 true cm %\def\leqslant{\le} \def\c{{\cal C}} \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} % \chaptitle \noindent \centerline{Elliptic Curves. Solutions to Sheet 2.} \rm \par \bigskip \noindent {\bf 1.} \par\noindent {\bf (a).} $(2,0) + (-(2,0)) = {\bf o}$, where $-(2,0)$ is the inverse of $(2,0)$; but we know that $-(2,0) = (2,-0) = (2,0)$ and so $(2,0) + (2,0) = {\bf o}$; that is, $(2,0)$ is of order~$2$, as required. \par\noindent {\bf (b).} In general, a point $(x,y)$ on $Y^2 = X^3 + f_2X^2 + f_1X + f_0$ has inverse: $(x,-y)$ and so $(x,y)$ is of order~2 exactly when $(x,y) + (x,y) = {\bf o} \iff (x,y) = (x,-y)$. But $y=-y$ precisely when $y=0$. So, we see that all points of order~2 are of the form~$(x,0)$ where~$x$ is a root of the cubic $X^3 + f_2X^2 + f_1X + f_0$. There are at most~3 such points over any field, since a cubic has at most 3~roots in any field. In the complex numbers, of course, there will always be exactly~3 such points. Applying this to the curve $Y^2 = X(X^2-3)$, we see that $X(X^2 - 3)$ has one $\bbQ$-rational root $X=0$, giving $(0,0)$ as the only $\bbQ$-rational point of order~2; the other two roots give the points $(\sqrt{3}, 0)$ and $(-\sqrt{3},0)$, which are the remaining two $\bbC$-rational points of order~2. The curve $Y^2 = X^3 - 7$ has no $\bbQ$-rational point of order~2 (since $X^3 - 7$ has no $\bbQ$-rational roots); over~$\bbC$, there are the 3~points of order~2 given by: $(\omega_1,0),(\omega_2,0),(\omega_3,0)$, where $\omega_k = 7^{1/3}e^{2\pi i k /3}$, for $k=0,1,2$. Finally, the curve $Y^2 = X(X-1)(X-7)$ has the points of order~2: $(0,0),(1,0),(7,0)$ all of which are both $\bbC$-rational and $\bbQ$-rational. Group structures of the $2$-torsion groups over~$\bbQ$ for the three curves are clearly: $C_2, C_1$ and $C_2 \times C_2$, respectively (since all members except~{\bf o} have order~2). \medskip \noindent {\bf 2.} Using $2YY' = 3X^2$, we see that the slope of the curve at~$(0,2)$ is $3\cdot 0^2 / (2\cdot 2) = 0$, and so the tangent to the curve at $(0,2)$ is the line: $Y=2$. The $x$-coordinate of the third point of intersection is then: $0^2 - 0 - 0 = 0$, with corresponding $y$-coordinate~$2$. In summary: The line $Y=2$ intersects the curve at $(0,2)$ with multiplicity~3. Hence: $(0,2) + (0,2) + (0,2) = {\bf o}$, and so: $3(0,2) = {\bf o}$. But, $1(0,2) = (0,2)$ and $2(0,2) = (0,-2)$, and so~$3$ is the smallest $k \ge 1$ such that $k(0,2) = {\bf o}$; that is, $(0,2)$ has order~$3$. \medskip \noindent {\bf 3.} \par\noindent {\bf (a).} As usual, $-(\alpha , 0) = (\alpha , 0)$, and so $(\alpha , 0) + (\alpha , 0) = {\bf o}$; that is, $(\alpha , 0)$ is a point of order~2, as required. We now compute $(x',y') = (x,y) + (\alpha , 0)$. First, the line through $(x,y)$ and $(\alpha , 0)$ is: $Y = (y/(x-\alpha )) (X-\alpha)$. Substituting this into the curve gives: $X^3 - (m^2 + \alpha - a) X^2 + \ldots = 0$, where $m = y/(x-\alpha )$. The sum of the roots of this cubic is: $m^2 + \alpha - a$, and so the $x$-coordinate of the third point of intersection is given by: $x' = m^2 + \alpha - a - x - \alpha = m^2 - x - a = y^2/(x-\alpha)^2 - x - a = (\alpha x + a\alpha + b)/(x-\alpha)$. So, we have: $$ T : x \mapsto {\alpha x + a\alpha + b\over x-\alpha}.$$ \noindent We now check $T^2(x) = T(T(x))$, which is: $$ T : x \mapsto {\alpha (\alpha x + a\alpha + b)/(x-\alpha) + a\alpha + b \over (\alpha x + a\alpha + b)/(x-\alpha)-\alpha}.$$ \noindent On multiplying numerator and denominator by $x-\alpha$ and simplifying, we get: $(\alpha^2 + a\alpha + b)x/(\alpha^2 + a\alpha + b)x$, which is~$x$, as required [Note that $\alpha^2 + a\alpha + b \not= 0$, since the cubic $(X-\alpha)(X^2 + aX + b)$ has distinct roots]. \par\noindent {\bf (b).} For~$Y^2 = (X-\alpha_1)(X-\alpha_2) (X-\alpha_3)$, the formula for~$T_1$ is the same as for~$T$ in part~(a), but with $\alpha,a,b$ replaced by $\alpha_1, -\alpha_2 - \alpha_3, \alpha_2\alpha_3$, respectively. That is: $$ T_1 : x \mapsto {\alpha_1 x - \alpha_1\alpha_2 - \alpha_1\alpha_3 + \alpha_2\alpha_3 \over x - \alpha_1}.$$ \noindent The formula for $T_2$ is the same as for~$T_1$, but with $\alpha_1,\alpha_2,\alpha_3$ replaced by $\alpha_2,\alpha_3,\alpha_1$, respectively. The formula for $T_3$ is the same as for~$T_2$, but (again) with $\alpha_1,\alpha_2,\alpha_3$ replaced by $\alpha_2,\alpha_3,\alpha_1$, respectively. These are then: $$ T_2 : x \mapsto {\alpha_2 x - \alpha_2\alpha_3 - \alpha_2\alpha_1 + \alpha_3\alpha_1 \over x - \alpha_2}.$$ $$ T_3 : x \mapsto {\alpha_3 x - \alpha_3\alpha_1 - \alpha_3\alpha_2 + \alpha_1\alpha_2 \over x - \alpha_3}.$$ On computing $T_1 T_2 (x) = T_1 (T_2 (x))$, and simplifying (by removing a common factor of $\alpha_1 - \alpha_2$ from the numerator and denominator), we get the formula for~$T_3$; that is: $T_1 T_2 = T_3$. The formula for $T_3$ is invariant under $\alpha_1 \leftrightarrow \alpha_2$ and so $T_2 T_1 = T_3$, also. Hence $T_1 T_2 = T_3 = T_2 T_1$, and so $T_1$ and $T_2$ commute. By symmetry, $T_1$ and $T_3$ commute, as do $T_2$ and $T_3$. Furthermore, $T_1 T_2 T_3 = T_3^2$ (since $T_1T_2 = T_3$), and we know from part~(a) that $T_3^2 : x\mapsto x$; hence $T_1 T_2 T_3 : x\mapsto x$, as required. \par Finally, the fixed points of $T_1$ are the solutions to $T_1(x) = x$, that is: $$ \alpha_1 x - \alpha_1\alpha_2 - \alpha_1\alpha_3 + \alpha_2\alpha_3 = (x - \alpha_1) x,$$ which has solutions: $$ x_1 = \alpha_1 + \sqrt{(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_3)}, \ \ x_2 = \alpha_1 - \sqrt{(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_3)}. $$ Computing $T_2(x_1)$ and $T_2(x_2)$ gives $x_2,x_1$, respectively. \medskip \noindent {\bf 4.} {\bf (a).} Let $L$ be the tangent line to the curve at~$P$, and let~$R$ be the third point of intersection; that is, the line and the curve meet at $P,P,R$. Then $P+P+R = {\bf o}$. Then, $3P = {\bf o} \iff R=P \iff$ $L$ intersects $\cal E$ only at~$P$ (3 times). \par\noindent {\bf (b).} The Hessian matrix is: $$ \pmatrix{ -6X_0 & 0 & -2AX_2 \cr 0 & 2X_2 & 2X_1 \cr -2AX_2 & 2X_1 & -2AX_0 - 6BX_2 \cr}. $$ \noindent The determinant is: \par $-6X_0\bigl( 2X_2 ( -2AX_0 - 6BX_2 ) - (2X_1)(2X_1) \bigr) -2AX_2 \bigl( 0 - (2X_2)(-2AX_2) \bigr)$ \par $= 8(3AX_0^2X_2 + 9BX_0X_2^2 + 3X_0X_1^2 - A^2 X_2^3)$. \par\noindent The only projective point $(X_0,X_1,X_2)$ on the curve with $X_2=0$ is $(0,1,0) = {\bf o}$, for which the statement is true, since $3{\bf o} = {\bf o}$ and $(X_0,X_1,X_2) = (0,1,0)$ makes the Hessian determinant~$0$. When $X_2 \not= 0$, we can write everything in affine form, with $x = X_0/X_2$ and $y = X_1/X_2$, when we see that the Hessian determinant (after dividing through by~$8 X_2^3$) is~0 exactly when: \medskip \hskip 5 cm $3Ax^2 + 9Bx + 3xy^2 - A^2 = 0$.\hfill (1) \medskip \noindent Also, the point $P = (x,y)\not= {\bf o}$, written in affine form, has order~3 exactly when $2(x,y) = (x,-y)$ which happens if and only if the $x$-coordinate of $2(x,y)$ is~$x$. But, as usual, the $x$-coordinate of $2(x,y)$ is $m^2 - 2x$, where $m = (3x^2 + A)/(2y)$ (note that $y\not= 0$ here, since $y=0$ would make $P$ be of order~$2$). %So, $P$ is of order~$3$ exactly when So, $P$ being of order~$3$ implies $\bigl( (3x^2 + A)/(2y) \bigr)^2 - 2x = x$, that is: \medskip \hskip 5 cm $-9 x^4 - 6 A x^2 + 12xy^2 - A^2 = 0$. \hfill (2) \medskip \noindent Finally, we use the fact the $(x,y)$ is a point on the curve, so that $y^2 = x^3 + Ax + B$, and so we can replace $y^2$ by $x^3 + Ax + B$ in equations~(1),(2). This makes both equations become the same equation: $3x^4 + 6Ax^2 + 12 B x - A^2 = 0$, as required. %\smallskip %\par{\bf (c).} The above showed that $x$ is the $x$-coordinate %of a point of order~$3$ precisely when $3x^4 + 6Ax^2 + 12 B x - A^2 = 0$. %Now, the discriminant of $q(x) = 3x^4 + 6Ax^2 + 12 B x - A^2 = 0$ %is the resultant $q(x)$ and $q'(x)$, which is %the determinant of the matrix: %$$ \pmatrix{ 0 & 0 & 3 & 0 & 6A & 12B & -A^2 \cr % 0 & 3 & 0 & 6A & 12B & -A^2 & 0 \cr % 3 & 0 & 6A & 12B & -A^2 & 0 & 0 \cr % 0 & 0 & 0 & 12 & 0 & 12A & 12B \cr % 0 & 0 & 12 & 0 & 12A & 12B & 0 \cr % 0 & 12 & 0 & 12A & 12B & 0 & 0 \cr % 12 & 0 & 12A & 12B & 0 & 0 & 0 } %$$ %giving: $2^8 3^4 (4 A^3 + 27 B^2 )^2$. %Since the original curve is an elliptic curve, we know %that $4 A^3 + 27 B^2 \not= 0$, and since $\hbox{Char }K \not= 2,3$, %we know that the above discriminant is nonzero, and so %the roots of the quartic $x_i$ (for $i=1,\ldots ,4$) are distinct. %Each $x_i$ gives rise to two distinct points $(x_i,y_i), (x_i,-y_i)$ %of order~$3$ [note that $y_i\not= 0$, since the points are %not of order~$2$], giving $8$ points of order~$3$ in~$\overline K$. %Together with {\bf o}, this gives $9$ points that are $3$-torsion. \smallskip \par{\bf (c).} By part~(b), the $x$-coordinate of any point of order~3 must be a root of the quartic $3x^4 + 6Ax^2 + 12 B x - A^2$, which has at most~4 roots $x_1,\ldots ,x_4$. Each~$x_i$ gives rise to at most two points $(x_i,y_i), (x_i,-y_i)$ on the curve, giving at most $8$ points of order~$3$. Together with {\bf o}, this gives at most $9$ points that are $3$-torsion. \medskip \vfil \eject %\end \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\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. %\def\sqr{$\vcenter{\hrule height .3mm %\hbox {\vrule width .3mm height 2mm \kern 1.4mm %\vrule width .3mm} \hrule height .3mm}$} % \null % %\vsize=19.5 true cm %\hsize=11.5 true cm %\vskip 5 true cm %\def\leqslant{\le} \def\c{{\cal C}} \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\Q{{\Bbb Q}} \def\F{{\Bbb F}} \def\Z{{\Bbb Z}} \def\R{{\Bbb R}} \def\C{{\Bbb C}} % \chaptitle \noindent \centerline{Elliptic Curves. Solutions to Sheet 3.} \rm \bigskip \noindent {\bf 1.} {\bf (a).} Let $x,y\in K$ be such that $|x| \not= |y|$, without loss of generality, say that $|x| < |y|$. Since $K$ is non-Archimedean, we know that $|x \pm y | \leqslant \hbox{max}( |x|, |y| ) = |y|$, and so it is sufficient to show that $|x \pm y | \not< |y|$. Imagine $|x \pm y | < |y|$; then $|y| = |(x \pm y) - x| \leqslant \hbox{max}(|x \pm y |, |x|) < |y|$, a contradiction, as required. [Note that an immediate consequence is the implication: $| u \pm v | < | u | \Rightarrow |u| = |v|$]. \par\noindent {\bf (b).} Let $x_1, \ldots , x_n \in K$ be such that $|x_\ell| > |x_i|$ for all $i\not= \ell$. Let $x = x_1 + \ldots + x_{\ell-1} + x_{\ell+1} + \ldots + x_n$ and let $y = x_\ell$. Then $|x| \leqslant \hbox{max}(|x_i| : 1\leqslant i \leqslant n, i\not= \ell\} < |y|$ and so: $|x_1 + \ldots + x_n| = |x + y| = |y| = |x_\ell|$, by part~(a). \smallskip \par\noindent {\bf (c).} Since $K,|\ |$ satisfies the triangle inequality $|x+y| \leqslant |x|+|y|$, we can apply the standard trick from first year Analysis: $|x| \leqslant |(x - y) + y| \leqslant |x-y| + |y|$ to give: $|x| - |y| \leqslant |x-y|$; similarly $|y| - |x| \leqslant |x-y|$, and so: $|\ |x| - |y|\ |_\infty \leqslant | x - y |$. If $s_n \rightarrow s$ in $K,|\ |$ then $|s_n - s| \rightarrow 0$ in $\R$, and so $|\ |s_n| - |s|\ |_\infty \leqslant | s_n - s | \rightarrow 0$, giving $|s_n| \rightarrow |s|$ in $\bbR, |\ |_\infty$, as required. \par %Suppose %that $s_n \rightarrow s \not= 0$ in~$\bbQ_p$. %Then $|s| = p^r$ for some $r \in \Z$. Let %$\epsilon = \hbox{min}(p^r - p^{r-1}, p^{r+1} - p^r) > 0$. Then %there exists $N$ such that, for all $n > N$, %$|\ |s_n|_p - |s|_p\ |_\infty < \epsilon$. But, since any $|s_n|_p$ %is an integer power of~$p$, there are no possible $|s_n|_p$ %within $\epsilon$ of $|s|_p$ apart from $|s|_p$ itself, %and so $|s_n|_p = |s|_p$, as required. {\it Alternatively:} Suppose $s \not= 0$; then $|s| > 0$ and taking $\epsilon = |s|$, there exists $N$ such that $|s_n - s| < |s|$ for all $n > N$, so that $|s_n| = |s|$ for all $n > N$ (since if $|s_n| \not= |s|$ then~(a) would give $|s_n - s| = \hbox{max}(|s_n|,|s|) \geqslant |s|$, a contradiction). %\medskip %\smallskip %\par\noindent {\bf (d).} %Let $\sum_{n=1}^\infty x_n$ %converge to $x \in K, |\ |$. From lectures, we know %that $|x_n| \rightarrow 0$ in $\R$. If all $x_n = 0$ then the %result is trivial. Otherwise there exists $|x_i| > 0$ and %there exists $N$ such that, for all $n > N$, $|x_n| < |x_i|$. %The finite set $|x_1|,\ldots ,|x_N|$ has a maximum element %$|x_m|$, say, which will also satisfy $|x_m| > |x_n|$ %for all $n > N$. %So, the set $\{ | x_i | : i \geqslant 1\} \subset \bbR$ indeed %has a maximum element, namely $|x_m|$. %\par Define $s_k = \sum_{n=1}^k x_n$, so that $s_k \rightarrow x$. %Then each $|s_k| \leqslant \hbox{max}(|x_1|,\ldots ,|x_k|) \leqslant |x_m|$; %since, by~(c), $|s_k| \rightarrow |x|$, this gives %that $|x| \leqslant |x_m| = \hbox{max}_i |x_i|$, also. %\par Suppose that there exists $\ell$ such that %$|x_\ell| > |x_i|$ for all $i\not= \ell$. %Then, for all $k > \ell$, we have: %$|s_k| = |x_\ell|$, by~(b). This means that $|s_k|$ %does not converge to~$0$ in~$\R$, and so %$\sum_{n=1}^\infty x_n$ does not converge to~$0$ %in $K,|\ |$. \medskip \noindent {\bf 2.} {\bf (a).} $3/50 = 5^{-2}\cdot 3/2$ (where $5$ has no common factor with either the numerator and denominator of $3/2$) and so $| 3/50 |_5 = 5^2$. Similarly, $3/50 = 3^{1}\cdot 1/50$ and so $| 3/50 |_3 = 3^{-1}$. Similarly, $3/50 = 7^0 \cdot 3/50$, and so $| 3/50 |_7 = 7^0 = 1$. \par $d_5(2/3 , 1/5) = | 2/3 - 1/5 |_5 = | 7/15 |_5 = 5$, $d_7(2/3 , 1/5) = | 7/15 |_7 = 7^{-1}$, $d_{11}(2/3, 1/5) = | 7/15 |_{11} = 1$. \par\noindent {\bf (b).} $ |3/7|_3 = 3^{-1}$, $ |3/7|_7 = 7$. For all other primes~$p$, $\hbox{gcd}(p,3) = \hbox{gcd}(p,7) = 1$ and so $ |3/7|_p = 1$. Note also that $ |3/7|_{\infty} = 3/7$. The given product $\prod | 3/7 |_i$ has all factors equal to~1 apart from the factors: $ |3/7|_3 = 3^{-1}$, $ |3/7|_7 = 7$ and $ |3/7|_3 = 3/7$, whose product is~$1$. Hence the given product $\prod | 3/7 |_i$ is~1. For any $x\in\Q$, write $x = \pm n/d$, where $n$ and $d$ are integers with $\hbox{gcd}(n,d) = 1$, and write $n,d$ in terms of their primes factorisations: $n = p_1^{s_1}\cdot \ldots p_k^{s_k}$ and $d = q_1^{t_1}\cdot \ldots q_\ell^{t_\ell}$, where $p_1,\ldots p_k, q_1,\ldots q_\ell$ are distinct primes. For any prime~$p$, if $p = p_i$ for some~$i$, then $|x|_p = p_i^{-s_i}$. If $p = q_j$ for some~$j$, then $|x|_p = q_j^{t_j}$. If $p=\infty$, then $|x|_p = n/d = (p_1^{s_1}\cdot \ldots p_k^{s_k})/ ( q_1^{t_1}\cdot \ldots q_\ell^{t_\ell})$. For all other primes~$p$, we have $|x|_p = 1$. It follows that the product $\prod | x |_i$ is equal to $(p_1^{-s_1}\cdot \ldots p_k^{-s_k})\cdot q_1^{t_1}\cdot \ldots q_\ell^{t_\ell}\cdot (p_1^{s_1}\cdot \ldots p_k^{s_k})/ ( q_1^{t_1}\cdot \ldots q_\ell^{t_\ell})$, which is again equal to~$1$. \medskip \noindent {\bf 3.} {\bf (a).} $| 1/5^n |_5 = 5^n \rightarrow \infty$; this means that $| 1/5^n |_5$ does not converge, and so $1/5^n$ does not converge in~$\Q_5$ (since, if $a_n \rightarrow \ell$ then $| a_n |_p \rightarrow | \ell |_p$). \par\noindent {\bf (b).} {\it Method 1.} $ | n |_5 \le 5^{-1}$ for $n = 5,10,15,\ldots$ and $ | n |_5 = 1$ otherwise; this means that $| n |_5$ does not converge (since there is a subsequence $| n |_5$ for $5 \not | n$ converging to~$1$ and a subsequence $| n |_5$ for $5 | n$ and $5^2 \not | n$ converging to~$1/5$), and so $n$ does not converge in~$\Q_5$ (since, if $a_n \rightarrow \ell$ then $| a_n |_p \rightarrow | \ell |_p$). \par\noindent {\bf (b).} {\it Method 2.} $n = \sum 1$, which does not converge, since $| 1 |_5 = 1$ which does not converge to~0 (using the result from lectures which says that $\sum c_i$ converges iff $| c_i |_p \rightarrow 0$). \par\noindent {\bf (c).} $n! = 5^r\cdot k$, where $r > n/5 - 1$, since every fifth factor of $1\cdot 2\cdot \ldots n$ (i.e. the factors $5,10,15,\ldots $) contributes at least one new factor of~5. Hence $ | n! - 0 |_5 = | n! |_5 < 5^{-(n/5 - 1)} \rightarrow 0$, and so $n!$ converges to~0 in $\Q_5$. \par \noindent {\bf (d).} $| (3+10^n) - 3 |_5 = |10^n|_5 = |5^n\cdot 2^n|_5 = 5^{-n} \rightarrow 0$, and so $3+10^n$ converges to~$3$ in $\Q_5$. \par \noindent {\bf (e).} $|10^n|_5 = |5^n\cdot 2^n|_5 = 5^{-n} \rightarrow 0$ in $\Q_5$ and so $\sum 10^n$ converges in $\Q_5$ (using the result from lectures which says that $\sum c_i$ converges iff $| c_i |_p \rightarrow 0$). \par \noindent {\bf (f).} $| 7^n |_5 = 1$ which does not converge to~0, and so $\sum 7^n$ does not converge in $\Q_5$ (using the same result from lectures as used in part (e)). %\par %Write $n = 5^r\cdot k$, where $\hbox{gcd}(5,k) = 1$, so that %$| n|_5 = 5^{-r} = 1/5^r > 1/n$. Hence $ |1/n |_5 < n$, %and so $ |5^n/ n|_5 = |5^n|_5 / |n|_5 = 5^{-n} / |n|_5 %< 5^{-n}\cdot n = n/5^n \rightarrow 0$. Hence (using the %which says that $\sum c_i$ converges iff $| c_i |_p \rightarrow 0$), %we have that $\sum {5^n\over n}$ %converges in $\Q_5$. %\medskip %\noindent {\bf 5.} {\bf (a).} Note first that $|1/2|_{257} = 1$ %and so the $257$-adic expansion %of $1/2$ is $a_0,a_1a_2a_3\ldots = a_0 + a_1p^1 + a_2p^2 + \ldots$ %(where $p=257$). %Using: $2 (a_0 + a_1p^1 + a_2p^2 + \ldots) = 1$, %we see that $2 a_0 \equiv 1$ mod~257, and so $a_0 = 129$. %Now choose~$x=129$, %so that $x-r = a_1\cdot 257^1 + \ldots$, %and so $|x-r|_{257} \leqslant 257^{-1}$, as required. %%$|x-1/2|_{257} \leqslant 257^{-1} %%\iff | 2x - 1 |_{257} \leqslant |2|_{257} 257^{-1}= 257^{-1} %%\iff 257 | (2x-1) \iff 2x \equiv 1 (\hbox{mod }257)$. %%So, we can take $x = 129$ (the inverse of 2 mod 257), for %%example. %\par\noindent {\bf (b).} If there were such an $x\in \Z$, then %we would have $|x|_3 \leqslant 1$ (since $|x|_p \leqslant 1$ %for any integer~$x$), whereas $| r |_3 = | 7/9 |_3 = 9 > 1$. %Hence, $| x - r |_3 = \hbox{max}( | r |_3 , |x|_3 ) = 9$ %(using the result that $|a+b|_p = \hbox{max}( |a|_p, |b|_p)$ %when $|a|_p \not= |b|_p$), contradicting $| x - r |_3 \leqslant 3^{-7}$. %Hence, no such~$x$ exists. %\par\noindent %{\bf (c).} Note first that $|1/4|_5 = 1$ and so the $5$-adic expansion %of $1/4$ is $a_0,a_1a_2a_3\ldots %= a_0 + a_1p^1 + a_2p^2 + a_3p^3 + \ldots$ %(where $p=5$). %Using: $4 (a_0 + a_1p^1 + a_2p^2 + a_3p^3 + \ldots) = 1$, %we see first that $4 a_0 \equiv 1$ mod~5, and so $a_0 = 4$. Proceeding %inductively, we find in turn that $a_1 = 3, a_2 = 3, a_3=3$, %and so: $1/4 = 4 + 3\cdot 5^1 + 3\cdot 5^2 + 3\cdot 5^3 + a_4\cdot 5^4 %+ \ldots$. If we now choose~$x$ to be $x=4 + 3\cdot 5^1 + % 3\cdot 5^2 + 3\cdot 5^3 = 469$, we see that $x-r = a_4\cdot 5^4 + \ldots$, %and so $|x-r|_5 \leqslant 5^{-4}$, as required. \medskip \noindent {\bf 4.} %N.B. $| r - x^2 |_p = |x^2 - r |_p$, so it makes %no difference whether we use $| r - x^2 |_p$ or $|x^2 - r |_p$. %\par\noindent {\bf (a).} We are looking for $x$ such that $|x^2 +1|_5 \leqslant 5^{-4}$. Take $x_0 = a_0 = 2$ as the initial approximation for which $|a_0^2 + 1|_5 = |5|_5 = 5^{-1}$. Take $x_1 = a_0 + 5a_1 = 2+5a_1$. Then we want $(2 + 5a_1)^2 \equiv -1$ mod~$5^2$, and so $20 a_1 \equiv -5$ mod~$5^2$, which is the same as $4a_1 \equiv -1$ mod~$5$, and so $a_1 = 1$. This gives a better approximation: $x_1 = 2 + 1\cdot 5 = 7$, which satisfies $|x_1^2 + 1|_5 = 5^{-2}$. Simlarly, we then find $a_2 = 2$ and so $x_2 = 7 + 2\cdot 5^2 = 57$. Then, finally, we similarly find $a_3 = 1$ and so $x_3 = 57 + 1\cdot 5^3 = 182$, satisfying $|x^2 + 1|_5 = 5^{-4}$, as required. \par \noindent {\bf (b).} If there were an integer~$x$ such that $|x^2 - 7/8|_3 \leqslant 3^{-7}$, then we would have $x^2 \equiv 7/8$ mod~$3^7$ and so $x^2 \equiv 7/8$ mod~$3$. But $7/8 \equiv 2$ mod~$3$, which is not a quadratic residue (since $0^2 \equiv 0, 1^2 \equiv 1, 2^2\equiv 1$ mod~3); this means that no such~$x$ can exist. \par\noindent {\bf (c).} If there were an integer~$x$ such that $|x^2 - 5/4 |_5 \leqslant 5^{-4}$ then consider the possibilities for $|x^2|_5$. If $|x^2|_5 > |5/4|_5$ then (using the result that $|a+b|_p = \hbox{max}( |a|_p, |b|_p)$ when $|a|_p \not= |b|_p$) we have $|x^2 - 5/4 |_5 = \hbox{max}(|x^2|_5 , |5/4|_5) = |x^2|_5 > |5/4|_5 = 5^{-1} > 5^{-4}$, contradicting $|x^2 - 5/4 |_5 \leqslant 5^{-4}$. If $|x^2|_5 < |5/4|_5$ then (using the same result) we have $|x^2 - 5/4 |_5 = \hbox{max}(|x^2|_5 , |5/4|_5) = |5/4|_5 = 5^{-1} > 5^{-4}$, again contradicting $|x^2 - 5/4 |_5 \leqslant 5^{-4}$. It must then be that $|x^2|_5 = |5/4|_5 = 5^{-1}$, but this is again impossible since $|x^2|_5 = 5^r$ where $r$ is an even integer. Hence no such~$x$ exists. \medskip \noindent {\bf 5.} $200$ mod $7$ is~$4$, so write: $200 = 4 + 7\cdot 28 = $ (similarly) $ 4 + 7\cdot (0 + 7\cdot 4) = 4 + 0\cdot 7^1 + 4\cdot 7^2 = 4,04$. \par For $3/14$, it is easier first to write: $3/14 = 7^{-1}\cdot 3/2$. First find the $7$-adic expansion of~$3/2$, with the idea that the $7$-adic expansion of $3/14$ will then just be the $7$-adic expansion of $3/2$ shifted one place to the left). Now, $|3/2|_7 = 1$ and so $3/2 = a_0,a_1a_2\ldots$. Using $2( a_0 + 7a_1 + 7^2a_2 + \ldots) = 3$, we first find $2a_0 \equiv 3$ mod~$7$ and so $a_0 = 5$. Continuing as usual, we find that $a_1 = 3, a_2=3, a_3=3\ldots$. At this point, we suspect that $3/2 = 5,\bar{3}$. We can prove this rigorously as follows. Let $\alpha = 5,\bar{3}$. Then $\alpha - 5 = 0,\bar{3}$, and so $7^{-1}(\alpha - 5) - 3 = 3,\bar{3} - 3 = 0,\bar{3} = \alpha - 5$. Hence, $\alpha - 5 - 21 = 7\alpha - 35$ and so $\alpha = 3/2$, proving that $3/2 = 5,\bar{3}$ in $\Q_7$. Finally, $3/14 = 7^{-1}\cdot 3/2 = 53,\bar{3}$. \par %The $5$-adic expansion $23,4$ is just %$2\cdot 5^{-1} + 3\cdot 5^0 + 4\cdot 5^1 = 117/5$. Let $\beta = 2,\overline{34} \in \Q_5$. Then $\beta - 2 = 0,\overline{34}$ and so $5^{-2}(\beta - 2) = 34,\overline{34}$, giving: $5^{-2}(\beta - 2) - 3\cdot 5^{-1} - 4\cdot 5^0 = 0,\overline{34} = \beta - 2$, and so: $\beta = -67/24$. \medskip\noindent {\bf 6.} Let $x\in \Q$. Write $x = \pm n/d$, where $n$ and $d$ are integers with $\hbox{gcd}(n,d) = 1$, and write $n,d$ in terms of their primes factorisations: $n = p_1^{s_1}\cdot \ldots p_k^{s_k}$ and $d = q_1^{t_1}\cdot \ldots q_\ell^{t_\ell}$, where $p_1,\ldots p_k, q_1,\ldots q_\ell$ are distinct primes. If $x\in \Z$ then $\ell = 0$ (i.e. there are no occurrences of $q_j$) and we see that when $p=p_i$ for some~$i$, $|x|_p = p_i^{-s_i} \leqslant 1$, and otherwise $|x|_p = 1$; hence $|x|_p \leqslant 1$ for all~$p$; that is to say: $x\in \Z_p \hbox{ for all }p$. \par Conversely, assume that $x\in \Z_p \hbox{ for all }p$. Then there could not be any occurrence of $q_j$ [since then at $p=q_j$ we would have $|x|_p = q_j^{t_j} > 1$, contradicting $x\in \Z_p$. Hence, $x$ has no denominator and is in~$\Z$. \medskip \noindent {\bf 7.} The number of positive multiples of an integer $k>0$ which are $\leqslant n$ is clearly $\bigl[ {n\over k} \bigr]$. To count the power of $p$ dividing $n!$, since $p$ is prime, it is enough to count the powers of $p$ dividing $1,2,3,\ldots ,n$ and add these powers up. Now, the number of multiples of $p$ among $1,2,3,\ldots ,n$ is $[ {n\over p} ]$. Each multiple of $p^2$ among $1,2,3,\ldots ,n$ gives an additional power of $p$ dividing into $n!$, giving $\bigl[ {n\over p} \bigr] + \bigl[ {n\over p^2} \bigr]$ so far. Continuing in this way we get that the total power of $p$ is as in the given formula. This gives that $|n!|_p = p^{-M}$, where $M = \bigl[ {n\over p} \bigr] + \bigl[ {n\over p^2} \bigr] + \ldots \leqslant {n\over p} + {n\over p^2} + \ldots = {n\over p-1}$. Hence $|n!| \geqslant p^{-{n\over p-1}}$, and so $|1/n!| \leqslant p^{n\over p-1}$. Suppose that $|x| < p^{-{1\over p-1}}$ and let $\tau = |x|/p^{-{1\over p-1}} < 1$. Then $\bigl| {x^n \over n!} \bigr| = \tau^n p^{-{n\over p-1}} \bigl| {1 \over n!} \bigr| \leqslant \tau^n p^{-{n\over p-1}} p^{n\over p-1} = \tau^n \rightarrow 0$. Hence, $\hbox{exp}_p(x)$ converges (using the result from lectures which says that $\sum c_i$ converges iff $| c_i |_p \rightarrow 0$). \par On the other hand, if $|x| \geqslant p^{-{1\over p-1}}$, note that, for any $n=p^\ell$, the above ${n\over p}, {n\over p^2},\ldots {n\over p^\ell} = p^{\ell-1}, p^{\ell-2}, \ldots , 1 \in \Z$ and so $M = p^{\ell-1} + p^{\ell-2}, \ldots + 1 = {p^\ell - 1\over p - 1}$; this means that the subsequence $\bigl| {x^{p^\ell}\over (p^\ell)!}\bigr| \geqslant (p^{-{1\over p-1}})^{p^\ell} p^{p^\ell - 1\over p - 1} = p^{-1 \over p-1}$, and so $\bigl| {x^n \over n!} \bigr| \not\rightarrow 0$. Hence $\hbox{exp}_p(x)$ does not converge when $|x| \geqslant p^{-{1\over p-1}}$. \par When $K = \Q_p$ ($p\not= 2$), any $|x|_p < 1$ must satisfy $|x|_p \leqslant p^{-1}$ [since $|x|_p = p^r$ for some $r\in \Z$] $< p^{-{1\over p-1}}$, since $p\geqslant 3$. Any $|x|_p \geqslant 1$ satisfies $|x| \geqslant p^{-{1\over p-1}}$. When $K = \Q_2$, any $|x|_2 < 1/2$ must satisfy $|x|_2 \leqslant 2^{-2}$ [since $|x|_p = p^r$ for some $r\in \Z$] $< p^{-{1\over p-1}} = 2^{-1}$. Any $|x|_p \geqslant 1/2$ satisfies $|x| \geqslant p^{-{1\over p-1}}= 2^{-1}$. \vfil \eject %\end \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\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. %\def\sqr{$\vcenter{\hrule height .3mm %\hbox {\vrule width .3mm height 2mm \kern 1.4mm %\vrule width .3mm} \hrule height .3mm}$} % \null % %\vsize=19.5 true cm %\hsize=11.5 true cm %\vskip 5 true cm %\def\leqslant{\le} \def\etq{{\cal E}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})} \def\ctq{{\cal C}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})} \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\q{{\Bbb Q}} \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\Q{{\Bbb Q}} \def\F{{\Bbb F}} \def\Z{{\Bbb Z}} \def\R{{\Bbb R}} \def\C{{\Bbb C}} % \chaptitle \noindent \centerline{Elliptic Curves. Solutions to Sheet 4.} \rm \bigskip \noindent {\bf 1.} First note that $-28 = 2^2\cdot (-7)$. In $\Q_2$, we know that any integer congruent to~1 mod~8 is a square, and so certainly $-7$ is a square; hence so is $-28 = 2^2\cdot (-7)$. \par In $\Q_3$, we observe that $|-28|_3 = 1$ and $-28 \equiv 2$ (mod~3) which is not a quadratic residue mod~3; hence~$-28$ is not a square in $\Q_3$. In $\Q_5$, we observe that $|-28|_5 = 1$ and $-28 \equiv 2$ (mod~5) which is not a quadratic residue mod~5; hence~$-28$ is not a square in $\Q_5$. In $\Q_7$, if there were an $x$ such that $x^2 = -28$, then $|x^2|_7 = |-28|_7 = 7^{-1}$, contradicting the fact that $|x^2|_7 = 7^r$ where~$r$ is an even integer; hence $-28$ is not a square in $\Q_7$. In $\Q_{11}$, we observe that $|-28|_{11} = 1$ and $-28 \equiv 5$ (mod~11) which is a quadratic residue mod~11 since $5 \equiv 4^2$ (mod~11); hence~$-28$ is a square in $\Q_{11}$. [The above repeatedly used the result that, when $p\not= 2$ and when $|a|_p = 1$ then $a$ is a square in $\Q_p$ iff it is a square mod~$p$.] %\medskip\noindent {\bf 2.} In $\Q_2$, let $x=4$. Then %$x^3 + x - 3 = 65$ which is an integer congruent to~1 mod~8, %and so, from lectures, is a square in $\Z_2$; that is, %$65 = \alpha^2$ for some $\alpha \in \Q_2$ [and, of course, %$|\alpha|_2^2 = |\alpha^2|_2 = |65|_2 = 1$, so that $|\alpha|_2 =1$, %and so $\alpha$ must be in $\Z_2$]. %Then $x=4, y=\alpha$ %give the required pair in $\Z_2$. %\par In $\Q_3$, let $x=2$. Then $x^3 + x - 3 = 7$, which %satisfies: $|7|_3 = 1$, and $7 \equiv 1^2$ is a quadratic %residue mod~3. Hence (by the same result from notes quoted %at the end of the soln to qn~8) we have that there exists %$\beta \in \Q_3$ s.t. $7 = \beta^2$ [and, of course, as above, %this $\beta \in \Z_3$, also]. %\par In $\Q_5$, let $x=1$. Then $x^3 + x - 3 = -1$, which %satisfies: $|-1|_5 = 1$, and $-1 \equiv 2^2$ is a quadratic %residue mod~5. Now apply the same argument as for~$\Q_3$. %\par In $\Q_7$, let $x=-1$. Then $x^3 + x - 3 = -5$, which %satisfies: $|-5|_7 = 1$, and $-5 \equiv 3^2$ is a quadratic %residue mod~5. Now apply the same argument as for~$\Q_3$. %\par Similarly for $\Q_{13}$, with~$x=1$, and $\Q_{19}$, with $x=0$. %%\par In $\Q_{13}$, let $x=1$. Then $x^3 + x - 3 = -1$, which %%satisfies: $|-1|_{13} = 1$, and $-1 \equiv 5^2$ is a quadratic %%residue mod~13. Now apply the same argument as for~$\Q_3$. %%\par In $\Q_{19}$, let $x=0$. Then $x^3 + x - 3 = -3$, which %%satisfies: $|-3|_{19} = 1$, and $-3 \equiv 4^2$ is a quadratic %%residue mod~19. Now apply the same argument as for~$\Q_3$. %\par Note that $\Delta = 4\cdot 1^3 + 27\cdot (-3)^2 = 247 = 13\cdot 19$, %and so the curve will still be an elliptic curve for any %$p$ not dividing $2\Delta$, that is, for any $p\not=2,13,19$. %For all primes $p \geqslant 11$, $p\not=13,19$ the number of %points on the elliptic curve mod~$p$ (not including the point at infinity) %over $\F_p$ is in the range %$[p- 2\sqrt{p},p + 2\sqrt{p}]$, and for $p\geqslant 11$, we %have $p- 2\sqrt{p} \geqslant 11 - 2\sqrt{11} > 4$. At most 3 %of these points over $\F_p$ are of the form $(v,0)$, %and so there must be at least one point of the form %$(v,w)$ over $\F_p$, with $w\not=0$. Now, let $x=v$; then %$v^3 + v - 3$ must be a quadratic residue mod~p and %nonzero mod~p. Now again apply the same argument as for $p=3$. \medskip \noindent {\bf 2.} In~$\R$, we have for example the root $\sqrt{2}$. In $\Q_2$, note that $17 \equiv 1$ mod~8, and so $17$ is a square in $\Q_2$. In $\Q_{17}$, note that $|2|_{17} = 1$ and $2$ is a quadratic residue mod~$17$, so that $2$ is a square in~$\Q_{17}$. \par Finally, let $p \not= 2,17$. Then each of $({2\over p}), ({17\over p}), ({34\over p})$ is $1$ or $-1$. They cannot all be $-1$ since $({34\over p}) = ({2\over p}) ({17\over p})$ [and $-1 \not= (-1)(-1)$], so at least one of them must be~1. Wlog, say that $({2\over p}) = 1$. Then $|2|_p = 1$ and $2$ is a quadratic residue mod~$p$, so that $2$ must be a square in $\Q_p$. \medskip\noindent {\bf 3.} Suppose there were an $x\in \Q_3$ such that $x^3 = 4$ in $\Q_3$. Then $|x|_3^3 = |4|_3 = 1$ and so $|x|_3 = 1$, which means that $x$ can be written $x = a_0,a_1a_2\ldots$. Reducing $x$ modulo~9 would then give the integer $a_0 + 3a_1$ whose cube is~$4$ modulo~9. But $0,\ldots 8$ square to give: $0,1,8,0,1,8,0,1,8$, respectively, so that $x^3 = 4$ is an impossible congruence mod~9. Hence~$4$ is not a cube in $\Q_3$. \par Let $f(x) = x^3 - 28$ and let $x_0 = 1$. Then $| f(x_0) |_3 = |-27 |_3 = 3^{-3}$, whereas $| f'(x_0) |_3 = | 3 |_3 = 3^{-1}$. Hence $| f(x_0) |_3 < | f'(x_0) |_3^2$, and so by Hensel's Lemma there must be a root of $f(x)$ in $\Q_3$; that is $28$ must be a cube in $\Q_3$. \par Let $f(x) = x^3 - 13$ and let $x_0 = -1$. Then $| f(x_0) |_7 = |-14 |_3 = 7^{-1}$, whereas $| f'(x_0) |_7 = | 3 |_7 = 1$. Hence $| f(x_0) |_7 < | f'(x_0) |_7^2$, and so by Hensel's Lemma again there must be a root of $f(x)$ in $\Q_7$; that is $13$ must be a cube in $\Q_7$. \medskip\noindent {\bf 4.} Let $\c$ be the curve $2 Y^2 = X^4 - 17$. Over~$\R$, the curve has the point $(4,\sqrt{239/2})$. In $\Q_2$, let $x=11$. Then $11^4 - 17 = 2^5 \cdot 457$; but 457 is an integer congruent to~1 mod~8, and so, from lectures, there exists $\gamma \in \Z_2 \subset \Q_2$ such that $457 = \gamma^2$; then $(11, 4\gamma)$ is a point on the curve over~$\Q_2$. For the next few primes, we shall repeatedly use the result from lectures that, if $p\not= 2$ and $|a|_p = 1$, then $a$ is a square in $\Q_p$ iff it is a square mod~$p$. Now, note that $-8,-34,-34,-8,10,-8$ are quadratic residues modulo~$p=3,5,7,11,13,17$, respectively, and so there exists $\gamma \in \Q_p$ such that $\gamma^2 = -8,-34,-34,-8,10,-8$, respectively; this gives the $\Q_p$-rational points on the curve: $(1,\gamma), (0,\gamma/2), (0,\gamma/2), (1,\gamma), (4,\gamma/2), (1,\gamma)$, respectively. Hence $\c$ has $\Q_p$-rational points for all primes up to and including $p=17$ [in fact, it was only necessary to do here $p=2,3,5,7,11,17$ here]. %\par Let $p$ be any prime $p\geqslant 19$. %If $-34$ is a quadratic residue mod~$p$, then (as above) %there exists $\gamma \in \Q_p$ such that $\gamma^2 = -34$, %in which case $(0,\gamma/2)$ is a $\Q_p$-rational point %on the curve. On the other hand, if -34 is not a quadratic %residue mod~$p$, then consider the variant curve: %$\c' : 2 Y^2 = -34(X^4 - 17)$ [that is: %$\c' : Y^2 = -17(X^4 - 17)$], which has the point $(0,17)$. %Furthermore, the quartic has no repeated roots mod~$p$ %[since the resultant is nonzero mod~$p$ for any $p\not= 2,17$] %and the curve is nonsingular mod~$p$. %Therefore, from lectures, $\c'$ is an elliptic curve, %and its number of affine points is at most %$p + 2\sqrt{p}$. At most~4 $x$-values give %$-17(x^4 - 17) \equiv 0$ mod~$p$, and so there are at least %$p-4$ values of~$x$ for which $-17(x^4 - 17)$ %is nonzero mod~$p$. If these were all quadratic residues mod~$p$, %the this would give rise to $2(p-4)$ points on~$\c'$ %[via $(x,y),(x,-y)$ for each such~$x$], %which is impossible for $p \geqslant 19$ [using the fact %that $2(p-4) \leqslant p + 2\sqrt{p} \Rightarrow (\sqrt{p} - 1)^2 \leqslant 9 %\Rightarrow \sqrt{p} \leqslant 4$]. %Hence, these exists %$x_0 \in \Z$ such that $-17(x_0^4 - 17)$ is a nonzero quadratic %nonresidue mod~$p$. Since $-34$ is also a quadratic nonresidue mod~$p$, %it follows that $(-34)\bigl(-17(x_0^4 - 17)\bigr)$, %and so $2(x_0^4 - 17)$ is a nonzero quadratic residue mod~$p$ %[using the rule that the product of two quadratic nonresidues mod~$p$ %always gives a quadratic residue mod~$p$]. \par Let $p$ be any prime $p\geqslant 13$ and $p\not= 17$. Our curve~$\c$ (after multiplying both sides by~$2$) is birationally equivalent to $V^2 = 2 X^4 - 34$, where $V = 2Y$; note that $2 X^4 - 34$ has discriminant $-2^{11} 17^3$. Recall Theorem~1.15 from lectures, that any curve $y^2 = Q(x)$, where $Q(x) = f_4 x^4 + \ldots + f_0$ has nonzero discriminant over~$\F_p$, has at least $p - 1 - 2\sqrt{p} > 4$ affine points over~$\F_p$, and so at least~$5$ affine points. At most~$4$ of these can have $2 x^4 - 34 \equiv 0$~(mod~$p$), and so there exists $x_0 \in \Z$ such that $2(x_0^4 - 17)$ is a nonzero quadratic residue mod~$p$. As above, there exists $\gamma \in \Q_p$ such that $\gamma^2 = 2(x_0^4 - 17)$, and so $(x_0,\gamma/2)$ is a $\Q_p$-rational point on our original curve~$\c$. Hence $\c$ has points in~$\R$ and every~$\Q_p$. \par {\it Alternative method for the above parts:} Note that for $p \notdiv 2\Delta$ (that is: $p \not= 2,17$, so that $\widetilde \c$ is still an elliptic curve mod~$p$) and $p \geqslant 5$, we have $\# {\widetilde \c} (\F_p) \geqslant p+1 - 2\sqrt{p} > 1$, and so the number of affine points is $ > 0$; these are non-singular, since ${\widetilde \c}$ is still an elliptic curve mod~$p$. Also, by a theorem from lectures, any non-singular point on ${\widetilde \c} (\F_p)$ lifts to a point on $\c (\Q_p)$; also, any affine non-singular point on ${\widetilde \c} (\F_p)$ lifts to an affine point on $\c (\Q_p)$ (since the point at infinity on $\c (\Q_p)$ maps to the point at infinity on ${\widetilde \c} (\F_p)$ under the reduction map mod~$p$). This shows the existence of such $x,y \in \Q_p$ for all~$p$ except $p = 2,3$ and bad primes, so only still need to consider $p = 2,3,17$. Also, note that for $p = 3$ there is the non-singular affine point $(1,1)$ on ${\widetilde \c} (\F_3)$, and for $p = 17$ there is the non-singular affine point $(1,3)$ on ${\widetilde \c} (\F_{17})$, and as before these must lift to affine points $(x,y) \in \c(\Q_p)$. Over~$\R$, the curve has the point $(4,\sqrt{239/2})$. In $\Q_2$, let $x=11$. Then $11^4 - 17 = 2^5 \cdot 457$; but 457 is an integer congruent to~1 mod~8, and so, from lectures, there exists $\gamma \in \Z_2 \subset \Q_2$ such that $457 = \gamma^2$; then $(11, 4\gamma)$ is a point on the curve over~$\Q_2$. This completes the alternative method of showing that $\c$ has points in~$\R$ and every~$\Q_p$. \par Now, imagine that $\c$ had a $\Q$-rational point; that is, imagine that there exist $X,Y\in \Q$ such that $2 Y^2 = X^4 - 17$. Let $X = t/r$, where $r,t\in \Z$ and $\hbox{gcd}(r,t) = 1$. Then $2(r^2 Y)^2 = t^4 - 17 r^4 \in \Z$. For any prime~$p$, this gives that $|2|_p |r^2 Y|_p^2 \leqslant 1$. When $p\not= 2$, we have $|2|_p = 1$ and so $|r^2 Y|_p^2 \leqslant 1$ and so $|r^2 Y|_p \leqslant 1$. When $p=2$, we have $2^{-1} |r^2 Y|_2^2 \leqslant 1$; but this still gives $|r^2 Y|_2 \leqslant 1$ [since $|r^2 Y|_2 > 1$ would mean $|r^2 Y|_2 \geqslant 2^1$ and so $2^{-1} |r^2 Y|_2^2 \geqslant 2^{-1} 2^2 > 1$]. Overall, we have shown that $|r^2 Y|_p \leqslant 1$ for all~$p$; since also $r^2 Y \in \Q$, this gives that $r^2 Y \in \Z$; let us say: $s = r^2 Y \in \Z$. Therefore $r,s,t\in\Z$ satisfy $2 s^2 = t^4 - 17 r^4$ and $\hbox{gcd}(r,t) = 1$. \par Let $q$ be any prime such that $q | s$. Note that $q\not= 17$ [since if $17 | s$ then our equation $2 s^2 = t^4 - 17 r^4$ would force $17 | t$, and so $17^2 | 2 s^2$ and $17^2 | t^4$, which would give $17^2 | 17 r^4$, forcing $17 | r$, which would contradict $\hbox{gcd}(r,t) = 1$]. Note also that we must not then have $q|r$ [since if $q|s$ and $q|r$, then our equation $2 s^2 = t^4 - 17 r^4$ would force $q | t$, which would contradict $\hbox{gcd}(r,t) = 1$]. Reducing our equation modulo~$q$ gives that $0 \equiv t^4 - 17 r^4$ and so $(t^2/r^2)^2 \equiv 17$ mod~$q$ [note that division by $r^2$ is allowable mod~$q$ since $r$ is not divisible by~$q$]. Hence $17$ is a nonzero quadratic residue mod~$q$. For $q\not= 2$, quadratic reciprocity then allows us to deduce that~$q$ is a quadratic residue mod~$17$ [since $17 \equiv 1$ (mod~$4$)]. Furthermore, one can check directly that $2$ is a quadratic residue mod~$17$, since $2 \equiv 6^2$ mod~$17$. Overall, we have shown that every prime~$q$ dividing~$s$ must be a quadratic residue mod~$17$, and we can take $s>0$ (if necessary, replacing~$s$ by~$-s$), so that~$s$ is the product of its prime factors. It follows that $s$ itself must be a quadratic residue mod~$17$, since it is a product of quadratic residues mod~$17$. Hence $s^2$ is a nonzero fourth power [also known as a {\it quartic residue}] mod~$17$. Note also that reducing our equation mod~$17$ gives $2 s^2 \equiv t^4$ (mod~$17$), so that $2 s^2$ is also a nonzero fourth power mod~$17$. Since $s, 2s^2$ are both nonzero fourth powers mod~$17$, it follows that $2 = (2s^2)/s^2$ is also a fourth power mod~$17$. On the other hand, one can compute directly that $0^4,1^4,2^4,3^4,4^4,5^4,6^4,7^4,8^4,9^4,10^4,11^4,12^4,13^4,14^4, 15^4,16^4$ are congruent mod~$17$ to: $0,1,16,13,1,13,4,4,16,16,4,4,13,1,13,16,1$, respectively, so that in fact $2$ is not a fourth power mod~$17$. This contradication show that our original curve $\c$ has no $\Q$-rational points. \par\noindent {\bf 5.} Since $p \equiv 2$~(mod~$3$), we have $\hbox{gcd}(3,p-1) = 1$, and so there exist $\lambda, \mu \in \bbZ$ such that $3\lambda + (p-1)\mu = 1$. For any $x,y\in\bbF_p^*$, we have $x^{p-1} = y^{p-1} = 1$ [by Fermat's Little Theorem], and so: $$ x^3 = y^3 \Rightarrow \bigl( x^3 \bigr)^\lambda \cdot 1^\mu = \bigl( y^3 \bigr)^\lambda \cdot 1^\mu \Rightarrow \bigl( x^3 \bigr)^\lambda \cdot \bigl(x^{p-1}\bigr)^\mu = \bigl( y^3 \bigr)^\lambda \cdot \bigl(y^{p-1}\bigr)^\mu $$ $$ \ \ \ \ \ \Rightarrow x^{ 3\lambda + (p-1)\mu } = y^{ 3\lambda + (p-1)\mu } \Rightarrow x = y. $$ Thus the map $x \mapsto x^3$ is injective on $\bbF_p^*$ and so is a bijection. \par Now, let $a\in \Z$ be such that $p\notdiv a$. From above, there must exist $x_0 \in \Z$ such that $x_0^3 \equiv a$ (mod~$p$); clearly $p\notdiv x_0$. Let $f(x) = x^3 - a$. Then $|f(x_0)|_p = |x_0^3 - a|_p < 1$, since $x_0^3 - a \equiv 0$ (mod~$p$). Also, $|f'(x_0)|_p = |3 x_0^2|_2 = 1$, since $|3|_p = 1$ and $p\notdiv x_0$. Hence $|f(x_0)|_p < |f'(x_0)|_p^2$, and so by Hensel's Lemma, there exists $x\in \Z_p$ with $f(x) = 0$, that is, $x^3 = a$, as required. \medskip\noindent {\bf 6.} Recall the standard fact about resultants (stated in Section~$0$, the preliminary reading), that there exist polynomials $p(x),q(x)\in R[x]$ such that: $p(x) f(x) + q(x) g(x) = \hbox{Res}\bigl( f(x), g(x) \bigr)$ Since the discriminant~$D$ is the resultant of $f(x)$ and $f'(x)$, there must exist polynomials $p(x),q(x)\in R[x]$ such that $p(x) f(x) + q(x) f'(x) = D$, and so: $p(a_0) f(a_0) + q(a_0) f'(a_0) = D$. Since $|a_0|\leqslant 1$, and since $p(x),q(x),f(x),f'(x)\in R[x]$, we must have $|p(a_0)|,|f(a_0)|,|q(a_0)|,|f'(a_0)| \leqslant 1$ and so $|D| \leqslant \hbox{max}( |p(a_0) f(a_0)|,|q(a_0) f'(a_0)| ) \leqslant 1$, which also implies $|D^2| = |D|^2 \leqslant |D|$. Now, we are given that $|f(a_0)| < |D|^2$ and so $| p(a_0) f(a_0) | \leqslant | f(a_0) | < |D^2| \leqslant |D|$. Hence $|D| \not= |p(a_0) f(a_0) |$, and so $|q(a_0) f'(a_0)| = |D - p(a_0) f(a_0)| = \hbox{max}( |D|, |p(a_0) f(a_0)|) = |D|$ [using the fact that, if $|u| \not= |v|$ then $|u\pm v| = \hbox{max}(|u|,|v|)$]. Since also $|q(a_0)| \leqslant 1$, this gives that $|f'(a_0)| \geqslant |D|$. Finally, $|f(a_0)| < |D|^2 \leqslant |f'(a_0)|^2$, and so $f(X)$ has a root $a\in R$ by Hensel's Lemma. \medskip\noindent {\bf 7.} Let $f(X) = 5 X^3 - 7 X^2 + 3 X + 6$, so that $f'(X) = 15 X^2 - 14 X + 3$. Let $x_0 = a_0 = 1$. Then $|f(x_0)|_7 = |7|_7 = 7^{-1}$ and $|f'(x_0)|_7 = |4|_7 = 1$. Then Hensel's Lemma tells us that there exists $\alpha \in \Z_7$ with $| \alpha - 1 |_7 < 1$, and the proof tells us to construct a sequence $x_{n+1} = x_n - f(x_n)/f'(x_n)$ which will converge to~$\alpha$, with each $f(x_n) \equiv 0$ (mod~$7^{n+1}$), so that $a = x_3$ will satisfy $|\alpha - a|_7 \leqslant 7^{-4}$. Now compute $x_1 = x_0 - f(x_0)/f'(x_0) = 1 - 7/4 = -3/4 \equiv 36$ (mod~$7^2$). Then compute $x_2 = x_1 - f(x_1)/f'(x_1) = 152494/6313 \equiv 36$ (mod~$7^3$). Finally compute $x_3 = x_2 - f(x_2)/f'(x_2) = 152494/6313 \equiv 379$ (mod~$7^4$). As a check, note that $f(x_3) = f(379) = 271195351 = 7^4\cdot 112951 \equiv 0$ (mod~$7^4$), and all $|f(x_n)|_7 \leqslant 7^{-4}$ for $n\geqslant 3$. So, for all $n\geqslant 3$, the process $x_{n+1} = x_n - f(x_n)/f'(x_n)$ will not affect $x_n$ mod~$7^4$; that is, $x_n \equiv 379$ mod~$7^4$ [and so $|x_n - 379|_7 \leqslant 7^{-4}$] for all $n\geqslant 3$. Since $x_n \rightarrow \alpha$ in $\Z_7$ it follows that $|\alpha - 379|_7 \leqslant 7^{-4}$, as required. \vfil\eject \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\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. %\def\sqr{$\vcenter{\hrule height .3mm %\hbox {\vrule width .3mm height 2mm \kern 1.4mm %\vrule width .3mm} \hrule height .3mm}$} % \null % %\vsize=19.5 true cm %\hsize=11.5 true cm %\vskip 5 true cm %\def\leqslant{\le} \def\etq{{\cal E}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})} \def\ctq{{\cal C}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})} \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\q{{\Bbb Q}} \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\Q{{\Bbb Q}} \def\F{{\Bbb F}} \def\Z{{\Bbb Z}} \def\R{{\Bbb R}} \def\C{{\Bbb C}} % \chaptitle \noindent \centerline{Elliptic Curves. Solutions to Sheet 5.} \rm \medskip\noindent {\bf 1.} We first (as suggested in the hint) show: \par\noindent {\bf Lemma (*).} For any $\alpha = a + b\sqrt{d} \in \Q_p(\sqrt{d})$, $|\alpha|_p \leqslant 1 \Rightarrow |\alpha + 1|_p \leqslant 1$. \par\noindent {\bf Proof.} Assume $|\alpha |_p \leqslant 1$, so that $| a^2 - b^2 d |_p \leqslant 1$. The result is trivial for $b=0$, so we can assume $b\not= 0$. Let: \par\noindent $f(x) = \bigl( x - (a + b\sqrt{d}) \bigr) \bigl( x - (a - b\sqrt{d}) \bigr) = x^2 - 2ax + a^2 - b^2 d$. \par Imagine $|2a|_p > 1$. Then $g(x) = f(x) / (2a) = {x^2\over 2a} - x + {a^2 - b^2 d \over 2a}$ would be defined over $\Z_p$, with $|g(0)|_p = | {a^2 - b^2 d \over 2a} |_p < 1 = |-1|_p^2 = | g'(0) |_p^2$, so by Hensel's Lemma, there would exist a root in $\Z_p$, contradicting the fact that $a + b\sqrt{d}, a - b\sqrt{d} \not\in \Z_p$ (since $d$ is non-square and $b\not= 0$). \par Hence $|2a|_p \leqslant 1$. So: \par\noindent $|\alpha + 1 |_p = |a+1 + b\sqrt{d}|_p = |(a + 1)^2 - b^2 d|_p^{1/2} = | (a^2 - b^2 d) + 2a + 1 |_p^{1/2} \leqslant 1$ (since $|a^2 - b^2 d|_p, |2a|_p, |1|_p \leqslant 1$), completing the proof of Lemma~(*). \par For any $\alpha = a + b\sqrt{d} \in \Q_p(\sqrt{d})$, let $\overline \alpha$ denote $a - b\sqrt{d}$; also, let: \par\noindent $\alpha_1 = a_1 + b_1\sqrt{d},\ \alpha_2 = a_2 + b_2\sqrt{d} \in \Q_p(\sqrt{d})$, where $a,b,a_1,b_1,a_2,b_2\in \Q_p$. \par Note that $|\alpha|_p = |a^2 - b^2 d|_p^{1/2} \geqslant 0$, and that: \par\noindent $|\alpha|_p = 0 \iff a^2 - b^2 d = 0 \iff a = b = 0 \iff \alpha = 0$ (since~$d$ is non-square). \par $|\alpha_1 \alpha_2 |_p = | \alpha_1 \alpha_2 \overline{\alpha_1 \alpha_2} |_p^{1/2} = | \alpha_1 \overline \alpha_1 \alpha_2 \overline \alpha_2 |_p^{1/2} = | \alpha_1 \overline \alpha_1 |_p^{1/2} | \alpha_2 \overline \alpha_2 |_p^{1/2} = | \alpha_1 |_p |\alpha_2 |_p$. \par Also, without loss of generality, $|\alpha_1|_p \leqslant |\alpha_2|_p$, so that $| {\alpha_1 \over \alpha_2} |_p \leqslant 1$. Then: \par\noindent $| \alpha_1 + \alpha_2 |_p = | {\alpha_1 \over \alpha_2} + 1 |_p | \alpha_2 |_p \leqslant $ [by Lemma~(*)] $| \alpha_2 |_p = \hbox{max}\bigl( |\alpha_1|_p, |\alpha_2|_p \bigr)$. \par Hence, $|\ |_p$ is a non-Archimedean valuation on $\Q_p(\sqrt{d})$. Finally, note that when $\alpha = a$ [with $b=0$], our definition of $|\ |_p$ on $\Q_p(\sqrt{d})$ gives $|\alpha|_p = | a^2 - 0^2 d |_p^{1/2} = | a^2 |_p^{1/2} = | a |_p$, which agrees with the usual $|\ |_p$ on $\Q_p$, so that our non-Archimedean valuation on $\Q_p(\sqrt{d})$ extends that on $\Q_p$. \medskip\noindent {\bf 2.} In projective form, the point is: $(-64/25,59/125,1)$. The coordinate with largest $5$-adic value is $59/125$, and on dividing all coordinates through by this, we can also represent the point as: $(-320/59,1,125/59)$, which is in $5$-adic standard form (that is, $1$ is the maximum of the $5$-adic valuations of the coordinates). Reducing mod~$5$ gives $(0,1,0)$ on $\widetilde \e$, which is the point at infinity. \medskip\noindent {\bf 3.} Suppose that $(x,y)$ on $\e$ reduces mod~$p$ to $(0,0)$ on~$\widetilde \e$. Then $|x|_p, |y|_p < 1$, indeed $|x|_p, |y|_p \leqslant p^{-1}$ (since any $p$-adic value is one of: $\ldots ..., p^{-2}, p^{-1}, 1, p^1, p^2, \ldots$, so that if a $p$-adic value is $< 1$ then it must be $\leqslant p^{-1}$), so that $|x|_p^3 \leqslant p^{-3}$. But $|p|_p = p^{-1}$ and so $|x|_p^3 \not= |p|_p$ which means that $|x^3 + p|_p = \hbox{max}(|x|_p , |p|_p) = p^{-1}$. Since $x,y$ satisfy $y^2 = x^3 + p$, we must therefore have: $|y|_p^2 = p^{-1}$, which contradicts the fact that $|y|_p^2 = p^{2r}$ for some integer~$r$. \medskip\noindent {\bf 4.} \par \noindent {\bf (a).} Let $\e : Y^2 = X^3 + p^2$. Then the point $(0,p)$ on $\e$ reduces modulo~$p$ to the cusp~$(0,0)$ on $\widetilde \e : Y^2 = X^3$, which is another way of saying that $(0,0)$ on $\widetilde \e$ lifts to the point $(0,p)$ in~$\e (\Q_p)$. \par \noindent {\bf (b).} Question~3 is an example of this. \par \noindent {\bf (c).} Let $\e : Y^2 = X^3 + X^2 + p^2$. Then the point $(0,p)$ on $\e$ reduces modulo~$p$ to the node~$(0,0)$ on $\widetilde \e : Y^2 = X^3 + X^2$, which is another way of saying that $(0,0)$ on $\widetilde \e$ lifts to the point $(0,p)$ in~$\e (\Q_p)$. \par \noindent {\bf (d).} Let $\e : Y^2 = X^3 + X^2 + p$ and $\widetilde \e : Y^2 = X^3 + X^2$. Then the node~$(0,0)$ on $\widetilde \e$ does not lift to any point in~$\e(\Q_p)$ by the same argument as in Question~2. \medskip\noindent {\bf 5.} Take $F(X,Y) = X + Y + t X Y^p$, clearly non-commutative. One only has to check associativity. $$ F(F(X,Y),Z) = F( X + Y + t X Y^p, Z ) = X + Y + t X Y^p + Z + t(X + Y + t X Y^p) Z^p $$ $$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = X + Y + t X Y^p + Z + t X Z^p + t Y Z^p,$$ since the remaining term $t^2 X Y^p Z^p \in I = t^2 \F_p[t]$ and so $t^2 X Y^p Z^p = 0$ in $R = \F_p[t]/I$. Also: $$ F(X, F(Y,Z)) = X + F(Y,Z) + t X F(Y,Z)^p = X + Y + Z + t Y Z^p + t X (Y + Z + t Y Z^p)^p $$ $$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = X + Y + Z + t Y Z^p + t X (Y^p + Z^p + (t Y Z^p)^p), $$ since all other terms in the binomial expansion of $(Y + Z + t Y Z^p)^p$ are divisible by~$p$ and so are equal to~$0$ in $R$, which has characteristic~$p$. But $(t Y Z^p)^p = t^2 (t^{p-2} Y^p Z^{p^2})\in I$ and so $(t Y Z^p)^p = 0$ in~$R$, so that the above are the same, giving associativity, as required. \medskip\noindent {\bf 6.} Recall from lectures that the first step in deriving the formal group of an elliptic curve is to write the equation $\e : Y^2 = X^3 + A X + B$ as: $\e' : w = f(z,w) = z^3 + A w^2 z + B w^3$, where $z = -x/y, w = -1/y$. We then Inductively define $f_n(z,w)$ by: $f_1(z,w) = f(z,w)$ and $f_{m+1}(z,w) = f_m( z, f(z,w) )$ and define $$ w(z) = \lim_{m\rightarrow \infty} f_m(z,0) \in \Z[A,B][[z]],$$ which satisfies $w(z) = f\bigl( z, w(z) \bigr)$. The terms of $f_1(z,w) = z^3 + A w^2 z + B w^3$ all have weighted degree~$\equiv 3$ [where we give $z$ weight~1, $w$ weight~$3$, $A$ weight~$-4$ and $B$ weight~$-6$]. Suppose that this is also true of $f_m(z,w)$; then $f_{m+1}(z,w) = f_m ( z, z^3 + A w^2 z + B w^3 )$, where $z$ has weighted degree~$1$, $z^3 + A w^2 z + B w^3$ is homogeneous of weighted degree~$3$, and so the same will be true of~$f_{m+1}$. So, by induction, all terms of every $f_m$ and so all terms of $w(z)$ will have weighted degree~$3$, and so $w(z) = \sum c_n z^n$, where $c_n = 0$ unless $n\equiv 3$ mod~$4$. We now perform the addition $ (z_1,w_1) + (z_2,w_2)$. As usual, we first write the line $z = \lambda w + \mu$ through the points, given by $\lambda = (w(z_1) - w(z_2))/(z_1 - z_2)$ and $\mu = (z_1 w(z_2) - z_2 w(z_1))/(z_1 - z_2)$, both in $\Z[A,B][[z_1,z_2]]$. Letting $z_1$ and $z_2$ each have weight~$1$, the numerator $w(z_1) - w(z_2)$ is homogeneous of weighted degree~$3$, and so $\lambda$ is homogeneous of weighted degree~$2$. Similarly, $\mu$ is homogeneous of weighted degree~$3$. As in lectures, substituting $w = \lambda z + \mu$ into $\e'$ gives $\lambda z + \mu = z^3 + A(\lambda z + \mu)^2 z + B (\lambda z + \mu)^3$, and so: $$(1 + A \lambda^2 + B \lambda^3) z^3 + ( 2 A \lambda \mu + 3 B \lambda^2 \mu ) z^2 + \ldots = 0.$$ Let $(z_3, w(z_3))$ be the third point of intersection of $\e'$ and the line $z = \lambda w + \mu$, so that $z_1,z_2,z_3$ are the roots of the above cubic, giving that $z_1 + z_2 + z_3 = -(\hbox{coeff of }z^2)/(\hbox{coeff of }z^3)$, so: $$ z_3 = -z_1 - z_2 - {2 A \lambda \mu + 3 B \lambda^2 \mu\over 1 + A \lambda^2 + B \lambda^3} \in \Z[A,B][[z_1,z_2]], $$ since the denominator is of the form $1 + \phi(z)$, where $\phi(z)$ has no constant term [and so is an invertible power series, with $1/(1 + \phi(z)) = 1 - \phi(z) + \phi(z)^2 + \ldots$]. The sum $(z_1,w_1) + (z_2,w_2) + (z_3,w_3) =$ the identity, and so $(z_1,w_1) + (z_2,w_2) = -(z_3,w_3)$. Negation $(x,y) \mapsto (x,-y)$ induces $(z,w)\mapsto (-z,-w)$ [since $z=-x/y, w=-1/y$], so that the $z$-coordinate of $(z_1,w_1) + (z_2,w_2)$ is given by $F_\e(z_1,z_2)$, where: $$ F_\e(z_1,z_2) = z_1 + z_2 + (\hbox{ terms of degree }\geqslant 2) \in \Z[A,B][[z_1,z_2]]. $$ But note that the expression for $z_3$ above consists of the terms $z_1,z_2$, both homogeneous of weighted degree~$1$, and the fraction whose numerator $2 A \lambda \mu + 3 B \lambda^2 \mu$ is homogeneous of weighted degree~$1$, and whose denominator $1 + A \lambda^2 + B \lambda^3$ is homogeneous of weighted degree~$0$. Therefore the final power series giving the formal group must be homogeneous of weighted degree~$1$. \par For the case when our curve is of the form $Y^2 = X^3 + AX$ (so that $B=0$), each term of the formal group must be an integer multiple of $A^k z_1^{n_1} z_2^{n_2}$. Since the weighted degree is~1 and since $A$ has weight~$-4$, we see that the degree purely in $z_1,z_2$ must by $\equiv 1$ mod~$4$, as required. \par When our curve is of the form $Y^2 = X^3 + B$ (so that $A = 0$) the fact that $B$ has weight~$-6$ similarly gives that each term of the formal group has degree $\equiv 1$ mod~$6$ in $z_1,z_2$. \vfil \eject %\end \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\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. %\def\sqr{$\vcenter{\hrule height .3mm %\hbox {\vrule width .3mm height 2mm \kern 1.4mm %\vrule width .3mm} \hrule height .3mm}$} % \null % %\vsize=19.5 true cm %\hsize=11.5 true cm %\vskip 5 true cm %\def\leqslant{\le} \def\etq{{\cal E}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})} \def\ctq{{\cal C}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})} \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\q{{\Bbb Q}} \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\Q{{\Bbb Q}} \def\F{{\Bbb F}} \def\Z{{\Bbb Z}} \def\R{{\Bbb R}} \def\C{{\Bbb C}} % \chaptitle \noindent \centerline{Elliptic Curves. Solutions to Sheet 6.} \rm \bigskip \noindent {\bf 1.} In all of the following, we use the result from lectures that (when the coefficients of $\cal E$ are in $\Bbb Z$) $\etq$ is isomorphic to a subgroup of $\widetilde {\cal E}$ mod~$p$, where $p\not=2$ is a prime not dividing the discriminant. Note that this typically gives a much faster way of computing $\etq$ than the Nagell-Lutz result. N.B. (a),(b),(c) are about the level that could be asked as part of a 3-hour exam. \par\noindent {\bf (a).} There are the obvious points $P = (0,1)$ of order~$3$ and $Q = (-1,0)$ of order~$2$ in $\etq$, which generate a subgroup of $\etq$ of size~$6$ (namely, the~$6$ points $mP+nQ$ for $0\leqslant m \leqslant 2$ and $0\leqslant n \leqslant 1$). Further, $\Delta = 4A^3 + 27B^2 = 27$, so we can reduce modulo any prime except~$2$ (which must always be avoided) and~$3$. Over ${\Bbb F}_5$, there are only six points: ${\bf o}, (0,\pm 1), (2,\pm 3), (4,0)$. So, we conclude that $\etq$ has size at most~$6$, which means that it consists of precisely the $C_6 = C_2\times C_3$ group of points we have found already. \par Note that, if $P=(0,1)$ and $Q=(-1,0)$, then the complete list of torsion points is: ${\bf o}, P = (0,1), 2P = (0,-1), Q = (-1,0), P+Q = (2,-3), 2P+Q = (2,3)$. \par\noindent {\bf (b).} Here, the (birational over~$\Bbb Q$) transformation $(X,Y) \mapsto (X-1,Y)$ takes the given curve to: $Y^2 = X(X+1)(X-1) = X^3 - X$, which has discriminant $-4$, and so we can reduce modulo the prime~$3$ Over~${\Bbb F}_3$ there are the points: ${\bf o}, (0,0), (1,0), (2,0)$ giving that $\etq$ has size at most~$4$. But, in fact, $\etq$ contains ${\bf o}, (0,0), (-1,0), (1,0)$, which means the this $C_2\times C_2$ group gives all of $\etq$. [{\it Alternatively, even without using a birational transformation to the form $Y^2=X^3 + AX + B$, we could just work entirely with the given equation of the curve, noting that the original cubic $X(X-1)(X-2)$ has no repeated roots~mod~$3$, so that $Y^2 = X(X-1)(X-2)$ is an elliptic curve mod~$3$, and then noting that the only points are: ${\bf o}, (0,0), (1,0), (2,0)$.}] \par\noindent {\bf (c).} The (birational over~$\Bbb Q$) transformation $(X,Y) \mapsto (3^2X, 3^3Y)$ takes the given curve to: $Y^2 = X^3 + 1$ which we have already seen in part~(a) to have a $C_2\times C_3$ group as its torsion group. \par Note that, if $P = (0,1/27)$ and $Q = (-1/9,0)$ then the complete list of torsion points is given by: ${\bf o}, P=(0,1/27), 2P=(0,-1/27), Q = (-1/9,0), P+Q = (2/9,-1/9), 2P+Q = (2/9,1/9)$. \par\noindent {\bf (d). [optional]} The discriminant of $X^3 - 219X + 1654$ is $\Delta = 4(-219)^3 + 27\cdot 1654^2 = 31850496 = 2^{17}3^5$, so we need only avoid the primes~$2,3$ (alternatively, it would be sufficient to compute that $\Delta \not\equiv 0$ mod~$5$, since that is all we shall require). Reducing modulo~$5$ gives the nine points: ${\bf o}, (0,\pm 2), (1,\pm 1), (2,\pm 2), (3,\pm 2)$, and so $\etq$ has size at most~$9$. A short integer search for possible $x$-coordinates reveals the point $P = (11,24)$ of order~$9$, as can be checked by computing $P,2P,3P,\ldots, 9P = {\bf o}$). Hence, $\etq$ is just the $C_9$ group consisting of these points. \par Note that, if $P = (11,24)$, then the complete list of torsion points is: ${\bf o}, P = (11,24), 2P = (-13,48), 3P = (3,-32), 4P = (35,-192), 5P = (35,192), 6P = (3,32), 7P = (-13,-48), 8P = (11,-24)$. %%\par\noindent {\bf (c).} There are the obvious points: ${\bf o}, %%(0,0), (-1,0),(-4,0)$, all of order~$2$ in $\etq$. There's a further %%obvious point: $(2,6)$, and note that $2(2,6) = (0,0)$ so that %%$(2,6) \in \etq$ also, and is of order~$4$. The points $P = (-1,0)$ %%and $Q = (2,6)$ generate a $C_2 \times C_4$ group in $\etq$ %%whose members are: $mP + nQ$ for $0\leqslant m \leqslant 1$ %%and $0\leqslant n \leqslant 3$. So, this group of size~$8$ %%is a subgroup of $\etq$. Now, over~$\Bbb Q$, we can transform $\cal E$ %%(after the map: $(x,y)\mapsto (9x+15,27y)$) to %%the curve: $Y^2 = X^3 - 351X + 1890$. This new cubic has %%discriminant $4(-351)^3 + 27(1890)^3 = -76527504 = 2^4 3^{14}$, %%and so we only need to avoid the primes~$2,3$. [N.B. If you don't %%like the idea of factorising a number like $76527504$, %%you can just observe that a map $x \mapsto kx + \ell$ only %%changes the discriminant of a polynomial by a power of $k$, %%and so we only need to avoid the primes dividing the discriminant %%of $X(X+1)(X+4)$ which is ~$144$ (i.e. the primes~$2,3$) plus %%any primes dividing the coefficient $9$ in ``$9x+15$''. %%This gives a computationally easier way to see that we only %%need to avoid the primes~$2,3$]. Reducing $Y^2 = X^3 - 351X + 1890$ %%modulo~$5$ gives: $Y^2 = X^3 - X$ [N.B. The fact that $4(-1)^3 + 27\cdot 0^2 %%\not\equiv 0$ mod~$5$ gives a further independent proof that~$5$ %%is a legitimate choice of prime]. %%This has the %%points: ${\bf o}, (0,0), (1,0), (2,\pm 1), (3,\pm 2), (4,0)$ %%over ${\Bbb F}_5$, and so $\etq$ has size at most~$8$. Hence, %%$\etq$ is precisely the $C_2\times C_4$ group of points %%given above. %\par\noindent {\bf (c).} There are the obvious $2$-torsion points %${\bf 0}, (0,0), (-81,0), (-256,0)$. %A short integer search for possible further points reveals: %$P = (24,840)$. Note that $2P = (144,-3600)$ and $4P = 2(144,-3600) %= (0,0)$, so that $8P = 2(0,0) = {\bf o}$, which means that %$P \in \etq$ and is of order~$8$. The point~$P$, together %with the independent point~$Q=(-81,0)$ of order~$2$, generate %a $C_2\times C_8$ group, which is a subgroup of $\etq$. %Finally, if we reduce our curve modulo~$2,3,5,7$, the corresponding reduced %curves are, respectively: $Y^2 = X(X+1)X,\ Y^2 = X^2(X+1),\ Y^2 = X(X+1)^2,\ %Y^2 = X(X+4)^2$, each of which has repeated roots and so %not an elliptic curve (N.B.\ {\it there is no need %to transform to the form %$Y^2 = X^3 + AX + B$}). %The first success is when $p=11$, when the %curve reduces to $Y^2 = X(X+4)(X+3) = X^3 + 7X^2 + X$, %which is an elliptic curve (mod~11), since there are no repeated %roots~(mod~11). So, we can compute %all points on the elliptic curve $Y^2 = X^3 + 7X^2 + X$ mod~$11$, %and we find %a total of~$16$ points, given by: %${\bf o}, (0,0), (1,3), (1,8), (2,4), (2,7), (3,4), (3,7)$, %$(4,2), (4,9), (6,1), (6,10), (7,0), (8,0), (10,4), (10,7)$. %This means that $\etq$ consists %precisely of the~$16$ points given above. %\par Note that, if $P=(24,840)$ and $Q = (-81,0)$, then the complete %list of torsion points is given by: ${\bf o}, P = (24,840), 2P = (144,-3600), %3P = (864,30240), 4P = (0,0), 5P = (864,-30240), 6P = (144,3600), %7P = (24,-840), Q = (-81,0), P+Q = (-216,1080), 2P+Q = (-144,-1008), %3P+Q = (-96,480), 4P+Q = (-256,0), 5P+Q = (-96,-480), %6P+Q = (-144,1008), 7P+Q = (-216,-1080)$. %\par\noindent {\bf (d).} Here, the (birational over~$\Bbb Q$) %transformation $(X,Y) \mapsto (X-1,Y)$ takes the given %curve to: $Y^2 = X(X+1)(X-1) = X^3 - X$, which has discriminant $-4$, %and so we can reduce modulo the prime~$3$ %Over~${\Bbb F}_3$ %there are the points: ${\bf o}, (0,0), (1,0), (2,0)$ giving %that $\etq$ has size at most~$4$. But, in fact, $\etq$ %contains ${\bf o}, (0,0), (-1,0), (1,0)$, which means %the this $C_2\times C_2$ group gives all of $\etq$. %[{\it Alternatively, even without using a birational transformation %to the form $Y^2=X^3 + AX + B$, we could just work entirely %with the given equation of the curve, noting that %the original cubic $X(X-1)(X-2)$ has no repeated %roots~mod~$3$, so that $Y^2 = X(X-1)(X-2)$ is an elliptic curve %mod~$3$, and then noting that the only points are: %${\bf o}, (0,0), (1,0), (2,0)$.}] %\par\noindent {\bf (e).} The (birational over~$\Bbb Q$) %transformation $(X,Y) \mapsto (3^2X, 3^3Y)$ takes the %given curve to: $Y^2 = X^3 + 1$ which we have already seen %in part~(a) to have a $C_2\times C_3$ group as its %torsion group. %\par Note that, if $P = (0,1/27)$ and $Q = (-1/9,0)$ then the %complete list of torsion points is given by: ${\bf o}, P=(0,1/27), %2P=(0,-1/27), Q = (-1/9,0), P+Q = (2/9,-1/9), 2P+Q = (2/9,1/9)$. %\par\noindent {\bf (f).} The (birational over~$\Bbb Q$) map %$(X,Y) \mapsto ( 4X , 8Y + 4)$ takes the given curve %to the curve: $Y^2 = X^3 - 16X + 848$. The discriminant %$4(-16)^3 + 27(848)^2 \equiv 1\cdot 2^3 + 0 \equiv 2$ mod~$3$, %so that $3$ does not divide~$4(-16)^3 + 27(848)^2$, and so it's %legitimate to reduce mod~$3$. %The reduced curve mod~$3$ is: %$Y^2 = X^3 + 2X + 2$ which has {\bf o} as its only point. %It follows that $\etq$ contains only {\bf o}, also. %%\par\noindent {\bf (h).} $Y^2 = X^3 - X^2 + 1/4$. \medskip\noindent {\bf 2.} Let $\phi : \c (\q ) \rightarrow \d (\q)$ be the usual isogeny, which is a group homomorphism with kernel: ${\bf o}, (0,0)$, and let $\hat\phi : \d (\q ) \rightarrow \c (\q)$ be the usual dual isogeny, which is also a group homomorphism with kernel: ${\bf o}, (0,0)$. Now, let $\cotq$ be the set of torsion elements of $\c (\q)$ which have odd order. First check that $\cotq$ is a subgroup: (1) The identity {\bf o} has order~1, which is odd, so ${\bf o}\in \cotq$, (2) If $P,Q$ have odd torsion orders $m,n$, respectively, then $mn(P+Q) = n(mP) + m(nQ) = {\bf o}$ and so the order of~$P+Q$ divides $mn$, giving that the order of $P+Q$ is odd, i.e.\ $P+Q \in \cotq$, (3) The order of $-P$ is the same as the order of $-P$, so $P\in\cotq \Rightarrow -P \in \cotq$. Similarly define $\dotq$, a subgroup of $\d (\q)$. \par Now, consider any $P\in \cotq$. Then $P$ must have odd order~$m$, say. Let $R = \phi (P)$. Then $mR = m\phi(P) = \phi (mP) = \phi ({\bf o}) = {\bf o}$, which means that the order of~$R$ divides~$m$, and so the order of~$R$ is odd; that is: $R\in \dotq$. Hence, $\phi$ gives a map from $\cotq$ to $\dotq$, which certainly satisfies the homomorphism property $\phi (P+Q) = \phi(P) + \phi(Q)$ for all $P,Q\in \cotq$, since $\phi$ satisfies this property on the larger set $\c (\q )$. Furthermore, the kernel of $\phi : \cotq \rightarrow \dotq$ must only contain {\bf o} [since $(0,0) \not\in \cotq$], and so $\phi : \cotq \rightarrow \dotq$ is an injection [any homomorphism with trivial kernel is an injection]. We have therefore established that there is an injective homomorphism (namely $\phi$) from $\cotq$ to $\dotq$, which implies that $\cotq$ is isomorphic to a subgroup of $\dotq$, namely the subgroup $\phi (\cotq))$ [note, since $\cotq$, $\dotq$ are finite, it follows from this that $\# \cotq = \# \phi (\cotq)) \leqslant \# \dotq$]. (We have just used here the general fact that if $\theta$ is a homomorphism from group~$G$ to group~$H$, then $G / \hbox{ker}\theta$ is isomorphic to $\hbox{im}\theta$, which is the same as $\theta(G)$; when $\theta$ is injective, $\hbox{ker}\theta$ contains only the identity element and $G / \hbox{ker}\theta$ can be replaced by~$G$). Applying the same argument to $\hat\phi : \dotq \rightarrow \cotq$ gives that $\dotq$ is isomorphic to a subgroup of $\cotq$, namely $\hat\phi ( \dotq )$ [and consequently $\# \dotq = \# \hat\phi ( \dotq ) \leqslant \# \cotq$]. So, all of $\cotq, \dotq, \hat\phi ( \dotq ), \hat\phi ( \dotq )$ must have the same number of elements. Combining the fact that $\phi (\cotq))$ is a subgroup of $\dotq$ and the fact that $\# \phi (\cotq)) = \# \dotq$ gives that $\phi (\cotq))$ must be equal to $\dotq$. But $\cotq$ is isomorphic to $\phi (\cotq))$ which is equal to $\dotq$; hence $\cotq$ is isomorphic to $\dotq$, as required. \medskip\noindent {\bf 3.} The preimages of~$(0,0)$ under $\hat\phi$ are given by the points of order~2 on $\d$ distinct from~$(0,0)$, namely $Q_1 = ((-a_1 + \sqrt{a_1^2 - 4b_1})/2,0) = (a + 2\sqrt{b},0)$ and $Q_2 = ((-a_1 - \sqrt{a_1^2 - 4b_1})/2,0) = (a - 2\sqrt{b},0)$. Recall the standard map from lectures $q : \d (\q) \rightarrow \qmods$ which takes ${\bf o} \rightarrow 1$, $(0,0) \rightarrow b_1$, and otherwise takes $(u,v) \rightarrow u$. We know from lectures that this map has kernel precisely $\phi ( \c (\Q ))$. Now, the multiplication by 2 map on $\c (\Q)$ is $\hat\phi \circ \phi$, and so $(0,0) \in 2\c (\Q)$ iff either $Q_1$ or $Q_2$ is a member of $\phi (\c (\Q) ) \iff q(Q_1) = 1 \hbox{ or } q(Q_2) = 1 \iff a+2\sqrt{b} \hbox{ or } a-2\sqrt{b}= 1 \in \qmods \iff a+2\sqrt{b} \hbox{ or } a-2\sqrt{b} \in \qss \iff b = m^2 \hbox{ and } a+2m = n^2$, for some $m,n \in \Q$; but in fact $m,n$ must be in $\Z$ since $a,b\in \Z$. \medskip\noindent {\bf 4.} First let us suppose that our elliptic curve $Y^2 = X^3 + A X + B$ has a point $(x_0,y_0)$ of order~$4$. Performing the birational transformation $(X,Y) \mapsto (X-x_0,Y-y_0)$ [which corresponds to replacing the variables $X,Y$ by $X + x_0, Y + y_0$] maps $(x_0,y_0)$ to $(0,0)$ on the birationally equivalent curve if the form: $Y^2 + 2 y_0 Y + y_0^2 =$ (monic cubic in~$X$), and so is of the form: $Y^2 + 2 y_0 Y = X^3 + f_2 X^2 + f_1 X$ [after absorbing $-y_0^2$ into the cubic in~$X$]. The fact that there is no constant term~$f_0$ is a consequence of the fact that our new curve contains the point~$(0,0)$. We perform the further birational transformation $(X,Y) \mapsto (X,Y - f_1 X/(2 y_0))$ [which corresponds to replacing the variable $Y$ by $Y + f_1 X/(2 y_0)$] which maps to the new birationally equivalent curve $Y^2 + b XY + a Y = X^3 + c X^2$, where $b = f_1/y_0, a = 2 y_0, c = f_2 - (f_1/(2 y_0))^2$]. Note that, for any $r$, the birational transformation $(X,Y) \mapsto ( r^2 X, r^3 Y )$ [which corresponds to replacing the variables $X,Y$ by $X/r^2, Y/r^3$] maps to the birationally equivalent curve: $Y^2 + b r XY + a r^3 Y = X^3 + c r^2 X^2$. Choosing $r = c/a$ forces the coefficients of $Y$ and $X^2$ both to be $c^3/a^2$. So, let $r = c/a, v = c^3 / a^2$ and $w = br = b^2/c$. Note that $y_0 \not= 0$, since $(x_0,y_0)$ is not of order~2, and so $a = 2 y_0 \not= 0$. Also, $c\not= 0$ (since if $c=0$ then $Y=0$ would meet the curve $Y^2 + b XY + a Y = X^3 + c X^2$ 3 times at~$(0,0)$ making $(0,0)$ of order~3, and so $(x_0,y_0)$ of order~3 on the original curve, a contradiction). Hence, it was legitimate to have divided by~$a$ and~$c$. %Note that $a=0 \iff (x_0,y_0)$ is of order~2, and $c=0 \iff (x_0,y_0)$ %is of order~3 on the original curve. Our curve is now of the form: $$ \e : Y^2 + w XY + v Y = X^3 + v X^2,$$ and $(0,0)$ is our point of order~$4$. So far, we have only used the fact that our original curve contained a rational point $(x_0,y_0)$ [not of order~2 or~3], which we have mapped to $(0,0)$. We now use the fact that $(0,0)$ is of order~$4$ on $\e$ to see what condition this gives on $v,w$. First perform the addition $(0,0) + (0,0)$. The line tangent to $\e$ at $(0,0)$ is $Y=0$. Substituting this into $\e$ gives: $X^3 + v X^2$, which has roots $0,0,-v$, and so $(0,0),(0,0),(-v,0)$ are the three points of intersection of $Y=0$ and $\e$, so that $(0,0) + (0,0) + (-v,0) = {\bf o}$, giving: $2(0,0) = -(-v,0)$. Now, since {\bf o} is the point at infinity, the negative of $(-v,0)$ is the other point on~$\e$ with $X$-coordinate~$-v$. Substituting $X=-v$ into~$\e$ gives: $Y^2 - v w Y + v Y = 0$ and so: $Y ( Y - v(w-1) )$, which has roots $0, v(w-1)$. So, the negative of $(-v,w)$ is $(-v,v(w-1))$, giving: $2(0,0) = (-v,v(w-1))$. Now, $(0,0)$ is of order~$4$ iff $2(0,0)$ is of order~$2$ iff $2(0,0)$ equals its own inverse (since it is not the identity). But $(-v,v(w-1))$ is the inverse of $(-v,0)$ and so equals its own inverse iff $v(w-1) = 0$. Now, we cannot have $v = 0$, since then $\e$ would be singular and so would not be an elliptic curve. So, $(0,0)$ having order~$4$ on~$\e$ is equivalent to $w = 1$. Substututing $w=1$ into $\e$ gives the required form. \par\noindent \vfil \eject %\end \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\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. %\def\sqr{$\vcenter{\hrule height .3mm %\hbox {\vrule width .3mm height 2mm \kern 1.4mm %\vrule width .3mm} \hrule height .3mm}$} % \null % %\vsize=19.5 true cm %\hsize=11.5 true cm %\vskip 5 true cm %\def\leqslant{\le} \def\etq{{\cal E}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})} \def\ctq{{\cal C}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})} \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\q{{\Bbb Q}} \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\Q{{\Bbb Q}} \def\F{{\Bbb F}} \def\Z{{\Bbb Z}} \def\R{{\Bbb R}} \def\C{{\Bbb C}} % \chaptitle \noindent \centerline{Elliptic Curves. Solutions to Sheet 7.} \rm \bigskip \noindent {\bf 1.} \par\noindent %{\bf (a).} %Let $\c : Y^2 = X(X^2 + aX + b) = X(X^2 + 3X + 5)$, %where $a=3, b=5$, %and isogenous curve $\d : Y^2 = X(X^2 + a_1X + b_1 ) = X(X^2 - 6X -11)$, %where $a_1 = -2a = -6, b_1= a^2 - 4b = -11$, %with the usual isogeny $\phi : \c (\Q ) \rightarrow %\d (\Q) : (x,y) \mapsto (y^2/x^2 , y - 5y/x^2)$, %and dual isogeny $\hat\phi : \d (\Q ) \rightarrow %\c (\Q) : (u,v) \mapsto ( {1\over 4} v^2/u^2 , {1\over 8}( v + 11v/u^2) )$. %\par %The map $q : \d (\Q) / \phi (\c (\Q)) \mapsto \qmods : (u,v) \mapsto u$, %for $(u,v) \not= (0,0)$, with $q : (0,0) \mapsto b_1$ %and $q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}q$ %contained in $\{ d : d\hbox{ is square free and } d | b_1\} %= \{ \pm 1 , \pm 11 \}$. Also, ${\bf o} \mapsto 1$, %$(0,0) \mapsto -11$ and the obvious point $(-1,2) \mapsto -1$, so that %$1,-11,-1 \in \hbox{im} q$; but $\hbox{im} q$ is a group, so %$11 \in \hbox{im} q$ also, and indeed we can just take: %$(0,0) + (-1,2) = (11,22)$, which maps to $11$ under $q$. %Hence, %$\{ \pm 1,\pm 11 \} \subset \hbox{im} q \subset \{ \pm 1 , \pm 11\}$, %that is $\hbox{im} q = \{ \pm 1 , \pm 11\}$, and so %$\d (\Q) / \phi (\c (\Q)) = \{ {\bf o}, (0,0), (-1,2), (11,22)\}$, %that is, $\d (\Q) / \phi (\c (\Q))$ is generated by~$(0,0),(-1,2)$. %\par %The map $\hat q : \c (\Q ) / \hat\phi (\d (\Q)) %\mapsto \qmods : (u,v) \mapsto u$, %for $(u,v) \not= (0,0)$, with $\hat q : (0,0) \mapsto a_1^2 - 4b_1 = b$ %and $\hat q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}\hat q$ %contained in $\{ d : d\hbox{ is square free and } d | b\} %= \{ \pm 1 , \pm 5\}$. %Also, ${\bf o} \mapsto 1$ and %$(0,0) \mapsto 5$, so that %$\{ 1,5 \} \subset \hbox{im} \hat q \subset \{ \pm 1 , \pm 5\}$. %\par There is only one coset to check, represented by $-1$, say. %We know that $-1 \in \hbox{im} \hat q$ iff there are integers %$\ell , m , n$, %not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: %$(-1)\cdot \ell^4 + a \ell^2m^2 + (b/(-1))\cdot m^4 = n^2$; %that is: $-\ell^4 + 3\ell^2m^2 - 5 m^4 = n^2$. %Multiply both sides by 4 and rewrite %as: $ - ( 2\ell^2 - 3 m^2 )^2 - 11 m^4 = 4 n^2$. We can see that %the LHS is $\leqslant 0$ and the RHS is $\geqslant 0$, %and the equation is satisfied iff $( 2\ell^2 - 3 m^2 ) = 11 m^4 = %4 n^2 = 0$. From this we see that $m=n=0$, which when combined %with $2\ell^2 - 3 m^2 = 0$ gives that $\ell = 0$, also, which %is a contradiction. %Hence $-1 \not\in \hbox{im} \hat q$. %\par We conclude that $\hbox{im} \hat q = \{ 1,5 \}$, and %so $\c (\Q) / \hat\phi (\d (\Q))$ is generated by~$(0,0)$. %\par Finally, since multiplication by 2 in $\c (\Q)$ %is $\hat\phi \circ \phi$, %we have that $\c (\Q) / 2\c (\Q)$ is generated by: generators for %$\c (\Q) / \hat\phi (\d (\Q))$ [namely: $(0,0)$] together with %the images under $\hat \phi$ of generators for $\d (\Q) / \phi (\c (\Q))$ %[namely, $\hat\phi \bigl( (0,0) \bigr) = {\bf o}$ and %$\hat\phi \bigl( (-1,2) \bigr) = (1,3)$]. Conclusion: %$\c (\Q ) / 2\c(\Q )$ is generated by $(0,0)$ and $(1,3)$ %and so is the group $C_2\times C_2$. %Now $\c (\q)/2\c (\q)$ is isomorphic to $\ctq/2\ctq \times C_2^{rank}$, and %$\ctq /2\ctq$ is isomorphic to the $2$-torsion group of $\ctq$ %which is $C_2$ (consisting only of {\bf o} and $(0,0)$), so that %$\ctq/2\ctq$ is isomorphic to $C_2$. Conclusion: rank $=1$. %\par\noindent %{\bf (a)} %Let $\c : Y^2 = X(X^2 + aX + b) = X(X^2 + 5X - 5)$, %where $a=5, b=-5$, %and isogenous curve %$\d : Y^2 = X(X^2 + a_1X + b_1 ) = X(X^2 - 10X + 45)$, %where $a_1 = -10, b_1=45$, %with the usual isogeny $\phi : \c (\Q ) \rightarrow %\d (\Q) : (x,y) \mapsto (y^2/x^2 , y + 5y/x^2)$, %and dual isogeny $\hat\phi : \d (\Q ) \rightarrow %\c (\Q) : (u,v) \mapsto ( {1\over 4} v^2/u^2 , %{1\over 8}( v - 45 v/u^2) )$. %%{\bf [2~marks]} %\par %The map $q : \d (\Q) / \phi (\c (\Q)) \rightarrow \qmods : (u,v) \mapsto u$, %for $(u,v) \not= (0,0)$, with $q : (0,0) \mapsto b_1$ %and $q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}q$ %contained in $\{ d : d\hbox{ is square free and } d | b_1\} %= \{ \pm 1 , \pm 3, \pm 5, \pm 15 \}$. Also, ${\bf o} \mapsto 1$, %$(0,0) \mapsto 45 = 5$. %Hence, %$\{ 1, 5 \} \subset \hbox{im} q \subset \{ \pm 1 , \pm 3, \pm 5, \pm 15 \}$. %%{\bf [3 marks]} %\par %There are only three cosets to check, represented by $-1,3,-3$, say. %We know that $-1 \in \hbox{im} q$ iff there are integers $\ell , m , n$, %not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: %$(-1)\cdot \ell^4 + a_1 \ell^2m^2 + (b_1/(-1))\cdot m^4 = n^2$ %that is: $-\ell^4 - 10 \ell^2m^2 - 45 m^4 = n^2$. %We can see that %the LHS is $\leqslant 0$ and the RHS is $\geqslant 0$, %and there is equality iff $\ell = m = n = 0$, a contradiction. %Hence $-1 \not\in \hbox{im} q$. Similarly $-3 \not\in \hbox{im} q$. %We know that $3 \in \hbox{im} q$ iff there are integers $\ell , m , n$, %not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: %$3\cdot \ell^4 + a_1 \ell^2m^2 + (b_1/3)\cdot m^4 = n^2$ %that is: $3\ell^4 - 10 \ell^2m^2 + 15 m^4 = n^2$. %Rewrite as: $(3\ell^2 - 5 m^2)^2 + 20 m^4 = 3 n^2$. %Reducing modulo~5 %gives: $ ( 3 \ell^2 - 5 m^2 )^2 \equiv 3 n^2$ (modulo~5). If $n$ were %coprime to~$5$, then this would give: $( 3 \ell^2 / n)^2 = 3$ %in $\F_5$, contradicting the fact that~$3$ %is not a quadratic residue modulo~5. So, %$5 | n$ and so $5 | ( 3\ell^2 - 5 m^2 )$, also. Hence $5^2$ divides %$( 3\ell^2 - 5 m^2 )^2$ and $3 n^2$ and so must divide $20 m^4$; %but $20$ is only divisible by $5$ (not by~$5^2$) so that $5$ must divide %$m^4$; hence $5$ divides $m$. Furthermore, the fact (already %found) that %$5 | ( 3 \ell^2 - 5 m^2 )$ gives that $5 | 3\ell^2$, %and so $5 | \ell$ also. We've shown %that $5$ divides both of $\ell,m$, which contradicts %the fact that $\hbox{gcd}(\ell,m) = 1$. %Hence $3 \not\in \hbox{im} \hat q$. %%{\bf [8 marks]} %\par We conclude that $\hbox{im} q = \{ 1,5 \}$ and that %$\d (\Q) / \phi (\c (\Q)) = \{ {\bf o}, (0,0)\}$, %and so $\d (\Q) / \phi (\c (\Q))$ is generated by $(0,0)$. %%{\bf [1 mark]} %\par %The map $\hat q : \c (\Q ) / \hat\phi (\d (\Q)) %\rightarrow \qmods : (x,y) \mapsto x$, %for $(x,y) \not= (0,0)$, with $\hat q : (0,0) \mapsto a_1^2 - 4b_1 = b$ %and $\hat q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}\hat q$ %contained in $\{ d : d\hbox{ is square free and } d | b\} %= \{ \pm 1, \pm 5 \}$. %Also, ${\bf o} \mapsto 1$, %$(0,0) \mapsto -5$, and the obvious point $(-1,3) \mapsto -1$, %which imply that $(0,0)+(-1,3) = (5, 15) \mapsto 5$, %so that %$\{ \pm 1 , \pm 5\} \subset \hbox{im} \hat q \subset \{ \pm 1 , \pm 5\}$. %%{\bf [5 marks]} %\par We conclude that $\hbox{im} \hat q = \{ \pm 1 , \pm 5\}$ and %$\c (\Q) / \hat\phi (\d (\Q)) = %\{ {\bf o}, (0,0), (-1,3), (5, 15)\}$, so that %$\c (\Q) / \hat\phi (\d (\Q))$ is generated by~$(0,0), (-1,3)$. %%{\bf [2 marks]} %\par Finally, since multiplication by 2 in $\c (\Q)$ %is $\hat\phi \circ \phi$, %we have that $\c (\Q) / 2\c (\Q)$ is generated by: generators for %$\c (\Q) / \hat\phi (\d (\Q))$ [namely: $(0,0), (-1,3)$] together with %the images under $\hat \phi$ of generators for $\d (\Q) / \phi (\c (\Q))$ %[namely, $\hat\phi \bigl( (0,0) \bigr) = {\bf o}$]. Conclusion: %$\c (\Q ) / 2\c(\Q )$ is generated by $(0,0), (-1,3)$, %and so is the group $C_2 \times C_2$. %Now $\c (\q)/2\c (\q)$ %is isomorphic to $\ctq/2\ctq \times C_2^{rank}$, and %$\ctq /2\ctq$ is isomorphic to the $2$-torsion group of $\ctq$ %which is $C_2$ (consisting only of {\bf o} and $(0,0)$, since %$x^2 + 5x - 5$ has no $\Q$-rational roots), %so that %$\ctq/2\ctq$ is~$C_2$. Conclusion: rank $=1$. \par\noindent{\bf (a).} Let $\c : Y^2 = X(X^2 + aX + b) = X(X^2 + 2X + 3)$, where $a=2, b=3$, and isogenous curve $\d : Y^2 = X(X^2 + a_1X + b_1) = X(X^2 - 4X - 8)$, where $a_1 = -4, b_1=-8$, with the usual isogeny $\phi : \c (\Q ) \rightarrow \d (\Q) : (x,y) \mapsto (y^2/x^2 , y - 3y/x^2)$, and dual isogeny $\hat\phi : \d (\Q ) \rightarrow \c (\Q) : (u,v) \mapsto ( {1\over 4} v^2/u^2 , {1\over 8}( v + 8v/u^2) )$. %{\bf [3~marks]} \par The map $q : \d (\Q) / \phi (\c (\Q)) \rightarrow \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $q : (0,0) \mapsto b_1$ and $q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}q$ contained in $\{ d : d\hbox{ is square free and } d | b_1\} = \{ \pm 1 , \pm 2 \}$. Also, ${\bf o} \mapsto 1$, $(0,0) \mapsto -8 = -2$, so that $\{ 1,-2 \} \subset \hbox{im} q \subset \{ \pm 1 , \pm 2\}$. %{\bf [3 marks]} \par There is only one coset to check, represented by $-1$, say. We know that $-1 \in \hbox{im} q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a_1 \ell^2m^2 + (b_1/(-1))\cdot m^4 = n^2$ that is: $-\ell^4 - 4 \ell^2m^2 + 8 m^4 = n^2$. Rewrite as: $-(\ell^2 + 2m^2)^2 + 12 m^4 = n^2$. Reducing modulo~3 gives $ - (\ell^2 + 2m^2 )^2 \equiv n^2$ (modulo~3). If $n$ were coprime to~$3$, then this would give: $(( \ell^2 + 2 m^2 )/n)^2 = -1$ in $\F_3$, contradicting the fact that~$-1$ is not a quadratic residue modulo~3. So, $3 | n$ and so $3 | (\ell^2 + 2m^2 )$ also. This means that $9 | (\ell^2 + 2m^2 )^2$ and $9 | n^2$, which can be combined with $-(\ell^2 + 2m^2)^2 + 12 m^4 = n^2$ to give: $9 | 12 m^4$ and so $3 | m$. Combining $3 | m$ with $3 | (\ell^2 + 2m^2 )$ gives that $3 | \ell$. This contradicts the fact that $\hbox{gcd}(\ell,m) = 1$. Hence our equation is impossible in~$\Q_3$, and so impossible in~$\Q$. Hence $-1 \not\in \hbox{im} q$. %{\bf [6 marks]} \par We conclude that $\hbox{im} q = \{ 1,-2 \}$, and so $\d (\Q) / \phi (\c (\Q))$ is generated by~$(0,0)$. %{\bf [1 mark]} \par The map $\hat q : \c (\Q ) / \hat\phi (\d (\Q)) \rightarrow \qmods : (x,y) \mapsto x$, for $(x,y) \not= (0,0)$, with $\hat q : (0,0) \mapsto b$ and $\hat q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}\hat q$ contained in $\{ d : d\hbox{ is square free and } d | b\} = \{ \pm 1 , \pm 3 \}$. Also, ${\bf o} \mapsto 1$ and $(0,0) \mapsto 3$, so that $\{ 1,3 \} \subset \hbox{im} \hat q \subset \{ \pm 1 , \pm 3\}$. %{\bf [3 marks]} \par There is only one coset to check, represented by $-1$, say. We know that $-1 \in \hbox{im} \hat q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a \ell^2m^2 + (b/(-1))\cdot m^4 = n^2$; that is: $-\ell^4 + 2\ell^2m^2 - 3 m^4 = n^2$. Rewrite as: $ - ( \ell^2 - m^2 )^2 - 2 m^4 = n^2$. This is impossible in~$\R$ (the left hand side is $\leqslant 0$ and the right hand side is $\geqslant 0$, and equality only occurs when $\ell^2 - m^2 = m^4 = n^2 = 0$, implying $\ell = m = n = 0$, which is not allowed). Hence $-1 \not\in \hbox{im} \hat q$. %{\bf [4 marks]} \par We conclude that $\hbox{im} \hat q = \{ 1,3 \}$, and so $\c (\Q) / \hat\phi (\d (\Q))$ is generated by~$(0,0)$. %{\bf [1 mark]} \par Finally, since multiplication by 2 in $\c (\Q)$ is $\hat\phi \circ \phi$, we have that $\c (\Q) / 2\c (\Q)$ is generated by: generators for $\c (\Q) / \hat\phi (\d (\Q))$ [namely: $(0,0)$] together with the images under $\hat \phi$ of generators for $\d (\Q) / \phi (\c (\Q))$ [namely, $\hat\phi \bigl( (0,0) \bigr) = {\bf o}$]. Conclusion: $\c (\Q ) / 2\c(\Q )$ is generated by $(0,0)$, and so is isomorphic to $C_2$. We also know that $\c (\Q ) / 2\c(\Q )$ is isomorphic to $\ctq / 2\ctq \times C_2^r$, which is isomorphic to $\c (\Q )[2] \times C_2^r$, where $\c (\Q )[2]$ is the $2$-torsion group and $r$ is the rank. The $2$-torsion points on $\c : Y^2 = X(X^2 + 2X + 3)$ are {\bf o} together with the points of the form $(x,0)$, where $x$ is a root of $X(X^2 + 2X + 3)$, that is: $(0,0)$, $(-1 + \sqrt{-2}, 0)$ and $(-1 + \sqrt{-2}, 0)$, of which only {\bf o} and $(0,0)$ are in $\c (\Q )[2]$, giving that $\c (\Q )[2]$ is isomorphic to $C_2$. Combining this with the facts (already found) that $\c (\Q ) / 2\c(\Q )$ is isomorphic both to $C_2$ and to $\c (\Q )[2] \times C_2^r$, give that $\c (\Q)$ has rank~0. \par\noindent {\bf (b).} Let $\c : Y^2 = X(X^2 + aX + b) = X(X^2 + 14X + 1)$, where $a=14, b=1$, and isogenous curve $\d : Y^2 = X(X^2 + a_1X + b_1 ) = X(X^2 - 28X +192)$, where $a_1 = -28, b_1=192$, with the usual isogeny $\phi : \c (\Q ) \rightarrow \d (\Q) : (x,y) \mapsto (y^2/x^2 , y - y/x^2)$, and dual isogeny $\hat\phi : \d (\Q ) \rightarrow \c (\Q) : (u,v) \mapsto ( {1\over 4} v^2/u^2 , {1\over 8}( v - 192 v/u^2) )$. \par The map $q : \d (\Q) / \phi (\c (\Q)) \mapsto \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $q : (0,0) \mapsto b_1$ and $q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}q$ contained in $\{ d : d\hbox{ is square free and } d | b_1\} = \{ \pm 1 , \pm 2, \pm 3, \pm 6 \}$. Also, ${\bf o} \mapsto 1$, $(0,0) \mapsto 192 = 3$ and the obvious point $(8,16) \mapsto 2$ [N.B. when searching for a point in~$\d (\Q)$ which might map to~$2$ under~$q$, one need only try points with $x$-coordinate equal to~$2$ modulo squares, e.g.~2,8,1/2,18,etc, so one should find the point $(8,16)$ quickly; also, its a good idea at the outset just to look for ``obvious'' members of $\d (\q )$ with $x$ coordinates being integers in the range from $-10$ to $10$; doing this at the outset would also reveal the point $(8,16)$.] This means that $1,3,2 \in \hbox{im} q$; but $\hbox{im} q$ is a group, so $6 \in \hbox{im} q$ also, and indeed we can just take: $(0,0) + (8,16) = (24,-48)$, which maps to $6$ under $q$. Hence, $\{ 1,2,3,6 \} \subset \hbox{im} q \subset \{ \pm 1 , \pm 2,\pm 3, \pm 6\}$. \par There is only one coset to check, represented by $-1$, say. We know that $-1 \in \hbox{im} q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a_1 \ell^2m^2 + (b_1/(-1))\cdot m^4 = n^2$ that is: $-\ell^4 - 28 \ell^2m^2 - 192 m^4 = n^2$. We can see that the LHS is $\leqslant 0$ and the RHS is $\geqslant 0$, and there is equality iff $\ell = m = n = 0$, a contradiction. Hence $-1 \not\in \hbox{im} q$. We conclude that $\hbox{im} q = \{ 1,2,3,6 \}$ and that $\d (\Q) / \phi (\c (\Q)) = \{ {\bf o}, (0,0), (8,16), (24,-48) \}$, and so $\d (\Q) / \phi (\c (\Q))$ is generated by $(0,0)$ and $(8,16)$. \par The map $\hat q : \c (\Q ) / \hat\phi (\d (\Q)) \mapsto \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $\hat q : (0,0) \mapsto a_1^2 - 4b_1 = b$ and $\hat q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}\hat q$ contained in $\{ d : d\hbox{ is square free and } d | b\} = \{ \pm 1 \}$. Also, ${\bf o} \mapsto 1$ and $(0,0) \mapsto 1$, so that $\{ 1 \} \subset \hbox{im} \hat q \subset \{ \pm 1 \}$. \par There is only one coset to check, represented by $-1$, say. We know that $-1 \in \hbox{im} \hat q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a \ell^2m^2 + (b/(-1))\cdot m^4 = n^2$; that is: $-\ell^4 + 14\ell^2m^2 - m^4 = n^2$. Rewrite as: $ - ( \ell^2 - 7 m^2 )^2 + 48 m^4 = n^2$. Reducing modulo~3 gives: $ - ( \ell^2 - 7 m^2 )^2 \equiv n^2$ (modulo~3). If $n$ were coprime to~$3$, then this would give: $(( \ell^2 - 7 m^2 )/n)^2 = -1$ in $\F_3$, contradicting the fact that~$-1$ is not a quadratic residue modulo~3. So, $3 | n$ and so $3 | ( \ell^2 - 7 m^2 )$, also. Hence $9$ divides $( \ell^2 - 7 m^2 )^2$ and $n^2$ and so must divide $48 m^4$; but $48$ is only divisible by $3$ (not by~$9$) so that $3$ must divide $m^4$; hence $3$ divides $m$. Combining this with the fact (already found) that $3 | ( \ell^2 - 7 m^2 )$ gives that $3 | \ell$ also. We've shown that $3$ divides all of $\ell,m,n$, a contradiction. Hence $-1 \not\in \hbox{im} \hat q$. \par We conclude that $\hbox{im} \hat q = \{ 1 \}$, and so $\c (\Q) / \hat\phi (\d (\Q))$ contains only {\bf o}. [N.B. $(0,0)$ should not also be included as a separate member of $\c (\Q) / \hat\phi (\d (\Q))$ even though it is a rational point; $(0,0)$ maps to~1 under~$\hat q$, and so $(0,0) \in \hat\phi (\d (\Q))$; that is $(0,0) = {\bf o}$ in $\c (\Q) / \hat\phi (\d (\Q))$; the same comment applies to the obvious point $(1,4)$; in general, for each distinct member $r$ of $\hbox{im} \hat q$, you should only include exactly one point in $\c (\Q)$ which maps to~$r$]. \par Finally, since multiplication by 2 in $\c (\Q)$ is $\hat\phi \circ \phi$, we have that $\c (\Q) / 2\c (\Q)$ is generated by: generators for $\c (\Q) / \hat\phi (\d (\Q))$ [namely: {\bf o}] together with the images under $\hat \phi$ of generators for $\d (\Q) / \phi (\c (\Q))$ [namely, $\hat\phi \bigl( (0,0) \bigr) = {\bf o}$ and $\hat\phi \bigl( (8,16) \bigr) = (1,-4)$]. Conclusion: $\c (\Q ) / 2\c(\Q )$ is generated by $(1,-4)$, and so is the group $C_2$. Now $\c (\q)/2\c (\q)$ is isomorphic to $\ctq/2\ctq \times C_2^{rank}$, and $\ctq /2\ctq$ is isomorphic to the $2$-torsion group of $\ctq$ which is $C_2$ (consisting only of {\bf o} and $(0,0)$), so that $\ctq/2\ctq$ is isomorphic to~$C_2$. Conclusion: rank~$=0$. [If this seems surprising, then note that $(1,4),(1,-4)$ are points of order~$4$ in $\c (\q)$]. %\par\noindent {\bf (d).} $Y^2 = X(X^2 + 2X + 9)$. %\par\noindent {\bf (e).} $Y^2 = X(X^2 + 9X - 1)$. %\par\noindent {\bf (f).} $Y^2 = X(X-12)(X-36)$. \medskip\noindent {\bf 2.} We are given that $A,+$ is an Abelian group, and that $h : A \rightarrow \R_{\ge 0}$ satisfies: \par\noindent \ \ \ \ (I) There exists a constant~$C$, independent of~$P,Q$, such that \par \ \ \ \ $ | h(P+Q) + h(P-Q) - 2 h(P) - 2 h(Q) | \le C $, for all $P,Q \in A$, \par\noindent \ \ \ \ (II) For any~$B\in\R$, the set $\{P\in A:h(P)\le B\}$ is finite. \par From~(I) we obtain: $h(P+Q) + h(P-Q) - 2 h(P) - 2 h(Q) \le C $ and so (since $h(P-Q) \ge 0$): $h(P+Q) \le h(P+Q) + h(P-Q) \le 2 h(P) + 2 h(Q) + C \le 2 h(P) + C_1(Q)$, where $C_1(Q) = 2 h(Q) + C$. Hence Property~(1) in the definition of height function is satisfied. \par Letting $Q=P$ in~(I), we obtain: \par \ \ \ \ $ | h(2P) + h(e) - 4 h(P) | \le C, \ \ \ \ (*) $ \par\noindent where~$e$ denotes the identity element of the group~$A$. This gives: $h(2P) + h(e) - 4 h(P) \ge -C$, and so: $h(2P) \ge 4 h(P) - C_2$, where $C_2 = C + h(e)$. Hence Property~(2) in the definition of height function is satisfied. Furthermore, (II) is the same as Property~(3) in the definition of height function. Hence all~3 required properties are satisfied, giving that~$h$ is a height function, as required. \par Replacing $P,Q$ in~(I) with $2P,P$, respectively, gives: \par \ \ \ \ $ | h(3P) - 2 h(2P) - h(P) | \le C $. \par\noindent Multiplying $(*)$ by~$2$ gives: \par \ \ \ \ $ | 2 h(2P) + 2 h(e) - 8 h(P) | \le 2C$. \par\noindent These last two equations then give: \par $ | h(3P) - 9 h(P) | = | h(3P) - 2 h(2P) - h(P) + 2 h(2P) + 2 h(e) - 8 h(P) - 2 h(e) |$ \par\ \ \ \ \ $\le | h(3P) - 2 h(2P) - h(P) | + | 2 h(2P) + 2 h(e) - 8 h(P) | + 2 | h(e) | \le C_3,$ \par\noindent where $C_3 = 3C + 2 | h(e) |$ (which is independent of~$P$). \medskip\noindent {\bf 3.} In all of the following, each step multiplies numbers $\leqslant N$ (followed by a possible reduction modulo~$N$), and so we are guaranteed that everything can be done on an $9$-digit calculator, since $N^2$ has only 9~digits. \par\noindent {\bf (a).} First compute (modulo $N=10481$): $2^1 \equiv 2$, $2^2 \equiv 4$, $2^4 \equiv 16$, $2^8 \equiv 256$, $2^{16} \equiv 2650$, $2^{32} \equiv 230$ (where each of these was obtained be squaring the previous one, and reducing modulo~$N$). Now, we write $46$ in base~2: $46 = 2 + 4 + 8 + 32$ and so $2^{46} \equiv 2^2 2^4 2^8 2^{32} \equiv 4\cdot 16 \cdot 256 \cdot 230 \equiv 64\cdot 6475 \equiv 5641$ modulo~$N$, so that $2^{46} - 1 \equiv 5640$ modulo~$N$. \par Now, compute $\hbox{gcd}(5640, N)$ by Euclid's Algorithm: $10481 = 1 \cdot 5640 + 4841$; $5640 = 1\cdot 4841 + 799$; $4841 = 6\cdot 799 + 47$, $799 = 17\cdot 47 + 0$. So, $47$ is a factor of $N$. Compute $10481/47 = 223$, giving the factorisation $N = 10481 = 47 \cdot 223$. %{\bf [7~marks]} \par\noindent {\bf (b).} The line tangent to~$\e$ at $P=(5,11)$ has slope $y'$ given by $2yy' = 3x^2 - 1$, with $x=5,y=11$; that is, the slope is $74/22 = 37/11$. This tangent line also goes through $(5,11)$ and so has equation: $Y = (37/11)X - 64/11$. The $x$-coordinate of $2P$ is therefore $(37/11)^2 - (5+5) = 159/121$. [It will turn out not to be necessary to evaluate this mod~$N$, although if this is done using EA, then it is~7364], and the $y$-coordinate is: $-((37/11)\cdot (159/121) - (64/11)) = 1861/1331$ [again, although unnecessary in this example, this can be computed by EA to be: 6679 mod~N], so that $Q = 2P = (159/121 , 1861/1331)$. We now wish to compute $3P = P + Q$, and so again the first step is to find the line joining $P$ and $Q$. This has slope given by $(1861/1331 - 11)/(159/121 - 5) = 6930/2453$, and so we need to compute $6930/2453$ (modulo~$N=10481$), for which the first step is to find the inverse of $2453$ (modulo~$N=10481$). Using Euclid's Algorithm: $10481 = 4\cdot 2453 + 669$; $2453 = 3 \cdot 669 + 446$; $669 = 1\cdot 446 + 223$; $446 = 2\cdot 223 + 0$. So, we cannot find the inverse of $2453$ (modulo~$N=10481$), and this step has given us a factor~$223$ of~$N$. As before, compute $10481/223 = 47$, giving the factorisation $N = 10481 = 47 \cdot 223$. %{\bf [10~marks]} \par\noindent {\bf (c).} Since $N = 47 \cdot 223$, we have $\phi (N) = 46 \cdot 222 = 10212$. Compute the gcd of $d=4085$ and $\phi(N)$, we see: $10212 = 2\cdot 4085 + 2042$; $4085 = 2\cdot 2042 + 1$, so that $\hbox{gcd}(10212,4085) = 1$. Reversing the steps: $1 = 4085 - 2\cdot 2042 = 4085 - 2\cdot (10212 - 2\cdot 4085) = 5\cdot 4085 - 2\cdot 10212$. Hence, $5$ is the inverse of $4085$ modulo~$10212$. The decoding operation is therefore $X \mapsto X^{5} \hbox{ mod }N$. Computing $6012^{5} = (6012^2)^2 \cdot 6012 \equiv 5656^2\cdot 6012 \equiv 2324\cdot 6012 \equiv 715$. (modulo~$N = 10481$). Also: $3236^{5} = (3236^2)^2 \cdot 3236 \equiv 1177^2 \cdot 3236 \equiv 1837 \cdot 3236 \equiv 1805$ (modulo~$N = 10481$). The decoded message is therefore: $0715,\, 1805$; that is: GORE. %In all of the following, each step multiplies %numbers $\leqslant N$ (followed by a possible reduction %modulo~$N$), and so everything can be done on an $8$-digit %calculator. %\par %Write $k$ in base~$2$ as: $k = 2^6 + 2^8 + 2^{10} + 2^{13} + 2^{14}$. %Now compute $2^rP$ for $r$ from $0$ to $14$. %The line tangent to~$\e$ at $P=(1,1)$ %has slope $y'$ given by $2yy' = 3x^2 + 1$, with $x=1,y=1$; %that is, the slope is $4/2 = 2$. This tangent line also goes %through $(1,1)$ and so has equation: $Y = 2X - 1$. %The $x$-coordinate of $2P$ is therefore $2^2 - (1+1) = 2$, %and the $y$-coordinate is: $-(2\cdot 2 - 1) = -3\equiv 8774$, %so that $Q = 2P \equiv (2 , 8774)$ (modulo~$N=8777$). We now wish %to double the point $Q = 2P$, and so again the first step %is to find the line tangent to~$\e$ at $Q$. This has %slope $y'$ given by $2\cdot 8774 \cdot y' = 3\cdot 2^2 + 1$, %and so we need to compute $(3\cdot 2^2 + 1) / (2\cdot 8774) %\equiv 13 / 8771$ %(modulo~$N=8777$), for which the first step is to find the %inverse of $8771$ (modulo~$N$). %Using Euclid's Algorithm: $8777 = 8771 + 6$; %$8771 = 1461 \cdot 6 + 5$; $6 = 5 + 1$; %$5 = 5 + 0$. So, reversing these steps: $1 = 6 - 5 %= 6 - (8771 - 1461\cdot 6) = 1462\cdot 6 - 8771 = 1462\cdot (8777 - 8771) %- 8771 = -1463 \cdot 8771 + 1462 \cdot 8777$. %So, the inverse of $8771$ is $-1463$ mod~$N$, %and so the slope of our line is~$13\cdot(-1463) \equiv 7312$ (mod~$N$). %Since $(2 , 8774)$ lies on the line, we deduce that the %equation of the line is: $Y= 7312 X - 5850$. The $x$-coordinate %of $2Q = 4P$ is then $7312^2 - 2 - 2 \equiv 4633$ (mod~$N$), %with $y$-coordinate $-(7312\cdot 4633 - 5850) \equiv 8574$. %Hence, $2^2P = 4P \equiv (4633, 8574)$ mod~$N$. Proceeding %in the way, we then get: $2^3 P \equiv (1610,740)$, %$2^4P \equiv (613, 4759)$, %$2^5P \equiv (2,265)$, %$2^6P \equiv (4633, 6229)$, %$2^7P \equiv (1610, 7775)$, %$2^8P \equiv (613, 6769)$, %$2^9P\equiv (2,8774) = 2^1P$, %$2^{10}P \equiv (4633, 8574) = 2^2P$, %$2^{11}P \equiv (1610,740) = 2^3P$, %$2^{12}P \equiv (613, 4759)= 2^4P$, %$2^{13}P \equiv (2,265) = 2^5P$, %$2^{14}P \equiv (4633, 6229)= 2^6P$ (mod~$N$). %Now, $k\cdot P = 2^6P + 2^8P + 2^{10}P +2^{13}P + 2^{14}P$, %and so a natural next computation is $2^6P + 2^8P$ (mod~$N$), %which is $(4633, 6229) + (613, 6769)$ (mod~$N$). This requires %finding the slope of the line through $(4633, 6229)$ and %$(613, 6769)$, that is: $(6229 - 6769)/(4633 - 613)$, %for which the first step is to find the inverse of %$4633-613 = 4020$ mod~$N$. Applying Euclid's Algorithm: %$8777 = 2\cdot 4020 + 737$; $4020 = 5\cdot 737 + 335$; %$737 = 2\cdot 335 + 67$; $335 = 5\cdot 67 + 0$. %Hence, we cannot %find the inverse of $4020$ (modulo~$N=8777$), and this %step has given us a factor~$67$ of~$N$. %Compute $8777/67 = 131$, giving the factorisation %$N = 8777 = 67 \cdot 131$. %\par %Pollard's $p-1$ method will not work if you only %multiply together prime powers for primes less than~$11$. %A natural attempt which will work is, for example, %$\ell=2^3\cdot 3^2\cdot 5\cdot 7\cdot 11 = 27720$ %[indeed, $2\cdot 3\cdot 11$ would work, but there's %no way of knowing that in advance!]. %First compute (modulo $N=8777$): %$2^{2^0} \equiv 2$, $2^{2^1} \equiv 4$, $2^{2^2} \equiv 16$, %$2^{2^3} \equiv 256$, %$2^{2^4} \equiv 4097$, $2^{2^5} \equiv 3785$, %$2^{2^6} \equiv 2161$, %$2^{2^7} \equiv 557$, %$2^{2^8} \equiv 3054$, %$2^{2^9} \equiv 5742$, %$2^{2^{10}} \equiv 4152$, %$2^{2^{11}} \equiv 1076$, %$2^{2^{12}} \equiv 7989$, %$2^{2^{13}} \equiv 6554$, %$2^{2^{14}} \equiv 278$ %(where each of these %was obtained be squaring the previous one, and reducing modulo~$N$; %we stop at $2^{14}$, which is the highest power of $2$ equal %or less than~$\ell$). %Now, we write $\ell$ in base~2: %$\ell = 2^3 + 2^6 + 2^{10} + 2^{11} + 2^{13} %+ 2^{14}$ and %so $2^\ell \equiv 2^{2^3} 2^{2^6} 2^{2^{10}} 2^{2^{11}} 2^{2^{13}} 2^{2^{14}} %\equiv ( 256 \cdot 2161 )\cdot (4152 \cdot 1076 )\cdot (6554 \cdot 278) %\equiv 265 \cdot 59 \cdot 5173 %\equiv 6858 \cdot 5173 %\equiv 8577$ modulo~$N$, %so that $2^\ell - 1 \equiv 8576$ modulo~$N$. %\par %Now, compute $\hbox{gcd}(8576, N)$ by Euclid's Algorithm: %$8777 = 8576 + 201$; $8576 = 42\cdot 201 + 134$; %$201 = 134 + 67$, $134 = 2\cdot 67 + 0$. %So, $67$ is a factor of $N$. %Compute $8777/67 = 131$, again giving the factorisation %$N = 8777 = 67 \cdot 131$. %\par Since $N = 67 \cdot 131$, we have %$\phi (N) =66 \cdot 130 = 8580$. Compute the gcd of $d=4903$ and %$\phi(N)$, we see: $8580 = 1\cdot 4903 + 3677$; %$4903 = 1\cdot 3677 + 1226$; %$3677 = 2\cdot 1226 + 1225$, $1226 = 1\cdot 1225 + 1$, %so that $\hbox{gcd}(8580,4903) = 1$. Reversing the steps: %$1 = 1226 - 1225 = 1226 - (3677 - 2\cdot 1226) %= 3\cdot 1226 - 3677 = 3\cdot (4903 - 3677) - 3677 %= 3\cdot 4903 - 4\cdot 3677 = 3\cdot 4903 - 4\cdot (8580 - 4903) %= 7\cdot 4903 - 4\cdot 8580$. %Hence, $7$ is the inverse of %$4903$ modulo~$8580$. %The decoding operation is therefore $X \mapsto X^{7} \hbox{ mod }N$. %\par %Computing $4195^7 = (4195^2)^2\cdot (4195^2 \cdot 4195)\equiv %140^2 \cdot (140 \cdot 4195) \equiv 2046 \cdot 8018 \equiv 615$ %(modulo~$N = 8777$). Similarly computing $7645^7, 1876^7, \ldots$ %(mod~$N$), we find that the complete decoded message is: %\par %$0615\vert 1520\vert 0201\vert 1212\vert 0009\vert 2001\vert 1209\vert 0100$, %%\par\noindent %\ \ \ \ \ which reads: Football Italia. \vfil \eject %\end % MARKS ALLOCATED AS FOLLOWS: % 1: 20. 2: 20. 3: 30. 4: 30. Total: 100. % In more detail: % 1: 20. 2: 20. 3: each part 10. 4: 30. % By the way, delete 3(c) (and soln). Refer them also to exams qn. 6. % Also, refer them after 4 to exams qn. 7. % Possible qn 6 for the future: y^2 = x*(x^2 + x + 7)? \input amssym.def \input amssym.tex \def\notdiv{{\not\hskip-.5pt |\ }} %\def\Bbb{\bf} \def\ge{\geqslant} \def\le{\leqslant} \nopagenumbers %\magnification=\magstep1 %\hoffset=1truecm %\voffset=2truecm \baselineskip = 5.2 true mm \font\eightsl=cmsl8 \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. %\def\sqr{$\vcenter{\hrule height .3mm %\hbox {\vrule width .3mm height 2mm \kern 1.4mm %\vrule width .3mm} \hrule height .3mm}$} % \null % %\vsize=19.5 true cm %\hsize=11.5 true cm %\vskip 5 true cm %\def\leqslant{\le} \def\etq{{\cal E}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})} \def\ctq{{\cal C}_{\lower 1pt\hbox{\eightsl tors}}({\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\Q{{\Bbb Q}} \def\F{{\Bbb F}} \def\Z{{\Bbb Z}} \def\R{{\Bbb R}} \def\C{{\Bbb C}} % \chaptitle \noindent \centerline{Elliptic Curves. Solutions to Sheet 8.} \bigskip \centerline{Solutions to the Exam-style Questions.} \rm \bigskip\noindent {\bf Question 1.} \par\noindent {\bf (i).} Let $x_0 = a_0 = 2$. Then $| x_0^2 + 3 |_7 = 7^{-1}$. Look for $x_1 = a_0 + 7a_1$ such that $| x_1^2 + 3 |_7 \leqslant 7^{-2}$. This is satisfied if: $(2 + 7a_1)^2 \equiv -3$ (mod $7^2$) $\iff 2\cdot 2\cdot 7 a_1 \equiv -3 - 2^2\ (\hbox{mod } 7^2) \iff 4 a_1 \equiv -1\ (\hbox{mod } 7) \iff a_1 \equiv 5\ (\hbox{mod } 7)$. So, now define: $a_1 = 5$ and $x_1 = a_0 + 7a_1 = 37$. Check that indeed $|x_1^2 + 3 |_7 =|1372|_7 = 7^{-3} < 7^{-1}$.~{\bf [5~marks]} \medskip\par\noindent {\bf (ii).} Compute $1,\overline{2} = 1 + 2p + 2p^2 + 2p^3 + \ldots = 1 + 2p(1 + p + p^2 + \ldots) = 1 + 2p/(1-p) = (1+p)/(1-p)$. So, $1,\overline{2} = -9/8 \iff (1+p)/(1-p) = -9/8 \iff 8(1+p) = -9(1-p) \iff p = 17$.~{\bf [3~marks]} %\par %Compute $11,\overline{1} = p^{-1} + 1 + p + p^2 + p^3 + \ldots %= p^{-1}(1 + p + p^2 + \ldots ) = p/(1-p)$. %So, $11,\overline{1} = -1/110 \iff p/(1-p) = -1/110 %\iff -p(1-p) = -110 \iff p^2 - p - 110 = 0 %\iff (p + 10)(p - 11) = 0 \iff p = -10$~or~$11$. %So we must have~$p=11$ since~$p$ is prime.~{\bf [3~marks]} \medskip\par\noindent {\bf (iii).} We are given that $p\equiv 1$~(mod~$3$). Then~$p\not= 2$, so $p$ is odd, and so satisfies either $p \equiv 1$~(mod~$4$) or $p \equiv 3$~(mod~$4$). In the first case, we have that $\bigl( {-3\over p} \bigr) = \bigl( {3\over p} \bigr) \bigl( {-1\over p} \bigr) = \bigl( {p\over 3} \bigr) \bigl( {-1\over 3} \bigr)$ [by quadratic reciprocity] $= \bigl( {1\over p} \bigr) \bigl( {-1\over 3} \bigr) = 1\cdot 1$ [using the result from notes that $\bigl( {-1\over p} \bigr) = 1$ when $p\equiv 1$~(mod~$4$)]. In the second case, $\bigl( {-3\over p} \bigr) = \bigl( {3\over p} \bigr) \bigl( {-1\over p} \bigr) = -\bigl( {p\over 3} \bigr) \bigl( {-1\over p} \bigr)$ [by quadratic reciprocity] $= -\bigl( {1\over 3} \bigr) \bigl( {-1\over p} \bigr) = -1\cdot (-1)$ [using the result from notes that $\bigl( {-1\over p} \bigr) = 1$ when $p\equiv 1$~(mod~$4$)]. In either case, we have that $\bigl( {-3\over p} \bigr) = 1$, as required. We may apply the general rule from lectures (which follows easily from Hensel's Lemma) that, if $p\not= 2$ and $|a|_p = 1$, then $a$ is a square in $\Q_p^*$ iff $a$ is a quadratic residue in $\F_p$. %From this, %it immediately follows that Hence $-3$ is a square in $\Q_p^*$.~{\bf [6~marks]} \medskip\par\noindent {\bf (iv).} For $p \equiv 2$~(mod~$3$), note that $3$ is coprime to $p-1$, and so there exist $\lambda , \mu \in \Z$ such that $(p-1) \lambda + 3 \mu = 1$. Now, suppose that $v^3 = w^3$ in $\F_p^*$. Then $v^{3 \mu} = w^{3 \mu}$. Also $(p-1) \lambda$ is a multiple of the order of the group $\F_p^*$ and so $v^{(p-1) \lambda} = w^{(p-1) \lambda}$ (since both are equal to~1). Multiplying these last two equations gives: $v^{(p-1) \lambda + 3 \mu} = w^{(p-1) \lambda + 3 \mu}$, and so $v = w$. We have shown that $v^3 = w^3$ implies $v=w$, and so the map $v \mapsto v^3$ is injective, and hence surjective (and hence bijective), from $\F_p^*$ to $\F_p^*$. Since also $0\mapsto 0^3$, the map $v \mapsto v^3$ must also be a bijection from $\F_p$ to $\F_p$. For any~$d\in\Z$ not divisible by~$3$, it follows that there exists~$x_0 \in \Z$ such that~$x_0^3 \equiv d$~(mod~$p$). This is the same as $| f(x_0) |_p < 1$, where $f(x) = x^3 - d$. But from the fact that $d$ is not divisible by~$p$ and $x_0^3 \equiv d$~(mod~$p$) it follows that $x_0$ is not divisible by~$p$, and so $|x_0|_p = 1$, giving that $| f'(x_0) |_p = | 3 x_0^2|_p = 1$. Therefore, $| f(x_0) |_p < | f'(x_0) |_p^2$ and so, by Hensel's Lemma, there exists a solution in~$\Q_p$ to $x^3 - d$; that is, $d$ is a cube in~$\Q_p$, for any $d\in\Z$ not divisible by~$3$. \par We have already established in~(c) that $X^2 + 3 = 0$ has a solution~$\Q_p$ for every $p\equiv 1$~(mod~3), and we have already established (immediately above), since the given $q$ is not divisible by~$3$, that $X^3 - q = 0$ has a solution~$\Q_p$ for every~$p\equiv 2$~(mod~3). For $p=3$, note that we are given $q \equiv 1$~(mod~$27$), so that $ | f(x_0) |_3 \le 3^{-3}$, where $x_0=1$ and $f(x) = x^3 - q$. Also, $| f'(x_0) |_3 = | 3\cdot 1^2 | = 3^{-1}$, so that $ | f(x_0) |_3 < | f'(x_0) |_3^2$. By Hensel's Lemma, it follows that there exists a solution in~$\Q_3$ to $X^3 - q = 0$. Finally note that, since $q>0$, there exists a solution to $X^3 - q = 0$ in~$\R$. Hence there is a solution to $(X^2 + 3)(X^3 - q) = 0$ in~$\R$ and every~$\Q_p$, as required.~{\bf [11~marks]} \eject \noindent {\bf Question 2.} \par\noindent {\bf (i).} Let~$\c$ be the elliptic curve~$Y^2 = X^3 + 3$. The discriminant of $X^3 + 3$ is~$4\cdot 0^3 + 27\cdot 3^2 = 3^5$, and so $\widetilde \c : Y^2 = X^3 + 3$ is an elliptic curve for all $p\not= 2,3$. The elements of $\widetilde \c (\F_5 )$ are: ${\bf o}, (1,\pm 2), (2,\pm 1), (3, 0)$, and so $\widetilde \c (\F_5 )$ has order~$6$. The elements of $\widetilde \c (\F_7 )$ are: ${\bf o}, (1,\pm 2), (2,\pm 2), (3, \pm 3), (4, \pm 2), (5, \pm 3), (6, \pm 3)$, and so $\widetilde \c (\F_7 )$ has order~$13$. Since the torsion subgroup of $\c (\Q)$ injects into both of these groups, its order must divide~$\hbox{gcd}(6,13) = 1$ and so equal~$1$. Hence the torsion subgroup of $\c (\Q)$ is precisely~$\{ {\bf o} \}$.~{\bf [5~marks]} \par\noindent {\bf (ii).} Let~$\d$ be the elliptic curve~$Y^2 = X^3 + 4X$. Then~{\bf o} and~$(0,0)$ [which has order~$2$] are both members of the torsion subgroup. Furthermore, consider the point $(2,4)\in \d(\Q)$; the tangent to~$\d$ at~$(2,4)$ has slope~$(3\cdot 2^2 + 4)/ (2\cdot 4) = 2$ and has equation~$Y = 2X$. So, the $x$-coordinate of $2(2,4)$ is $2^2 - 2 - 2 = 0$, with corresponding $y$-coordinate~$0$, giving: $2(2,4) + (0,0) = {\bf o}$ and so $2(2,4) = -(0,0) = (0,0)$. Hence~$4(2,4) = 2(0,0) = {\bf o}$; similarly~$4(2,-4) = {\bf o}$. We have found~$4$ torsion elements: ${\bf o}, (0,0), (2,4), (2,-4)$, and so the torsion subgroup is of order at least~$4$. The discriminant of $X^3 + 4X$ is~$4\cdot 4^3 + 27\cdot 0^2 = 2^8$, and so $\widetilde \d : Y^2 = X^3 + 4X$ is an elliptic curve for all $p\not= 2$. The elements of $\widetilde \d (\F_3 )$ are: ${\bf o}, (0, 0), (2,\pm 1)$, and so $\widetilde \d (\F_3 )$ has order~$4$. Since the torsion subgroup of $\d (\Q)$ injects into this groups, its order must divide~$4$. Hence the torsion subgroup of $\d (\Q)$ consists precisely of the~$4$ torsion elements already found, namely:~$\{ {\bf o}, (0,0), (2, \pm 4) \}$.~{\bf [6~marks]} \par\noindent {\bf (iii).} Let~$\cal F$ be the elliptic curve~$Y^2 = X(X+1)(X+n^2) = X^3 + (n^2+1)X^2 + n^2 X$. The tangent to~$\d$ at~$(n, n(n+1))$ has slope~$(3n^2 + 2(n^2+1)n + n^2)/ (2n(n+1)) = n+1$, and so equation~$Y = (n+1)X +c$ for some~$c$; the line must pass through~$(n, n(n+1))$ so that~$c=0$ and the equation of the tangent is: $Y = (n+1)X$. The $x$-coordinate of~$P = 2(n, n(n+1))$ is $(n+1)^2 - (n^2+1) - 2n = 0$, with $y$-coordinate~$(n+1)\cdot 0 = 0$. Hence $2(n, n(n+1)) + (0,0) = {\bf o}$, so that $2(n, n(n+1)) = -(0,0) = (0,0)$, giving $4(n, n(n+1)) = {\bf o}$. So $(n, n(n+1))$ is torsion, as must be $(n, -n(n+1)) = -(n, n(n+1))$. Replacing every occurrence of~$n$ with~$-n$ gives that the tangent to~$\d$ at~$(-n, n(n-1))$ has equation~$Y = (-n+1)X$ and~$2(-n, n(n-1)) = (0,0)$; we deduce, as before, that $(-n, \pm n(n-1))$ are torsion elements. Therefore there are at least~$8$ members of the torsion subgroup of~${\cal F}(\Q)$, namely: ${\bf o}, (0,0), (-1,0), (-n^2,0)$ and the above~$4$ points of order~$4$. Finally, let~$n \equiv 2$~(modulo~$5$), so that~$n^2 \equiv 4$~(modulo~$5$). Then $\widetilde {\cal F}$ modulo~$5$ is: $Y^2 = X(X+1)(X+4)$, and the cubic has no repeated roots, so that~$f$ has good reduction at~$5$. The elements of $\widetilde {\cal F} (\F_5 )$ are: ${\bf o}, (0, 0), (1, 0), (2, \pm 1), (3, \pm 2), (4,0)$, and so $\widetilde {\cal F} (\F_5 )$ has order~$8$. Since the torsion subgroup of ${\cal F} (\Q)$ injects into this groups, its order must divide~$8$. Hence the torsion subgroup of ${\cal F} (\Q)$ consists precisely of the~$8$ torsion elements already found, namely:~$\{ {\bf o}, (0,0), (-1,0), (-n^2,0), (n, \pm n(n+1)) , (-n, \pm n(n-1)) \}$. {\bf [7~marks]} \par\noindent{\bf (iv).} The disciminant of $X^3 - k^2 X + k^3$ is $4\cdot (-k^2)^3 + 27 \cdot (k^3)^2 = 23 k^6$. If $(x,y)$ is a $\Q$-rational torsion point on $\e : Y^2 = X^3 - k^2 X + k^3$, then we know by Nagell-Lutz that $x,y\in \Z$ with $y^2 | 23 k^6$ and so $| y |^2 \le 23 |k|^6$; that is: $| y | \le \sqrt{23} |k|^3 = (4.79583\ldots) |k|^3 \le 5 |k|^3$. \par We cannot have $k=0$ since then the discriminant would be~0 and $\e$ would not be an elliptic curve. % The cases $k = \pm 1$ have already % been dealt with in~$(b)$, when it the torsion group of $\e(\Q)$ % was shown to be trivial. So, we can take $| k | > 1$, Since $k\in \Z$, we have $|k| \geqslant 1$ and so: $| k |^2 \leqslant | k |^4$ and $| k |^3 \leqslant | k |^6$. \par\noindent Imagine $ | x | > 3 | k |^2$. Then: $| x^2 - k^2 | \ge | x^2 | - | k^2 | > 9 | k |^4 - | k^2 | \geqslant 9 | k |^4 - | k^4 | = 8 | k |^4$, so: \par\noindent $ | y |^2 = %| x^3 - k^2 x + k^3 | = | x(x^2 - k^2) + k^3 | \ge | x(x^2 - k^2) | - | k^3 | = | x | | x^2 - k^2 | - | k^3 | > 3 | k |^2 8 | k |^4 - | k |^6 = 23 | k |^6$, \par\noindent contradicting $| y |^2 \le 23 |k|^6$ (above). Hence $| x | \le 3 | k |^2$, as required.~{\bf [7~marks]} \par\noindent{\bf Question 3.} \medskip\par\noindent {\bf (i).} Let $r \in \qmods, r \in \hbox{im}\, q, r \in \Z, r$ square free. We want to prove that $r | b_1$. Suppose $r = q(u,v)$, where $(u,v)\in \d (\Q)$, which must exist since $r \in \hbox{im}\, q$. Then: $r = q(u,v) = u = u^2 + a_1 u + b_1$ in $\qmods$ [since $u(u^2 + a_1 u + b_1) = v^2$]. So, $r, u, u^2 + a_1 u + b_1$ are all the same modulo squares, which means we can write: $$ u^2 + a_1 u + b_1 = rs^2,\ \ \ u = rt^2,\ \ \ \hbox{ for some } s,t \in \Q. $$ Hence: $ (rt^2)^2 + a_1 (rt^2) + b_1 = rs^2$. Let $t = \ell/m$, where $\ell, m \in \Z$ and gcd$(\ell,m) = 1$. Then: $ r^2 \ell^4 / m^4 + a_1 r \ell^2 / m^2 + b_1 = rs^2$, and so: $ r^2 \ell^4 + a_1 r \ell^2 m^2 + b_1 m^4 = r(m^2s)^2$. Now, $a_1, b_1, r,\ell, m \in \Z$, so the LHS of this last equation is in~$\Z$, and so the RHS is also in~$\Z$; that is: $r(m^2s)^2\in \Z$. Since $r$ is square free, we must therefore have $m^2 s \in \Z$. Define: $n = m^2 s \in \Z$. Then our equation becomes: $$ r^2 \ell^4 + a_1 r \ell^2 m^2 + b_1 m^4 = r n^2,\ \ \hbox{ for some } \ell, m, n \in \Z, \hbox{ gcd}(\ell,m) = 1. \eqno (\dagger ) $$ We want to show that $r | b_1$, and we know that $r$ is square free. It is sufficient to show, for any prime~$p$, that $p | r \Rightarrow p | b_1$. \par \item{}Imagine $p | r$ and $p \notdiv b_1$, for some prime~$p$. Then $p | r^2 \ell^4, a_1 r \ell^2 m^2, r n^2$ and so by $(\dagger)$, $p | b_1 m^4$, which in turn gives: $p | m$ [since $p \notdiv b_1$]. Hence, since now $p | r$ and $p | m$, we have: $p^2 | r^2 \ell^4, a_1 r \ell^2 m ^2, b_1 m^4$, and so by $(\dagger)$, $p^2 | r n^2$, which in turn gives: $p | n$ [since~$r$ is square free]. Hence, since now $p | r,m,n$, we have: $p^3 | a_1 r \ell^2 m^2, b_1 m^4, r n^2$, and so by $(\dagger)$, $p^3 | r^2 \ell^4$, which in turn gives: $p | \ell$ [since~$r$ is square free]. This is a contradiction, since $p | \ell$ and $p | m$ but $\hbox{ gcd}(\ell,m) = 1$. \par\noindent The above assumption that $p | r$ and $p \notdiv b_1$ let to a contradiction, and so it is impossible for any prime~$p$ to satisfy $p | r$ and $p \notdiv b_1$. This is the same as saying that $ p | r \Rightarrow p | b_1 $ for any prime~$p$. Since $r$ is square free, we conclude that $r | b_1$, as required~{\bf [13~marks]} \medskip\par\noindent{\bf (ii)} Let $\c : Y^2 = X(X^2 + aX + b) = X(X^2 + 2X + 3)$, where $a=2, b=3$, and isogenous curve $\d : Y^2 = X(X^2 + a_1X + b_1) = X(X^2 - 4X - 8)$, where $a_1 = -4, b_1=-8$, with the usual isogeny $\phi : \c (\Q ) \rightarrow \d (\Q) : (x,y) \mapsto (y^2/x^2 , y - 3y/x^2)$, and dual isogeny $\hat\phi : \d (\Q ) \rightarrow \c (\Q) : (u,v) \mapsto ( {1\over 4} v^2/u^2 , {1\over 8}( v + 8v/u^2) )$. \par The map $q : \d (\Q) / \phi (\c (\Q)) \rightarrow \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $q : (0,0) \mapsto b_1$ and $q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}q$ contained in $\{ d : d\hbox{ is square free and } d | b_1\} = \{ \pm 1 , \pm 2 \}$. Also, ${\bf o} \mapsto 1$, $(0,0) \mapsto -8 = -2$, so that $\{ 1,-2 \} \subset \hbox{im} q \subset \{ \pm 1 , \pm 2\}$. \par There is only one coset to check, represented by $-1$, say. We know that $-1 \in \hbox{im} q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a_1 \ell^2m^2 + (b_1/(-1))\cdot m^4 = n^2$ that is: $-\ell^4 - 4 \ell^2m^2 + 8 m^4 = n^2$. Rewrite as: $-(\ell^2 + 2m^2)^2 + 12 m^4 = n^2$. Reducing modulo~3 gives $ - (\ell^2 + 2m^2 )^2 \equiv n^2$ (modulo~3). If $n$ were coprime to~$3$, then this would give: $(( \ell^2 + 2 m^2 )/n)^2 = -1$ in $\F_3$, contradicting the fact that~$-1$ is not a quadratic residue modulo~3. So, $3 | n$ and so $3 | (\ell^2 + 2m^2 )$ also. This means that $9 | (\ell^2 + 2m^2 )^2$ and $9 | n^2$, which can be combined with $-(\ell^2 + 2m^2)^2 + 12 m^4 = n^2$ to give: $9 | 12 m^4$ and so $3 | m$. Combining $3 | m$ with $3 | (\ell^2 + 2m^2 )$ gives that $3 | \ell$. This contradicts the fact that $\hbox{gcd}(\ell,m) = 1$. Hence our equation is impossible in~$\Q_3$, and so impossible in~$\Q$. Hence $-1 \not\in \hbox{im} q$. \par We conclude that $\hbox{im} q = \{ 1,-2 \}$, and so $\d (\Q) / \phi (\c (\Q))$ is generated by~$(0,0)$. \par The map $\hat q : \c (\Q ) / \hat\phi (\d (\Q)) \rightarrow \qmods : (x,y) \mapsto x$, for $(x,y) \not= (0,0)$, with $\hat q : (0,0) \mapsto b$ and $\hat q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}\hat q$ contained in $\{ d : d\hbox{ is square free and } d | b\} = \{ \pm 1 , \pm 3 \}$. Also, ${\bf o} \mapsto 1$ and $(0,0) \mapsto 3$, so that $\{ 1,3 \} \subset \hbox{im} \hat q \subset \{ \pm 1 , \pm 3\}$. \par There is only one coset to check, represented by $-1$, say. We know that $-1 \in \hbox{im} \hat q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a \ell^2m^2 + (b/(-1))\cdot m^4 = n^2$; that is: $-\ell^4 + 2\ell^2m^2 - 3 m^4 = n^2$. Rewrite as: $ - ( \ell^2 - m^2 )^2 - 2 m^4 = n^2$. This is impossible in~$\R$ (the left hand side is $\leqslant 0$ and the right hand side is $\geqslant 0$, and equality only occurs when $\ell^2 - m^2 = m^4 = n^2 = 0$, implying $\ell = m = n = 0$, which is not allowed). Hence $-1 \not\in \hbox{im} \hat q$. \par We conclude that $\hbox{im} \hat q = \{ 1,3 \}$, and so $\c (\Q) / \hat\phi (\d (\Q))$ is generated by~$(0,0)$. \par Finally, since multiplication by 2 in $\c (\Q)$ is $\hat\phi \circ \phi$, we have that $\c (\Q) / 2\c (\Q)$ is generated by: generators for $\c (\Q) / \hat\phi (\d (\Q))$ [namely: $(0,0)$] together with the images under $\hat \phi$ of generators for $\d (\Q) / \phi (\c (\Q))$ [namely, $\hat\phi \bigl( (0,0) \bigr) = {\bf o}$]. Conclusion: $\c (\Q ) / 2\c(\Q )$ is generated by $(0,0)$, and so is isomorphic to $C_2$. We also know that $\c (\Q ) / 2\c(\Q )$ is isomorphic to $\ctq / 2\ctq \times C_2^r$, which is isomorphic to $\c (\Q )[2] \times C_2^r$, where $\c (\Q )[2]$ is the $2$-torsion group and $r$ is the rank. The $2$-torsion points on $\c : Y^2 = X(X^2 + 2X + 3)$ are {\bf o} together with the points of the form $(x,0)$, where $x$ is a root of $X(X^2 + 2X + 3)$, that is: $(0,0)$, $(-1 + \sqrt{-2}, 0)$ and $(-1 + \sqrt{-2}, 0)$, of which only {\bf o} and $(0,0)$ are in $\c (\Q )[2]$, giving that $\c (\Q )[2]$ is isomorphic to $C_2$. Combining this with the facts (already found) that $\c (\Q ) / 2\c(\Q )$ is isomorphic both to $C_2$ and to $\c (\Q )[2] \times C_2^r$, give that $\c (\Q)$ has rank~0. \par\hfill{\bf [12 marks]} \bigskip\noindent{\bf Question 4.} \par\noindent {\bf (i).} First compute (modulo $N=10001$): $2^1 \equiv 2$, $2^2 \equiv 4$, $2^4 \equiv 16$, $2^8 \equiv 256$, $2^{16} \equiv 5530$, $2^{32} \equiv 7843$, $2^{64} \equiv 6499$ (where each of these was obtained be squaring the previous one, and reducing modulo~$N$). Now, we write $68$ in base~2: $68 = 4 + 64$ and so $2^{68} \equiv 2^4 2^{64} \equiv 16\cdot 6499 \equiv 3974$ modulo~$N$, so that $2^{68} - 1 \equiv 3973$ modulo~$N$. \par Now, compute $\hbox{gcd}(3973, N)$ by Euclid's Algorithm: $10001 = 2 \cdot 3973 + 2055$; $3973 = 1\cdot 2055 + 1918$; $2055 = 1\cdot 1918 + 137$, $1918 = 14\cdot 137 + 0$. So, $137$ is a factor of $N$. Compute $10001/137 = 73$, giving the factorisation $N = 10001 = 73 \cdot 137$. {\bf [5~marks]} \medskip\par\noindent {\bf (ii).} Since $N = 73 \cdot 137$, we have $\phi (N) = 72 \cdot 136 = 9792$. Computing the gcd of $d=4451$ and $\phi(N)$, we see: $9792 = 2\cdot 4451 + 890$; $4451 = 5\cdot 890 + 1$, so that $\hbox{gcd}(9792,4451) = 1$. Reversing the steps: $1 = 4451 - 5\cdot 890 = 4451 - 5\cdot (9792 - 2\cdot 4451) = - 5\cdot 9792 + 11\cdot 4451$ Hence, $11$ is the inverse of $4451$ modulo~$9792$. The decoding operation is therefore $X \mapsto X^{11} \hbox{ mod }N$. Computing $6847^{11} = ((6847^2)^2)^2 \cdot (6847^2 \cdot 6847) \equiv (6722^2)^2 \cdot (6722 \cdot 6847) \equiv 766^2\cdot 932 \equiv 6698\cdot 932 \equiv 1912$. (modulo~$N = 10001$). Also: $2577^{11} = ((2577^2)^2)^2 \cdot (2577^2 \cdot 2577) \equiv (265^2)^2 \cdot (265 \cdot 2577) \equiv 218^2 \cdot 2837 \equiv 7520 \cdot 2837 \equiv 2107$ (modulo~$N = 10001$). The decoded message is therefore: $1912,\, 2107$; that is: SLUG. {\bf [5~marks]} \medskip\par\noindent {\bf (iii).} We are given that $A/2A$ is finite, so let $A/2A = S = \{ Q_1,\ldots Q_r \} \subset A$. Let~$P$ be any element of~$A$. Then $P = Q_{i_1}$ in $A/2A$ for some $Q_{i_1} \in S$ and so we can write: $P = 2P_1 + Q_{i_1}$, for some $P_1 \in A$. Indusctively, continue to write: $P_1 = 2P_2 + Q_{i_2}, P_2 = 2P_3 + Q_{i_3}, \ldots$, where each $P_j \in A$ and each $Q_{i_j}\in S$. Now: \noindent $h(P_j) \le {1\over 4}\bigl( h(2P_j) + C_2 \bigr)$ [by~(2)] $= {1\over 4}\bigl( h(P_{j-1}-Q_{i_j}) + C_2 \bigr) \le {1\over 4}\bigl( 2h(P_{j-1}) + C_1' + C_2 \bigr)$ [by~(1)], where: \noindent $C_1' = \hbox{max}\{ C_1(-Q) : Q\in S\}$. So, if $h(P_{j-1}) > (C_1' + C_2)/2$ then $h(P_j) < {1\over 4} \bigl( 2h(P_{j-1}) + 2h(P_{j-1}) \bigr) = h(P_{j-1})$. Imagine that $h(P) > (C_1' + C_2)/2$ and $h(P_j) > (C_1' + C_2)/2$ for all~$j$. Then $h(P), h(P_1), h(P_2),\ldots$ would be a strictly decreasing sequence, giving infinitely many distinct members of~$A$ with height $\le h(P)$, which would contradict~(3). This contradiction show that there must exists an~$n$ such that $h(P_n) \le (C_1' + C_2)/2$. So, we can write: $P = 2P_1 + Q_{i_1} = 2(2P_2 + Q_{i_2}) + Q_{i_1} = \ldots$, and after~$n$ steps $P$ will be written as a linear combination of~$P_n$ and members of~$S$. Let $T = \{ Q \in A : h(Q) \le (C_1' + C_2)/2\}$. We have shown (since $P_n\in T$) that any $P\in A$ is a linear combination of members of~$S \cup T$. Furthermore, $T$ is finite, by~(3). In conclusion: $A$ is generated by the finite set $S \cup T$, and so is finitely generated.~{\bf [10~marks]} \medskip\par\noindent {\bf (iv).} Suppose that $h(P) > {1\over 3}C_2$. Then $C_2 < 3h(P)$ and so by~(2): $h(2P) \ge 4h(P) - C_2 > 4h(P) - 3h(P) = h(P)$. Inductively: $h(P) < h(2P) < h(4P) < \ldots$, so that $P,2P,4P,\ldots$ are all distinct, which forces~$P$ to have infinite order (since otherwise $P$ would generate a finite group). Hence, if $P$ is a torsion element then $h(P) \le {1\over 3}C_2$.~{\bf [5~marks]} \bigskip \bigskip \hrule \bigskip \bigskip \chaptitle \noindent \centerline{Solutions to the 2006 Examination Questions.} \rm \par\bigskip \noindent {\bf Question 1.} \par\noindent (i) [{\it Proof from lectures}]. Define $f_j(x)$ by: \par $f(x + y) = f_0(x) + f_1(x) y + f_2(x) y^2 + \ldots,$ \par\noindent so that $f_0(x) = f(x), f_1(x) = f'(x)$. Define $b_0 = -f(a_0)/f'(a_0)$. By $(*)$, $|b_0| < 1$. \par Define $a_1 = a_0 + b_0 = a_0 - f(a_0)/f'(a_0)$. Then: \par $| f'(a_1) - f'(a_0) | = | f'(a_0+b_0) - f'(a_0) | = | (\hbox{poly in }a_0) b_0 + (\hbox{poly in }a_0) b_0^2 + \ldots |$ \par \ \ \ \ \ \ \ \ \ \ \ $\leqslant | b_0 | < | f'(a_0) |$, \par\noindent so that $| f'(a_1) | = | f'(a_0) |$. \par Also, $| f(a_1) | = | f(a_0 + b_0) | | f_0(a_0) + f_1(a_0)b_0 + f_2(a_0)b_0^2 + \ldots |$ \par\ \ \ $= | f_2(a_0)b_0^2 + \ldots |$\ \ [since $f_0(a_0) + f_1(a_0)b_0 = 0$] \par\ \ \ $\leqslant \hbox{max}_{j\geqslant 2} |f_j(a_0)| |b_0|^j \leqslant | b_0 |^2 = {{ |f(a_0 )|^2}\over {|f'(a_0 )|^2}} = \rho | f(a_0) | < | f(a_0) |$, \par\noindent where $\rho = {{ |f(a_0 )|}\over {|f'(a_0 )|^2}} < 1$. \par Summarising: $ |f'(a_1) | = |f'(a_0)|$ and $| f(a_1) | \leqslant \rho | f(a_0) | < | f(a_0) |$, where $\rho = {{ |f(a_0 )|}\over {|f'(a_0 )|^2}} < 1$. \par\noindent Given $a_n\in R$, define $b_n = -f(a_n)/f'(a_n)$ and $a_{n+1} = a_n + b_n = a_n -f(a_n)/f'(a_n)$. %\par As above: $|f'(a_{n+1})| = \ldots |f'(a_1)| = |f'(a_0)|$. %\ \ \ \ \ \ (1) \par\noindent Assume, as induction hypothesis, that: \par $|f'(a_n)| = \ldots = |f'(a_1)| = |f'(a_0)|$ and $ | f(a_n) | \leqslant \rho |f(a_{n-1})| \leqslant \ldots \leqslant \rho^n |f(a_0)|$.\ \ \ \ \ \ (1) \par\noindent Then, as above: $|f'(a_{n+1})| = \ldots = |f'(a_1)| = |f'(a_0)|$. \par\noindent Then $| f(a_{n+1}) | \leqslant |b_n|^2$ \ \ [justified as for the case~$n=0$ above] \par \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $= {{ | f(a_n) |^2 }\over { | f'(a_n) |^2 }} = {{ | f(a_n) |^2 }\over { | f'(a_0) |^2 }}$\ \ [by~(1), the induction hypothesis] \par \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\leqslant {{ | f(a_0) |^2 }\over { | f'(a_0) |^2 }} |f(a_0)|$ \ \ [since $|f(a_n)| \leqslant |f(a_0)|$ by (1), the induction hypothesis] \par \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $= \rho | f(a_0) | \leqslant \rho^{n+1} | f(a_0) |$ \ \ [by (1), the induction hypothesis]. \par\noindent By induction, $\forall n$, $|f'(a_n)| = |f'(a_0)|$ and $ | f(a_{n}) | \leqslant \rho^n | f(a_0) |$ which $\rightarrow 0$ as $n\rightarrow \infty$. \ \ \ \ \ \ (2) \par\noindent Now, $| b_n | = | f(a_n) |/|f'(a_n)| = | f(a_n) |/|f'(a_0)| \rightarrow 0$, so [by the theorem from lectures that a series converges in a non-Archimedean field iff its terms converge to~$0$]: \par $a_n = a_0 + b_0 + b_1 + \ldots + b_n$ converges to~$a$, say. \par\noindent By continuity of polynomials, $f(a) = \lim f(a_n) = 0$\ \ [by~(2)]. \par\noindent Furthermore: $| a - a_0 | = | \sum b_n | \leqslant \max | b_n | = \max {{| f(a_n) |}\over {|f'(a_n)|}} = \max {{| f(a_n) |}\over {|f'(a_0)|}} = {{| f(a_0) |}\over {|f'(a_0)|}}$, as required. \par\noindent Imagine that $\hat a \not= a$ also satisfied $f(\hat a) = 0$ and $| \hat a - a_0 | \leqslant | f(a_0) |/ | f'(a_0) |$. Let $\hat b = \hat a - a$. \par Then $0 = f(\hat a) - f(a) = f(a + \hat b) - f(a) = {\hat b} f_1(a) + {\hat b}^2 f_2(a) + \ldots$ \ \ \ \ \ \ (3) \par\noindent But $| \hat b | = | \hat a - a_0 + a_0 - a | \leqslant \max( | \hat a - a_0 |, | a - a_0 | ) \leqslant | f(a_0) |/ | f'(a_0) | $ \par \ \ \ $< | f'(a_0) | = | f_1(a_0) | = | f_1(a) |$ \ \ [by~(1) and continuity of $|f'(x)|$]. \par\noindent This gives $|{\hat b}^j f_j(a)| \leqslant |{\hat b}^j| \leqslant |{\hat b}^2| < |{\hat b} f_1(a)|$ for~$j\geqslant 2$, so that the leading term of the sum in~(3) has valuation strictly greater than the valuations of the other terms, which is inconsistent with the sum being~$0$. Hence~$a$ is unique. \hfill {\bf [10~marks]} \par\noindent (ii) [{\it They have seen a variation on an exercise sheet, although this one is substantially harder}]. For~$p=2$, can see the equation becomes $-y^2 \equiv 2$~(mod~$4$), which is impossible, since $y \equiv 0,1,2,3$ give $y^2 \equiv 0,1,0,1$, respectively. Hence, there are no such $x,y\in \Z_2$. For $p=3$, take $x=1$, in which case we want~$y\in\Z_3$ such that~$3y^2 = -6$, that is: $f(y) = y^2 + 2 = 0$. But $a_0 = 1$ gives $|f(a_0)|_3 = 3^{-1}$ and $|f'(a_0)| = 1$, so by Hensel's Lemma, there exists~$y\in\Z_3$ which is a root of~$f$. For~$p=5$, take~$x=2$, in which case we want~$y\in\Z_5$ such that~$g(y) = 3y^2 - 22 = 0$. But $a_0 = 2$ gives $|g(a_0)|_5 = 5^{-1}$ and $|g'(a_0)| = 1$, so by Hensel's Lemma, there exists~$y\in\Z_5$ which is a root of~$g$. For~$p=7$, take~$y=0$, in which case we want~$x\in\Z_7$ such that~$h(x) = (4x^3 - 10)/3 = 0$. But $a_0 = -1$ gives $|h(a_0)|_7 = 7^{-1}$ and $|h'(a_0)| = 1$, so by Hensel's Lemma, there exists~$x\in\Z_7$ which is a root of~$h$. For $p\geqslant 11$, the curve is an elliptic curve mod~$p$ [since the only bad primes are $2,3,5$] and $\# {\widetilde{\cal E}} (\bbF_p) \geqslant p+1 - 2\sqrt{p} > 5$. At most 4 of these are $2$-torsion, so there exists~$x_0$ such that $h(x_0)$ is a nonzero residue mod~$p$; applying Hensel's Lemma to $y^2 - h(x_0)$ gives the existence of a corresponding $y\in\Z_p$. In summary, there exist such $x,y\in\Z_p$ for all~$p$ except~$2$. \hfill {\bf [7~marks]} \par\noindent (iii) [{\it Unseen}]. We may apply the general rule from lectures (which follows easily from Hensel's Lemma) that, if $p\not= 2$ and $|\alpha|_p = 1$, then $\alpha$ is a square in $\Q_p^*$ iff $\alpha$ is a quadratic residue in $\F_p$. An element $\alpha \in ({\Q}_p^*)^2$ iff (both $| \alpha |_p = p^r$ for $r$ even and the leading digit $a_r$ of $\alpha$ is a quadratic residue mod~$p$) [N.B. if $| \alpha |_p = p^r$ for $r$ odd, then $\alpha \not= \beta^2$ for any $\beta$, since $|\beta|_p = p^s$ for some integer~$s$]. Let $H_1 = ({\Q}_p^*)^2$, and fix a quadratic nonresidue mod~$5$, say~$\gamma$. Let $H_2 = \gamma H_1$, $H_3 = p H_1$, $H_4 = \gamma p H_1$. Since any nonresidue = $\gamma$(residue) and vice versa, these give all the cases when ($a_r$ is a residue and $r$ is even)~[H1], ($a_r$ is a nonresidue and $r$ is even)~[H2], ($a_r$ is a residue and $r$ is odd)~[H3], ($a_r$ is a nonresidue and $r$ is odd)~[H4], which partition~${\Q}_p^*$. Hence there are precisely 4 distinct elements in ${\Q}_p^*/({\Q}_p^*)^2$, when $p\not=2$. \par Similarly, 1,3,5,7 are all distinct in ${({\Q_2}^*)^2}$, since for $|\alpha|_2 = 1$, Hensel's Lemma gives that $\alpha\in {({\Q_2}^*)^2}$ iff $\alpha \equiv 1$~(mod~$8$). Furthermore, any $\alpha \equiv 1$ is equal to prrecisely one of~$1,3,5,7$ in ${{\Q_2}^*/({\Q_2}^*)^2}$. For any~$\alpha$, if $|\alpha|_2 = 2^r$ with~$r$ even, then $\alpha = \alpha \cdot 2^{r} = 1,3,5$~or~$7$ in ${{\Q_2}^*/({\Q_2}^*)^2}$, since $| \alpha \cdot 2^{r}|_2 = 1$. If~$r$ is odd, then $\alpha/2 = 2^r \alpha = 1,3,5$~or~$7$ in ${{\Q_2}^*/({\Q_2}^*)^2}$, and so $\alpha = 2,6,10$~or~$14$ in ${{\Q_2}^*/({\Q_2}^*)^2}$. So, there are 8~elements when $p=2$, represented, say, by $1,3,5,7,2,6,10,14$ (generated by $3,5,2$). \hfill {\bf [8~marks]} \par\bigskip \noindent {\bf Question 2.} Let~$R$ be any ring, and let~$F,G$ be formal groups over~$R$. \par\noindent (i) [{\it Proof from lectures}]. Let $P(T) = F_X(0, T)^{-1}$, so that $\omega(T) = P(T){\hbox{d}} T$. Note that $F_X(0, T) = 1 + \ldots$ is invertible, so that $P(T)$ is indeed a member of~$R[[T]]$. Furthermore, $P(0) = 1$, so that it is normalised. \par We need to show that $\omega$ is an invariant differential; that is, $\omega \circ F(T,S) = \omega(T)$, which is equivalent to: $P\bigl( F(T,S) \bigr) F_X (T,S) = P(T)$ so, in out case, it is sufficient to show: $$ F_X\bigl( 0, F(T,S) \bigr)^{-1} F_X(T,S) = F_X(0,T)^{-1}, $$ which is true iff: $$ F_X\bigl( 0, F(T,S) \bigr) = F_X(T,S) F_X(0,T). $$ But this last statement is immediate from differentiating $F\bigl( U, F(T,S) \bigr) = F\bigl( F(U,T), S \bigr)$ [associativity] with respect to~$U$ to get: $F_X\bigl( U, F(T,S) \bigr) = F_X\bigl( F(U,T), S\bigr) F_X(U,T)$ and setting~$U = 0$. Hence~$\omega$ is an invariant differential. \par Suppose that $\hat \omega = Q(T){\hbox{d}} T \in R[[T]] {\hbox{d}} T$ is also an invariant differential, so that $Q(T)$ satisfies $Q\bigl( F(T,S) \bigr) F_X(T,S) = Q(T)$. Substituting~$T=0$ gives $Q(S) F_X(0,S) = Q(0)$, so that $Q(S) = Q(0) F_X(0,S)^{-1}$. It follows that $\hat \omega = a \omega$, where $a = Q(0)$. \hfill {\bf [9~marks]} \par\noindent (ii) [{\it Proof from lectures}]. First, note that $\omega_G \circ f\bigl( F(T,S) \bigr) = \omega_G \bigl( G( f(T), f(S) \bigr) = \omega_G \circ f (T)$, so that $\omega_G \circ f$ is an inviariant differential on~$G$. From part~(a), it follows that $\omega_G \circ f = a\ \omega_F$, for some~$a\in R$. Since~$\omega_F,\omega_G$ are normalised, $(1 + \ldots ) {\hbox{d}} f(T) = a (1 + \ldots ){\hbox{d}} T$, and so $(1 + \ldots ) f'(T) {\hbox{d}} T = a (1 + \ldots ){\hbox{d}} T$; equating constant terms gives $a = f'(0)$, as required. \par Let~$\omega$ be the normalised invariant differential on~$F$. Since $[p](T) = p T + \dots$, it satisfies $[p]'(0) = p$. Applying the previous result to~$[p]$, a homomorphism from~$F$ to itself, gives: $\omega \circ [p] = [p]'(0) \omega = p \omega$, and so $$ p \omega(T) = \omega \circ [p](T) = (1 + \ldots ) {\hbox{d}} ( [p](T) ) = (1 + \ldots ) [p]'(T) {\hbox{d}} T.$$ Hence $[p]'(T) \in p\, R_T$. Each term $a_n T^n$ in $[p](T)$ msut then satisfy $ p | n a_n$ and so $p | n$ or $p | a_n$, as required. \hfill {\bf [8~marks]} \par\noindent (iii) [{\it Unseen}]. Let ${\e}$ be the given curve, which is of the form $Y^2 = X^3 - M^2 X + N^2$, where~$M = m^2 + 1$ and~$N = 3n$. First note that for $m\equiv 0,1,2$~(mod~$3$), $M \equiv 1,2,2$~(mod~$3$), so that~$M$ is never divisible by~$3$. Hence $\Delta = 4(- M^2)^3 + 27 (N^2)^2 = -4 M^6 + 27 N^4$ is never~$0$, and so the curve is non-singular and is an elliptic curve. It has the obvious point~$(0,N)$. Let~$P = 2(0,N) \in {\e}(\Q)$. The tangent to~${\e}$ at~$(0,N)$ has slope~$(3\cdot 0^2 - M^2)/ (2\cdot N) = -M^2/2N$, and so the $x$-coordinate of~$P = 2(0,N)$ is $\bigl( -M^2/2N \bigr)^2 - 2\cdot 0$, which is not an integer, since the numerator is not divisible by~$3$, but the denominator $N = 3n$ is divisible by~$3$. Therefore~$P$ is not a torsion point (using the result from lectures that any torsion point in~${\e}(\Q)$ must have both coordinates in~$\Z$), and so must be of infinite order, giving that~${\e}(\Q)$ is infinite, as required. \hfill {\bf [8~marks]} \par\bigskip \noindent {\bf Question 3.} \par\noindent (i) [{\it Seen similar on an exercise sheet}]. The line tangent to~$\e$ at $P=(1,1)$ has slope $y'$ given by $2yy' = 3x^2 + 7$, with $x=1,y=1$; that is, the slope is $10/2 = 5$. This tangent line also goes through $(1,1)$ and so has equation: $Y = 5 X - 4$. The $x$-coordinate of $2P$ is therefore $5^2 - (1+1) = 23$, and the $y$-coordinate is: $-(5\cdot 23 - 4) = -111 \equiv 1406$, so that $Q = 2P \equiv (23 , 1406)$ (modulo~$N=1517$). We now wish to compute $4P = 2Q = 2(23 , 1406)$, and so the first step is to find the line tangent to~$\e$ at~$Q$. This has slope~$y'$ given by $2yy' = 3x^2 + 7$, with $x=23,y=1406$, that is, $y' = 1594/2812$. In order to compute this modulo~$N = 1517$, we must first find the inverse of~$2812$ modulo~$1517$. Using Euclid's Algirithm: $2812 = 1\cdot 1517 + 1295$, $1517 = 1\cdot 1295 + 222$, $1295 = 5\cdot 222 + 185$, $222 = 1\cdot 185 + 37$, $185 = 5\cdot 37 + 0$. So, we cannot find the inverse of $2812$ (modulo~$N=1517$), and this step has given us a proper factor~$37$ of~$N$. \hfill {\bf [8~marks]} \par\noindent (ii) [{\it Seen similar on an exercise sheet}]. The elliptic curve has $\Delta = 4A^3 + 27B^2 = 4\cdot (-2)^3 + 27\cdot 0^2 = -32$, and so has good reduction at all primes~$p$ except~$p=2$. Taking $p=3$, we find that $\e(\F_3) = \{{\bf o}, (0,0),(2,1),(2,2)\}$, and since $\etq$ is isomorphic to a subgroup of $\e(\F_3)$, this gives that $\# \etq \ | \ 4$. Taking $p=5$, we find that $\e(\F_5) = \{{\bf o}, (0,0),(1,2),(1,3),(2,2),(2,3),(3,1),(3,4),(4,1),(4,4)\}$, and since $\etq$ is isomorphic to a subgroup of $\e(\F_5)$, this gives that $\# \etq \ | \ 10$. Hence, $\# \etq \ | \ \hbox{gcd}(4,10) = 2$. However, there are the obvious torsion points: ${\bf o}$ and $(0,0)$ and so $\# \etq = \{ {\bf o}, (0,0) \}$. \par\hfill {\bf [7~marks]} \par\noindent (iii) [{\it Unseen}]. Using $2YY' = 3X^2$, we see that the slope of the curve at any~$(x,y)$ is $3 x^2 / (2 y)$, and so the tangent to the curve at $(x,y)$ is the line: $Y= \bigl( {{{3 x^2}\over {2 y}}} \bigr) X + c$, for some~$c$. The $x$-coordinate of the third point of intersection is then: $\bigl( {{{3 x^2}\over {2 y}}} \bigr)^2 - x - x = {{{ 9 x^4 - 8 x y^2 }\over { 4 y^2 }}} = {{{ 9 x^4 - 8 x (x^3 + k) }\over { 4 (x^3 + k) }}} = {{{ x (x^3 - 8k) }\over { 4 (x^3 + k) }}}$. Furthermore $3(x,y) = {\bf o} \iff 2(x,y) = -(x,y) = (x,-y) \iff \bigl( {{{ x (x^3 - 8k) }\over { 4 (x^3 + k) }}} , ? ) = (x,-y) \iff {{{ x (x^3 - 8k) }\over { 4 (x^3 + k) }}} = x$ [the $\Rightarrow$ of the last step is trivial; for $\Leftarrow$, note that if $2(x,y)$ has that same $x$-coordinate as~$(x,y)$ then $2(x,y) = (x,-y)$~or~$(x,y)$; but the latter would imply~$(x,y)={\bf o}$, which we have excluded]. Note further that this last condition on~$x$ is satisfied when $ x (x^3 - 8k) = 4 x (x^3 + k) $, that is: $x ( x^3 + 4k ) = 0$. This gives on~$\e_k({\C})$ the points~$(0,\sqrt{k})$ and~$(-\root3 \of{4k}, i\sqrt{3k})$. But it is impossible for both~$\sqrt{k}$ and~$i\sqrt{3k}$ to be in~$\Q$ (indeed they cannot both be in~$\R$), so that at least one of these points of order~$3$ is not in~$\e_k(\Q)$. \hfill {\bf [10~marks]} \par\bigskip \noindent {\bf Question 4.} \par\noindent (i) [{\it Seen similar on an Exercise Sheet}]. Let $\c : Y^2 = X(X^2 + aX + b) = X(X^2 + 3X + 5)$, where $a=3, b=5$, and isogenous curve $\d : Y^2 = X(X^2 + a_1X + b_1 ) = X(X^2 - 6X -11)$, where $a_1 = -2a = -6, b_1= a^2 - 4b = -11$, with the usual isogeny $\phi : \c (\Q ) \rightarrow \d (\Q) : (x,y) \mapsto (y^2/x^2 , y - 5y/x^2)$, and dual isogeny $\hat\phi : \d (\Q ) \rightarrow \c (\Q) : (u,v) \mapsto ( {{1}\over {4}} v^2/u^2 , {{1}\over {8}}( v + 11v/u^2) )$. \par The map $q : \d (\Q) / \phi (\c (\Q)) \mapsto \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $q : (0,0) \mapsto b_1$ and $q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}q$ contained in $\{ d : d\hbox{ is square free and } d | b_1\} = \{ \pm 1 , \pm 11 \}$. Also, ${\bf o} \mapsto 1$, $(0,0) \mapsto -11$ and the obvious point $(-1,2) \mapsto -1$, so that $1,-11,-1 \in \hbox{im} q$; but $\hbox{im} q$ is a group, so $11 \in \hbox{im} q$ also, and indeed we can just take: $(0,0) + (-1,2) = (11,22)$, which maps to $11$ under $q$. Hence, $\{ \pm 1,\pm 11 \} \subset \hbox{im} q \subset \{ \pm 1 , \pm 11\}$, that is $\hbox{im} q = \{ \pm 1 , \pm 11\}$, and so $\d (\Q) / \phi (\c (\Q)) = \{ {\bf o}, (0,0), (-1,2), (11,22)\}$, that is, $\d (\Q) / \phi (\c (\Q))$ is generated by~$(0,0),(-1,2)$. \par The map $\hat q : \c (\Q ) / \hat\phi (\d (\Q)) \mapsto \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $\hat q : (0,0) \mapsto a_1^2 - 4b_1 = b$ and $\hat q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}\hat q$ contained in $\{ d : d\hbox{ is square free and } d | b\}$ $= \{ \pm 1 , \pm 5\}$. Also, ${\bf o} \mapsto 1$ and $(0,0) \mapsto 5$, so that $\{ 1,5 \} \subset \hbox{im} \hat q \subset \{ \pm 1 , \pm 5\}$. \par There is only one coset to check, represented by $-1$, say. We know that $-1 \in \hbox{im} \hat q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a \ell^2m^2 + (b/(-1))\cdot m^4 = n^2$; that is: $-\ell^4 + 3\ell^2m^2 - 5 m^4 = n^2$. Multiply both sides by 4 and rewrite as: $ - ( 2\ell^2 - 3 m^2 )^2 - 11 m^4 = 4 n^2$. We can see that the LHS is $\leqslant 0$ and the RHS is $\geqslant 0$, and the equation is satisfied iff $( 2\ell^2 - 3 m^2 ) = 11 m^4 = 4 n^2 = 0$. From this we see that $m=n=0$, which when combined with $2\ell^2 - 3 m^2 = 0$ gives that $\ell = 0$, also, which is a contradiction. Hence $-1 \not\in \hbox{im} \hat q$. \par We conclude that $\hbox{im} \hat q = \{ 1,5 \}$, and so $\c (\Q) / \hat\phi (\d (\Q))$ is generated by~$(0,0)$. \par Finally, since multiplication by 2 in $\c (\Q)$ is $\hat\phi \circ \phi$, we have that $\c (\Q) / 2\c (\Q)$ is generated by: generators for $\c (\Q) / \hat\phi (\d (\Q))$ [namely: $(0,0)$] together with the images under $\hat \phi$ of generators for $\d (\Q) / \phi (\c (\Q))$ [namely, $\hat\phi \bigl( (0,0) \bigr) = {\bf o}$ and $\hat\phi \bigl( (-1,2) \bigr) = (1,3)$]. Conclusion: $\c (\Q ) / 2\c(\Q )$ is generated by $(0,0)$ and $(1,3)$ and so is the group $C_2\times C_2$. Now $\c (\Q)/2\c (\Q)$ is isomorphic to $\ctq/2\ctq \times C_2^{rank}$, and $\ctq /2\ctq$ is isomorphic to the $2$-torsion group of $\ctq$ which is $C_2$ (consisting only of {\bf o} and $(0,0)$), so that $\ctq/2\ctq$ is isomorphic to $C_2$. Conclusion: rank $=1$. \hfill {\bf [13~marks]} % See Ex Sheet 7, 3(a). On Ex Sheet, replace this % with 1997, 6. \par\noindent (ii) [{\it Unseen}]. Let $\c : Y^2 = X(X^2 + aX + b) = X(X^2 + p^2)$, where $a=0, b=p^2$, and isogenous curve $\d : Y^2 = X(X^2 + a_1X + b_1 ) = X(X^2 - 4p^2)$, where $a_1 = -2a = 0, b_1= a^2 - 4b = -4p^2$, with the usual isogeny $\phi : \c (\Q ) \rightarrow \d (\Q) : (x,y) \mapsto (y^2/x^2 , y - p^2y/x^2)$, and dual isogeny $\hat\phi : \d (\Q ) \rightarrow \c (\Q) : (u,v) \mapsto ( {{1}\over {4}} v^2/u^2 , {{1}\over {8}}(v + 4p^2v/u^2))$. \par The map $q : \d (\Q) / \phi (\c (\Q)) \mapsto \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $q : (0,0) \mapsto b_1$ and $q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}q$ contained in $\{ d : d\hbox{ is square free and } d | b_1\} = \{ \pm 1 , \pm 2, \pm p, \pm 2p \}$. Also, ${\bf o} \mapsto 1$, $(0,0) \mapsto -1$, $(2p,0)\mapsto 2p$, $(-2p,0)\mapsto -2p$, so that $\{ \pm 1,\pm 2p \} \subset \hbox{im} q \subset \{ \pm 1 , \pm 2, \pm p, \pm 2p \}$. There is only one coset to check, represented by~$-2$, say. We know that $-2 \in \hbox{im} q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-2)\cdot \ell^4 + a_1 \ell^2m^2 + (b_1/(-2))\cdot m^4 = n^2$; that is: $-2\ell^4 + 2 p^2 m^4 = n^2$. Since $p\equiv 5$~(mod~$8$), $\bigl( {{-2}\over {p}} \bigr) = -1$, and so $p | \ell$, $p | n$. Let $\ell = p \ell'$, $n = p n'$. Then $-2 p^2 (\ell')^4 + 2 m^4 = (n')^2$. Since also $\bigl( {{2}\over {p}} \bigr) = -1$, this implies $p | m$, $p | n'$. This gives a contradiction, since $p| \ell, p | m$ and $\hbox{gcd}(\ell,m) = 1$. Hence~$-2\not\in \hbox{im} q$, giving that $\hbox{im} q = \{ \pm 1 , \pm 2p \}$, and $\d (\Q) / \phi (\c (\Q)) = \{ {\bf o}, (0,0), (2p,0), (-2p,0)\}$. \par The map $\hat q : \c (\Q ) / \hat\phi (\d (\Q)) \mapsto \qmods : (u,v) \mapsto u$, for $(u,v) \not= (0,0)$, with $\hat q : (0,0) \mapsto a_1^2 - 4b_1 = b$ and $\hat q: {\bf o} \mapsto 1$, is an injection with $\hbox{im}\hat q$ contained in $\{ d : d\hbox{ is square free and } d | b\}$ $= \{ \pm 1 , \pm p\}$. Also, ${\bf o} \mapsto 1$ and $(0,0) \mapsto p^2 = 1$, so that $\{ 1 \} \subset \hbox{im} \hat q \subset \{ \pm 1 , \pm p\}$. \par We know that $-1 \in \hbox{im} \hat q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $(-1)\cdot \ell^4 + a \ell^2m^2 + (b/(-1))\cdot m^4 = n^2$; that is: $-\ell^4 - p^2 m^4 = n^2$, clearly impossible in~$\R$. Hence~$-1\not\in \hbox{im} \hat q$. Similarly, $-p\not\in\hbox{im} \hat q$. We know that $p \in \hbox{im} \hat q$ iff there are integers $\ell , m , n$, not all~0, and with $\hbox{gcd}(\ell,m) = 1$, such that: $p\cdot \ell^4 + a \ell^2m^2 + (b/p)\cdot m^4 = n^2$; that is: $p\ell^4 + p m^4 = n^2$, and so $p | n$. Letting $n = p n'$, the equation becomes $\ell^4 + m^4 = p^2 (n')^2$, which in turn implies $p | \ell, p | m$, since $-1 \not\equiv a^4$ for any~$a$ [as can be seen by taking both sides of $-1 \equiv a^4$ to the power of $(p-1)/4$]. This again contradicts $\hbox{gcd}(\ell,m) = 1$, and so $p\not\in \hat q$. \par We conclude that $\hbox{im} \hat q = \{ 1 \}$, and so $\c (\Q) / \hat\phi (\d (\Q)) = \{ {\bf o} \}$. \par Finally, since multiplication by 2 in $\c (\Q)$ is $\hat\phi \circ \phi$, we have that $\c (\Q) / 2\c (\Q)$ is generated by: generators for $\c (\Q) / \hat\phi (\d (\Q))$ [namely: ${\bf o}$] together with the images under $\hat \phi$ of generators for $\d (\Q) / \phi (\c (\Q))$ [namely, $\hat\phi \bigl( (0,0) \bigr) = {\bf o}$, and $\hat\phi \bigl( (2p,0) \bigr) = (0,0)$]. Conclusion: $\c (\Q ) / 2\c(\Q )$ is generated by $(0,0)$, and so is the group $C_2$. Now $\c (\Q)/2\c (\Q)$ is isomorphic to $\ctq/2\ctq \times C_2^{rank}$, and $\ctq /2\ctq$ is isomorphic to the $2$-torsion group of $\ctq$ which is $C_2$ (consisting only of {\bf o} and $(0,0)$), so that $\ctq/2\ctq$ is isomorphic to $C_2$. Conclusion: rank $=0$. \hfill {\bf [12~marks]} % See Ex Sheet 7, 3(a). On Ex Sheet, replace this % with 1997, 6. \bigskip \bigskip \hrule \vfil \eject \end