We were talking about how to prove Fermat’s last theorem. We constructed an elliptic curve $E$ and for a prime $\ell \gt 3$ looked at the Galois action $$ \bar{r}_ {E,\ell} \colon G_\mathbb{Q} \to \GL_2(\mathbb{F}_ \ell). $$ Because $E$ was everywhere semistable, we found out that $\bar{r}_ {E,\ell}$ was everywhere unramified away from $2$ and $\ell$, where $$ \bar{r}_ {E,\ell} \vert_{G_{\mathbb{Q}_ 2}} \cong \begin{pmatrix} \bar{\epsilon}_ \ell & \ast \br 0 & 1 \end{pmatrix} \otimes \delta^2 $$ where $\delta$ is unramified with $\delta^2 = 1$, and $$ \bar{r}_ {E,\ell} \vert_{G_{\mathbb{Q}_ \ell}} \cong \operatorname{Ind}_ {G_{\mathbb{Q}_ {\ell^2}}}^{G_{\mathbb{Q}_ \ell}} \bar{\theta} \text{ or } \begin{pmatrix} \bar{\chi}^{-1} \bar{\epsilon}_ \ell & \ast \br 0 & \bar{\chi} \end{pmatrix} $$ where $\bar{\theta} \colon I_\ell \twoheadrightarrow \mathbb{F}_ {\ell^2}^\times$ and $\bar{\chi}$ is unramified and with $\ast$ being integral under Kummer theory (this was called peu ramifiée).
Theorem 1. The representation $\bar{r}_ {E,\ell}$ is absolutely irreducible.
Proof.
If not, it is an extension of characters over $\bar{\mathbb{F}}_ \ell$, but because $\det \bar{r}_ {E,\ell}(c) = -1$ we have $\bar{\chi}_ 1(c), \bar{\chi}_ 2(c)$ are $\pm 1$. It follows that the eigenspaces are defined over $\mathbb{F}_ \ell$, so we have $$ \bar{r}_ {E,\ell} \sim \begin{pmatrix} \bar{\chi}_ 1 & \ast \br 0 & \bar{\chi}_ 2 \end{pmatrix} $$ over $\mathbb{F}_ \ell$. Now we have $\bar{\chi}_ 1$ and $\bar{\chi}_ 2$ are unramified away from $\ell$, and we have $\bar{\chi}_ i \vert_{I_\ell}$ are either $1$ or $\bar{\epsilon}_ \ell$. So either $\bar{\chi}_ i$ or $\bar{\chi}_ i \bar{\epsilon}_ \ell^{-1}$ is unramified everywhere, which means that it is trivial. So either $$ \bar{r}_ {E,\ell} \sim \begin{pmatrix} \bar{\epsilon}_ \ell & \ast \br 0 & 1 \end{pmatrix} \text{ or } \begin{pmatrix} 1 & \ast \br 0 & \bar{\epsilon}_ \ell \end{pmatrix}. $$ In the second case, we see that $\bar{r}_ {E,\ell} \vert_{I_2} = 1 \oplus \bar{\epsilon}_ \ell$ and so we get a contradiction. In the first case, we have $\mu_\ell \hookrightarrow E$, and so we choose $m$ maximal such that $\mu_{\ell^m} \hookrightarrow E$. (We can’t have $\mu_{l^\infty} \hookrightarrow E$ because then the Frobenius at $p$ has eigenvalue $p$ on $T_p E$, which is not possible by the Riemann hypothesis.) If we consider $$ E^\prime = E / \mu_{\ell^m} $$ then $E^\prime[\ell]$ has the same properties, and so we see that $$ \bar{r}_ {E^\prime, \ell} \cong \begin{pmatrix} 1 & \ast \br 0 & \bar{\epsilon}_ \ell \end{pmatrix} $$ using the same argument. Moreover, $E^\prime\lbrack 2 \rbrack(\mathbb{Q}) = 4$.
Theorem 2 (Mazur). If $E^\prime / \mathbb{Q}$ is an elliptic curve with $\lvert E^\prime\lbrack 2 \rbrack(\mathbb{Q}) \rvert = 4$ and if $\ell \gt 3$ is a prime then $E^\prime$ does not have a $\mathbb{Q}$-point of exact order $\ell$.
Idea of proof.
We take care of the cases $\ell = 5, 7, 13$ by some explicit arguments. In the other cases, suppose that $$ \bar{r}_ {E^\prime,\ell} \sim \begin{pmatrix} 1 & \ast \br 0 & \bar{\epsilon}_ \ell \end{pmatrix}. $$ Note that (1) $E$ has everywhere semistable reduction and (2) $\bar{r}_ {E^\prime,\ell}$ is not split. Let us call $0 \neq P \in E^\prime\lbrack \ell \rbrack(\mathbb{Q})$.
If it has bad reduction at $p$, then we have $$ \bar{r}_ {E^\prime,\ell} \vert_{G_{\mathbb{Q}_ p}} \sim \begin{pmatrix} \bar{\epsilon}_ \ell & \ast \br 0 & 1 \end{pmatrix} \otimes \delta $$ where $\delta$ is unramified and $\delta^2 = 1$. So we have two cases: either (I) $\bar{r}_ {E^\prime,\ell} \vert_{G_{\mathbb{Q}_ p}} \sim \bar{\epsilon}_ \ell \oplus 1$ or (II) $\ell \mid p-1$ and $P$ reduces to a root of unity in $(\mathbb{Q}_ p^\times / q^\mathbb{Z})(\delta)$.
If we are in case I for every $p$ then we have that $\bar{r}_ {E^\prime,\ell}$ is unramified everywhere after restricting to $\mathbb{Q}(\zeta_\ell)$ and moreover its kernel corresponds to a degree $\ell$ extension of $\mathbb{Q}(\zeta_\ell)$. By class field theory, it corresponds to an element $C \in \mathrm{Cl}(\mathbb{Z}[\zeta_\ell])$ and ${}^\sigma C = C^{\bar{\epsilon}_ \ell(\sigma)^{-1}}$ for all $\sigma \in \Gal(\mathbb{Q}(\zeta_\ell) / \mathbb{Q})$. This is a classical algebraic number theory question and is a contradiction by Herbrand’s theorem.
What about case II? There is a rational point $$ X_0(\ell)(\mathbb{Q}) \ni x = (E^\prime, \langle P \rangle), $$ where $X_0(\ell)$ has two cusps $0$ and $\infty$. Now the claim is that $x \mod p$ is $\infty$ in case II and $0$ in case I. That means that in the Jacobian $$ J_0(\ell)(\mathbb{Q}) $$ we have $[x] - [0]$ reducing to $[\infty] - [0]$ in case II and $0$ in case I. On the other hand, $[\infty] - [0]$ is torsion of exact order the numerator of $(\ell-1)/12$, which is greater than $1$ if $\ell = 11$ or $\ell \gt 13$. Mazur then found the Eisenstein quotient $$ J_0(\ell) \twoheadrightarrow J^\prime $$ such that $[\infty] - [0]$ have the same order, and $J^\prime(\mathbb{Q})$ is finite. Because $p \gt 2$, the only finite flat group scheme over $\mathbb{Z}_ p$ whose generic fiber is constant is constant, so we have $$ J^\prime(\mathbb{Q}) \hookrightarrow J^\prime(\mathbb{F}_ p). $$ But $[x] - [0]$ in $J^1(\mathbb{Q})$ reduces to $[\infty] - [0]$ in case II and $0$ in case I. So this shows that we are always in case I or always in case II.
On the other hand we can check that $3$ is a prime with bad reduction (because if not $\mathbb{Z}/\ell$ injects into $E^\prime(\mathbb{F}_3)$ which has order at most $1 + 3 + 2 \sqrt{3} \le 8$, and it is in case I because $\ell$ does not $p - 1 = 2$.
So it suffices to prove the following theorem.
Theorem 3 (Theorem A). if $\ell \ge 3$ and $\bar{r} \colon G_\mathbb{Q} \to \mathrm{GL}_ 2(\mathbb{F}_ \ell)$ is a continuous representation with
- $\det \bar{r} = \bar{\epsilon}_ \ell$,
- $\bar{r}$ unramified away from $2\ell$,
- $\bar{r} \vert_{G_{\mathbb{Q}_ 2}} = \begin{pmatrix} \bar{\epsilon}_ \ell & \ast \br 0 & 1 \end{pmatrix} \otimes \delta$ for $\delta$ unramified with $\delta^2 = 1$,
- $\bar{r} \vert_{G_{\mathbb{Q}_ \ell}}$ is either an unramified quadratic induction of an unramified character, or peu ramifiée,
then $\bar{r}$ is reducible.
Before proving this, our first goal is to prove the following, which is for $\ell = 3$ but more general.
Theorem 4 (Theorem B). Suppose $\bar{r} \colon G_\mathbb{Q} \to \mathrm{GL}_ 2(\bar{\mathbb{F}}_ 3)$ is a continuous representation satisfying (1)–(4) of Theorem A, then $$ \bar{r} \sim \begin{pmatrix} \bar{\epsilon}_ 3 & \ast \br 0 & 1 \end{pmatrix}. $$
For the second theorem, what we do is to look at $L$ the fixed points of the kernel of $\bar{r}$ and then $L/\mathbb{Q}$ is a finite extension satisfying $$ \lvert D_{L/\mathbb{Q}} \rvert^{1/[L:\mathbb{Q}]} \lt 8.25, $$ and then there are some results that tell us that $L/\mathbb{Q}$ has small degree.