text:
\begin{document}

\title[A local-global principle for rational isogenies of prime degree]{A local-global principle for\\rational isogenies of prime degree}
\author{\sc Andrew V. Sutherland}
\address{Andrew V. Sutherland\\Department of Mathematics\\Massachusetts Institute of Technology\\77 Massachusetts Avenue\\Cambridge, MA  02139}
\email{drew@math.mit.edu}
\urladdr{\url{http://math.mit.edu/~drew}}
\subjclass[2010]{11G05}
\keywords{elliptic curve, isogeny, local-global principle}

\maketitle

\begin{resume}
Soit $K$ un corps de nombres. Nous \'etudions un principe local-global
pour les courbes elliptiques $E/K$ admettant ou non une isog\'enie
rationnelle de degr\'e premier $\ell$. Pour des corps $K$ convenables
(dont $K=\mathbb Q$), nous d\'emontrons ce principe pour tout $\ell\equiv
1\bmod 4$ et tout $\ell<7$ mais exhibons une courbe elliptique
d'invariant modulaire $2268945/128$ comme contre-exemple pour $\ell=7$.
Nous montrons alors qu'il s'agit du seul contre-exemple \`a isomorphisme
pr\`es lorsque $K=\mathbb Q$.
\end{resume}

\begin{abstr}
Let $K$ be a number field.  We consider a local-global principle for elliptic curves $E/K$ that admit
(or do not admit) a rational isogeny of prime degree $\ell$.
For suitable $K$ (including $K=\Q$), we prove that this principle holds for all $\ell\equiv 1\bmod 4$, and for $\ell < 7$,
but find a counterexample when $\ell=7$ for an elliptic curve with $j$-invariant $2268945/128$.
For $K=\Q$ we show that, up to isomorphism, this is the only counterexample.
\end{abstr}

\section*{Introduction}

Let $E$ be a non-singular elliptic curve defined over a number field $K$, and let~$\ell$ be a prime number.
We say that $E$ admits an $\ell$-isogeny over $K$ if there is an isogeny $\phi\colon E\to E'$ of degree $\ell$
defined over $K$.  If $\fp$ is a prime of~$K$ where $E$ has good reduction, we say that $E$ admits
an $\ell$-isogeny locally at~$\fp$ if the reduction of $E$ modulo $\fp$ admits an isogeny of degree $\ell$
defined over the residue field.

If $E$ admits an $\ell$-isogeny over $K$, then $E$ necessarily admits an $\ell$-isogeny locally everywhere,
that is, at every prime of good reduction.  We ask the converse:

\begin{center}
\emph{If $E$ admits an $\ell$-isogeny locally everywhere,\\ must $E$ admit an $\ell$-isogeny over $K$?}
\end{center}

We can immediately answer this question with a counterexample.
Consider the elliptic curve $E/\Q$ defined by the Weierstrass equation
\begin{equation}\label{E2450A}
y^2 + xy = x^3 - x^2 - 107x - 379.
\end{equation}
This curve admits a $7$-isogeny locally at every prime of good reduction (and over $\R$) but it does not admit a $7$-isogeny over~$\Q$.
We note that $E$ does admit a 7-isogeny over a quadratic extension of~$\Q$.
Our first theorem implies that this is necessarily the case, as is the fact that we used a prime $\ell\equiv 3 \bmod 4$.

\begin{thm1}
Assume $\sqrt{\inkron{-1}{\ell}\ell}\notin K$, and suppose $E/K$ admits an $\ell$-isogeny locally at a set of primes with density one.
Then~$E$ admits an $\ell$-isogeny over a quadratic extension of $K$, and if $\ell\equiv 1 \bmod 4$ or $\ell < 7$, then $E$ admits an $\ell$-isogeny over~$K$.
\end{thm1}

Whether an elliptic curve  admits an $\ell$-isogeny over $K$ (or not) depends only on its
$j$-invariant $j(E)$, which uniquely identifies its isomorphism class over any algebraic
closure of~$K$.
Let us call a pair $(\ell,j(E))$ exceptional if $E/K$ admits an
$\ell$-isogeny locally everywhere, but not over~$K$.
The pair $(7,\ 2268945/128)$ corresponds to the counterexample above.
\mbox{Theorem}~1 admits the possibility
of infinitely many exceptional pairs, and this occurs, for example,
when $K=\Q(i)$.  However, it does not happen when $K=\Q$.

\begin{thm2}
The pair $(7,\ 2268945/128)$ is the only exceptional pair for $\Q$.
\end{thm2}

The analogous local-global question for $\ell$-torsion was addressed by Katz in~\cite{Katz:TorsionPoints}.
In general, an elliptic curve $E/K$ with non-trivial $\ell$-torsion locally everywhere may have
trivial $\ell$-torsion over $K$.  However, Katz shows that such an $E$ must be
rationally isogenous to an elliptic curve which has non-trivial $\ell$-torsion over $K$.  He proves this
by reducing the problem to a purely group-theoretic statement, an approach advocated by Mazur \cite[p.~483]{Katz:TorsionPoints}.

We take a similar line in our treatment of Theorem 1, using a Galois representation to reduce the problem
to a question regarding the structure of certain subgroups of $\glmodl$, which we are then able to address
with purely elementary methods.  The proof of Theorem 2 requires more, and here we also use the theory of complex multiplication and,
crucially, a theorem of Parent~\cite{Parent:Triviality} that characterizes the rational points on the modular
curve $X_0^+(\ell^2)(\Q)$ for certain values of $\ell$.

\section{Preliminaries}

\subsection{Galois Representations}
We follow the notation in \cite{Mazur:GaloisRepresentations}.  Let us fix a number field~$K$
with algebraic closure $\overline{K}$.  If $S$ is a finite set of non-archimedean primes of~$K$,
let $\overline{K}_S$ denote the maximal algebraic extension of $K$ in $\overline{K}$ unramified outside of~$S$.
Given an elliptic curve $E/K$ and a prime number $\ell$, the absolute Galois group $G_K=\Gal(\overline{K}/K)$
acts on the group of $\ell$-torsion points $E[\ell]$.  Provided $S$ contains the primes that divide~$\ell$, and
the primes where $E$ has bad reduction, this action factors through $G_{K,S}=\Gal(\overline{K}_S/K)$, yielding a representation
\[
\rhobarl\colon G_{K,S}\rightarrow\Aut(E[\ell])\cong{\rm GL}_2(\Z/\ell\Z)\cong \glmodl.
\]

If $\fp$ is a non-archimedean prime of $K$ not in $S$ with residue field $k_\fp$, let~$\frob$ denote the
$\leftvert k_\fp\vert$-power Frobenius automorphism, which we may view as an element of $G_{K,S}$ with the understanding that it
is determined only up to conjugacy.  We identify the image of $\frob$ under $\rhobarl$ as a
conjugacy class $\frobbarl$ of $\glmodl$, and we have
\[
\det(\frobbarl)\equiv \leftvert k_\fp\vert \bmod\ell,\qquad \tr(\frobbarl)\equiv \leftvert k_\fp\vert+1-\leftvert E(k_\fp)\vert\bmod \ell.
\]
The Chebotarev density theorem implies that each conjugacy class $\frobbarl$ in~$G$ arises for a set of primes with density $|\frobbarl|/|G| > 0$.

\subsection{Subgroups of $\glmodl$}
We recall some of the classical subgroups of $\glmodl$.

A \emph{Cartan} subgroup is an absolutely semisimple maximal abelian subgroup, of which there are two types.
A \emph{split} Cartan subgroup is isomorphic to $\F_\ell^*\times\F_\ell^*$ and
conjugate to the group of diagonal matrices.  A \emph{non-split} Cartan subgroup is isomorphic to $\F_{\ell^2}^*$
and conjugate to a group $C_{ns}$ we may define as follows:
for $\ell=2$ let $C_{ns}$ be the unique subgroup of order 3, and for $\ell>2$ fix a quadratic non-residue $\delta\in\Fl^*$
and let
\[
C_{ns} = \left\{\left(\afrac{x}{y}\thinspace\afrac{\delta y}{x}\right):x,y\in\Fl, (x,y)\ne(0,0)\right\}.
\]
Every Cartan subgroup has index 2 in its normalizer and contains the group of scalar matrices.

The semisimple elements of $\glmodl$ are those of order prime to $\ell$.  The following proposition
characterizes the semisimple subgroups of $\glmodl$ in terms of their images in $\pglmodl$.

\begin{proposition}\label{prop:SSS}
Let $G$ be a subgroup of $\glmodl$ of order prime to $\ell$, and let $H$ be the image of $G$ in $\pglmodl$.
Then exactly one of the following holds:
\begin{enumerate}
\item[(a)]
$H$ is cyclic and $G$ lies in a Cartan subgroup;
\item[(b)]
$H$ is dihedral and $G$ lies in the normalizer of a Cartan subgroup but not in the Cartan subgroup itself;
\item[(c)]
$H$ is isomorphic to $A_4$, $S_4$, or $A_5$.
\end{enumerate}
Here $A_n$ and $S_n$ denote the alternating and symmetric groups on $n$ letters, respectively.
\end{proposition}
\begin{proof}
See \cite[Thm.~XI.2.3]{Lang:ModularForms} or \cite{Serre:GaloisRepresentations}.
\end{proof}

Let $\Omega$ denote the set of linear subspaces of $\Fl^2$.  The group $\glmodl$
acts on~$\Omega$, and the induced action of $\pglmodl$ is faithful, since only scalar matrices act trivially.
For an element or subgroup~$g$ of $\glmodl$ or $\pglmodl$,
we let $\Omega/g$ denote the set of $g$-orbits of $\Omega$, and define $\Omega^g$ to be
the set of elements fixed by~$g$.  We recall the orbit-counting lemma,
\[
|\Omega/G| = \frac{1}{|G|}\sum_{g\in G}|\Omega^g|,
\]
and understand that the size of each orbit in $\Omega/G$ must divide $|G|$.

We can use Proposition~\ref{prop:SSS} to characterize the action of each element of $\glmodl$ on $\Omega$.
This yields Proposition~\ref{prop:action}, which encapsulates the group-theoretic content of two results credited to Atkin \cite[\S 6]{Schoof:ECPointCounting2}.  For each $h$ in $\pglmodl$, let $\sigma(h)$
denote the sign of $h$ as a permutation of $\Omega$ (thus for $\ell>2$ we have $\sigma(h)=1$ if and only if $h\in\pslmodl$).

\begin{proposition}\label{prop:action}
Let $g\in\glmodl$ have image $h$ in $\pglmodl$ with order $r$, let $k=|\Omega^h|$, and let $s=|\Omega/h|$.
Then $k$ is $0, 1, 2$, or $\ell+1$, and the $s-k$ non-trivial $h$-orbits have size $r$.
When $\ell>2$ we also have $\sigma(h)=(-1)^s$.
\end{proposition}
\begin{proof}
The proposition clearly holds for $\ell=2$, so we assume $\ell > 2$.
When $r=1$, we have $k=s=\ell+1$ and $\sigma(h)=1=(-1)^s$.
When $r=\ell$, there are exactly two $h$-orbits, of sizes $1$ and~$\ell$, and we have $k=1$, $s=2$,
and $\sigma(h) = 1 = (-1)^s$.  The proposition holds in both cases.

Otherwise the cyclic group $H=\langle h \rangle$ has order $r$ prime to $\ell$ and we are in case (a) of Proposition~\ref{prop:SSS}.  Thus $g$ lies in a Cartan subgroup $C$, whose image in $\pglmodl$ has order $\ell-1$ or $\ell+1$, as $C$ is split or non-split (resp.).  We consider the two cases.

If $C$ is split then $g$ is diagonalizable, so $k=|\Omega^h|=|\Omega^g|=2$, and the same is true for every non-trivial element of $H$. The orbit-counting lemma yields
\[
s = \frac{1}{r}\sum_{h'\in H}\left|\Omega^{h'}\right| = \frac{1}{r}\bigl((r-1)2+\ell+1\bigr) = 2 + \frac{\ell-1}{r}.
\]
The sizes of the $(\ell-1)/r$ non-trivial orbits all divide $r$ and sum to $\ell-1$, hence they are all equal to~$r$.
If $r$ is odd then $s$ is even and $\sigma(h)=1=(-1)^s$, and if $r$ is even then $\sigma(h)=(-1)^{s-2}=(-1)^s$.
Thus the proposition holds when $C$ is split.

If $C$ is non-split then $g$ has no eigenvalues in $\Fl$, hence $k=|\Omega^h|=|\Omega^g|=0$,
and the same is true for every non-trivial element of $H$.
The orbit-counting lemma yields $s=(\ell+1)/r$, and every orbit
must have size~$r$.
If $r$ is odd, then $s$ is even and $\sigma(h)=1=(-1)^s$, otherwise $r$ is even and $\sigma(h)=(-1)^s$.
Thus the proposition also holds when $C$ is non-split.
\end{proof}

\section{Proof of Theorem 1}

We first prove a group-theoretic lemma from which Theorem~1 will follow.
In terms of the elliptic curve $E$ of Theorem~1, the group $G$ in Lemma~\ref{lemma:main} is the image of the Galois representation $\rhobarl$.
The hypothesis of the lemma is met precisely when $E$ admits an $\ell$-isogeny locally everywhere but not globally; the lemma then imposes specific constraints on $G$ and $\ell$.

\begin{lemma}\label{lemma:main}
Let $G$ be a subgroup of $\glmodl$ whose image $H$ in $\pglmodl$ does not lie in $\ker \sigma$.
Suppose $|\Omega^g|>0$ for all $g\in G$, but $|\Omega^G|=0$.
The following hold:
\begin{enumerate}
\item $H$ is dihedral of order $2n$, where $n>1$ is an odd divisor of $(\ell-1)/2$;\label{lemma:main1}
\item $G$ is properly contained in the normalizer of a split Cartan subgroup;\label{lemma:main2}
\item $\ell\equiv 3\bmod 4$;\label{lemma:main3}
\item $\Omega/G$ contains an orbit of size $2$.\label{lemma:main4}
\end{enumerate}
\end{lemma}
\begin{proof}
No subgroup of ${\rm GL}_2(\F_2)$ satisfies the hypothesis of the lemma, so we assume $\ell > 2$.
Let $G$ and $H$ be as in the lemma, and note that $\Omega^H=\Omega^G$, $\Omega/H=\Omega/G$,
and if $h\in H$ is the image of $g\in G$ then $\Omega^h=\Omega^g$.

We first show that $\ell$ does not divide $m=|H|$.  The orbit-counting lemma yields
\[
|\Omega/H| = \frac{1}{m}\sum_{h\in H}|\Omega^h| \ge \frac{1}{m}(\ell+m) > 1,
\]
since $|\Omega^h| > 0$ for all $h\in H$ and $|\Omega^h|=\ell+1$ when $h$ is the identity.
If $\ell\mid m$ then $H$ contains an element $h$ of order $\ell$ and $\Omega/h$
consists of two orbits, of sizes 1 and $\ell$.  These must also be the
orbits of $\Omega/H$, since $|\Omega/H|>1$.  But this contradicts our assumption that $|\Omega^H|=0$, thus $\ell\nmid m$.

Since $\ell\nmid m$ we must have $|\Omega^h|\ne 1$ for all $h\in H$, as may be seen from the
proof of Proposition~\ref{prop:action}, thus $|\Omega^h|=2$ for every non-trivial $h\in H$.
We note that $H$ cannot be a cyclic group~$\langle h \rangle$, for this would imply $\Omega^h=\Omega^H$, contrary to our hypothesis.

We now show that $H$ is not isomorphic to $A_4$, $A_5$, or $S_4$.  The kernel of $\sigma\colon H\to\{\pm 1\}$ must
be an index 2 subgroup of $H$.
Neither $A_4$ nor $A_5$ contain such a subgroup.
Proposition~\ref{prop:action} implies that for $h\in H$, the value $\sigma(h)=(-1)^s$ depends
only on the order $r$ of $h$, since either $r=1$ and $s=\ell+1$, or $r >1$ and $s=2+(\ell-1)/r$ (since $k=|\Omega^h|=2$).
But there is no non-trivial homomorphism from $S_4$ to $\{\pm 1\}$ with this property: the sequence 1,9,8,6 that
counts the elements of order 1,2,3,4 (resp.) in~$S_4$ has no subsequence whose sum is 12.  It follows that $H\not\cong S_4$.

Proposition~\ref{prop:SSS} then implies that $H$ is a dihedral group of order $2n$
for some $n > 1$.  If $n$ is even then $H$ contains $n+1$ elements of order 2,
but $\sigma$ cannot be constant on a subset of $H$ with size $n+1>|H|/2$, so $n$ is odd.
For an element $h\in H$ of order $n$,
Proposition~\ref{prop:action} implies that $s=2+(\ell-1)/n$ is an integer, thus $n$
divides $\ell-1$ and in fact divides $(\ell-1)/2$.
This completes the proof of (\ref{lemma:main1}).

We are in case (b) of Proposition~\ref{prop:SSS}, so $G$ lies in the normalizer $N(C)$ of a Cartan subgroup $C$,
which must be split because $n$ does not divide $2(\ell+1)$.
The group $G$ must be properly contained in $N(C)$, since $n$ properly divides $\ell-1$.  This proves (\ref{lemma:main2}).

If $\ell\equiv 1\bmod 4$, then for each of the $n$
elements of $H$ with order 2 we have $s=2+(\ell-1)/2$ with even parity in Proposition~\ref{prop:action},
and $s$ also has even parity when $h$ is the identity.  But this implies that $\sigma$ is constant on a subset of
$H$ with size $n+1 >|H|/2$, which is again a contradiction.  Therefore $\ell\equiv 3\bmod 4$, proving (\ref{lemma:main3}).

Let $h\in H$ have order $n$.  Proposition~\ref{prop:action} implies that
$\Omega/h$ consists of two trivial orbits and $(\ell-1)/n$ orbits of size $n$.
It follows that if $t$ is the size of an orbit of $\Omega/H$, then $t$ is congruent to
0, 1, or 2 modulo $n$, and we also know that $t$ divides $|H|=2n$ and $t\ne 1$ (since $|\Omega^H|=0$).  Therefore either $t=2$
or $t$ is a multiple of $n$.  But $n$ does not divide $\ell+1$, so the size of at least one orbit in $\Omega/H$
is not divisible by $n$.  This orbit must have size $t=2$, which proves (\ref{lemma:main4}).
\end{proof}

\begin{thm1}
Assume $\sqrt{\inkron{-1}{\ell}\ell}\notin K$, and suppose $E/K$ admits an $\ell$-isogeny locally at a set of primes with density one.
Then~$E$ admits an $\ell$-isogeny over a quadratic extension of $K$, and if $\ell\equiv 1 \bmod 4$ or $\ell < 7$, then $E$ admits an $\ell$-isogeny over~$K$.
\end{thm1}
\begin{proof}
Let the finite set $S$ consist of the primes where $E$ has bad reduction and the primes that divide $\ell$.
Let $G$ be the image of $\rho_{E,\ell}\colon G_{K,S}\to\glmodl$, and let~$g\in G$.  By the Chebotarev density theorem,
the conjugacy class of $g$ is equal to $\frobbarl$ for a set of primes $\fp$ with positive density, thus
we may choose $\fp$ so that $g=\frobbarl$ and $E$ admits an $\ell$-isogeny locally at $\fp$.  The Frobenius
endomorphism $\frob$ fixes a linear subspace of $E[\ell]$, hence $\frobbarl$ fixes an element of $\Omega$.
Thus $|\Omega^g|>0$ for all $g\in G$.

If $|\Omega^G|>0$, then $G_{K,S}$ fixes a linear subspace of $E[\ell]$ which is the kernel of an $\ell$-isogeny defined over $K$ (and any quadratic extension).  The theorem holds in this case, so we assume $|\Omega^G|=0$.  No subgroup $G\subset {\rm GL}_2(\F_2)$ has $|\Omega^G|=0$ and $|\Omega^g|>0$ for all $g\in G$, so $\ell\ne 2$.

The hypothesis on $K$ implies that some element of $G$ has a non-square determinant, otherwise $G_{K,S}$ fixes the quadratic Gauss sum $\sum_{n=0}^{\ell-1}\zeta_\ell^{n^2}$, which is equal to $\pm\sqrt{\inkron{-1}{\ell}\ell}$; see, e.g., \cite{Ireland:ClassicalModern}.
It follows that the image of $G$ in $\pglmodl$ does not lie in the kernel of $\sigma$, and we may apply Lemma~\ref{lemma:main}.
Thus $\ell\equiv 3\bmod 4$ and $\ell \ne 3$ (by part (\ref{lemma:main1}) of the lemma), and $\Omega/G$ contains an orbit of size 2.  Let $x\in\Omega$ be an element of this orbit.  The stabilizer of~$x$ in $G_{K,S}$ is a subgroup of index 2, corresponding to a quadratic extension of $K$ over which $E$ admits an isogeny of degree $\ell$.
\end{proof}

We now show that subgroups of the form permitted by Lemma~\ref{lemma:main} do in fact exist.

\begin{proposition}\label{prop:construct}
Let $\ell\equiv 3 \bmod 4$ be a prime number greater then $3$.
There exists a subgroup $G$ of $\glmodl$ with the following properties:
\begin{enumerate}
\item
The determinant map from $G$ to $\Fl^*$ is surjective;\label{prop:construct1}
\item
$|\Omega^g|\ge 2$ for all $g\in G$;\label{prop:construct2}
\item
$|\Omega^G|=0$.\label{prop:construct3}
\end{enumerate}
\end{proposition}
\begin{proof}
Let $\alpha$ be a generator for $\Fl^*$ and for integers $i$ and $j$ define
\[
A(i,j) =  \left( \afrac{\alpha^i}{0}\thinspace\afrac{0}{\alpha^j} \right);\qquad B(i,j) = \left( \afrac{0}{\alpha^j}\thinspace\afrac{\alpha^i}{0} \right). \]
Let $G$ be the set of matrices $A(i,j)$ and $B(i,j)$ for which $i$ and $j$ have the same parity.
It is easily checked that $G$ is a group.  It contains the scalar matrices $A(i,i)$, and the matrix $B(0,0)$ whose determinant $-1$ is not a quadratic residue in $\Fl^*$, since $\ell\equiv 3\bmod 4$. Therefore (\ref{prop:construct1}) holds.  Clearly $A(i,j)$ is diagonalizable, and so is $B(i,j)$, with distinct eigenvalues $\alpha^{(i+j)/2}$ and $-\alpha^{(i+j)/2}$.  This implies (\ref{prop:construct2}).
The matrices $A(0,2)$ and $B(0,2)$ have no common eigenspace, which proves (\ref{prop:construct3}).
\end{proof}

The group in Proposition \ref{prop:construct} has a dihedral image in $\pglmodl$ of order~$2n$, where $n=(\ell-1)/2$.
A similar construction works for any odd $n\ge 3$ dividing $(\ell-1)/2$.

\section{A counterexample}

The existence of the subgroups prescribed by Lemma~\ref{lemma:main} and constructed in Proposition \ref{prop:construct} does not imply that they actually arise as the image of~$\rho_{E,\ell}$ for some elliptic curve $E/K$.  Indeed, if~$E$ does not have complex multiplication, then $\rho_{E,\ell}$ is known to be surjective for all sufficiently large~$\ell$, as shown by Serre in \cite{Serre:GaloisRepresentations}.  Surjectivity certainly precludes an exceptional image of the type permitted by Lemma~\ref{lemma:main}, and, as we will show in the proof of Theorem 2, so does complex multiplication.

But there is an elliptic curve $E/\Q$ and a prime $\ell$ for which the image of $\rho_{E,\ell}$ satisfies the hypothesis of Lemma~\ref{lemma:main}: the curve defined by equation~(\ref{E2450A}) of the introduction, with $j$-invariant $j(E)=2268945/128$, and $\ell=7$.

Let $\Phi_N(X,Y)$ denote the classical modular polynomial \cite[\S 5]{Lang:EllipticFunctions}.
For a field $F$ of characteristic not dividing $N$, an elliptic curve $E/F$ admits a cyclic isogeny of degree~$N$ defined over~$F$ if and only if $\Phi_N(X,j(E))$ has a linear factor in $F[X]$, as proven in~\cite{Igusa:ModularPolynomial}.  We note that an isogeny of prime degree is necessarily cyclic.
\vspace{2pt}

The irreducible factorization of $\phi(X) = \Phi_7(X,2268945/128)$ over $\Q[X]$ is given below:
\small
\begin{align*}
\phi(X) =\  &(X^2 -  1306315496294666865/2^{49}\thinspace X + 1567296563714555573025/2^{50})\\
                  &(X^3 - 405372852936775146868447415805\thinspace X^2\\
                   &\quad\medspace\medspace\thinspace\thinspace+ 55385584722536349330202265781434325/4\thinspace X - 17996436663345^3/8)\\
                  &(X^3 - 4209728442885/2\thinspace X^2 + 3961627130765133274725/2\thinspace X - 16205314545^3/8).
\end{align*}
\normalsize
The absence of a linear factor shows that $E$ does not admit a 7-isogeny over~$\Q$, but it does admit a 7-isogeny (two in fact) over a quadratic extension of $\Q$, as required by Theorem~1.  The discriminants of the three factors of $\phi(X)$ are each of the form $-7a^2/4^b$ for some positive integers $a$ and $b$.  It follows that the reduction of $\phi(X) \bmod p$ has two linear factors in $\Fp[X]$ for every odd prime $p$: either $-7$ is a quadratic residue mod $p$ and the quadratic factor splits in $\Fp[X]$, or it is not and both cubic factors split into a linear and a quadratic factor in $\Fp[X]$.  Thus $E$ admits a 7-isogeny locally at every prime of good reduction (all $p\nmid 70$).

It is easily verified that the cubic factors of $\phi(X)$ have the same splitting field, in which the quadratic factor also splits.  Its Galois group is isomorphic to $S_3$, which we may view as a subgroup of ${\rm PGL}_2(\F_7)$ via its action on the roots of $\phi(X)$.  Thus we have a dihedral subgroup of ${\rm PGL}_2(\F_7)$ with order $2n$, where $n=3$ divides $(\ell-1)/2$, as required by Lemma~\ref{lemma:main}.  Up to conjugacy, the image of $\rho_{E,7}$ in ${\rm GL}_2(\F_7)$ is precisely the group $G$ constructed in Proposition \ref{prop:construct}.

To determine whether there are any other exceptional pairs of the form $(7,j(E))$, we consider the moduli space of elliptic curves with the required level 7 structure.  As explained by Elkies in \cite[\S 4.2]{Elkies:EightfoldWay}, the corresponding modular curve may be constructed as a quotient of $X(7)$ by a suitable subgroup of ${\rm PSL}_2(\F_7)$.  We are interested in
elliptic curves $E$ for which the image $H$ of $\rho_{E,7}$ in ${\rm PGL}_2(\F_7)$ is a dihedral group of order 6 not contained in the kernel of $\sigma$.  Up to conjugacy there is precisely one such $H$, and its intersection with ${\rm PSL}_2(\F_7)$ is a cyclic group of order 3.

As shown in \cite{Elkies:EightfoldWay}, the quotient of $X(7)$ by such a group is isomorphic (over~$\overline{\Q}$) to $X_0(49)$, an elliptic curve.
We are interested in the twist of $X_0(49)$ by $\Gal(\Q(\sqrt{-7})/\Q)$, relative to the Fricke involution~$w_{49}$, since if $(7,j(E))$ is an exceptional pair for $\Q$, then $\Q(\sqrt{-7})$ is the unique quadratic extension of $\Q$ over which $E$ admits a 7-isogeny (by Theorem~1 there is such an extension, and from the proof of Theorem~1 it follows that if $\sqrt{-7}\not\in K$ then $(7,j(E))$ will remain exceptional for $K$).
If we make the cusp of $X_0(49)$ at infinity the origin, the cusp at zero is a rational point of order~2, and there are two non-cuspidal irrational 2-torsion points defined over $\Q(\sqrt{-7})$.
Our twist makes the two rational cusps irrational, and the irrational 2-torsion points are made rational and swapped by the $w_{49}$ involution.  An explicit computation by Elkies \cite{Elkies:X49twist} finds that this twist has the equation
\begin{equation}\label{X49twist}
-7y^2 = x^4 + 2x^3 - 9x^2 - 10x - 3,
\end{equation}
which is a genus 1 curve with rational points $(-1/2,\pm 1/4)$.  If $(x,y)$ is a rational solution to~(\ref{X49twist}),
then the rational map
\small
\[
f(x) = -(x-3)^3(x-2)(x^2+x-5)^3(x^2+x+2)^3(x^4-3x^3+2x^2+3x+1)^3 / (x^3-2x^2-x+1)^7
\]
\normalsize
yields the $j$-invariant of an elliptic curve that admits a 7-isogeny locally everywhere but not over~$\Q$.
Applying $f$ to either of the points $(-1/2,\pm 1/4)$ yields the $j$-invariant $2268945/128$ that we have already exhibited.

Taking either of $(-1/2, \pm1/4)$ as the origin, the curve in (\ref{X49twist}) is isomorphic over~$\Q$ to the elliptic curve with Weierstrass equation
\begin{equation}\label{49a3}
y^2 + xy = x^3 - x^2 - 107x + 552.
\end{equation}
This is curve 49a3 in Cremona's tables \cite{Cremona:Database}.  It has just two $\Q$-rational points, and these correspond
to the two solutions of (\ref{X49twist}) that we already know.  Thus $(7,\ 2268945/128)$ is the only exceptional pair for $\Q$ with $\ell=7$.

However, over a finite extension $K$ of $\Q$ the curve 49a3 may have infinitely many rational points, corresponding to infinitely many exceptional pairs $(7,j(E))$ for $K$.
Over $K=\Q(i)$, for example, the projective point $(-14:7+29i:1)$ on the curve 49a3 has infinite order, yielding infinitely many solutions to $(\ref{X49twist})$.
The first of these has $x$-coordinate $(7i-29)/58$, and in general, if $(u,v)$ is a solution to $(\ref{49a3})$ then there is a solution to $(\ref{X49twist})$ with $x$-coordinate $(3u-v+42)/(u+2v)$.

\section{Proof of Theorem 2}

Recall that a pair $(\ell,j(E))$ is said to be exceptional for $K$ when $E/K$ admits an $\ell$-isogeny locally everywhere but not over $K$.

\begin{thm2}
The pair $(7,\ 2268945/128)$ is the only exceptional pair for $\Q$.
\end{thm2}
\begin{proof}
By Theorem 1 we need only consider exceptional pairs with $\ell \ge 7$ and $\ell\equiv 3\bmod 4$.
The case $\ell=7$ is treated above, so we assume $\ell > 7$.

We first show that if $(\ell,j(E))$ is an exceptional pair then $E$ cannot have complex multiplication (CM). Suppose the contrary. Then $E$ has CM by an imaginary quadratic order $\cO$, since we are in characteristic zero, and it must have class number $h(\cO)= 1$, since $E$ is defined over~$\Q$.  By Theorem~1 there is an $\ell$-isogenous elliptic curve $E'$ that is defined over a quadratic extension of $\Q$ but not over $\Q$ (since we are in an exceptional case).  The curve~$E'$ must have CM by an imaginary quadratic order $\cO'$ with class number $h(\cO')= 2$.  Since $E$ and $E'$ are $\ell$-isogenous and $h(\cO') > h(\cO)$, the order $\cO'$ must be properly contained in $\cO$ with index $\ell$.
Via \cite[Thm.~7.24]{Cox:ComplexMultiplication}, we may compute the ratio $h(\cO')/h(\cO)$ as
\[
\frac{h(\cO')}{h(\cO)} = \frac{1}{[\cO^*:\cO'^*]}\left(\ell - \kron{\disc(\cO)}{\ell}\right) \ge \frac{1}{3}\bigl(\ell-1\bigr).
\]
But we already know that $h(\cO')/h(\cO) = 2/1 = 2$, which yields a contradiction for $\ell > 7$.  Therefore $E$ does not have CM.

By Lemma 1, it suffices to consider pairs $(j(E),\ell)$ for which the image of $\rho_{E,\ell}$ in ${\rm GL}_2(\mathbb{F}_\ell)$ is contained in the normalizer of a split Cartan group, as explained in the proof of Theorem~1.  For such a pair, the $j$-invariant $j(E)$ corresponds to a rational non-cuspidal point on the modular curve $X_{\rm split}(\ell)(\Q)$, which is isomorphic over $\Q$ to $X_0^+(\ell^2)$, the quotient of $X_0(\ell^2)$ by the Fricke involution $w_{\ell^2}$.  By Theorem 1.1 of \cite{Parent:Triviality}, for all $\ell > 7$ congruent to $3\bmod 4$, the only rational non-cuspidal points on $X_0^+(\ell^2)(\Q)$ correspond to elliptic curves with CM.\footnote{This is also implied by the recent (and stronger) result in \cite{BiluParentRebolledo:Xsplit}, which addresses all $\ell\ne 13$.}  But we have shown that no curve with CM can arise in an exceptional pair, and the theorem follows.
\end{proof}

The argument used to rule out CM in the proof above works over any number field, and when $K\ne \Q$ we have $[\cO^*:\cO'^*]=1$, which covers the case $\ell=7$ as well.
\smallskip

\section{Acknowledgments}
I thank Nicholas Katz, Barry Mazur, and Ken Ribet for several helpful discussions, and Pierre Parent for directing my attention to his result in~\cite{Parent:Triviality}.  I am especially grateful to Noam Elkies for his explicit computations in the case $\ell=7$, and to John Cremona for identifying the constraint on~$K$ required by Theorem~1.

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}


TITLE: True or False: If $A^2=0$ for a 10x10 matrix $A$, then rank($A$) $\le 5$
QUESTION [3 upvotes]: True or False:
  If $A^2=0$ for a $10$ by $10$ matrix $A$, then the inequality rank($A$) $\le 5$ must hold.

I am guessing that this is false. Apparently proving or disproving this uses knowledge of basis, change of coordinates, or dimension (the chapter of the problem is about these). I tried proof by contradiction, which didn't work so far.

REPLY [0 votes]: For a matrix with $A^2=0$ assuming it be different from null matrix because for the null matrix the case the result is trivial.
Consider $A^2=0$ , $m_A(x)=x^2$
There exists a Jordan block corresponding to the eigenvale $0$ of size $2×2$. And every jordan block corresponding to the eigenvalue $0$ of size $n×n$ has rank $n-1$. So in order to maximize your rank of the matrix you can have jordan blocks of size $2×2$ and not more than this size due to the restriction by the minimal polynomial. So you can have $5$ blocks ,each increasing the rank of the matrix by $1$.so maximum rank can be $5$
In general a matrix $A$ of size $n×n$ such that $A^2=0$
has rank atmost $\lfloor \frac{n}{2}\rfloor$


TITLE: Hom(V,W) is a vector space
QUESTION [0 upvotes]: WTS: Hom(V,W) which is the set of all linear maps is a vector space.
The question I have is that, isn't Hom(V,W) a subset of W? Because Hom(V,W) is just the set of all linear maps from V to W, so why can't I just show that it is a subspace?

REPLY [1 votes]: Hom(V,W) is a set of homomorphisms(vector space homomorphism)/linear maps from V to W, which forms vector space! whilst what you're thinking is just image of V with particular element f of Hom(V,W)(i.e. f(V) is subspace of W). These two are different things!


TITLE: Help with a problem with matrices
QUESTION [0 upvotes]: Hi everyone I've been messing with a problem with matrices and I can't get it to work. I think I am not seeing it the right way because else I get an 8 variable quadratic system of equations.
If $A$ and $B$ are $2\times 2$ matrices and $A^2 = B^2 = I$ and $$AB =
\begin{pmatrix}
0 & -1 \\
1 & 2 \end{pmatrix}$$
and $$BA =
\begin{pmatrix}
2 & 1 \\
-1 & 0 \end{pmatrix}$$
calculate
$(A+B)^2$.
Any help will be very appreciated

REPLY [1 votes]: Please, type in TeX. Consider $(A+B)^2=(A+B)(A+B)=AA+AB+BA+BB$ and do the math.


TITLE: How can I determine the force of a blow that launches an object?
QUESTION [0 upvotes]: If I have a blow of force N that strikes an object and launches that object of weight W at angle A into the air moving said object a distance D from where it was struck.
What other considerations (efficiency of energy transfer?) and formula can I use to determine what N is if I know A, D and W?

REPLY [1 votes]: From the information that you have got you can find the impulse $\displaystyle \int F (t) \, dt$ over the period for which the force was acting.
What you need to find is the time interval over which the force was acting and how the force was changing during that time interval which is usually quite a difficult thing to do without specialised apparatus.
There are devices which measure the force as a function of time directly and here is a description of an experiment, Measuring Impulse and Momentum Change in 1 Dimension, which uses such a device.
In terms of your measured projectile motion you would have to consider the effect of air resistance.


TITLE: How is DA equal to 1/x.
QUESTION [0 upvotes]: Image of shape for question context
The question asked about this shape was...
"Let BP = x, and show that AB = AP = DP = 1, and DA = 1/x"
From what I extract, triangle ADB is isosceles where DA = DB, where DB = DP + BP, and from the question BP = x, and DP = 1 and therefore DA = 1 + x, not 1/x.
Is this question wrong.

REPLY [0 votes]: Because the parts are in a golden ratio, $\frac{1}{x} = 1+x$ in this case. That is, provided the side length of the (regular) pentagon is $1$.