The Exact Sequence

Because $ D_\mathfrak{p}$ preserves $ \mathfrak{p}$, there is a natural reduction homomorphism

$\displaystyle \varphi :D_\mathfrak{p}\to \Gal (k_{\mathfrak{p}}/\mathbf{F}_p).

Theorem 9.3.5   The homomorphism $ \varphi $ is surjective.

Proof. Let $ \tilde{a} \in k_{\mathfrak{p}}$ be an element such that $ k_{\mathfrak{p}} = \mathbf{F}_p(\tilde{a})$. Lift $ \tilde{a}$ to an algebraic integer $ a\in\O_K$, and let $ f=\prod_{\sigma\in D_p}(x-\sigma(a))\in K^D[x]$ be the characteristic polynomial of $ a$ over $ K^D$. Using Proposition 9.3.4 we see that $ f$ reduces to a multiple of the minimal polynomial $ \tilde{f}=\prod (x-\widetilde{\sigma(a)})\in \mathbf{F}_p[x]$ of $ \tilde{a}$ (by the Proposition the coefficients of $ \tilde{f}$ are in $ \mathbf{F}_p$, and $ \tilde{a}$ satisfies $ \tilde{f}$). The roots of $ \tilde{f}$ are of the form $ \widetilde{\sigma(a)}$, and the element $ \Frob _p(a)$ is also a root of $ \tilde{f}$, so it is of the form $ \widetilde{\sigma(a)}$. We conclude that the generator $ \Frob _p$ of $ \Gal (k_{\mathfrak{p}}/\mathbf{F}_p)$ is in the image of $ \varphi $, which proves the theorem. $ \qedsymbol$

Definition 9.3.6 (Inertia Group)   The inertia group associated to $ \mathfrak{p}$ is the kernel $ I_\mathfrak{p}$ of $ D_\mathfrak{p}\to \Gal (k_{\mathfrak{p}}/\mathbf{F}_p)$.

We have an exact sequence of groups

$\displaystyle 1 \to I_\mathfrak{p}\to D_\mathfrak{p}\to \Gal (k_{\mathfrak{p}}/\mathbf{F}_p)\to 1.$ (9.2)

The inertia group is a measure of how $ p$ ramifies in $ K$.

Corollary 9.3.7   We have $ \char93  I_\mathfrak{p}= e(\mathfrak{p}/p)$, where $ \mathfrak{p}$ is a prime of $ K$ over $ p$.

Proof. The sequence (9.3.1) implies that $ \char93 I_\mathfrak{p}= (\char93 D_\mathfrak{p})/f(K/\mathbf{Q})$. Applying Propositions 9.3.3-9.3.4, we have

$\displaystyle \char93 D_\mathfrak{p}= [K:L] = \frac{[K:\mathbf{Q}]}{g} = \frac{efg}{g} = ef.$

Dividing both sides by $ f=f(K/\mathbf{Q})$ proves the corollary. $ \qedsymbol$

We have the following characterization of $ I_\mathfrak{p}$.

Proposition 9.3.8   Let $ K/\mathbf{Q}$ be a Galois extension with group $ G$, and let  $ \mathfrak{p}$ be a prime of $ \O_K$ lying over a prime $ p$. Then

$\displaystyle I_\mathfrak{p}= \{\sigma\in G   :  \sigma(a) \equiv a\pmod{\mathfrak{p}}$ for all $\displaystyle a\in\O_K\}.

Proof. By definition $ I_\mathfrak{p}= \{\sigma\in D_\mathfrak{p}: \sigma(a) \equiv
a\pmod{\mathfrak{p}}$ for all $ a\in\O_K\}$, so it suffices to show that if $ \sigma\not\in D_\mathfrak{p}$, then there exists $ a\in\O_K$ such that $ \sigma(a)\not\equiv a\pmod{\mathfrak{p}}$. If $ \sigma\not\in D_\mathfrak{p}$, then $ \sigma^{-1}\not\in D_\mathfrak{p}$, so $ \sigma^{-1}(\mathfrak{p})\neq \mathfrak{p}$. Since both are maximal ideals, there exists $ a\in\mathfrak{p}$ with $ a\not\in\sigma^{-1}(\mathfrak{p})$, i.e., $ \sigma(a)\not\in\mathfrak{p}$. Thus $ \sigma(a)\not\equiv a\pmod{\mathfrak{p}}$. $ \qedsymbol$

William Stein 2012-09-24