\documentclass{oxmathexam} \input{standardmacros.sty} \newcommand{\F}{{\mathbb F}} \renewcommand{\ss}{s} %% the prime p - 2 in solution %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % USER DEFINED MACROS - IF YOU MUST USE THEM, INSERT THEM HERE \def\g{{\cal G}} \def\h{{\cal H}} \def\o{\underline{\bf o}} \renewcommand{\C}{C} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% \school{Part C} %% \title{Modular Forms} %% \date{Version December 7, 2016. Checker: Victor Flynn} \begin{document} \begin{questions} \question \begin{parts} \part[11] State and prove {\it Hensel's Lemma}. \part[6] Let $p$ be a prime and $a$ an integer. \begin{subparts} \subpart Show that if $p$ divides $a$ then $x^p - x + a$ splits into linear factors over $\Z_p$, and otherwise $x^p - x + a$ has no roots in $\Z_p$. \subpart Compute an integer $x$ such that $x \equiv 1 \bmod{3}$ and $x^3 - x + 3 = 0 \bmod{3^{4}}$. \subpart Hence, or otherwise, find integers $x$ and $y$ such that $x^3 - x = 3^y - 3$ and $x > 3$. \end{subparts} \part[8] Let $p$ be a prime. Show that $x^2 + 7 y^3 + 5 z^5 = 0$ has a solution $x,y,z \in \Z_p$ with $(x,y,z) \ne (0,0,0)$. [{\it Hint: It may help to observe $\frac{1}{2} + \frac{1}{3} + \frac{1}{5} > 1$.}] \end{parts} \question \begin{parts} \part[6] Compute the torsion subgroup of the elliptic curve $y^2 = x^3 + 5x^2 + 3x + 7$. \part[6] Let $n$ be an integer with $n > 2$. Show that the equation $y^2 = x^3 + x + (n^2 - 2)$ has infinitely many rational solutions. \part[13] Let $R$ be a field of characteristic zero and let \[ F(X,Y) := \frac{X + Y}{1 + XY} \in R[[X,Y]].\] \begin{subparts} \subpart Prove that $F$ defines a formal group over $R$. \subpart Compute explicitly the formal logarithm $\log_F(T)$. \subpart Hence find an explicit isomorphism $f(T) \in T R[[T]]$ from $F$ to the formal multiplicative group \[ \widehat{G}_m(X,Y) = X + Y + XY\] with $f(T)$ a rational function in $T$. \end{subparts} \end{parts} \question \begin{parts} \part[14] Let ${\mathcal C} : Y^2 = X(X^2 + aX + b)$, where $a,b\in \Z, b \not= 0, a^2 - 4b \not= 0$, and let ${\mathcal D} : V^2 = U(U^2 + a_1 U + b_1)$, where $a_1 = -2a$ and $b_1 = a^2 - 4b$. Let the map $q$ be defined by \[ q: {\mathcal D}(\Q) \rightarrow \Q^* / (\Q^*)^2, \,\,\, (u,v) \mapsto u \mbox{ when } u \ne 0,\] \[ q: (0,0) \mapsto b_1,\,\,\, q: \underline{\bf o} \mapsto 1.\] \begin{subparts} \subpart Show that $q(P + Q) = q(P) q(Q)$ in the case when none of $P,Q,P + Q$ are $(0,0)$ or $\underline{\bf o}$. \subpart Let $P = (u,v) \in {\mathcal D}(\Q)$ with $P \ne (0,0)$. Explicitly compute the sum $P + (0,0)$, %% Prove that %% \[ P + (0,0) = \left( \frac{b_1}{u}, - \frac{b_1 v}{u^2} \right).\] and deduce that $q(P + (0,0)) = q(P)q((0,0))$. \subpart By considering the remaining cases, prove that $q$ is a homomorphism. \end{subparts} \part[11] Compute the rank of the elliptic curve $y^2 = x(x^2 - 3 x + 10)$. %% [{\it You may use any results from lectures provided they are clearly stated.}] \end{parts} \newpage {\bf Solutions:} 1. (a) [Bookwork] (Hensel's Lemma). Let~$K$ be a field, complete with respect to a non-Archimedean valuation $| \cdot |$, with valuation ring~$R = \{ x\in K : |x| \leqslant 1\}$. \par Let $f(x) \in R[x]$ and let $a_0\in R$ satisfy: \ \ \ \ $| f(a_0) | < | f'(a_0) |^2. \ \ \ \ \ \ \ (*)$ \par\noindent Then there exists a unique $a\in R$ such that $f(a) = 0$ and $| a - a_0 | \leqslant | f(a_0) |/ | f'(a_0) |$. [2 marks]\\ % N.B. I might verbally add: note that this is in contrast % with Newton-Raphson in R, where we need to know something % about f(x) for all x in an interval about the initial approximation. Proof: Define $f_j(x)$ by: \ \ \ \ $f(x + y) = f_0(x) + f_1(x) y + f_2(x) y^2 + \ldots,$ \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 \ \ \ \ \ \ \ \ \ \ \ $\leqslant | b_0 | < | f'(a_0) |$ \ \ (by $(*)$), \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 $f_0(a_0) + f_1(a_0)b_0 = 0$] % N.B. I might omit the following line: \par\ \ \ $\leqslant \hbox{max}_{j\geqslant 2} |f_j(a_0)| |b_0|^j \leqslant | b_0 |^2 = \frac{ |f(a_0 )|^2}{|f'(a_0 )|^2} = \rho | f(a_0) | < | f(a_0) |$, where $\rho = \frac{ |f(a_0 )|}{|f'(a_0 )|^2} < 1$. \par Summarising: $ |f'(a_1) | = |f'(a_0)|$ and $| f(a_1) | \leqslant \rho | f(a_0) | < | f(a_0) |$, where $\rho = \frac{ |f(a_0 )|}{|f'(a_0 )|^2} < 1$. [3 marks] \par\noindent For all~$n$, 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 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $= \frac{ | f(a_n) |^2 }{ | f'(a_n) |^2 } = \frac{ | f(a_n) |^2 }{ | f'(a_0) |^2 }$\ \ [by (1), the induction hypothesis] \par \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\leqslant \frac{ | f(a_0) | }{ | f'(a_0) |^2 } |f(a_n)|$ \ \ [since $|f(a_n)| \leqslant |f(a_0)|$ by (1), the induction hypothesis] \par \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $= \rho | f(a_n) | \leqslant \rho^{n+1} | f(a_0) |$ \ \ [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 a standard theorem in non-archimedean analysis) \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)]. Furthermore: \par\noindent $| a - a_0 | = | \sum b_n | \leqslant \max | b_n | = \max \frac{| f(a_n) |}{|f'(a_n)|} = \max \frac{| f(a_n) |}{|f'(a_0)|} = \frac{| f(a_0) |}{|f'(a_0)|}$ [by~(2)], as required. [4 marks] \par\noindent For uniqueness, imagine $\hat a \not= a$ also satisfied $f(\hat a) = 0$ and $| \hat a - a_0 | \leqslant | f(a_0) |/ | f'(a_0) |$. Let $\hat b = \hat a - a \not= 0$. \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 \ \ \ $< | f'(a_0) | \hbox{ [by (*)] } = | f_1(a_0) | = | f_1(a) |$ \ \ [by~(2) and continuity of $|f'(x)|$]. \par\noindent This gives $|{\hat b}^j f_j(a)| \leqslant |{\hat b}^j| \leqslant |{\hat b}^2| < |{\hat b} f_1(a)|$ % N.B. I might omit the next line: (since $|{\hat b}| \not= 0$ \& $|{\hat b}| < |f_1(a)|$) 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. [2 marks]\\ (b) (i) [2 marks] Let $f(x) = x^p - x + a$. If $p|a$ then $f(x) = x^p - x = \prod_{\alpha = 0}^{p-1} (x - \alpha) \bmod{p}$. Also for any $\alpha \in \Z$, $f^\prime(\alpha) = p \alpha^{p-1} - 1 + a \equiv - 1 \not = 0 \bmod{p}$. So we may apply Hensel's lemma to each $0 \leq \alpha \leq p-1$ to get a root $\hat{\alpha}$ and factorisation $f(x) = \prod_{\alpha = 0}^{p-1} (x - \hat{\alpha})$. [1 mark] Suppose now $a \bmod{p} \ne 0$. Then for any $\alpha \in \Z$ we have $f(\alpha) = \alpha^p - \alpha + a \equiv a \ne 0 \bmod{p}$, so there are no roots modulo $p$ and thus none in $\Z_p$. (ii) [2 marks] Let $a_0 = 1$ and $f(x) = x^3 - x + 3$, so $f^\prime(x) = 3 x^2 - 1$. Then iterating Newton we find \[ a_1 = 1 - \frac{3}{2} = - \frac{1}{2},\, a_2 = - \frac{1}{2} - \frac{-\frac{1}{8}+ \frac{1}{2} + 3}{\frac{3}{4} - 1} = 13.\] We know as the lifting is in fact quadratically convergent this should be a solution. And indeed $13^3 - 13 + 3 = 2187 = 3^7$. (iii) [1 mark: {\it this also gives a hint for the calculation in (ii)}] $x = 13$ and $y = 7$. (c) [{\it This is new, but a similar idea is used in question 1 of the 2017 paper.}] Let $p$ be a prime and consider the sets \[ A_2:= \{ x^2 : x \in \F_p^\star\},\, A_3 :=\{ 7 y^3 : y \in \F_p^\star\},\, A_5 :=\{ 5 z^5: z \in \F_p^\star\}.\] Assume now $p \ne 5, 7$ (to make sure none of these sets are just $\{0\}$). Since always $x \mapsto x^m$ is at most $m$ to $1$ on $\F_p^\star$ these sets have size at least $(p-1)/2$, $(p-1)/3$ and $(p-1)/5$ respectively. As \[ \frac{1}{2} + \frac{1}{3} + \frac{1}{5} > 1\] they must overlap. [1 mark] If $A_2 \cap A_3 \ne \emptyset$ we can find non-zero $x,y\in \F_p$ so that $x^2 = 7 y^3 \bmod{p}$ and then $x^2 + 7 (-y)^3 + 5 \cdot 0^5 \equiv 0 \mod{p}$. We can now apply Hensel, using for $p \ne 2$ the polynomial $f(X) = X^2 + 7 (-y)^3$ with starting value $X = x$, to get a solution $(\hat{x},-y,0)$. And using for $p = 2$ the polynomial $f(Y) = 7 Y^3 + x^2$ with starting value $Y = -y$, to get a solution $(x,\hat{y},0)$. [2 marks] If $A_2 \cap A_5 \ne 0$ we can find non-zero $x,z\in \F_p$ so that $x^2 = 5 z^5 \bmod{p}$ and then $x^2 + 7 \cdot 0^3 + 5 (-z)^5 \equiv 0 \mod{p}$. Now apply Hensel, using $f(X) = X^2 + 5 (-z)^5$ for $p \ne 2$ and $f(Z) = 5 Z^5 + x^2$ for $p = 2$. [2 marks] If $A_3 \cap A_5 \ne 0$ we can find non-zero $y,z\in \F_p$ so that $7 y^3 = 5 z^5 \bmod{p}$ and then $0^2 + 7 (-y)^3 + 5 z^5 \equiv 0 \mod{p}$. Now apply Hensel, using $f(Y) = 7 Y^3 + 5 (-z)^5 $ for $p \ne 3$ and $f(Z) = 5 Z^5 + 2(-y)^3$ for $p= 3$. [2 marks] Finally when $p = 5$ as $x \mapsto x^3$ is a bijection on $\F_p$ we have $A_2 \cap A_3 \ne \emptyset$, and the argument before works. And for $p = 7$, since $x \mapsto x^5$ is a bijection on $\F_p$ we have $A_2 \cap A_5 \ne 0$, and we find a solution as before. [1 mark] \newpage 2 (a) [Similar: I think this is a little tricky though, as it is somewhat different from the other computations of this sort they have done.] Reducing modulo $3$ we get the equation $y^2 = x^3 + 2x^2 + 7$. Now the polynomial on the RHS has no solutions in $\F_3$, and thus must be squarefree (any repeated root must be defined over the ground field). So the reduction is smooth, and once checks the points over $\F_3$ are \[ \{ \o, (0,1),(0,2),(1,1),(1,2) \}.\] So by the results in lectures the torsion subgroup has order dividing $5$. [2 marks] Now there is the obvious point $P = (1,4)$ on the curve. So all we need to do is check whether this order $5$ (rather than infinite order). The tangent at a point $(x,y)$ has gradient \[ \frac{3x^2 + 10 x + 3}{2 y}.\] So the tangent at $P$ is $y = 2x + 2$. Putting into the curve we get \[ 4 x^2 + \cdots = x^3 + 5 x^2 + \cdots \] \[ x^3 + x^2 + \cdots = (x - x(2P))(x-1)(x-1).\] So $x(2P) = -3$, and $2P = (-3,4)$. Doubling again we see that the tangent at $2P$ has gradient \[ \frac{3 \cdot 9 + 10 \cdot (-3) + 3}{8} = 0.\] So this tangent is $y = 4$. Putting in we get \[ (x^3 + 5 x^2 + 3x + 7) - 16 = (x + 3)(x + 3)(x - x(4P)).\] So looking at constants we get $x(4P) = 1$, and so $4P = (1,-4)$. Without further computation one now sees $P + 4P = \o$. [4 marks] (b) [Similar / New: they have seen that a non-rational solution implies positive rank.] This equation defines an elliptic curve exactly when $\Delta_n := 4 + 27 (n^2 - 2)^2 \ne 0$. So let us first assume $\Delta_n \ne 0$. [1 mark] The curve has the obvious point $P = (1,n)$. The strategy is to double it and observe $2P$ is not integral, when $n \ne \pm 2$, so $2P$ has infinite order (by an important result from lectures) giving our infinite set of solutions. [1 mark] The tangent at a point $(x,y)$ has gradient \[ \frac{3 x^2 + 1}{2 y} \] so at $P$ this is $\frac{4}{2n} = \frac{2}{n}$. Hence the tangent line is $y = \frac{2}{n} x + n - \frac{2}{n}$. Thus \[ \left( \frac{2}{n} x + \cdots\right)^2 = x^3 + 0 x^2 + \cdots\] and \[ x^2 - \frac{4}{n^2} x + \cdots = (x - 1)(x - 1)(x - x(2P)).\] Thus we find \[ x(2P) = \frac{4}{n^2} - 2. \] This is not integral provided $n \ne \pm 1, \pm 2$. [3 marks] On the discriminant, if we suppose $\Delta_n= 0$ then reducing mod $5$ we get $-1 + 2(n^2 - 2)^2 \equiv 0 \bmod{5}$ and so $(n^2 - 2)^2 = \frac{1}{2} \equiv 3 \bmod{5}$, which is a contradiction. Or alternatively, note $\Delta_n$ is in fact always positive. [1 mark] (c) (i) [4 marks: similar] We need to check the axioms. Commutativity is obvious, and also \[ F(X,Y) = X + Y + \cdots \] since $(1 + XY)^{-1} = 1 + (XY) + (XY)^2 + \cdots$. For associativity we have \[ F(X,F(Y,Z)) = \frac{X + \frac{Y + Z}{1 + YZ}}{1 + X\left(\frac{Y + Z}{1 + YZ}\right)} \] and \[ F(F(X,Y),Z) = \frac{\frac{X + Y}{1 + XY} + Z}{1 + \frac{X + Y}{1 + XY} Z} \] and both are \[ \frac{X + Y + Z + XYZ}{1 + YZ + XY + XZ}. \] (ii) [4 marks: similar] The canonical differential is $F_X(0,T)^{-1} dT$. One computes \[ F_X(X,Y) = \frac{(1 + XY) - (X + Y)Y}{(1 + XY)^2} = \frac{1 - Y^2}{(1+XY)^2}.\] Hence $F_X(0,T) = 1 - T^2$. So \[ \omega = \frac{dT}{1 - T^2} \] and the formal logarithm is \[ \int \omega = \int \frac{dT}{(1 + T)(1 - T)} = \frac{1}{2} \log \left(\frac{1 + T}{1 - T}\right) \,\,\left( = \tanh^{-1}(T) \right).\] (iii) [New] Thus by the lectures \[ \log_F(T) = \frac{1}{2} \log \left(\frac{1 + T}{1 - T}\right) \] is an isomorphism from $F$ to the formal additive group $\widehat{G}_a$. [1 mark] Now $\log(1 + T)$ is an isomorphism from $\widehat{G}_m$ to $\widehat{G}_a$, with inverse $\exp(T) - 1$. [1 mark] Thus the composition \[ \exp\left( \frac{1}{2} \log \left(\frac{1 + T}{1 - T}\right) \right) - 1 \] is an isomorphim. But this is not a rational function in $T$. [2 marks] However, recall that multiplication by $2$ is an isomorphism from $\widehat{G}_a$ to itself. This allows us to remove the factor $\frac{1}{2}$ and take \[ \exp\left(\log \left(\frac{1 + T}{1 - T}\right) \right) - 1 =\frac{2T}{1 - T}.\] [2 marks] (And indeed one can check by a computation this is indeed a homomorphism, and the series is certainly functionally invertible.) \newpage 3 (a) (i) [Bookwork] [6 marks] Let~$(u_1,v_1), (u_2,v_2),(u_3,v_3)$ be~3 points on~${\mathcal H} = {\mathcal D}(\Q)$ which sum to~$\o$, [so that $(u_1,v_1) + (u_2,v_2)= (u_3,-v_3)$]. Then these are the~3 points of intersection between~${\mathcal D}$ and some line defined over~$\Q$: $V = \ell U + m$, say. Substituting~$V = \ell U + m$ into ${\mathcal D}$ gives: $U(U^2 + a_1 U + b_1) - (\ell U + m)^2$, whose 3~roots must be~$u_1,u_2,u_3$. That is: $U(U^2 + a_1 U + b_1) - (\ell U + m)^2 = (U-u_1)(U-u_2)(U-u_3)$. Equating constant terms gives: $u_1 u_2 u_3 = m^2 = 1$ in $ \Q^* / (\Q^*)^2$, and so $u_1 u_2 = 1/u_3 = u_3$ in $ \Q^* / (\Q^*)^2$. (Note $u_1 u_2 u_3 \ne 0$, by our assumption.) Therefore, by the definition of~$q$ we have: $q\bigl( (u_1,v_1) \bigr) q\bigl( (u_2,v_2) \bigr) = q\bigl( (u_3,-v_3) \bigr)= q\bigl( (u_1,v_1) + (u_2,v_2) \bigr)$.\\ (ii) [The first part is in the lectures, but on the curve ${\mathcal C}$. The second part is new.] [5 marks] When $u\not= 0$, we first find the line through $(0,0)$ and $(u,v)$, which is: $V = \frac{v}{u} U$. Substituting this into~${\mathcal D}$ gives: \[ \Bigl( \frac{v}{u} \Bigr)^2 U^2 = U(U^2 + a_1 U + b_1) \] \[ v^2 U^2 = u^2 U^3 + a_1 u^2 U^2 + b_1 u^2 U\] \[u(u^2 + a_1 u + b_1) U^2 = u^2 U^3 + a_1 u^2 U^2 + b_1 u^2 U \mbox{ [since $(u,v)$ is on ${\mathcal D}$]} \] \[0 = u U^3 - (u^2 + b_1) U^2 + b_1 u U, \mbox{ [since $u\not= 0$]} \] and so $U(U - u)(u U - b_1) = 0$. The roots of this cubic are: $U = 0, U = u, U = b_1/u$. The line $V = \frac{v}{u} U$ and~${\mathcal D}$ intersect at: $$ (0,0), (u,v)\hbox{ and } \Bigl( \frac{b_1}{u}, \frac{b_1 v}{u^2} \Bigr) \hbox{ [since $U = \frac{b_1}{u}$ gives $V = \frac{v}{u}\frac{b_1}{u} = \frac{b_1 v}{u^2}$]} $$ and so $(u,v) + (0,0) = \Bigl( \frac{b_1}{u}, -\frac{b_1 v}{u^2} \Bigr)$ Thus $q(P + (0,0)) = b_1/u \equiv b_1 \cdot u = q((0,0)) q(P)$. (iii) [3 marks: new, and I think a little tricky.] If $P$ or $Q$ is $\underline{\bf o}$ the result is immediate. If $P + Q = \underline{\bf o}$ then we have $Q = - P$ and then always $q(Q) = q(P)$ and so $q(P)q(Q) = q(P)^2 \equiv 1 = q( \underline{\bf o})$. If $P = Q = (0,0)$ then $P + Q = \underline{\bf o}$ and $b_1 \cdot b_1 \equiv 1$. It remains then to consider the case $P + Q = (0,0)$. Then $Q = -P + (0,0)$ so by (ii) we have $q(Q) = q(-P + (0,0)) = q(-P)q((0,0))$ which is $q(P)q((0,0))$. Thus $q(P)q(Q) = q(P)^2 q((0,0)) \equiv q((0,0)) = q(P + Q)$. (b) [Similar: this is a standard type of exam question.] Using the notation from lectures we have $a = -3, b = 10, a_1 = 6, b_1 = -31$, and $C: y^2 = x(x^2 - 3x + 10)$, $D:v^2 = u(u^2 + 6 u - 31)$. [1 mark] We first consider $\h / \phi(\g)$. Under $q$ this is isomorphic to a subgroup of the squarefree divisors of $b_1 = -31$. So ${\rm im}(q) \leq \langle -1,-31 \rangle$. Taking $r = -1$ we easily spot the point $(-1,6) \in D(\Q)$ with $q((-1,6)) = -1$. And $q((0,0)) = -31$. %% For $r = 31$, this being quite a big number, we first examine the homogeneous space %% \[ W_{31}: 31 \ell^4 + 6 \ell^2 m^2 - m^2 = n^2.\] %% This has the obvious solution $(\ell,m,n) = (1,1,6)$. %% (This corresponds to the point $(31,6 \cdot 31)$: there %% is a formula for this point in the notes given $(\ell,m,n)$. Note finding this point by inspection would be much harder.) So ${\rm im}(q) = \langle -1,31 \rangle \cong C_2^2$. [4 marks] We next consider $\g / \hat{\phi}(\h)$. Under $\hat{q}$ this is isomorphic to a subgroup of the squarefree divisors of $b = 10$. So ${\rm im}(\hat{q}) \leq \langle -1,2,5 \rangle$. Again we easily spot points $(2,4)$ and $(5,10)$ on $C(\Q)$ and so $2$ and $5$ are in the image. For $r = -1$ the homogeneous space is \[ W_{-1} : - \ell^4 - 3 \ell^2 m^2 - 10 m^4 = n^2.\] This has no non-trivial solutions in $\R$, since always LHS is $\leq 0$ and RHS is $\geq 0$, so none in $\Z$. Hence ${\rm im}(\hat{q}) \leq \langle 2,5 \rangle> \cong C_2$. [5 marks] So in conclusion using the formula from Sheet 4 Question 3, the rank is $2 + 2 - 2 = 2$. [1 mark] \end{questions} \end{document}