Lecture 18: Criterion for nonexistence of torsion points
We prove Theorem A from Lecture 1, which states that if one can find an appropriate map from $X_0(N)$ to an abelian variety, then no elliptic curve over the rational numbers has a rational point of order $N$. Combined with Theorem B from Lecture 1, this gives an axiomatic approach to proving Mazur's theorem.
Statement of criterion
Our purpose today is to prove the following result (Theorem A from Lecture 1):
Theorem (Theorem 1). Let $N \gt 7$ be a prime number. Suppose there exists an abelian variety $A/\bQ$ and a map of varieties $f \colon X_0(N) \to A$ satisfying the following conditions:

$A$ has good reduction away from $N$.

$A(\bQ)$ has rank 0.

$f(0) \ne f(\infty)$.
Then no elliptic curve defined over $\bQ$ has a rational point of order $N$.
Combined with Theorem B from Lecture 1 [PP69BM], we have the following criterion:
Theorem (Theorem 2). Let $N \gt 7$ be a prime number and let $p \ne N$ be a second prime number. Suppose there exists an abelian variety $A/\bQ$ and a map $f \colon X_0(N) \to A$ satisfying the following:

$A$ has good reduction away from $N$.

$A$ has completely toric reduction at $N$.

The JordanHolder constituents of $A[p](\ol{\bQ})$ are 1dimensional and either trivial or cyclotomic.

$f(0) \ne f(\infty)$.
Then no elliptic curve defined over $\bQ$ has a rational point of order $N$.
Remark. The above theorem, and the proof presented here, comes from III.5 of Mazur’s paper “Modular curves and the Eisenstein ideal” (MR488287). It is not stated there explicitly, however.
Initial reduction
We will prove the following theorem in what follows. In this section, we explain how it implies Theorem 1.
Theorem (Theorem 3). Suppose that $A$ and $f$ are as in the statement of the theorem. Suppose that $E/\bQ$ has a point of order $N$. Then $E[N] \cong \bZ/N\bZ \oplus \mu_N$.
Note that $E$ having a point of order $N$ implies that $\bZ/N\bZ$ is a sub group scheme of $E[N]$; the Weil pairing implies that the quotient is $\mu_N$. The content of the above proposition is that this extension is split.
Before proving Theorem 1, we need two lemmas:
Lemma. Suppose we have $A$ and $f$ as in the statement of the theorem. Then $X_0(N)(\bQ)$ is finite.
Proof. By assumption, $A(\bQ)$ is finite. Since $f(0) \ne f(\infty)$, the map $f$ is nonconstant, and so the fibers of the map $f \colon X_0(N)(\bQ) \to A(\bQ)$ are finite. The result follows. ◾
Lemma. Let $E/\bQ$ be an elliptic curve. Then $\End(E)=\bZ$. (I.e., even if $E$ is CM, its extra endomorphisms are not defined over $\bQ$.)
Proof. We have a ring homomorphism $\End(E) \otimes \bQ \to \End(T_0E)$, where $T_0E$ is the tangent space at the identity. Of course, $\End(T_0E)=\bQ$. We know that over any field of characteristic 0, $\End(E)$ is either $\bZ$ or an order in an imaginary quadratic field. Thus $\End(E) \otimes \bQ$ is a number field. Since it admits a homomorphism to $\bQ$, it must be $\bQ$ itself, and so $\End(E) \cong \bZ$. ◾
We now prove the Theorem 1, assuming Theorem 3. Let $E_1$ be an elliptic curve over $\bQ$ having a point $P_1$ of order $N$. By Theorem 3, $E$ contains $\mu_N$ as a subgroup. Define $E_2=E_1/mu_N$. The image $P_2$ of $P_1$ in $E_1$ is a point of order $N$. Thus, Theorem 3 yields a $\mu_N$ in $E_1$. Continuing, we obtain a sequence of isogenies
$$ E_1 \to E_2 \to E_3 \to \cdots $$each of degree $N$. Furthermore, if $P_i$ denotes the image of $P_1$ in $E_i$ then the isogeny $E_i \to E_{i+1}$ does not kill $P_i$.
By the first Lemma, two of these curves, say $E_i$ and $E_j$ (with $i \lt j$), must be isomorphic. Let $f \colon E_i \to E_j$ be an isomorphism, and let $g \colon E_i \to E_j$ be the degree $N^{ji}$ isogeny defined above. Then $f^{1} g$ is an endomorphism of $E_i$ of degree $N^{ji}$ and does not kill $P_i$, a point of order $N$. It follows that $f^{1} g$ cannot be multiplication by an integer (as it would have to be multiplication by $N^{(ij)/2}$ by degree considerations, but this kills $P_i$), and so $\End(E) \ne \bZ$. But this contradicts the second Lemma. We conclude that no such $E_1$ exists. This proves Theorem 1.
Proof of the Theorem 3
We fix an elliptic curve $E/\bQ$ having a point $P$ of order $N$. Our goal is to show that the sequence
$$ 0 \to \bZ/N\bZ \to E[N] \to \mu_N \to 0 $$splits. Let $\cE/\bZ$ be the Néron model of $E$, and let $\cP$ be the $\bZ$point of $\cE$ extending $P$. We proceed in four steps.
Step 1
Proposition. $E$ has everywhere semistable reduction.
Proof. Suppose $E$ has additive reduction at a prime $p$. By the classification of the special fibers of Néron models, we know that $\cE_{\bF_p}$ has at most four components. Since $N \gt 7$ is prime, the image of $\cP_{\bF_p}$ in the component group must therefore vanish. It follows that $\cP_{\bF_p}$ is contained in the identity component. However, the identity component is $p$torsion, since $E$ has additive reduction. We conclude that $p=N$.
By the semistable reduction theorem, there is an extension $K/\bQ_p$ such that $E$ has semistable reduction over $K$. In fact, we can take $K$ to have degree at most 6 over $\bQ_p$ (since $p=N \gt 3$). Let $\cO$ be the ring of integers in $K$, let $k$ be the residue field of $\cO$, and let $\cE'$ be the Néron model of $E$ over $\cO$. The Néron mapping property yields a map $f \colon \cE_{\cO} \to \cE'$. We claim that $f$ maps $\cE_k^{\circ}$ to the identity. Indeed, $\cE_k^{\circ}$ is $\bG_a$, while $(\cE')_k^{\circ}$ is an elliptic curve or a torus. Since there are no nonconstant maps from $\bG_a$ to an elliptic curve or torus, the claim follows.
Let $\cP' \in \cE'(\cO)$ be the section extending $P$. We note $\cP'=f(\cP)$ since the two sections agree over $K$. In particular, $\cP'$ reduces to the identity element of $\cE'(k)$. But this is a contradiction, as we have previously proved [RC46CS] the reduction map $\cE'[N](\cO) \to \cE'[N](k)$ to be injective when $e \lt p1$ (and note here that $e \le 6$ and $p=N \gt 7$). ◾
Remark. The key result above (injectivity of reduction) makes essential use of Raynaud's theorem.
Step 2
Proposition. Suppose $p \in \{2,3\}$. Then $E$ has multiplicative reduction at $p$ and the reduction of $\cP$ at $p$ is not contained in the identity component of $\cE_{\bF_p}$.
Proof. Suppose $E$ has good reduction at $p$. Since $N$ is prime to $p$, the reduction map on $N$torsion is injective, and so the reduction of $\cP$ at $p$ is a point of order $N$ on $\cE_{\bF_p}$. But by the Hasse bound, $\# \cE(\bF_p) \le p+1+2\sqrt{p} \le 7.5$ and $N \gt 7$. This is a contradiction, and so $E$ must have bad reduction. We have previously ruled out additive reduction.
Since $E$ has multiplicative reduction, $\cE_{\bF}^{\circ}$ is a one dimensional torus over $\bF_p$. Up to isomorphism, there are two such tori: $\bG_m$, and the norm 1 piece of the restriction of scalars of $\bG_m$ from $\bF_{p^2}$. These have $p1$ and $p+1$ points over $bF_p$. It follows that $\cP$ cannot reduce into $\cE_{\bF}^{\circ}$, since $N \gt p+1$. ◾
Corollary. We have $E[N]=\bZ/N\bZ \oplus \mu_N$ over $\bQ_p$.
Proof. Let $G \subset E[N]$ be the subgroup scheme that reduces into the identity component of $\cE_{\bF_p}$. Then $G$ has order $N$ but does not intersect the $\bZ/N\bZ$ coming from $P$, and is thus a complementary space to this $\bZ/N\bZ$. ◾
Step 3
Proposition. Let $p \not\in \{2,3\}$ be a prime of bad reduction for $E$. Then the reduction of $\cP$ at $p$ is not contained in the identity component of $\cE_{\bF_p}$.
Proof. First note: (1) the reduction of $\cP$ is a point of order $N$ on $\cE_{\bF_p}$, since the reduction map is injective on torsion; and (2) $E$ has multiplicative reduction at $p$ (by Step 1). If $p=N$ then $\cE^{\circ}(\bF_N)$ has either $N1$ or $N+1$ points and therefore does not contain a point of order $N$. We therefore assume $p \ne N$ in what follows.
We are interested in three $\bZ[1/N]$ points of $X_0(N)$:

The point $\infty$ corresponding to the generalized elliptic curve which is a 1gon. The smooth locus is $\bG_m$, and the $\Gamma_0(N)$structure is $\mu_N$, which is contained in the identity component.

The point 0 corresponding to the generalized elliptic curve which is an $N$gon. The smooth locus is $\bG_m \times \bZ/N\bZ$ and the $\Gamma_0(N)$structure is $\bZ/N\bZ$, which is not contained in the identity component.

The point $x$ corresponding to $(\cE, \bZ/N\bZ)$, where $\cE$ is the curve we have at hand. Note that this defines a $\bZ[1/N]$point of $X_0(N)$ since $E$ has everywhere semistable reduction, and thus its minimal regular model is a generalized elliptic curve.
Let $\cA/\bZ[1/N]$ be the abelian scheme extending $A$. The reduction map $\cA(\bZ[1/N])_{\tors} \to \cA(\bF_p)$ is injective for $p \ne 2$, since for such $p$ we have $e=1 \lt p1$. But by hypothesis, every $\bZ[1/N]$point of $\cA$ is torsion, and so we can say that the reduction map $\cA(\bZ[1/N]) \to \cA(\bF_p)$ is injective. Note that the map $f$ extends to a map $f \colon X_0(N) \to \cA$ over $\bZ[1/N]$ by the Néron mapping property.
By Step 2, $x$ and 0 reduced to the same point of $X_0(N)(\bF_3)$. Thus $f(x)$ and $f(0)$ reduce to the same point of $\cA(\bF_3)$. It follows from the above reasoning that $f(x)=f(0)$ in $\cA(\bZ[1/N])$.
Suppose now that the reduction of $\cP$ were contained in $\cE_{\bF_p}^{\circ}$. Then $x$ reduces to $\infty$ in $X_0(N)(\bF_p)$. Thus $f(x)$ and $f(\infty)$ reduce to the same point of $\cA(\bF_p)$, and so $f(x)=f(\infty)$. But this contradicts $f(0) \ne f(\infty)$. The result follows. ◾
Corollary. Under the above hypotheses, $E[N] \cong \bZ/N\bZ \oplus \mu_N$ over $\bQ_p$.
Step 4
Let $\Gamma$ be the absolute Galois group of $\bQ$, and let $\rho \colon \Gamma \to \GL_2(\bF_N)$ be the representation given by $E[N]$. Let $K=\bQ(\mu_N)$. We begin with the following observation:
Proposition. $\rho \vert_K$ is everywhere unramified.
Proof. We consider several cases:

$p \ne N$ is a prime of good reduction. Then $\rho \vert_{\bQ_p}$ is already unramified by the easy direction of NéronOggShafarevich.

$p=N$ is a prime of good reduction. Then $\cE[N]$ is a finite flat group scheme over $\bZ_p$ containing $\bZ/N\bZ$ as a sub and $\mu_N$ as a quotient. The connectedétale sequence goes the opposite direction, and thus implies that $\cE[N]=\mu_N \oplus \bZ/N\bZ$ over $\cO$. Thus $\rho \vert_{\bQ_p} = \bZ/N\bZ \oplus \bZ/N\bZ(1)$, and is thus unramified over $K$.

$p$ is a prime of bad reduction. By Steps 2 and 3 above, we have $E[N]=\bZ/N\bZ \oplus \mu_N$, and so the result follows.
◾
Now, the representation $\rho$ looks like
$$ \rho = \mat{1}{f_0}{}{\chi} $$where $\chi$ is the mod $N$ cyclotomic character and $f_0$ is a 1cocyle $\Gamma \to \bF_N$ for $\chi^{1}$. The map $f_0$ restricts to a homomorphism $f \colon \Gamma_K \to \bF_N$ which is everywhere unramified. By class field theory, we can therefore regard $f$ as a homomorphism $\Cl(K) \to \bF_N$. Let $H=\Cl(K) \otimes \bZ/N\bZ$, so that $f \in H^*$. Now, $\Gal(K/\bQ)=\bF_N^{\times}$ acts on $H$. Let $H^i$ be the piece on which it acts by through the $i$th power map. Using the cocycle identity for $f_0$, we see that $f^{\sigma}=\chi(\sigma) f$ for $\sigma \in \Gal(K/\bQ)$, and so $f_0$ belongs to $(H^*)^1=(H^{1})^*$. We now appeal to Herbrand's theorem:
Theorem (Herbrand). Let $j \gt 1$ be odd. Then $H^j \ne 0$ only if $N$ divides $B_{Nj}$.
In particular, we see that $H^{1}=H^{N2}$ is nonzero only if $N$ divides $B_2=1/6$. Since this never happens, we see that $H^{1}=0$, and so $f_0=0$. But this means that $\rho \vert_K$ is the trivial representation, and so we can regard $\rho$ as a representation of $\Gal(K/\bQ)$. It follows that $\rho$ is semisimple, since the order of this group is prime to $N$. This proves Theorem 3.