Remark 15.112.9. Let $A$ be a discrete valuation ring with fraction field $K$. Let $L/K$ be a finite Galois extension. Let $\mathfrak m \subset B$ be a maximal ideal of the integral closure of $A$ in $L$. Let

$P \subset I \subset D \subset G$

be the wild inertia, inertia, decomposition group of $\mathfrak m$. Consider the diagram

$\xymatrix{ \mathfrak m \ar@{-}[d] \ar@{-}[r] & \mathfrak m^ P \ar@{-}[d] \ar@{-}[r] & \mathfrak m^ I \ar@{-}[d] \ar@{-}[r] & \mathfrak m^ D \ar@{-}[d] \ar@{-}[r] & A \cap \mathfrak m \ar@{-}[d] \\ B & B^ P \ar[l] & B^ I \ar[l] & B^ D \ar[l] & A \ar[l] }$

Observe that $B^ P, B^ I, B^ D$ are the integral closures of $A$ in the fields $L^ P$, $L^ I$, $L^ D$. Thus we also see that $B^ P$ is the integral closure of $B^ I$ in $L^ P$ and so on. Observe that $\mathfrak m^ P = \mathfrak m \cap B^ P$, $\mathfrak m^ I = \mathfrak m \cap B^ I$, and $\mathfrak m^ D = \mathfrak m \cap B^ D$. Hence the top line of the diagram corresponds to the images of $\mathfrak m \in \mathop{\mathrm{Spec}}(B)$ under the induced maps of spectra. Having said all of this we have the following

1. the extension $L^ I/L^ D$ is Galois with group $D/I$,

2. the extension $L^ P/L^ I$ is Galois with group $I_ t = I/P$,

3. the extension $L^ P/L^ D$ is Galois with group $D/P$,

4. $\mathfrak m^ I$ is the unique prime of $B^ I$ lying over $\mathfrak m^ D$,

5. $\mathfrak m^ P$ is the unique prime of $B^ P$ lying over $\mathfrak m^ I$,

6. $\mathfrak m$ is the unique prime of $B$ lying over $\mathfrak m^ P$,

7. $\mathfrak m^ P$ is the unique prime of $B^ P$ lying over $\mathfrak m^ D$,

8. $\mathfrak m$ is the unique prime of $B$ lying over $\mathfrak m^ I$,

9. $\mathfrak m$ is the unique prime of $B$ lying over $\mathfrak m^ D$,

10. $A \to B^ D_{\mathfrak m^ D}$ is étale and induces a trivial residue field extension,

11. $B^ D_{\mathfrak m^ D} \to B^ I_{\mathfrak m^ I}$ is étale and induces a Galois extension of residue fields with Galois group $D/I$,

12. $A \to B^ I_{\mathfrak m^ I}$ is étale,

13. $B^ I_{\mathfrak m^ I} \to B^ P_{\mathfrak m^ P}$ has ramification index $|I/P|$ prime to $p$ and induces a trivial residue field extension,

14. $B^ D_{\mathfrak m^ D} \to B^ P_{\mathfrak m^ P}$ has ramification index $|I/P|$ prime to $p$ and induces a separable residue field extension,

15. $A \to B^ P_{\mathfrak m^ P}$ has ramification index $|I/P|$ prime to $p$ and induces a separable residue field extension.

Statements (1), (2), and (3) are immediate from Galois theory (Fields, Section 9.21) and Lemma 15.112.5. Statements (4) – (9) are clear from Lemma 15.112.1. Part (12) is Lemma 15.112.8. Since we have the factorization $A \to B^ D_{\mathfrak m^ D} \to B^ I_{\mathfrak m^ I}$ we obtain the étaleness in (10) and (11) as a consequence. The residue field extension in (10) must be trivial because it is separable and $D/I$ maps onto $\text{Aut}(\kappa (\mathfrak m)/\kappa _ A)$ as shown in Lemma 15.112.4. The same argument provides the proof of the statement on residue fields in (11). To see (13), (14), and (15) it suffices to prove (13). By the above, the extension $L^ P/L^ I$ is Galois with a cyclic Galois group of order prime to $p$, the prime $\mathfrak m^ P$ is the unique prime lying over $\mathfrak m^ I$ and the action of $I/P$ on the residue field is trivial. Thus we can apply Lemma 15.112.5 to this extension and the discrete valuation ring $B^ I_{\mathfrak m^ I}$ to see that (13) holds.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).