The Exact Sequence

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

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

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

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

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

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

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

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 $\sigma(a)\not\equiv a\pmod{\mathfrak{p}}$. $\qedsymbol$