Lemma 58.13.1. In the situation described above, via the isomorphism $\pi _1(U) = \text{Gal}(K^{sep}/K)$ the diagram (58.13.0.1) translates into the diagram

\[ \xymatrix{ I \ar[r] \ar[rd]_1 & D \ar[d] \ar[r] & \text{Gal}(K^{sep}/K) \ar[d] \\ & \text{Gal}(\kappa (\mathfrak m^{sh})/\kappa ) \ar[r] & \text{Gal}(M/K) } \]

where $K^{sep}/M/K$ is the maximal subextension unramified with respect to $A$. Moreover, the vertical arrows are surjective, the kernel of the left vertical arrow is $I$ and the kernel of the right vertical arrow is the smallest closed normal subgroup of $\text{Gal}(K^{sep}/K)$ containing $I$.

**Proof.**
By construction the group $D$ acts on $(A^{sep})_{\mathfrak m^{sep}}$ over $A$. By the uniqueness of $A^{sh} \to (A^{sep})_{\mathfrak m^{sep}}$ given the map on residue fields (Algebra, Lemma 10.155.10) we see that the image of $A^{sh} \to (A^{sep})_{\mathfrak m^{sep}}$ is contained in $((A^{sep})_{\mathfrak m^{sep}})^ I$. On the other hand, Lemma 58.12.5 shows that $((A^{sep})_{\mathfrak m^{sep}})^ I$ is a filtered colimit of étale extensions of $A$. Since $A^{sh}$ is the maximal such extension, we conclude that $A^{sh} = ((A^{sep})_{\mathfrak m^{sep}})^ I$. Hence $K^{sh} = (K^{sep})^ I$.

Recall that $I$ is the kernel of a surjective map $D \to \text{Aut}(\kappa (\mathfrak m^{sep})/\kappa )$, see More on Algebra, Lemma 15.110.10. We have $\text{Aut}(\kappa (\mathfrak m^{sep})/\kappa ) = \text{Gal}(\kappa (\mathfrak m^{sh})/\kappa )$ as we have seen above that these fields are the algebraic and separable algebraic closures of $\kappa $. On the other hand, any automorphism of $A^{sh}$ over $A$ is an automorphism of $A^{sh}$ over $A^ h$ by the uniqueness in Algebra, Lemma 10.155.6. Furthermore, $A^{sh}$ is the colimit of finite étale extensions $A^ h \subset A'$ which correspond $1$-to-$1$ with finite separable extension $\kappa '/\kappa $, see Algebra, Remark 10.155.4. Thus

\[ \text{Aut}(A^{sh}/A) = \text{Aut}(A^{sh}/A^ h) = \text{Gal}(\kappa (\mathfrak m^{sh})/\kappa ) \]

Let $\kappa '/\kappa $ be a finite Galois extension with Galois group $G$. Let $A^ h \subset A'$ be the finite étale extension corresponding to $\kappa \subset \kappa '$ by Algebra, Lemma 10.153.7. Then it follows that $(A')^ G = A^ h$ by looking at fraction fields and degrees (small detail omitted). Taking the colimit we conclude that $(A^{sh})^{\text{Gal}(\kappa (\mathfrak m^{sh})/\kappa )} = A^ h$. Combining all of the above, we find $A^ h = ((A^{sep})_{\mathfrak m^{sep}})^ D$. Hence $K^ h = (K^{sep})^ D$.

Since $U$, $U^ h$, $U^{sh}$ are the spectra of the fields $K$, $K^ h$, $K^{sh}$ we see that the top lines of the diagrams correspond via Lemma 58.6.3. By Lemma 58.8.2 we have $\pi _1(X^ h) = \text{Gal}(\kappa (\mathfrak m^{sh})/\kappa )$. The exactness of the sequence $1 \to I \to D \to \text{Gal}(\kappa (\mathfrak m^{sh})/\kappa ) \to 1$ was pointed out above. By Proposition 58.11.3 we see that $\pi _1(X) = \text{Gal}(M/K)$. Finally, the statement on the kernel of $\text{Gal}(K^{sep}/K) \to \text{Gal}(M/K) = \pi _1(X)$ follows from Lemma 58.11.4. This finishes the proof.
$\square$

## Comments (0)