The Exact Sequence

There is a natural reduction homomorphism

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

Theorem 14.1.5   The homomorphism $ \varphi $ is surjective.

Proof. Let $ \tilde{a} \in \mathbf{F}_\mathfrak{p}$ be an element such that $ \mathbf{F}_\mathfrak{p}=\mathbf{F}_p(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 14.1.4 we see that $ f$ reduces to the minimal polynomial $ \tilde{f}=\prod (x-\tilde{\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}$, and the degree of $ \tilde{f}$ equals the degree of the minimal polynomial of $ \tilde{a}$). The roots of $ \tilde{f}$ are of the form $ \tilde{\sigma}(a)$, and the element $ \Frob _p(a)$ is also a root of $ \tilde{f}$, so it is of the form $ \tilde{\sigma(a)}$. We conclude that the generator $ \Frob _p$ of $ {\mathrm{Gal}}(\mathbf{F}_\mathfrak{p}/\mathbf{F}_p)$ is in the image of $ \varphi $, which proves the theorem. $ \qedsymbol$

Definition 14.1.6 (Inertia Group)   The is the kernel $ I_\mathfrak{p}$ of $ D_\mathfrak{p}\to{\mathrm{Gal}}(\mathbf{F}_\mathfrak{p}/\mathbf{F}_p)$.

Combining everything so far, we find an exact sequence of groups

$\displaystyle 1 \to I_\mathfrak{p}\to D_\mathfrak{p}\to {\mathrm{Gal}}(\mathbf{F}_\mathfrak{p}/\mathbf{F}_p)\to 1.$ (14.1)

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

Corollary 14.1.7   We have $ \char93  I_\mathfrak{p}= e(\mathfrak{p}/p)$, where $ \mathfrak{p}$ is a prime of $ K$ over $ p$.

Proof. The sequence (14.1.1) implies that $ \char93 I_\mathfrak{p}= \char93 D_\mathfrak{p}/f(K/\mathbf{Q})$. Applying Propositions 14.1.3-14.1.4, 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 14.1.8   Let $ K/\mathbf{Q}$ be a Galois extension with group $ G$, let $ \mathfrak{p}$ be a prime lying over a prime $ p$. Then

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

Proof. By definition $ I_\mathfrak{p}=
\{\sigma\in D_\mathfrak{p}: \sigma(a) = 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)=a\pmod{\mathfrak{p}}$. If $ \sigma\not\in D_\mathfrak{p}$, we have $ \sigma^{-1}(\mathfrak{p})\neq \mathfrak{p}$, so 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$

Figure 14.1.2 is a picture of the splitting behavior of a prime $ p\in\mathbf{Z}$.

Figure 14.1.1: The Splitting of Behavior of a Prime in a Galois Extension

William Stein 2004-05-06