Lecture 22: 13 torsion
We prove the theorem of Mazur and Tate: no elliptic curve over the rational numbers has a point of order 13.
Setup
Statement of the main theorem
Theorem (Main theorem). No elliptic curve over $\bQ$ has a rational point of order 13.
To prove this theorem, we'll show that $X_1(13)$ has no rational points other than the cusps. The main work lies in showing that $J_(13)$ has MordellWeil rank 0.
Notation
We use the following notation in this lecture:

$X=X_1(13)$ and $J=J_1(13)$. Both are considered over $\bZ[1/13]$.

$G=\Gal(\ol{\bQ}/\bQ)$.

$K=\bQ(\sqrt[13]{1})$.

$K^+$ is the maximal real subfield of $K$.
References
We follow the article of Mazur and Tate (MR0347826) in our proof of the main theorem. We also need some results of Ogg, established in MR0337974.
Preliminaries on X_1(13)
Definition
We begin by giving the definition of $X=X_1(13)$. We work throughout over $\bZ[1/13]$. Recall that $Y_1(13)$ parametrizes pairs $(E, P)$ where $E$ is an elliptic curve and $P$ is a point of order 13 on $E$. The space $X$ parametrizes pairs $(E, P)$ where $E$ is a generalized elliptic curves and $P$ is a point on $E$ of order 13 such that the group generated by $E$ meets all irreducible components of $E$ over the algebraic closure.
There is a natural map $X \to X_0(13)$ taking $(E, P)$ to $(E, G)$, where $G$ is the group generated by $P$. A simple calculation with the genus formula shows that $X_0(13)$ has genus 0 and $X_1(13)$ has genus 2.
Cusps
There are two cusps on $X_0(13)$: for one, $E$ is a 1gon and $G$ is $\mu_{13}$, and for the other, $E$ is a 13gon and $G=\bZ/13 \bZ$. Both of these are defined over $\bQ$. There are six points of $X$ lying over each of these two cusps: there are 12 choices of generator for the group $G$, but $(E,P)$ and $(E,P)$ are isomorphic. When $E$ is the 1gon, picking a generator involves picking a root of unity, and so these points are not rational: they are defined over $K^+$. When $E$ is a 13gon, the generators of $G$ are rational, and so the corresponding points are rational as well. We thus see that $X$ has 12 cusps, 6 of which are defined over $\bQ$.
The twisted dihedral action
The group $(\bZ/13\bZ)^{\times}$ acts on $X$ by scaling the point $P$, that is, $a \in (\bZ/13\bZ)^{\times}$ takes $(E,P)$ to $(E,aP)$. The element $1 \in (\bZ/13\bZ)^{\times}$ acts trivially on $X$, since $(E,P)$ and $(E,P)$ are isomorphic. Let $\Gamma$ be the quotient of $(\bZ/13\bZ)^{\times}$ by $\{\pm 1\}$, a cyclic group of order 6. Then $\Gamma$ acts faithfully on $X$. For $a \in \Gamma$, we let $\gamma_a$ be the corresponding automorphism of $X$.
There are also AtkinLehner type automorphisms of $X$. Let $\zeta$ be a primitive 13th root of unity, and let $(E,P)$ be a point of $Y_1(13)$. There is then a 13torsion point $Q$ on $E$ which pairs with $P$ under the Weil pairing to $P$. This point is unique up to translation by $P$, i.e., if $Q_1$ is a second such point then $QQ_1$ is a multiple of $P$. Let $E'$ be the quotient of $E$ by the group generated by $P$, and let $P'$ be the image of $Q$ in $E'$, which is welldefined. Then $(E',P')$ is a new point on $Y_1(13)$. We let $\tau_{\zeta}$ be the map $Y_1(13) \to Y_1(13)$ taking $(E,P)$ to $(E',P')$. This can be extended to a map on $X$.
One readily verifies that $\tau_{\zeta}$ and $\tau_{\zeta^{1}}$ induce the same map of $X$. We let $\Gamma'$ be the set of primitive 13th roots of unity modulo $\zeta=\zeta^{1}$, a set of cardinality 6. Thus $\tau_{\zeta}$ is welldefined for $\zeta \in \Gamma'$.
One has the following relations between the $\gamma$'s and $\tau$'s:
$$ \gamma_m \tau_{\zeta} = \tau_{\zeta^m}, \qquad \tau_{\zeta} \gamma_m \tau_{\zeta}^{1} = \gamma_m^{1}, \qquad \tau_{\zeta}^2=1. $$It follows that $\Delta = \Gamma \cup \Gamma'$ is a group of automorphisms of $X_1(13)$, and isomorphic to the dihedral group of order 12.
There is a natural action of $G$ on $\Delta$ which fixes $\Gamma$ and acts on $\Gamma'$ in the obvious manner. This action is compatible with the action on $X$. In other words, via the Galois action we can regard $\Delta$ as an étale group scheme over $\bZ[1/13]$, and the action of this group scheme on $X$ is defined over $\bZ[1/13]$.
The group $\Delta$ acts simply transitively on the cusps, while the subgroup $\Gamma$ acts simply transitively on the rational cusps.
Ogg's results
The subgroup generated by the cusps
Proposition. Let $P_i$ for $1 \le i \le 6$ be the 6 rational cusps on $X$. Then for $i \ne j$, the divisor $[P_i][P_j]$ defines a point of order 19 in $J(\bQ)$. Furthermore, these points all generate the same subgroup of $J(\bQ)$.
Proof. (Idea.) The idea of the proof is similar to the computation of the order of $[0][\infty]$ in $J_0(N)$ given in Lecture 20, but a bit more complicated. For $1 \le a \le 6$, let
$$ E_{2,a}(z) = \sum_{n,m} \frac{1}{(mz+n)^2}, $$where the sum is over integers $n$ and $m$ such that $n=0 \pmod{13}$ and $m=a \pmod{13}$. Then $\varphi_{i,j}=E_{2,i}E_{2,j}$ is a weight 2 modular form for $\Gamma_1(13)$. As such, one can show that it must have 14 zeros in the fundamental domain for $\Gamma_1(13)$. On the other hand, one can compute the order of vanishing of $\varphi_{i,j}$ at the cusps (by computing the first few terms of $q$series), and one finds that it has 14 zeros at the cusps. It follows that $\varphi_{i,j}$ only vanishes at the cusps, and so the full divisor of $\varphi_{i,j}$ is known; call it $D_{i,j}$. Since $\varphi_{i,j}/\varphi_{k,\ell}$ is a meromorphic function on $X$, its divisor represents the point 0 in $J$. It follows that $D_{i,j}=D_{k,\ell}$ as points of the $J$. This gives enough relations between the $[P_i][P_j]$ to establish the proposition. ◾
The torsion subgroup of J(Q)
Proposition. The torsion subgroup of $J(\bQ)$ is cyclic of order 19 and generated by the difference of any two cusps.
Proof. Observe the following:

The only points on $X$ defined over $\bF_4$ are the 6 rational cusps. Indeed, an elliptic curve over $\bF_2$ or $\bF_4$ cannot have a point of order 13 (by the Hasse bound), and the other 6 cusps are not rational over $\bF_2$ or $\bF_4$.

Similarly, the only points on $X$ defined over $\bF_3$ are the 6 rational cusps.

There is a unique elliptic curve $E$ over $\bF_9$ with a point of order 13, and $\Aut(E)=\bZ/6\bZ$. This yields 2 points on $Y_1(\bF_9)$. The only other points on $X$ defined over $\bF_9$ are the 6 rational cusps.
We thus see that $\# X(\bF_2)=\# X(\bF_4)=6$, and so $\# J(\bF_2)=19$ by the following lemma. We also see that $\# X(\bF_3)=6$ and $\# X(\bF_9)=8$, and so $\# J(\bF_3)=19$ by the following lemma. Since the kernel of $J(\bQ)_{\tors} \to J(\bF_2)$ consists only of 2torsion and the kernel of $J(\bQ)_{\tors} \to J(\bF_3)$ consists only of 3torsion, the result follows. (In fact, $J(\bQ) \to J(\bF_3)$ is injective using the stronger results coming from Raynaud's theorem, so the analysis at 2 is unnecessary.) ◾
Lemma. Let $X/\bF_q$ be a curve of genus 2 and let $J$ be its Jacobian. Then
$$ \# J(\bF_q)=q+ \tfrac{1}{2} \# X(\bF_{q^2}) + \tfrac{1}{2} (\# X(\bF_q))^2. $$Proof. Let $V=\rH^1_{\et}(X_{\ol{\bF}_q}, \bQ_{\ell})$, a four dimensional $\bQ_{\ell}$ vector space, and let $F$ be Frobenius. Then
$$ \# X(\bF_q) = 1 + q  \tr(F \mid V), $$and a similar formula holds for $\# X(\bF_{q^2})$ using $F^2$ in place of $F$. Also,
$$ \# J(\bF_q) = \sum_{i=0}^4 (1)^i \tr(F \mid \bigwedge^i{V}). $$Finally, note that $q{}^tF^{1}$ is conjugate to $F$, by duality. The result is now a linear algebra exercise. ◾
The image of X in J
Proposition. The image of $X(\bC)$ in $J(\bC)$ (via $P \mapsto [P][P_6]$) intersects $J(\bQ)_{\tors}$ only at the 6 points $[P_i][P_6]$.
Proof. (Idea) If $[P][P_6]$ belonged to $J(\bQ)_{\tors}$ for some $P \in X(\bC)$ then we would have an equality $[P][P_6]=n([P_1][P_6])$ in $J$, and so $[P]n[P_1]+(n1)[P_0]$ would be the divisor of a function on $X$. Ogg shows that this cannot happen unless $n([P_1][P_6])=[P_i][P_6]$ for some $i$. ◾
Corollary. To prove the main theorem, it suffices to show that $J(\bQ)$ has rank 0.
Ogg notes that this is predicted by the BSD conjecture.
Structure of J
Proposition. $J$ is a simple abelian variety over $\bQ$.
Proof. Suppose that
$$ 0 \to J_1 \to J \to J_2 \to 0 $$is an exact sequence of abelian varieties, with $J_1$ and $J_2$ elliptic curves. Then either $J_1$ or $J_2$ would have a point of order 19, which we have already shown to be impossible. (In this case, it is easy to see, as $J_1$ and $J_2$ have good reduction away from 13, and so reducing modulo 2 would give an elliptic curve over $\bF_2$ with a point of order 19, which is impossible.) ◾
Observation: The element $\gamma_2$ generates $\Gamma$.
Proposition. The action of $\gamma_2$ on $J$ satisfies the polynomial $1x+x^2=0$.
Proof. This follows from the following observations:

$\gamma_2$ satisfies the polynomial $x^61$ but not any polynomial of the form $x^d1$ with $d$ a proper divisor of 6. This is because $\gamma_2$ has exact order 6 on $X$, and hence $J$.

If $\gamma_2$ satisfies some polynomial then it satisfies some irreducible factor of that polynomial. This is because $J$ is simple.

The polynomial $x^61$ factors as $(1x)(1+x)(1+x+x^2)(1x+x^2)$, and $1x$, $1+x$, and $1+x+x^2$ all divide polynomials of the form $x^d1$ with $d$ a proper divisor of 6.
It follows from the above proposition that the ring $D=\bZ[\Delta]/(1\gamma_2+\gamma_2^2)$ acts on $J$. This action is necessarily faithful since $D$ is an order in a simple algebra (in fact, $D \otimes \bQ=M_2(\bQ)$). Note that $\bZ[\gamma_2] \subset D$ is isomorphic to $\bZ[\sqrt[3]{1}]$, and $\gamma_2=\sqrt[3]{1}$, a 6th root of unity. ◾
Remark. The action of $D$ on $J$ is defined over $K^+$, and so $\End(J_{K^+}) \otimes \bQ$ contains $M_2(\bQ)$. It follows that $J$ is not simple over the field $K^+$.
The 19torsion in J
Let $V=J[19](\ol{\bQ})$ be the 19torsion in $J$, which is a 4dimensional vector space over the field $\bF_{19}$. This space admits compatible actions of $G$ and $\Delta$. In particular, it is a module over the ring $\bZ[\gamma_2]=\bZ[\sqrt[3]{1}]$.
The prime 19 splits in $\bZ[\sqrt[3]{1}]$; let $19=\pi \ol{\pi}$ be its factorization. Let $V_{\pi}$ and $V_{\ol{\pi}}$ be the kernels of $\pi$ and $\ol{\pi}$ on $V$, so that $V=V_{\pi} \oplus V_{\ol{\pi}}$. The spaces $V_{\pi}$ and $V_{\ol{\pi}}$ are stable by $G$ and $\Gamma$, since $\gamma_2$ commutes with these groups, but interchanged under $\tau_{\zeta}$, since conjugation by $\tau_{\zeta}$ induces a nontrivial automorphism of $\bZ[\gamma_2]$, and thus interchanges $\pi$ and $\ol{\pi}$.
Proposition. The Weil pairing on $V$ induces a Cartier duality between $V_{\pi}$ and $V_{\ol{\pi}}$.
Proof. It suffices to show that $V_{\pi}$ is selforthogonal under the Weil pairing. Denote the pairing by $(,)$. We have $(\gamma_2 x, \gamma_2 y)=(x, y)$, since $\gamma_2$ induces the identity on $\rH_2$ of $X$. On the other hand, $V_{\pi}$ and $V_{\ol{\pi}}$ are eigenspaces of $\gamma_2$ with eigenvalues $\zeta^{\pm 1}$, where $\zeta \in \bF_{19}$ is a primitive 6th root of unity. Thus for $x,y \in V_{\pi}$ we have
$$ (x,y)=(\gamma_2 x, \gamma_2 y)=(\zeta x, \zeta y)=\zeta^2 (x,y), $$and so $(x,y)=0$, since $\zeta^2 \ne 1$. ◾
We now define some more subspaces of $V$:

Let $V(1) \subset V$ be the 1dimensional space $J(\bQ)_{\tors}$ spanned by the cusps, on which $G$ acts trivially. Since $V(1)$ is stable by $\gamma_2$, it is contained in $V_{\pi}$ or $V_{\ol{\pi}}$. Relabeling if necessary, we can assume it's contained in $V_{\ol{\pi}}$.

Let $V(\gamma) \subset V$ be the space of vectors $v$ satisfying $a v=\gamma_a v$ for $a \in \Gamma$, where here we identify $\Gamma$ with $\Gal(K^+/\bQ)$. One easily sees that the action of $\Gamma'$ interchanges $V(1)$ and $V(\gamma)$, and so $V(\gamma)$ is a one dimensional subspace of $V_{\pi}$.

Let $V(\chi)$ be the one dimensional $\bF_{19}$ space on which $G$ acts through the mod 19 cyclotomic character. The inclusion $V(1) \to V_{\ol{\pi}}$ combined with the above proposition yields a surjection $V_{\pi} \to V(1)$.
Proposition. The sequence
$$ 0 \to V(\gamma) \to V_{\pi} \to V(\chi) \to 0 $$is exact.
Proof. The action of $G$ on $V(\gamma)$ factors through $\Gal(K^+/\bQ)$ and no smaller quotient, while the action of $G$ on $V(\chi)$ factors through $\Gal(\bQ(\sqrt[19]{1})/\bQ)$ and no smaller quotient. Thus $V(\gamma)$ and $V(\chi)$ are nonisomorphic representations of $G$, and so the composite is 0. This proves the proposition. ◾
Rank 0
First reduction
To prove that $J(\bQ)$ has rank 0, it is enough to show that the map $\pi$ on $J(\bQ)$ is surjective, as a finitely generated $\bZ[\sqrt[3]{1}]$module on which $\pi$ acts surjectively is necessarily finite. We have the following diagram:
$$ \xymatrix{ J(\bZ[1/13]) \ar[r]^{\pi} \ar[d] & J(\bZ[1/13]) \ar[r] \ar[d] & \rH^1_{\rm fppf}(\bZ[1/13], J[\pi]) \ar[d]^{\rho} \\ J(\bQ_{13}) \ar[r]^{\pi} & J(\bQ_{13}) \ar[r] & \rH^1_{\rm fppf}(\bQ_{13}, J[\pi]) } $$To prove that $\pi$ is surjective on $J(\bZ[1/13])=J(\bQ)$, it is enough to show (i) that $\pi$ is surjective on $J(\bQ_{13})$; and (ii) that $\rho$ is injective.
Proof of (i)
Let $\cJ$ be the Néron model of $J$ over $\bZ_{13}$; note that $\cJ(\bZ_{13})=J(\bQ_{13})$. Let $N$ be the kernel of the reduction map $\cJ(\bZ_{13}) \to \cJ(\bF_{13})$. We have the following diagram
$$ \xymatrix{ 0 \ar[r] & N \ar[r] \ar[d]^{\pi} & \cJ(\bZ_{13}) \ar[r] \ar[d]^{\pi} & \cJ(\bF_{13}) \ar[d]^{\pi} \ar[r] & 0 \\ 0 \ar[r] & N \ar[r] & \cJ(\bZ_{13}) \ar[r] & \cJ(\bF_{13}) \ar[r] & 0 } $$Now, $N$ is pro13 and so 19, and thus $\pi$, acts bijectively on it. By the snake lemma, the cokernel of $\pi$ on $\cJ(\bZ_{13})$ is isomorphic to the cokernel of $\pi$ on $\cJ(\bF_{13})$, and so it suffices to show that this is 0. But $\cJ(\bF_{13})$ is a finite set, and so the cokernel of $\pi$ on it vanishes if and only if the kernel of $\pi$ on it vanishes. Again, the snake lemma shows that the kernel of $\pi$ on $\cJ(\bF_{13})$ is isomorphic to the kernel of $\pi$ on $\cJ(\bZ_{13})$. We are thus reduced to showing that the kernel of $\pi$ on $J(\bQ_{13})$ vanishes.
Now, $J(\bQ_{13})[\pi]$ is exactly $V_{\pi}^D$, where $D \subset G$ is the decomposition group at 13. As we have an exact sequence
$$ 0 \to V(\gamma) \to V_{\pi} \to V(\chi) \to 0, $$it suffices to show that $V(\gamma)^D=0$ and $V(\chi)^D=0$. The former assertion follows since 13 ramifies in $K^+$, while the latter follows since 13 does not split in $\bQ(\sqrt[19]{1})$ (as $13 \ne 1 \pmod{19}$).
Proof of (ii)
Proposition. We have a short exact sequence of group schemes over $\bZ[1/13]$:
$$ 0 \to E \to J[\pi] \to \mu_{19} \to 0 $$where $E$ is finite étale and becomes constant over $\bZ[1/13, \sqrt[13]{1}]$.
Proof. Let $E$ be the scheme theoretic closure of $V(\gamma)$ in $J[\pi]$. The restriction of $E$ to the finite étale cover $\bZ[1/13, \sqrt[13]{1}]$ is trivial by Raynaud's theorem, as the restriction of $V(\gamma)$ to $\bQ(\sqrt[13]{1})$ is trivial. The quotient $J[\pi]/E$ is generically isomorphic to $\mu_{19}$, as its Galois representation is $V(\chi)$, and so isomorphic to $\mu_{19}$ over $\bZ[1/13]$ by Raynaud's theorem. ◾
Consider the diagram
$$ \xymatrix{ \rH^1_{\rm fppf}(\bZ[1/13], E) \ar[r] & \rH^1_{\rm fppf}(\bZ[1/13], J[\pi]) \ar[r] \ar[d]^{\rho} & \rH^1_{\rm fppf}(\bZ[1/13], \mu_{19}) \ar[d]^{\rho'} \\ & \rH^1_{\rm fppf}(\bQ_{13}, J[\pi]) \ar[r] & \rH^1_{\rm fppf}(\bQ_{13}, \mu_{19}) } $$To show that $\rho$ is injective, it suffices to show (iia) that $\rho'$ is injective; and (iib) that $\rH^1_{\rm fppf}(\bZ[1/13], E)=0$. We do this in the following two propositions.
Proposition. $\rho'$ is injective.
Proof. By Kummer theory, $\rho'$ is equivalent to the map
$$ \bZ[1/13]^{\times}/(\bZ[1/13]^{\times})^{19} \to \bQ_{13}^{\times}/(\bQ_{13}^{\times})^{19}. $$This map is injective because $\bZ[1/13]^{\times}=\{\pm 13^n\}$ and 13 is not a 19th power in $\bQ_{13}$ (it has valuation 1). ◾
Proposition. $\rH^1_{\rm fppf}(\bZ[1/13], E)=0$.
Proof. We have $\rH^1_{\rm fppf}(\bZ[1/13], E)=\rH^1_{\rm fppf}(\bZ[1/13, \sqrt[13]{1}], \bZ/19\bZ)^{\Gal(K/\bQ)}$, and so it suffices to show that $\rH^1_{\rm fppf}(\bZ[1/13, \sqrt[13]{1}], \bZ/19\bZ)=0$. This amounts to showing that $K$ has no abelian extension of degree 19 unramified everywhere away from the prime $\lambda$ of $K$ above 13. Such abelian extensions correspond to quotients of the ray class group for the modulus $\lambda$. The ray class group is an extension of the class group by a quotient of the group of $\lambda$adic units. Both of these groups are prime to 19, which proves the result. ◾