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\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}