\documentclass[a4paper]{article} \usepackage{amsmath,amssymb,amsthm,enumerate,eucal,tikz} %\usepackage{geometry} %\geometry{ % includeheadfoot, % margin = 2.54cm %} %\setlength\topmargin{-.1in} %\setlength\headheight{0in} %\setlength\headsep{0.2in} %\setlength\textheight{8.5in} %\setlength\textwidth{5.9in} %\setlength\oddsidemargin{0.4in} %\setlength\evensidemargin{0.4in} %% \usepackage[T1]{fontenc} %% \usepackage[scaled]{beramono,berasans} %% \usepackage[charter]{mathdesign} \usepackage{microtype} \newfont{\Bb}{msbm10 scaled\magstep0} \newfont{\Bbl}{msbm10 scaled\magstep1} \newfont{\Bbs}{msbm10 scaled 800} \def \qmodsquares {{\Q}^*/({\Q}^*)^2} \def\sqr{\ifmmode\square\else{$\square$}\fi} \def\square{\vcenter{ \hrule height.1mm \hbox{\vrule width.1mm height2.2mm\kern2.18mm\vrule width.1mm} \hrule height.1mm}} % This is a slimmer sqr. \null \def\le{\leqslant} \def\ge{\geqslant} \def\etq{{\cal E}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q})} \def\etqp{{\cal E}_{\lower 1pt\hbox{\eightrm tors}}({\Bbb Q}_p)} \def\c{{\cal C}} \def\d{{\cal D}} \def\e{{\cal E}} \def\pk{\phi _\kappa} \def\im{{\hbox{\sl im}}} \def\hs{H_{\varsigma}} \def\hpk{\hat \phi _\kappa} \font\sc=cmssqi8 \def\scc#1{\hbox{\sc #1}} \def\sf{{\scc F}} \def\pnbq{{\Bbb P}^n(\overline {\Bbb Q} )} \def\hk{{\hat \kappa}} \def\bq{{\overline {\Bbb Q}}} \def\hq{{\hat q}} \def\pv{\prod\limits_v } \def\pnk{{\Bbb P}^n(K)} \def\mnkvw{{\Bbb M}^n(K[{\bf v}^2,{\bf w}^2])} \def\pnkv{{\Bbb P}^n(K[{\bf v}^2])} \def\kj{\kappa (J)} \def \qmods {{\Bbb Q}^*/({\Bbb Q}^*)^2} \def \qmodss { {\Bbb Q}^*/({\Bbb Q}^*)^2 \times {\Bbb Q}^*/({\Bbb Q}^*)^2 } \def \qs{{\Bbb Q}^*} \def \qss{({\Bbb Q}^*)^2} \def\bbQ{\Bbb Q} \def\bbF{\Bbb F} \def\bbZ{\Bbb Z} \def\bbR{\Bbb R} \def\bbC{\Bbb C} \def\notdiv{{\not\hskip-.5pt |\ }} \def\Q{{\Bbb Q}} \def\F{{\Bbb F}} \def\Z{{\Bbb Z}} \def\R{{\Bbb R}} \def\C{{\Bbb C}} % \begin{document} \centerline{\bf Elliptic Curves. MT 2025. Sheet 0 solutions.} \rm \bigskip % {\it This sheet is not intended to be handed in or discussed in classes. It is % for you to use to reinforce the background material discussed in the preliminary % reading file.} % \bigskip \noindent {\bf 1.} Determine whether the following are groups. \par\noindent {\bf (a).} The set of all $2\times 2$ matrices under matrix multiplication. \textbf{Solution:} \textit{No: no inverses for singular matrices.} \par\smallskip \noindent {\bf (b).} The set of all $2\times 2$ matrices under matrix addition. \textbf{Solution:} \textit{Yes!} \par \medskip \noindent {\bf 2.} For each of the following, decide whether $\phi$ is a homomorphism. When $\phi$ is a homomorphism, decide whether~$\phi$ is injective, surjective, bijective, and find the kernel of~$\phi$. \par\noindent {\bf (a).} $\phi : \mathbb{Z} , + \rightarrow \mathbb{Q}^* , \times : x \mapsto x^2+1$. \textbf{Solution:} \textit{No: for example, $\phi(2) \neq \phi(1)^2$.} \par\smallskip \noindent {\bf (b).} $\phi : \mathbb{Q} , + \rightarrow \mathbb{R} , + : w \mapsto \sqrt{2}\, w$. \textbf{Solution:} \textit{Can check directly that this is a homomorphism. It is bijective ($\sqrt{2}$ is invertible, so multiplication by it is bijective), so the kernel is zero.} \par\smallskip \noindent {\bf (c).} $\phi : \mathbb{Z} , + \rightarrow \mathbb{Z} / 3\mathbb{Z} , + : x \mapsto 2x$. \textbf{Solution:} \textit{This is a surjective homomorphism, since $2$ is coprime to $3$. The kernel is $3\Z$.} \par \medskip \noindent {\bf 3.} \par\noindent {\bf (a).} In $\qmodsquares$, decide whether the following are true or false: $3=1/27$, $-4=4$, $3=5/6$. \textbf{Solution:} \textit{$3\times 27 = 3^4$ is a square, so $3 = 1/27$ mod $(\Q^*)^2$. } \textit{$4 = 1$ and $-4 = -1$ mod $(\Q^*)^2$. But $-1$ is not a rational square, so $-4 \neq 4$ mod $(\Q^*)^2$.} \textit{$5/18$ is not a rational square (it has prime factors appearing with odd powers).} \par\smallskip \noindent {\bf (b).} In $\qmodsquares$, write each of the following as a square free integer: $-2/27$, $16$, $12$, $1/3$. \textbf{Solution:} \textit{$-2\cdot 3^{-3} = -2\cdot 3 = -6$ mod $(\Q^*)^2$. } \textit{$16 = 1$ mod $(\Q^*)^2$.} \textit{$12 = 3$ mod $(\Q^*)^2$.} \textit{$1/3 = 3$ mod $(\Q^*)^2$.} \par\smallskip \noindent {\bf (c).} Perform each of the following in $\qmodsquares$, writing your answer as a square free integer: $6\times 10$, $10 / 21$, $15^{101}$, $3^{-1}$. \textbf{Solution:} \textit{$6\times 10 = 2^2\cdot 3 \cdot 5 = 15$ mod $(\Q^*)^2$.} \textit{$10/21 = 2\cdot 5 \cdot 3^{-1}\cdot 7^{-1} = 2\cdot 5 \cdot 3\cdot 7 = 210$ mod $(\Q^*)^2$.} \textit{$15^{101} = 15$ mod $(\Q^*)^2$.} \textit{$3^{-1} = 3$ mod $(\Q^*)^2$.} \par\smallskip \noindent {\bf (d).} How many elements are in each of the groups: $\qmodsquares$, ${\mathbb{R}}^*/({\mathbb{R}}^*)^2$, ${\mathbb{C}}^*/({\mathbb{C}}^*)^2$? \textbf{Solution:} \textit{The elements of $\qmodsquares$ are in bijection with square free integers. So there are (countably) infinitely many.} \textit{Every positive real is a square, so the sign map gives an isomorphism} \[{\mathbb{R}}^*/({\mathbb{R}}^*)^2 \cong \{\pm 1\}.\] \textit{Every complex number can be written as a square of another complex number, so the group ${\mathbb{C}}^*/({\mathbb{C}}^*)^2$ is trivial.} \par\medskip \newpage \noindent {\bf 4.} \par\noindent {\bf (a).} Find all singular points on the curve (defined over $\C$) $$ \mathcal{C} : f(X,Y) = X^4 + Y^3 - 3 X^2 Y = 0. $$ \textbf{Solution:} \textit{For $(x,y)$ to be a singular point, we need $f(x,y) = \frac{\partial f}{\partial X}(x,y) = \frac{\partial f}{\partial Y}(x,y) = 0$. In particular, we have $4x^3 - 6xy = 0$ and $3y^2-3x^2 = 0$. We deduce from these two equations that $y^2 = x^2$, hence $y = \pm x$, and then $4x^3 \mp 6x^2 = 0$. This gives the possibilities $(x,y) = (0,0), (\pm3/2,3/2)$. Only the first is a point on the curve, so the unique singular point is $(0,0)$.} \medskip\noindent Find all tangents to $\mathcal{C}$ at the point $(0,0)$. \textbf{Solution:} \textit{See Comment 0.100 for how to do this computation. We write} \[f(X,Y) = Y^3 - 3X^2Y + (\text{higher order terms})\] \textit{and then factorise $Y^3 - 3X^2Y = Y(Y-\sqrt{3}X)(Y+\sqrt{3}X)$. So we have three tangents: $Y = 0, Y = \sqrt{3}X, Y = -\sqrt{3}X$. Try sketching the graph (e.g.~with Wolfram Alpha.)} \par\smallskip \noindent {\bf (b).} Find all singular points on the curve (defined over $\C$) $$ \mathcal{C} : f(X,Y) = Y^2 - X(X^2-1)^2 = 0.$$ \textbf{Solution:} \textit{Computing the partial derivative with respect to $Y$, we see that $y = 0$ is necessary for a singular point. So the possible singular points are $(0,0), (1,0), (-1,0)$. We have $\frac{\partial f}{\partial X} = -(X^2-1)^2 -2X(X^2-1)(2X)$, so the two singular points are $(x,y) = (\pm1, 0)$. } \medskip \noindent Find all tangents to $\mathcal{C}$ at the points $(0,0)$ and $(1,0)$. \textbf{Solution:} \textit{The unique tangent at $(0,0)$ is $X = 0$. At $(1,0)$ we compute} \[f(1+X,Y) = Y^2 - (1+X)(X^2+2X)^2 = Y^2 - 4X^2 + (\text{higher order terms}).\] \textit{So we have two tangent lines at $(1,0)$, $Y = \pm 2(X-1)$. } \par \medskip \noindent {\bf 5.} Show that $\mathcal{C} : Y^2 = X^3 + AX + B$ is smooth if $4A^3 + 27B^2 \not= 0$ and we work over a field with characteristic $\neq 2$. What happens in characteristic $2$? \textbf{Solution:} \textit{We set $f(X,Y) = Y^2 - X^3 - AX - B$. So $\frac{\partial f}{\partial Y}(x,y) = 0$ implies $y = 0$ (if $2 \neq 0$). So the possible singular points are $(x,0)$ where $x$ is a root of the cubic $X^3 + AX + B$. The vanishing $\frac{\partial f}{\partial X}(x,0) = 0$ is then equivalent to $x$ being a repeated root of the cubic. The discriminant of the cubic polynomial is $4A^3 + 27B^2$, so that gives the desired criterion for smoothness. } \textit{In characteristic $2$, we have $\frac{\partial f}{\partial Y}(x,y) = 0$ for all points $(x,y)$. The equation $\frac{\partial f}{\partial X}(x,y) = 0$ gives us $x^2 = A$. So we have singular points $(x,y)$ when $x^2 = A$ and $y^2 = B$. } \par \medskip\noindent {\bf 6.} For each of the following curves, find the irreducible components over~$\mathbb{Q}$ and the irreducible components over~$\mathbb{C}$. \par\noindent {\bf (a).} $\mathcal{C} : Y^2 = X^5$. \textbf{Solution:} \textit{We have to factorise the polynomial $Y^2 - X^5$ over $\mathbb{Q}$ and $\mathbb{C}$. We claim that $Y^2 - X^5$ is irreducible over $\mathbb{C}$. Here is a long-winded proof (a more efficient argument might exist!). View $Y^2 - X^5$ as an element of $(\C[X])[Y]$, i.e.~a polynomial in $Y$ with coefficients in $\C[X]$. We cannot factor it as a product of polynomials in $Y$ with positive degree, since $X^5$ does not have a square root in $\C[X]$. So we deduce that if $Y^2 - X^5 = f_1(X,Y)f_2(X,Y)$, then one of the factors, say $f_1$ is actually just a polynomial in $X$. But then $f_1$ must actually be a constant, otherwise there would be a (complex) root $x_0$ of $f_1$ which would satisfy $y^2 - x_0^5 = 0$ for all $y \in \C$. } \par\smallskip \noindent {\bf (b).} $\mathcal{C} : Y^3 = X^3$. \textbf{Solution:} \textit{We factorise $Y^3 - X^3 = (Y-X)(Y^2+XY+X^2)$. So we get $Y = X$ as one component, and $Y^2 + XY + X^2 = 0$ as another. The latter is irreducible over $\Q$ but reducible over $\C$. We factorise } \[Y^2 + XY + X^2 = (Y-\omega X)(Y-\bar{\omega}X)\] \textit{where $\omega = \frac{-1 + \sqrt{-3}}{2}$, a primitive third root of unity, satisfies $\omega + \bar{\omega} = -1$ and $\omega\bar{\omega} = 1$. So over $\C$ the components are $Y=X$, $Y = \omega X$ and $Y = \bar{\omega}X$. } \par\smallskip \noindent {\bf (c).} $\mathcal{C} : Y^2 = X^3 + 1$. \textbf{Solution:} \textit{As for part (a), we observe that the polynomial $Y^2 - X^3 - 1$ is irreducible viewed as a polynomial in $Y$ with coefficients in $\C[X]$. Similarly to part (a), it is also not divisible by a non-constant element of $\C[X]$. So this curve is irreducible. } %\par\medskip\noindent{\bf 7.} %\par\noindent {\bf (a).} Find a birational transformation over~$\mathbb{Q}$ %between the curves $2X^2 - Y^2 = 1$ and $X^2 + Y^2 - 6XY = 1$. % %\textbf{Solution:} \textit{We find rational points on each curve. This tells us that each curve is birational to $\mathbb{P}^1$, and hence birational to each other. For the first, we have $(1,1)$. For the second, we have $(1,0)$. So the points of the first conic are parameterised by $t = \frac{y-1}{x-1} \in \mathbb{P}^1(\Q)$. To find a corresponding point on the second curve, we intersect $Y = t(X-1)$ with the curve. We get the equation $X^2 + t^2(X-1)^2 - 6tX(X-1) = 1$. The coefficient of $X^2$ is $1+t^2 -6t$ and coefficient of $X$ is $-2t^2 + 6t$. So if the intersection point is $(x_1,y_1)$, we have $x_1 + 1 = \frac{2t^2 - 6t}{t^2-6t+1}$, and hence $x_1 = \frac{t^2 - 1}{t^2-6t+1}, y_1 = \frac{2t(3t-1)}{t^2-6t+1}$. Substituting $t = \frac{y-1}{x-1}$, we get the rather unpleasant birational transformation from the first curve to the second} %\[(x,y) \mapsto \left(\frac{(y-1)^2 - (x-1)^2}{(y-1)^2-6(y-1)(x-1)+(x-1)^2}, \frac{2(y-1)(3(y-1)-(x-1))}{(y-1)^2-6(y-1)(x-1)+(x-1)^2} \right).\] % % %\par\smallskip %\noindent {\bf (b).} Find a birational transformation over~$\mathbb{Q}$ %between the curves $Y^2=(X+2)^6(X^3+1)$ and $Y^2 = X^3 + 1$. % %\textbf{Solution:} \textit{We can rewrite the first equation as $(\frac{Y}{(X+2)^3})^2 = X^3 + 1$. So we can take the birational transformation} %\[(x,y) \mapsto (x,\frac{y}{(x+2)^3}).\] % %\par\smallskip %\noindent {\bf (c).} Find a birational transformation over~$\mathbb{C}$ %between the curves~$Y^2=2X^2$ and~$Y^2=X^2$. Is there a birational %transformation over~$\mathbb{Q}$? % % %\textbf{Solution:} %\textit{Over $\C$, we have the birational transformation $(x,y) \mapsto (\sqrt{2}x,y)$. Over $\Q$, the second curve is reducible, with irreducible components $Y = \pm X$. The first curve is irreducible. So the two curves are not birational over $\Q$. Alternatively, the first curve's only rational point is $(0,0)$, whilst the second has infinitely many, so again they cannot be birational over $\Q$.} % %\par\medskip \noindent {\bf 7.} \par\noindent {\bf (a).} Find the discriminant of~$X^4-2$. \textbf{Solution:} \textit{Write down the resultant matrix for $(X^4-2,4X^3)$. Repeatedly doing Laplace expansion down the columns (from right to left) gives determinant $(-2)^34^4 = -2^{11}$. } \par\smallskip \noindent {\bf (b).} Find the resultant of $X^3 - a$ and $X^2 - b$, where $a,b$ are constants. \textbf{Solution:} {$b^3 - a^2$.} \par\medskip\noindent {\bf 8.} Find all intersection points (with multiplicities) over~$\mathbb{C}$ of the curves: $X^3 + Y^3 = Z^3$ and $X^2 + Y^2 = Z^2$. \textbf{Solution:} \textit{See Comment 0.122. We first compute intersection points with $Z \ne 0$. We compute the resultant of $f(x,y) = x^3 + y^3 - 1$ and $g(x,y) = x^2 + y^2 - 1$, viewed as polynomials in the variable $y$ over $\C[x]$. By 8(b) we get resultant $(1-x^2)^3 - (1-x^3)^2 = -(x - 1)^2 x^2 (2 x^2 + 4 x + 3)$. Let's consider the multiple roots $x = 0, x = 1$. We get, respectively, $y^3 = 1, y^2 = 1$ and $y^3 = 0, y^2 = 0$. So we have intersection points $(0:1:1)$ and $(1:0:1)$, both with multiplicity 2, and two (complex conjugate) intersection points with multiplicity 1: $(-1+\frac{\sqrt{2}}{2}i:-1-\frac{\sqrt{2}}{2}i:1),(-1-\frac{\sqrt{2}}{2}i:-1+\frac{\sqrt{2}}{2}i:1).$ That gives all the sections, since we've found 6 with multiplicity. We can also check directly that there are no intersection points with $Z=0$.} \par \medskip \noindent {\bf 9.} \par\noindent {\bf (a).} Decide whether each of $2,3,5,10,15$ are quadratic residues modulo~1009 (if you use quadratic reciprocity, this should not involve any lengthy computations). \textbf{Solution:} \textit{Note that 1009 is prime. We have $\left(\frac{2}{1009}\right) = +1$, since $1009 = 1$ mod $8$. } \textit{We have $\left(\frac{3}{1009}\right) = \left(\frac{1009}{3}\right) = \left(\frac{1}{3}\right) = +1.$} \textit{We have $\left(\frac{5}{1009}\right) = \left(\frac{1009}{5}\right) = \left(\frac{4}{5}\right) = +1.$} \textit{We have $\left(\frac{10}{1009}\right) = \left(\frac{2}{1009}\right)\left(\frac{5}{1009}\right)= +1.$} \textit{We have $\left(\frac{15}{1009}\right) = \left(\frac{3}{1009}\right)\left(\frac{5}{1009}\right)= +1.$} \par\smallskip \noindent {\bf (b).} Describe all primes~$p$ such that $3$ is a quadratic residue modulo~$p$. Describe all primes~$p$ such that $5$ is a quadratic residue modulo~$p$. Describe all primes~$p$ such that $10$ is a quadratic residue modulo~$p$. \textbf{Solution:} \textit{For $p > 3$, we have $\left(\frac{3}{p}\right) = (-1)^{(p-1)/2}\left(\frac{p}{3}\right)$. So $3$ is a QR mod $p$ if and only if $p = \pm 1$ mod $12$. } \textit{For an odd prime $p \neq 5$, we have $\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right)$. So $5$ is a QR mod $p$ if and only if $p = \pm 1$ mod $5$. } \textit{For an odd prime $p \neq 5$, we have $\left(\frac{10}{p}\right) = \left(\frac{2}{p}\right)\left(\frac{p}{5}\right)$. So $10$ is a QR mod $p$ if and only if one of the following holds:} \begin{itemize} \item $p \equiv \pm 1$ mod $5$ and $\pm 1$ mod $8$ \item $p \equiv \pm 3$ mod $5$ and $\pm 3$ mod $8$ \end{itemize} \textit{Equivalently, $10$ is a QR mod $p$ if and only if $p$ mod $40 \in \{\pm1,\pm3,\pm9,\pm13\}$. Note that this covers 8 of the 16 congruence classes in $(\Z/40\Z)^\times$.} \par\medskip \noindent {\bf 10.} Are there integers $a,b,c$, not all~$0$, such that $2a^2 + 5b^2 = c^2$? \textbf{Solution:} \textit{We can reduce to looking for solutions which are pairwise coprime. Then consider the equation mod $5$. It says $2a^2 = c^2$ mod $5$, which implies that $a = c = 0$ mod $5$ (since $2$ is not a QR mod $5$). This contradicts coprimality of $a$ and $c$. So there are no non-trivial integer solutions. } \par\medskip \noindent {\bf 11.} For any $n\in{\mathbb{N}}$ define, as usual, Euler's $\phi$-function by: $$ \phi(n) = \# \{ x : 1 \leqslant x \leqslant n \hbox{ and gcd} (x,n) = 1 \}. $$ For any prime~$p$, what is $\phi(p^r)$? For any distinct primes $p_1,p_2$, what is $\phi(p_1 p_2)$? \textbf{Solution:} \textit{There are $p^{r-1}$ multiplies of $p$ in the interval $[1,p^r]$. So $\phi(p^r) = p^r-p^{r-1} = p^{r-1}(p-1)$.} \par \medskip For each of the following examples of the type $a^b \ (\hbox{mod }n)$, reduce $a^b \ (\hbox{mod }n)$ to a member of $\{ 0, \ldots , n-1 \}$. \par\noindent $2^{12} \ (\hbox{mod }13)$, $3^{12} \ (\hbox{mod }13)$, $3^{24} \ (\hbox{mod }13)$, $3^{12000} \ (\hbox{mod }13)$, $3^{12002} \ (\hbox{mod }13)$, \hfill\par\noindent $4^{24} \ (\hbox{mod }35)$, $4^{48} \ (\hbox{mod }35)$, %$4^{48000} \ (\hbox{mod }35)$, $4^{48000001} \ (\hbox{mod }35)$, \hfill\par\noindent $7^{24} \ (\hbox{mod }35)$, $7^{48} \ (\hbox{mod }35)$, %$7^{48000} \ (\hbox{mod }35)$, $7^{48000001} \ (\hbox{mod }35)$. \textbf{Solution:} \textit{The first eight follow easily from Fermat--Euler: $1, 1, 1, 1, 9, 1, 1, 4$.} \textit{We have $7^{24} = 0$ mod $7$ and $1$ mod $5$. So $7^{24} = 21$ mod $35$.} \textit{Squaring, we also have $7^{48} = 0$ mod $7$ and $1$ mod $5$. So $7^{48} = 21$ mod $35$.} \textit{In fact, the same argument shows that $7^{24k} = 21$ mod $35$ for any positive integer $k$. So $7^{48000001} = 7\times 21 = 7$ mod $35$.} \end{document}