# Lecture 18: Criterion for non-existence of torsion points

$$\DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\Cl}{Cl} \newcommand{\tors}{\mathrm{tors}} \newcommand{\GL}{\mathrm{GL}} \def\mat#1#2#3#4{\left( \begin{array}{cc} #1 & #2 \\ #3 & #4 \end{array} \right)} \newcommand{\bQ}{\mathbf{Q}} \let\ol\overline \newcommand{\bZ}{\mathbf{Z}} \DeclareMathOperator{\End}{End} \newcommand{\cE}{\mathcal{E}} \newcommand{\cP}{\mathcal{P}} \newcommand{\bF}{\mathbf{F}} \newcommand{\cO}{\mathcal{O}} \newcommand{\bG}{\mathbf{G}} \newcommand{\cA}{\mathcal{A}}$$

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 Jordan--Holder constituents of $A[p](\ol{\bQ})$ are 1-dimensional 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 non-constant, 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 $N^{j-i}$ isogeny defined above. Then $f^{-1} g$ is an endomorphism of $E_i$ of degree $N^{j-i}$ 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 $N^{(i-j)/2}$ 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 semi-stable 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 semi-stable reduction theorem, there is an extension $K/\bQ_p$ such that $E$ has semi-stable 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 non-constant 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 $e \lt p-1$ (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 $p-1$ 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 $N-1$ 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 1-gon. 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 semi-stable 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 $e=1 \lt p-1$. 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 Néron--Ogg--Shafarevich.

• $p=N$ 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 connected--étale 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 1-cocyle $\Gamma \to \bF_N$ 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 $B_{N-j}$.

In particular, we see that $H^{-1}=H^{N-2}$ is non-zero 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 semi-simple, since the order of this group is prime to $N$. This proves Theorem 3.