\documentclass[11pt]{article} \usepackage{amssymb,amsmath} \def\ms{\medskip} \def\Q{\mathbb{Q}} \def\cE{{\cal E}} \def\Z{\mathbb{Z}} \begin{document} \begin{flushleft} Question 1. \ms (a) To see this, note that if $|y|>1$, then from the equation $$|y|^2=|x^3+Ax+B|>1,$$ we must have $|x|>1$. On the other hand, if $|x|>1$, then $|x^3+Ax+B|=|x^3|=|y|^2$, so $|y|>1$. That is, we have $|x|>1$ and $|y|>1$ if and only if $|x|>1$ or $|y|>1$. Now if $(x,y)=(x:y:1)$ reduces to $(0:1:0)$ modulo $p$, then either $|x|>1$ or $|y|>1$, since otherwise, the reduction would be $(\tilde{x}:\tilde{y}:1)$. Now suppose that $|x|>1$ or $|y|>1$. Then from $|y|^2=|x^3+Ax+B|=|x|^3$, we see that $|y|>1$, $|x|>1$, and $|y|>|x|$. So the reduction is $(\widetilde{x/y}:1:\widetilde{1/y})=(0:1:0).$ [7 marks. Similar to bookwork.] \ms (b) If $a\in \Q^*$ has a square root in $\Q_p^*$ for all $p$ then $a$ has a square root in $\Q$. \ms {\em Proof.} Write $a=\pm\prod_p p^{n_p}$ for distinct primes $p$ and integers $n_p$. For each $p$, since $a=x^2$ in $\Q_p$, we have that $n_p$ is even. Thus, we can write $a=\pm \prod_p p^{2m_p}$ for integers $m_p$. But then, since $\prod_pp^{-n_p}$ has a square root $\prod_pp^{-m_p}$ in $\Q$, $a\prod_pp^{-n_p}=\pm 1$ would have a square root in each $\Q_p$. But $-1$ does not have a square root in, say, $\Q_3$, since any square root would be in $\Z_3$, giving rise to a square root of $-1$ in $\Z/3$, which doesn't exist. Thus, we must have $a\prod_pp^{-n_p}= 1$, and hence, $a=\prod_pp^{2m_p}$, which has the square root $\prod_pp^{m_p}$ in $\Q$. [7 marks. Unseen, but similar seen.] \ms (c) Hensel: Let $K$ be a field, complete w.r.t. a non-archimedean valuation $|\cdot|$, with valuation ring $R$. Let $f(x)\in R[x]$, and let $a\in R$ satisfy \[|f(a)|<|f'(a)|^2.\] Then there is a unique $a^*\in R$ with $f(a^*)=0$ and \[|a^{*}-a|\le |f(a)|/|f'(a)|.\] [2 marks. Bookwork] \ms (d) Let $p\neq 2$. Suppose $a$ has a square root. Then clearly $n=2m$, so that $u=ap^{-2m}$ has a square root. Conversely, suppose $a=p^nu$ with $n$ even such that $u\in \Z_p^*$ has a square root modulo $p$. Then clearly, $p^n$ has a square root, so it suffices to show that $u$ has a square root. For this, consider $f(x)=x^2-u$ and let $a\in \Z_p$ such that $f(a)\equiv 0 \mod p$. Then $a\in \Z_p^*$, so $f'(a)=2a\in \Z_p^*$. Hence, $|f'(a)|^2=1>|f(a)|$. Hence, by Hensel's Lemma, there is a $b\in \Z_ps$ such that $b\equiv a\mod p$ and $f(b)=0$, which then is a square root of $u$. When $p=2$, $a\in \Q_2^*$ has a square root if and only if $a=2^nu$ where $n$ is even and where $u\in \Z_2^* $has a square root modulo 8. That the condition is necessary is clear exactly as in the $p\neq 2$ case. To show that it is sufficient, note again that it suffices to show that $u\in \Z_2^*$ has a square root. Again, put $f(x)=x^2-u$ and let $a$ be a square root of $u$ modulo 8. Then $a\in \Z_s^*$ and $f'(a)=2a$. So $|f'(a)|^2=2^{-2}> |f(a)|=2^{-3}.$ Therefore, Hensel's Lemma applies again to give us a $b\in \Z_2$ such that $f(b)=0$. [4 marks. Seen in example in lecture.] \ms (e) First, since 11 has a square root in $\mathbb{R}$, it is clear that the polynomial has a root in $\mathbb{R}$. In $\Q_2$, we have $-7\equiv 1 \mod 8$, so that $-7$ has a square root in $\Z/8$, and hence, in $\Z_2$ by Hensel's lemma. In $\Q_7$, $11\equiv 4=2^2\mod 7$, so 11 has a square root in $\Z_7$ by Hensel's lemma. Also, $-7\equiv 4\equiv 2^2 \mod 11$, so that $-7$ has a square root in $\Z_{11}$. Now, let $p$ be any prime other than 2, 7 and 11. Then $-7$ and 11 are units in $\Z_p$, and $$(-77/p)=(-7/p)(11/p),$$ so that at least one of $-77$, $-7$ and 11 must have a square root mod $p$. Hence, by Hensel's lemma, one of them must have a square root in $\Z_p$. [5 marks. Unseen, but similar seen.] \bigskip \bigskip Question 2. \ms (a) A formal group over $R$ is a power series $F(X,Y)\in R[[X,Y]]$ such that $$F(X,Y)=X+Y+\mbox{terms of degree $\geq 2$};$$ $$F(F(X,Y),Z)=F(X,F(Y,Z));$$ $$F(X,Y)=F(Y,X).$$ Given an elliptic curve $y^2=x^3+Ax+B$, there is a formal group $F_{\cE}(X,Y)\in \Z[A,B][[X,Y]]$ such that $p\Z_p$ with the group law induced by $F_{\cE}$ is isomorphic to $\cE_1(\Q_p)$ with the elliptic curve group law. [2 marks.] \ms (b) Let $\omega=F_X(0,T)^{-1}dT$, where the subcript refers to the derivative with respect to the first variable. Then $$\omega\circ F(T,S)=F_X(0, F(T,S))^{-1}F_X(T,S)dT.$$ So we must show $$F_X(0,F(T,S))^{-1}F_X(T,S)=F_X(0,T)^{-1},$$ or $$F_X(0,T)F_X(T,S)=F_X(0,F(T,S)).$$ But we have $$F(U, F(T,S))=F(F(U,T),S).$$ Differentiation with respect to $U$ yields $$F_X(U, F(T,S))=F_X(F(U,T),S)F_X(U,T).$$ Setting $U=0$, we get $$F_X(0, F(T,S))=F_X(T,S)F_X(0,T),$$ which is what we wanted. Hence, $\omega$ is an invariant differential form. Since $F(X,Y)\equiv X+Y$ up to terms of higher order, it is clear $\omega $ is normalized, i.e., of the form $$\omega=a(T)dT,$$ where $a(T)=1+Tb(T)$ and $b(T)\in \Z_p[T]$. \ms Let $\alpha =P(T)dT$ be an invariant differential form. Then $$P(F(T,S))F_X(T,S)dT=P(T)dT.$$ So $$P((F(T,0))F_X(T,0)=P(0),$$ or $$P(T)=P(0)F_X(T,0)^{-1}.$$ That is $\alpha=P(0)\omega.$ Hence, if $\alpha $ is normalized, then $\alpha =\omega$. [3 marks] \ms (c) We put $[0](T)=0$, $[1](T)=T$ and $[m](T)=F([m-1](T),T)$ for $m>1$. Also, $[m](T)=F([m+1](T), i(T))$ for $m<1$, where $i(T)\in T\Z_p[[T]]$ is the unique power series such that $F(T,i(T))=0$. We see by induction that $[m](T)=mT+h(T)$ where $T^2|h(T)$. Now, from $$F([m](X), [m](Y))=[m](F(X,Y)),$$ we see that $$F_X([m](X), [m](Y))[m]'(X)=[m]'(F(X,Y))F_X(X,Y).$$ Setting $X=0$, we get $$F_X(0, [m](Y))[m]'(0)=[m]'(Y)F_X(0,Y),$$ so that $$[m]'(Y)F_X(0,[m](Y))^{-1}=[m]'(0)F_X(0,Y)^{-1}.$$ That is, $$\omega\circ [m]=[m]'(0)\omega.$$ Applying this to $[p]$, we get $$\omega\circ [p]=p\omega.$$ If we put $\omega=P(T)dT$, where $P(T)=1+Th(T)$, this implies $p|[p]'(T)$. Writing $$[p](T)=pT+\sum_{n=2}^{\infty} a_nT^n,$$ we get $p|na_n$ for all $n$. Thus, we can group the terms for which $p|n$ into $g(T^p)$ to get the form we want. [5 marks] \ms (d) We have the isomorphism $$(p\Z_p, F_{\cE}(X,Y))\simeq (\cE, +_{\cE}).$$ So the torsion subgroups will be isomorphic. If $m$ is not divisible by $p$, then $$[m](T)=mT+T^2g(T),$$ is an isomorphism (since $m$ is a unit in $\Z_p$), so no non-trivial element is killed by $[m]$. If $p|m$, and $p^a$ is the largest power of $p$ that divides $m$, then $m=p^am_0$ for $(m_0,p)=1$. Therefore, for any point $x$ of order $m$, $m_0x$ has order $p^a$. So by induction on $a$, it suffices to show there is no $p$-torsion. If $p=2$, then the 2 torsion points in $\cE(\Q_p)$ satisfy $y=0$, so they cannot reduce mod $p$ to the origin. Hence, we assume $p>2$. Let $z\in p\Z_p$ correspond to a $p$-torsion point in $\cE_1(\Q_p)$. If $[p](z)=0$, then we have $$pz=z^pr(z)$$for some power series $r\in \Z_p[[T]]$. Thus, $$p^{-1}|z|=|z|^p|r(z)|\leq |z|^p.$$ If $|z|\neq 0$, then we get $1/p^{1/(p-1)}\leq |z|\leq 1/p$, which is a contradiction for $p>2$. So we must have $z=0$. [7 marks. Bookwork to here.] \ms (e) Note that $(1,2)$ is a rational point. We compute $2(1,2)=(a,b)$. The slope of the tangent line is $y'=3x^2/2y=3/4$ at$ (1,2)$. Thus, the tangent line is $y=3x/4+5/4.$ To find the point of intersection, we substitute $$x^3-(3x/4+5/4)^2+3.$$ The $x^2$ term has coefficient $-9/16$. Thus $1+1+a=9/16$. Therefore, $a=-23/16$ has non-integral coordinates. But then, $(a,b)\in \cE_1(\Q_2)$ and $(a,b)$ is not torsion. Hence, $\cE(\Q)$ is infinite. [8 marks. Unseen.] \ms \ms Question 3. \bigskip (a) As coset representatives for $\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^2$ we may take non-zero square-free values $r\in\mathbb{Z}$. Such an $r$ arises from $q(x,y)$ if and only if $x=ru^2$ and $x^2+ax+b=rv^2$ for some rational $u,v$. If $u=l/m$ in lowest terms, then $r^2l^4+ral^2m^2+bm^4=rn^2$ with integers $l,m$ for which $(l,m)=1$ and $m\neq 0$. Since $rn^2$ is an integer and $r$ is square-free it follows that $n$ is also an integer. We claim that $r|b$. Otherwise there would be a prime $p|r$ with $p\nmid b$. Then $p|r(n^2-rl^4-al^2m^2)=bm^4$, whence $p|m$. This entails $p^2|r^2l^4+ral^2m^2+bm^4$, whence $p|rn^2$, from which we deduce that $p|n$. Finally we would have $p^3|rn^2-ral^2m^2-bm^4=r^2l^4$, which imples that $p|l$. This contradicts our assumption that $(l,m)=1$, and therefore shows that no such $p$ can exist. It therefore follows that $r|b$, so that $q$ must have finite image. [7 marks] \bigskip (b) The maps $\phi:\mathcal{C}\rightarrow\mathcal{D}$ and $\hat{\phi}:\mathcal{D}\rightarrow\mathcal{C}$ are homomomorphisms with $\phi\hat{\phi}$ and $\hat{\phi}\phi$ being $[2]$ on the respective groups $\mathcal{D}$ and $\mathcal{C}$. We are told that ${\rm ker}(q)=\hat{\phi}(\mathcal{D})$, and we have a similar map $\hat{q}$ on $\mathcal{D}$ with ${\rm ker}(\hat{q})=\phi(\mathcal{C})$. We then obtain induced injective homomorphisms from $\mathcal{C}/\hat{\phi}(\mathcal{D})$ and $\mathcal{D}/\phi(\mathcal{C})$ into $\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^2$, each with finite image, as above. Thus $\mathcal{C}/\hat{\phi}(\mathcal{D})$ and $\mathcal{D}/\phi(\mathcal{C})$ are finite. Let $\{g_1,\ldots,g_k\}$ be coset representatives for $\mathcal{C}/\hat{\phi}(\mathcal{D})$, and similarly $\{h_1,\ldots,h_l\}$ for $\mathcal{D}/\phi(\mathcal{C})$. Given $g\in \mathcal{C}$ there is a $g_i$ and an $h\in\mathcal{D}$ such that $g=g_i+\hat{\phi}(h)$. Similarly there is an $h_j$ and a $g'\in \mathcal{C}$ so that $h=h_j+\phi(g')$. Since $\hat{\phi}$ is a homomorphism, and $\hat{\phi}\phi=[2]$ we have \[g=g_i+\hat{\phi}(h)=g_i+\hat{\phi}(h_j+\phi(g')) =g_i+\hat{\phi}(h_j)+\hat{\phi}\phi(g')=g_i+\hat{\phi}(h_j)+2g'.\] Thus the cosets $g_i+\hat{\phi}(h_j)+2\mathcal{C}(\mathbb{Q})$ cover $\mathcal{C}$, whence $\mathcal{C}/2\mathcal{C}$ is finite. [7 marks --- Bookwork to here] \bigskip (c) For $y^2=x(x^2-p^2)$ we have $s=2$. The image of $q$ corresponds to solvable equations $r^2l^4-p^2m^4=rn^2$, with $r|p$; and since $\mathcal{E}'$ is given by $y^2=x(x^2+4p^2)$ the image of $\hat{q}$ corresponds to solvable equations $r^2l^4+4p^2m^4=rn^2$, with $r|2p$. For $r^2l^4-p^2m^4=rn^2$ there are solutions $(1,0,1),(0,1,p), (1,1,0)$ and $(1,1,0)$ for $r=1,-1,p$ and $-p$ respectively. For $r^2l^4+4p^2m^4=rn^2$ we must clearly have $r>0$. Thus there are at most 4 possibilities for $r$, namely $1,2,p$ and $2p$. Hence, from the above, \[2^{R+2}=\#\mathcal{C}/2\mathcal{C}\le \big(\#\mathcal{C}/\hat{\phi}(\mathcal{D})\big) \big(\#\mathcal{D}/\phi(\mathcal{C})\big)\le 4\times 4,\] so that the rank is always at most 2. [5 marks] \bigskip (d) If $x^4+p^2y^4=2z^2$ has a non-trivial integral solution we may choose a minimal one. If $p|z$ we would have $p|x$ and so $y^4+p^2(x/p)^4=2(z/p)^2$, contradicting minimality. Hence $p\nmid z$. However $2z^2\equiv x^4\pmod{p}$, and therefore $(\tfrac{2}{p})=+1$. [2 marks] \bigskip (e) Thus if $r^2l^4+4p^2m^4=rn^2$ and $r=2$ then $p\equiv\pm 1\pmod{8}$. Thus for $p\equiv 3$ or $5\pmod{8}$ the case $r=2$ cannot occur, giving $\#\mathcal{D}/\phi(\mathcal{C})\le 3$, whence $R<2$. [2 marks] \bigskip For $p=5$ the curve $C$ has a point $(x,y)=(25/4,75/8)$, which is non-integral and therefore of infinite order. Thus $R\ge 1$, and hence $R=1$. [2 marks --- Unseen, but similar to previous exam questions] \end{flushleft} \end{document}