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\begin{document}
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\textbf{18.786 Number Theory II\hfill Spring~2024}\\\vspace{2pt}
\textbf{Problem Set \#7\hfill Due: 5/14/2024}\vspace{-8pt}
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\subsection*{Description}

Your solutions should be written up in latex and submitted as a pdf-file on \href{https://www.gradescope.com/courses/730173}{Gradescope} before midnight on the date due.
Collaboration is permitted/encouraged, but you must identify your collaborators or the name of your group on \href{https://psetpartners.mit.edu}{pset partners}, and any references you consulted. If there are none write ``\textbf{Sources consulted: none}'' at the top of your solution.
Pick any combination of problems that sum to 100 points, then complete the survey problem.

\subsection*{Problem 1. Commutator subgroups and adelic images of $E/\Q$ (50 points)}

Let $E/\Q$ be an elliptic curve without potential complex multiplication, let
\[
\rho_E\colon \Gal_\Q\to \GL_2(\Zhat)
\]
be its adelic Galois representation, and let $G_E:=\rho_E(\Gal_\Q)$ denote its image.

Serre's open image theorem implies that $G_E$ is an open subgroup of $\GL_2(\Zhat)$, and in particular, has finite index.  The goal of this problem is to prove another result of Serre, which states that the index $[\GL_2(\Zhat):G_E]$ is divisible by $2$, and we thus never have $G_E=\GL_2(\Zhat)$; the proof also gives an explicit non-obvious constraint on the open subgroups of $\GL_2(\Zhat)$ that can arise as Galois images of elliptic curves $E/\Q$ that allows us to rule out some otherwise plausible looking candidates.

Recall that every group $G$ has a \defi{commutator subgroup}
\[
[G:G]:= \{ ghg^{-1}h^{-1}:g,h\in G\},
\]
a normal subgroup of $G$ that determines the \defi{maximal abelian quotient} $G^\ab:= G/[G:G]$.
For any $G\le \GL_2(\Zhat)$ the commutator subgroup $[G,G]$ lies in the intersection $G\cap \SL_2(\Zhat)$.
Indeed, $G\cap \SL_2(\Zhat)$ is a normal subgroup contained in the kernel of the determinant map $\det\colon G\to \Zhat^\times$, and we have an abelian quotient $G/(G\cap \SL_2(\Zhat))\subseteq \Zhat^\times = \GL_2(\Zhat)/\SL_2(\Zhat)$.

One might na\"ively guess that $[G,G]=G\cap\SL_2(\Zhat)$, but this is not always true.
Indeed, it fails for $G=\GL_2(\Zhat)$, but it always holds for $G_E$ (as you will prove).  This depends crucially on the fact that $E$ is defined over $\Q$ rather than an extension of $\Q$; for elliptic curves $E$ over number fields $K$ that do not contain any roots of unity other than $\pm 1$ we will have $\rho_E(G_K)=\GL_2(\Zhat)$ for most $E/K$, except when $K=\Q$.

For an open $H\le \SL_2(\Z_\ell)$ the \defi{level} of $H$ is the least $\ell^n$ for which $H=\pi_{\ell^n}^{-1}(\pi_{\ell^n}(H))$, where $\pi_{\ell^n}\colon \SL_2(\Z_\ell)\twoheadrightarrow \SL_2(\Z/\ell^n\Z)$ is the reduction-modulo-$\ell^n$ map.

Let $\chi_\cyc\colon \Gal_\Q\to \Zhat^\times$ be the \defi{cyclotomic character} defined by $\sigma(\zeta_n)=\zeta_n^{\chi_\cyc(\sigma)\bmod n}$.

\begin{enumerate}[{\bf(a)}]
\item Show that for every $m\in \Z_{\ge 1}$ the reduction-modulo-$m$ map $\SL_2(\Z)\to \SL_2(\Z/m\Z)$ is surjective.
\item Show that $\#\SL_2(\Z)^\ab|12$ and $\SL_2(\Z/2\Z)^\ab$, $\SL_2(Z/3\Z)^\ab$, $\SL_2(Z/4\Z)^\ab$ are cyclic of order $2,3,4$, respectively.
\item Show that $\SL_2(\Z)^\ab\simeq \Z/12\Z$ and reduction modulo $12$ induces an isomorphism $\SL_2(\Z)^\ab \simeqarrow \SL_2(\Z/12\Z)^\ab$.
Then show that for any positive integer $m$ the group $\SL_2(\Z/m\Z)^\ab$ is cyclic of order $\gcd(m,12)$.
\item Show that $\SL_2(\Z_2)^\ab$ is cyclic of order $4$, $\SL_2(\Z_3)^\ab$ is cyclic of order 3, and $\SL_2(\Z_\ell)^\ab$ is trivial for every prime $\ell \ge 5$, and that the commutator subgroup of $\SL_2(\Z_2)$ has level $4$ in $\SL_2(\Z_2)$, while the commutator subgroup of $\SL_2(\Z_3)$ has level $3$ in $\SL_2(\Z_3)$.
\item Show that the commutator subgroup of $\GL_2(\Z_\ell)$ is $\SL_2(\Z_\ell)$ for all $\ell \ge 3$.
\item Show that the commutator subgroup of $\GL_2(\Z_2)$ has level~2 and index 2 in $\SL_2(\Z_2)$ and conclude that the commutator subgroup of $\GL_2(\Zhat)$ is an open subgroup of $\SL_2(\Zhat)$ of level $2$ and index $2$, and in particular, is not equal to $\GL_2(\Zhat)\cap \SL_2(\Zhat)$.
\item Show that if $G$ is an open subgroup of $\GL_2(\Zhat)$ or $\SL_2(\Zhat)$ then $[G,G]$ is an open subgroup of $\SL_2(\Zhat)$.
\item Show that $\det\circ\rho_E=\chi_\cyc^{-1}$, and conclude that $\det(G_E)=\Zhat^\times$.
\end{enumerate}

Recall the Kronecker-Weber theorem: the cyclotomic field $\Q^\cyc=\bigcup_{n\ge 1}\Q(\zeta_n)$ is the maximal abelian extension $\Q^\ab$ of $\Q$.

\begin{enumerate}[{\bf(a)}]
\setcounter{enumi}{8}
\item Show that $G_E\cap \SL_2(\Zhat)=\rho_E(\Gal(\Qbar/\Q^\ab)=[G_E,G_E]$.  Deduce that $G_E\ne \GL_2(\Zhat)$, and moreover, that $2$ divides the index $[\GL_2(\Zhat):G_E]$.
\end{enumerate}


\subsection*{Problem 2. Quadratic twists and refinements (50 points)}

Let $E\colon y^2=x^3+Ax+B$ be an elliptic curve over $\Q$ with $A,B\in \Z$, let $d$ denote a nonsquare squarefree integer, and let $E_d\colon dy^2=x^3+Ax+B$ denote the quadratic twist of $E$ by $\Q(\sqrt{d})/\Q$. Let $\chi_\cyc\colon \Gal_\Q\to\Zhat$ denote the cyclotomic character defined in Problem~1,
and let $\chi_d\colon \Gal_\Q\to \{\pm 1\}$ be the unique homomorphism that factors through the isomorphism $\Gal(\Q(\sqrt{d})/\Q))\simeqarrow \{\pm 1\}$.

We write $H\sim K$ when $H$ and $K$ are conjugate subgroups of $\GL_2(\Zhat)$ or $\GL_2(\Z/N\Z)$.
When $H$ is a subgroup of $\GL_2(\Zhat)$ or $\GL_2(\Z/N\Z)$ that contains $-1$, we call the index-2 subgroups of $H$ that do not contain $-1$ \defi{refinements} of $H$.

\begin{enumerate}[{\bf(a)}]
\item Show there is a unique homomorphism $\psi\colon \Zhat\to \{\pm 1\}$ for which $\chi_d=\psi\circ\chi_\cyc^{-1}$ and
let $H_d:= \{\psi(\det g)g:g\in G_E\}$.  Prove $H_d\sim\rho_{E_d}(\Gal_\Q)$ and $\pm \rho_E(\Gal_\Q)\sim \pm \rho_{E_d}(\Gal_\Q)$.
\item Let $H'$ be an index-2 subgroup of $\pm \rho_E(\Gal_\Q)$ that does not contain $-1$.  Show that there is a unique $d$ for which $\rho_{E_d}(\Gal_\Q)\sim H_d\sim H'$.  Conclude that every refinement of $\pm \rho_E(\Gal_\Q)$ can be realized as $\rho_{E'}(\Gal_\Q)$ for some quadratic twist $E'/\Q$ of $E$ that is unique up to $\Q$-isomorphism.
\item For each $N\le \Z_{\ge 1}$ show that all $\rho_{E,N}(\Gal_\Q)\sim\rho_{E_d,N}(\Gal_\Q)$  for all but finitely many~$d$, and that every refinement of $\pm \rho_{E,N}(\Gal_\Q)$ arises as $\rho_{E_d,N}(\Gal_\Q)$ for a unique $d$ that divides $\disc(\Q(E[N])$.
\end{enumerate}

Let $H$ be an open subgroup of $\GL_2(\Zhat)$ of level $N$ that contains $-1$. For each $M\in \Z_{\ge 1}$ let $\pi_M\colon\GL_2(\Zhat)\to \GL_2(\Z/M\Z)$ denote the reduction-modulo-$m$ map.

\begin{enumerate}[{\bf(a)}]
\setcounter{enumi}{3}
\item Let $N|M\in \Z_{\ge 1}$.  Show that $H$ has a refinement of level dividing~$M$ if and only if $\pi_M(H)$ has a refinement, and the level of every refinement of $H$ is a multiple of $N$.
\item Show by example that $H$ may admit a refinement of level $2N$ even when it has no refinements of level $N$.
\item Show that $H$ has a refinement whose level divides $2N$ if and only $-1$ is not a square in $\pi_{2N}(H)^\ab$, the maximal abelian quotient of $\pi_{2n}(H)$.
\item Show that if $H$ has any refinements then it has infinitely many.
\item Let $p\perp M\in \Z_{\ge 1}$ be an odd prime.  Show that if $H$ has a refinement of level $NMp^e$ for some $e>0$ then $H$ has a refinement of level $NM$ (hint: use Goursat's lemma).
Conclude that if $H$ has a refinement then it has one of level $N2^e$ for some $e\ge 0$.
\item Show that if $H$ has a refinement of level $N2^e$ for some $e\ge 1$ then $H$ has a refinement of level $2N$, and if $N$ is odd then $H$ has a refinement of level $N$.

Conclude that either $H$ has no refinements or it has infinitely many, including one of level $N$ or $2N$, depending on whether $-1\in \pi_{2N}(H)^\ab$ is a square or not.
\end{enumerate}

\subsection*{Problem 3. Complex conjugation and curves of genus zero (50 points)}

Recall that the categories of elliptic curves $E/\C$ (up to isomorphism) and complex tori $\C/L$ (up to homethety) are equivalent.
Here $L\subseteq \C$ is a free $\Z$-module of rank 2 and thus has a $\Z$-module basis $[\omega_1,\omega_2]$ for some $\R$-linearly independent $\omega_1,\omega_2\in \C^\times$.  After applying the homethety $\pm \omega_2/\omega_1$ we can assume $L=[1,\tau]$ with $\tau\in \H$,
in which case the corresponding elliptic curve $E_L$ is defined by
\[
E_L\colon y^2 = 4x^3-g_2(\tau)x-g_3(\tau),
\]
where $g_2(\tau)=60\sum'\frac{1}{(m+n\tau)^4}$ and $g_3(\tau)=140\sum'\frac{1}{(m+n\tau)^6}$, with the sums $\Sigma'$ taken over the nonzero lattice points $m+n\tau\in L$.  The map $\Phi\colon \C/L\to E_L(\C)$ that sends $L$ to the identity element of $E_L(\C)$ and $z+L$ to the point $(\wp(z),\wp'(z))$ is then an isomorphism of topological groups that is also an isomorphism of compact Riemann surfaces, where
\[
\wp(z):= \frac{1}{z^2} + \sum_{\omega\in L-\{0\}}\left(\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right).
\]
Every elliptic curve $E/\C$ is isomorphic an elliptic curve defined by an equation of the form above, and this includes the base change to $\C$ of any elliptic curve defined over~$\Q$.  Moreover, if $E\colon y^2 = x^3+Ax+B$ is an elliptic curve over $\Q$, there is a $\tau\in \H$ for which $g_2(\tau)=-4A$ and $g_3(\tau)=-4B$, and for $L=[t,\tau]$ the corresponding curve $E_L$ will have rational coefficients.

\begin{enumerate}[{\bf(a)}]
\item Show that if $g_2(L),g_3(L)\in \R$ (so $E_L$ is defined over $\R$ not just $\C$), then $L$ is stable under complex conjugation, in which case $\Phi(\R)\subseteq E_L(\R)$ but $\Phi(\C)\not\subseteq E_L(\R)$.
\end{enumerate}

Now let $E$ be an elliptic curve over $\Q$, let $\rho_E\colon \Gal_\Q\to \GL_2(\Zhat)$ be its adelic Galois representation, fix an embedding $\Qbar\subseteq \C$, and let $c\in \Gal_\Q$ denote the automorphism corresponding to the action of complex conjugation on $\Qbar\subseteq \C$.

\begin{enumerate}[{\bf(a)}]
\setcounter{enumi}{1}
\item Show that $E(\R)$ has a subgroup isomorphic to $\R/\Z$, and we can choose a compatible system of bases for $E[N]$ so that $\rho_E(c)$ is upper triangular with a $1$ in the upper left.  Then show that $\det(\rho_E(c))=-1$ and $\tr(\rho_E(c))=0$.  Conclude that for $N>2$ we cannot have $E[N]\subseteq E(\Q)$.
\item Show that for odd $N$ we can chose a basis for $E[N]$ so that $\rho_E(c)=\smat{1}{0}{0}{-1}$, but this is not necessarily true for $N=4$.
Then show that for every $N$ we can choose a basis for $E[N]$ so that $\rho_E(c)$ is either $\smat{1}{0}{0}{-1}$ or $\smat{1}{1}{0}{-1}$.
\item Let $H\le \GL_2(\Zhat)$ be an open subgroup of level $N$.  Show that if the reduction of $H$ to $\GL_2(\Z/N\Z)$ contains no element conjugate to  $\smat{1}{0}{0}{-1}$ or $\smat{1}{1}{0}{-1}$ then $X_H(\R)=\emptyset$, and in particular, $X_H(\Q)=\emptyset$.
\end{enumerate}

Let $X$ be a \defi{nice} (smooth, projective, geometrically integral) curve over $\Q$ of genus~0.  Then $X$ can be defined by a quadratic form  $ax^2+by^2+cz^2 = 0$ with $a,b,c\in \Z$ and $abc\ne 0$. Recall that if $X$ has a rational point $P$, then $X\simeq \P^1_\Q$ (the isomorphism identifies $Q\in X(\Q)$ with the slope of the line $\overline{PQ}$) and $X$ has infinitely many rational points.  The Hasse-Minkowski theorem implies that $X$ has a rational point if and only if it has a rational point over every completion of $\Q$.  It follows from Hilbert reciprocity that the number of places $p\le \infty$ of $\Q$ for which $X(\Q_p)=\emptyset$ is finite and even.

\begin{enumerate}[{\bf(a)}]
\setcounter{enumi}{5}
\item Show that if $p$ is an odd prime that does not divide $abc$ then $X(\Q_p)\ne \emptyset$.
\end{enumerate}

Let $H\le \GL_2(\Zhat)$ be an open subgroup of level $N$ with $\det(H)=\Z^\times$. Then $X_H$ is a nice curve over $\Q$ that has good reduction modulo every prime $p\nmid N$ (this means that the reduction of $X_H$ to $\F_p$ is also a nice curve).  If $X_H$ has genus zero this means that we can choose an integral model $ax^2+by^2+cz^2$ for $X_H$ with $p\nmid abc$.

\begin{enumerate}[{\bf(a)}]
\setcounter{enumi}{6}
\item Suppose $N$ is a prime power and $X_H$ has genus zero.  Show that $X_H(\Q)=\emptyset$ if and only if the image of $H$ to $\GL_2(\Z/N\Z)$ contains no conjugate of $\smat{1}{0}{0}{-1}$ or $\smat{1}{1}{0}{-1}$.
\item By applying the genus formula for $X_H$ with $H=\{\pm 1\}\le \GL_2(\Z/N\Z)$ (see the notes for Lecture 21), derive a formula for $g(X(N))$ as a function of $N$ and list the positive integers $N$ for which $g(X(N))=0$; your list should include $N\le 3$.
Conclude that for the $N$ in your list, for every $H$ of level $N$ we have $g(X_H)=0$.

\item Up to conjugation there are 6 subgroups of $\GL_2(\Z/3\Z)$ with $\det(H)=(\Z/3\Z)^\times$ that contain $-1$; their indexes are 1, 3, 4, 6, 6, 12.  Use \textbf{(g)} to determine which of the corresponding modular curves $X_H$ have rational points.

\item Up to conjugation there are 3 subgroups of $\GL_2(\Z/3\Z)$ with $\det(H)=(\Z/3\Z)^\times$ that do not contain $-1$; there indexes are 8, 8, 24.  Determine which of the corresponding modular curves $X_H$ have rational points.
\end{enumerate}

\subsection*{Problem 4. Survey}
Complete the following survey by rating each problem you attempted on a scale of 1 to~10 according to how interesting you found it (1 = ``mind-numbing," 10 = ``mind-blowing"), and how difficult you found it (1 = ``trivial," 10 = ``brutal").  Also estimate the amount of time you spent on each problem to the nearest half hour.

\begin{center}
\begin{tabular}{l|r|r|r|}
& Interest & Difficulty & Time Spent\\\hline
Problem 1 & & & \\\hline
Problem 2 & & & \\\hline
Problem 3 & & & \\\hline
\end{tabular}
\end{center}


\noindent
Please rate each of the following lectures that you attended, according to the quality of the material (1=``useless", 10=``fascinating"), the quality of the presentation (1=``epic fail", 10=``perfection"), the pace (1=``way too slow", 10=``way too fast", 5=``just right") and the novelty of the material to you (1=``old hat", 10=``all new").

\begin{center}
\begin{tabular}{l|l|r|r|r|r|r}
Date & Lecture Topic & Material & Presentation & Pace & Novelty\\\hline
5/6 & Modular curves $X_H$ & & & & \\\hline
5/8 & Counting points on $X_H$ & & & & \\\hline 
\end{tabular}
\end{center}

\noindent
Please feel free to record any additional comments you have on the problem sets and the lectures, in particular, ways in which they might be improved.
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