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Ramification theory

Theorem 1. Suppose we have a continuous representation $\bar{r} \colon G_\mathbb{Q} \to \mathrm{GL}_ 2(\bar{\mathbb{F}}_ 3)$ satisfying

  1. $\det \bar{R} = \bar{\epsilon}_ 3$,
  2. $\bar{r}$ is unramified outside $2$ and $3$,
  3. $\bar{r} \vert_{G_{\mathbb{Q}_ 2}} \sim \begin{pmatrix} \bar{\epsilon}_ 3 & \ast \br 0 & 1 \end{pmatrix} \otimes \delta$ for $\delta^2 = 1$ unramified,
  4. $\bar{r} \vert_{G_{\mathbb{Q}_ 3}}$ is either $\operatorname{Ind}_ {G_{\mathbb{Q}_ 9}}^{G_{\mathbb{Q}_ 3}} \bar{\theta}$ for some character $\bar{\theta} \colon G_{\mathbb{Q}_ 9} \twoheadrightarrow \mathbb{F}_ 9^\times$ totally ramified, or $\begin{pmatrix} \bar{\chi}^{-1} \bar{\epsilon}_ 3 & \ast \br 0 & \bar{\chi} \end{pmatrix}$ for $\bar{\chi}$ unramified and $\ast$ integral.

Then $\bar{r}$ is reducible, and moreover is of the form $\begin{pmatrix} \bar{\epsilon}_ 3 & \ast \br 0 & 1 \end{pmatrix}$.

Let $L = \bar{\mathbb{Q}} ^{\ker \bar{r}}$ over $\mathbb{Q}$. Then the lack of ramification will tell us that $\lvert D_{L/\mathbb{Q}} \rvert$ is small, and then $[L:\mathbb{Q}]$ is small.

Local ramification

Let $K/\mathbb{Q}_ p$ be a finite extension. We have $\mathfrak{p}_ K \subseteq \mathcal{O}_ K$ with residue field $k$. We can do the same for the algebraic closure $\mathfrak{p}_ \bar{K} \subseteq \mathcal{O}_ \bar{K}$ with residue field $\bar{k}$. Then we have a short exact sequence $$ 1 \to I_K \to G_K \to G_k \to 1. $$ We have a geometric Frobenius $\mathrm{Frob}_ k$ with $(\mathrm{Frob}_ k \alpha)^{\lvert k \rvert} = \alpha$. The inertia has a Sylow pro-$p$-subgroup $P_K \vartriangleleft I_K$ with quotient $I_K / P_K \cong \prod_{\ell \neq p} \mathbb{Z}_ \ell$.

Consider a finite Galois extension $L/K$. For each $\sigma \in \Gal(L/K)$, there is a number $i_{L/K}(\sigma)$ satisfying $$ \mathfrak{a}_ \sigma = \langle \sigma \alpha - \alpha : \alpha \in \mathcal{O}_ L \rangle = \mathfrak{p}_ L^{i_{L/K}(\sigma)}. $$ We see that $$ i_{L/K}(\tau \sigma \tau^{-1}) = i_{L/K}(\sigma), \quad i_{L/K}(\sigma \tau) \ge \min(i_{L/K}(\sigma), i_{L/K}(\tau)). $$ So if we define $$ \Gal(L/K)_ i = \lbrace \sigma \in \Gal(L/K) : i_{L/K} \ge i \rbrace $$ then this is a subgroup. We then have $$ \Gal(L/K)_ 1 = I_{L/K} \twoheadleftarrow I_K, \quad \Gal(L/K)_ 2 = P_{L/K} \twoheadleftarrow P_K. $$ We also see that $$ \Gal(L/K)_ i / \Gal(L/K)_ {i+1} \begin{cases} \cong \Gal(k_L/k_K) & i = 0, \br \hookrightarrow k_L^\times & i = 1, \br \hookrightarrow k_L & i \ge 2, \end{cases} $$ where for $i = 1$ we send $\sigma \mapsto \sigma(\pi_L) / \pi_L$ and for $i \ge 2$ we send $\sigma \mapsto (\sigma(\pi_L) - \pi_L) / \pi_L^i$.

We also write $e = \lvert I_{L/K} \rvert$ and define $$ u_{L/K}(\sigma) = \frac{1}{e} \sum_{\tau \in \Gal(L/K)_ 1} \min(i_{L/K}(\tau), i_{L/K}(\sigma)) \in \frac{1}{e}\mathbb{Z}. $$ Then again, we have $u_{L/K}(\sigma \tau \sigma^{-1}) = u_{L/K}(\tau)$ and $u_{L/K}(\sigma \tau) \ge \min(u_{L/K}(\sigma), u_{L/K}(\tau))$ and so we can define $$ \Gal(L/K)^u = \lbrace \sigma \in \Gal(L/K) : u_{L/K}(\sigma) \ge u \rbrace \vartriangleleft \Gal(L/K) $$ for every $u \in \mathbb{R}_ {\ge 0}$. Then we have $$ \begin{align} \Gal(L/K)^0 &= \Gal(L/K), \quad \Gal(L/K)^u = I_{L/K} \text{ for } u \in (0, 1], \br \Gal(L/K)^u &= P_{L/K} \text{ for } u \in (1, \tfrac{e+1}{e}]. \end{align} $$

Proposition 2. If $M/L/K$ are finite Galois extensions, then $\Gal(M/K)^u \twoheadrightarrow \Gal(L/K)^u$.

So we can now define $$ G_K^u = \varprojlim_{L/K} \Gal(L/K)^u. $$

Consider the fractional ideal $$ \mathcal{D}_ {L/K}^{-1} = \lbrace \alpha \in L : \tr(\alpha \mathcal{O}_ L) \subseteq \mathcal{O}_ K \rbrace \supseteq \mathcal{O}_ L. $$ Its inverse $\mathcal{D}_ {L/K} \subseteq \mathcal{O}_ L$ is an ideal, and we define $D_{L/K} = N_{L/K} \mathcal{D}_{L/K}$.

Proposition 3. We have $\displaystyle v_L(\mathcal{D}_ {L/K}) = e \max_{1 \neq \sigma \in \Gal(L/K)} u_{L/K}(\sigma) - \max_{1 \neq \sigma \in \Gal(L/K)} i_{L/K}(\sigma)$.

Global ramification

Now let us look at the situation where $K/\mathbb{Q}$ is finite and $L/K$ is finite Galois. The same definitions of $\mathcal{D}_ {L/K}$ and $D_{L/K}$ make sense. If $v$ is a place of $K$ and $w \mid v$ is a place of $L$, we have its stabilizer $$ \Gal(L/K) \supseteq \Gal(L/K)_ w \cong \Gal(L_w / K_v) $$ which is called the decomposition group. So we have $$ D_{L/K,v} = \prod_{w \mid v} D_{L_w/K_w}. $$

Corollary 4. We have the same formula $$ \frac{1}{[L:K]} v(D_{L/K}) = \max_{1 \neq \sigma \in \Gal(L_w/K_v)} u_{L_w/K_v}(\sigma) - \frac{1}{e} \max_{1 \neq \sigma \in \Gal(L_w/K_v)} i_{L_w/K_v}(\sigma). $$ So for example if $v$ is tamely ramified this number is $1 - 1/e$.

Let us now go back to the situation of $L = \bar{\mathbb{Q}}^{\ker \bar{r}}$ and consider $D_{L/\mathbb{Q}}$. There were three cases.

  1. Assume that $\bar{r} \vert_{G_{\mathbb{Q}_ 3}}$ is the unramified induction of $\bar{\theta}$ of order $8$. Then this is tamely ramified. Locally at $2$, the inertia acts unipotently so it has order $1$ or $3$, so in either case it is tamely ramified. So we see that $$ \lvert D_{L/\mathbb{Q}} \rvert^{1/n} \le 2^{1-1/3} 3^{1-1/8} \lt 4.16 $$ where $n = [L:\mathbb{Q}]$. On the other hand, $L/\mathbb{Q}$ is totally complex because $\det \bar{r}(c) = -1$, and so Minkowski tells us that $$ \lvert D_{L/\mathbb{Q}}^{1/n} \rvert \ge \frac{\pi}{4} \frac{n^2}{(n!)^{2/n}} $$ and this implies $n \le 14$. Now $16 = [L_w : \mathbb{Q}_ 3]$ divides $[L:\mathbb{Q}]$ so we get a contradiction.
  2. Assume that $\bar{r} \vert_{G_{\mathbb{Q}_ 3}} \sim \begin{pmatrix} \bar{\chi}^{-1} \bar{\epsilon}_ 3 & \ast \br 0 & \bar{\chi} \end{pmatrix}$ is tamely ramified. Then we similarly get $\lvert D_{L/\mathbb{Q}} \rvert^{1/n} \lt 2.75$ and by Minkowski $n \lt 6$. Because $n$ is even, we have either $n = 2$ or $n = 4$. Then because $\Gal(L/\mathbb{Q})$ is abelian and $\bar{r}$ is semi-simple, we have $\bar{\rho} = \bar{\chi}_ 1 \oplus \bar{\chi}_ 2$. Because it is unramified way from $3$, we have $\bar{\chi}_ i = \bar{\epsilon}_ 3$ or $1$. This shows that $\bar{r} = 1 \oplus \bar{\epsilon}_ 3$.
  3. Now assume that we have $\bar{r} \vert_{G_{\mathbb{Q}_ 3}} \sim \begin{pmatrix} \bar{\chi}^{-1} \bar{\epsilon}_ 3 & \ast \br 0 & \bar{\chi} \end{pmatrix}$ where $\bar{\chi}$ is unramified and $\ast$ is ramified but peu ramifée. In this case, we can do some calculation and show that $G_\mathbb{Q}^{3/2}$ fixes $L_w$ for $w \mid 3$. Then we get $$ \lvert D_{L/\mathbb{Q}} \rvert^{1/n} \le 2^{2/3} 3^{3/2} \lt 8.25. $$ If we use the naive Minkowski bound, we don’t get anything useful. But if we use an improved bound of Poitou, then we get $n \le 14$ and $6 \mid n$ and so $n = 6$ or $n = 12$. Then we can analyze them.