\documentclass[a4paper]{article} \usepackage{amsmath,amssymb,amsthm,enumerate,eucal,tikz,url} \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\sevrm=cmr7 \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} \def\R{\Bbb R} \def\Q{\Bbb Q} \def\Z{\Bbb Z} \def\F{\Bbb F} \def\e{{\mathcal E}} % \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})} \newcommand{\M}{\mathfrak{m}} \begin{document} \noindent \centerline{{\bf Elliptic Curves MT25: Solutions to Sheet 3}} \bigskip \begin{enumerate}[{\bf (1)}] \item Following the hint, there 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 \item 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 %\item %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 $p\geqslant 13$ 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~$\Q_p$. %\newpage % %\par {\it Alternative method for the above parts:} %Note that for $p \notdiv 2\Delta$ %(that is: $p \not= 2,17$) %$p \geqslant 7$, we have (from a theorem from lectures) %the number of affine points in %${\widetilde \c} (\F_p) \geqslant p-1 - 2\sqrt{p} > 0$, and so the %number of affine points is $ > 0$; these are non-singular, since %$2\Delta \not= 0$ in~$\F_p$. 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 $p = 2,3,5$ and bad primes, so only still need %to consider $p = 2,3,5,17$. Also, note that for $p = 3$ there %is the non-singular affine point $(1,1)$ on ${\widetilde \c} (\F_3)$, %for $p = 5$ there %is the non-singular affine point $(0,2)$ on ${\widetilde \c} (\F_5)$, %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 $\hbox{gcd}(r,t) = 1$. %%% \newpage % %\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~$17$ %(alternatively: if $2 \equiv w^4$ (mod~17) then $-1 \equiv 2^4 %\equiv w^{16} \equiv 1$ (mod~17) by FLT, a contradiction, %avoiding the need for an enumeration). %This contradication show that our original curve $\c$ %has no $\Q$-rational points. % %\medskip % % % %\item %\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 % %\item %By definition of a homomorphism, we need to show %\[ [m] F(X,Y) \stackrel{?}{=} F([m](X),[m]Y).\] %This is true when $m = 0$, since $[0](T) := 0$. %Now for $m \geq 1$ we can use the recursive definition to rewrite this as %\[ F([m-1]F(X,Y),F(X,Y)) \stackrel{?}{=} F(F([m-1](X),X), F([m-1]Y,Y)).\] %By induction we can assume %$[m-1] F(X,Y) = F([m-1](X),[m-1]Y)$. %Starting from the LHS we compute %\[ F([m-1]F(X,Y),F(X,Y)) = F(F([m-1](X),[m-1](Y)), F(X,Y))\] \[ = %F([m-1](X),F([m-1](Y),F(X,Y))) = F([m-1](X),F(F(X,Y),[m-1](Y))) \] \[ = F([m-1](X),F(X,F(Y,[m-1](Y)))) %= F(F([m-1](X),X),F(Y,[m-1](Y)))\] %which by commutativity equals the RHS. %Here we have used, in order, induction, associativity, commutativiity, associativity, associativity. % %\item {\bf (a)} We can regard $F$ as a formal group law over $K$, and then $\log_F$ is a homomorphism from $F$ to $\widehat{\mathbb{G}}_a$. In particular, we have an identity of formal power series (with coefficients in $K$): %$\log_F(X)+\log_F(Y) = \log_F(F(X,Y))$. When $x,y \in \M$, both sides converge, since the $n$th term in $\log_F$ contributes something of size $\le \frac{|\varpi|^n}{|n|}$ and this goes to $0$ as $n \to \infty$. This also shows that we have $\log_F(x)+\log_F(y) = \log_F(F(x,y))$: the identity of formal power series gives us an identity between terms of degree $\le n-1$ for each $n$, and the terms of degree $\ge n$ contribute something of size $\le \frac{|\varpi|^n}{|n|}$. Note that you do have to be careful when composing power series --- see Example 8.14 in \url{https://kconrad.math.uconn.edu/blurbs/gradnumthy/infseriespadic.pdf}. % %\item %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: ${\bf o}, P=(0,1/27), %2P=(0,-1/27), Q = (-1/9,0), P+Q = (2/9,-1/9), 2P+Q = (2/9,1/9)$. \item[{\bf (8)}] 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. \item[{\bf (9)}] Hint: consider the order $4$ automorphism (of elliptic curves over $\mathbb{C}$) $(x,y) \mapsto (-x,iy)$. This induces an automorphism of the formal group associated to $Y^2 = X^3 + AX$. See the full solutions file for the full solution! \item[{\bf (10)}] First we show convergence of the right hand side. For each $i$, set $g_i = G_i(x_1,\ldots,x_l)$. We have $|g_i| \le \max(|x_i|) < 1$. Choose $\rho < 1$ such that $\max(|x_i|) = \rho\delta$ with $\delta < 1$. Since $\{|a_n|\rho^{\sum n_i}\}$ is bounded, say by $C$, we have \[\left|\sum_{n: \sum n_i \ge N} a_n g_1^{n_1}\cdots g_k^{n_k}\right| \le C \delta^N.\] It follows that the right hand side converges, and is the limit of the sequence \[\sum_{n: \sum n_i \le N-1} a_n g_1^{n_1}\cdots g_k^{n_k}\] as $N \to \infty$. Write $G_i = \sum b_{i,m} X_1^{m_1} \ldots X_l^{m_l}$ and for a positive integer $M$ consider the finite truncations $G_{i,M} = \sum_{\sum m_i \le M-1} b_{i,m} X_1^{m_1} \ldots X_l^{m_l}$ with values $g_{i,M}$ when we substitute the $x_i$. We have \[\left|a_n g_1^{n_1}\cdots g_k^{n_k}-a_n g_{1,M}^{n_1}\cdots g_{k,M}^{n_k}\right| =\left|a_n(g_1^{n_1}\cdots g_k^{n_k}- g_{1,M}^{n_1}\cdots g_{k,M}^{n_k})\right| \le |a_n|\rho^M\delta^M,\] since all the terms which appear in $g_1^{n_1}\cdots g_k^{n_k}$, but not in $g_{1,M}^{n_1}\cdots g_{k,M}^{n_k}$, have valuation $\le \rho^M\delta^M$. We deduce that \[F(g_1,\ldots,g_k) - \sum_{n: \sum n_i \le N-1} a_n g_{1,N}^{n_1}\cdots g_{k,N}^{n_k}\] \[\le \max(C\delta^N,|a_n| \rho^N\delta^N : \sum n_i \le N-1) = C\delta^N.\] On the other hand, if we consider the formal power series $F\circ G$ then the difference \[(F\circ G) - \sum_{n: \sum n_i \le N-1} a_n G_{1,N}^{n_1}\cdots G_{k,N}^{n_k}\] only contains terms of degree $\ge N$. From this and the condition on the $|a_n|$, it follows easily that $(F\circ G)(x_1,\ldots,x_l)$ converges and its value is the limit of \[\sum_{n: \sum n_i \le N-1} a_n g_{1,N}^{n_1}\cdots g_{k,N}^{n_k}\] as $N \to \infty$. \end{enumerate} \end{document}