TANT 14 - Galois extensions of complete, valued fields

Hello there! These are notes for the fourteenth class of the course "Topics in algebra and number theory" held in Block 4 of the academic year 2017/18 at the University of Copenhagen.

In the previous lecture we have seen how every extension $(K,\phi) \hookrightarrow (L,\psi)$ of complete, non-Archimedean, discretely valued fields can be split into an unramified and a totally ramified extension. We have seen moreover that unramified extensions correspond bijectively to separable extensions of the residue field and totally ramified extensions correspond to Eisenstein polynomials.

In this lecture we will see how the presence of the absolute values allows us to define a filtration on the Galois group of any Galois extension $(K,\phi) \hookrightarrow (L,\psi)$ of complete, non-Archimedean, discretely valued fields. This filtration will be really useful in the following lectures, to prove the theorem of Kronecker and Weber.

The ramification filtration

Let $(K,\phi)$ be a complete, discretely valued field, and let $K \subseteq L$ be an algebraic extension. Then there exists a unique absolute value $\psi \colon L \to \mathbb{R}_{\geq 0}$ which extends $\phi$. Indeed we can use Theorem 1 of the twelfth lecture to extend $\phi$ to every Galois sub-extension $K \subseteq E \subseteq L$ such that $K \subseteq E$ is finite, and then we can recall from Galois theory that $$L = \varinjlim_{\substack{K \subseteq E \subseteq L \\ [E \colon K] < +\infty}} \, E$$ and thus all these extensions "glue together" to a unique absolute value on $L$. Observe moreover that if $K \subseteq L$ is finite then $(L,\psi)$ is also discretely valued.

 Definition 1  Let $(K,\phi) \hookrightarrow (L,\psi)$ be a finite, Galois extension of complete, discretely valued fields, and let $\pi_L \in L$ be a uniformizer. Then we define the ramification filtration $$\dots \subseteq G_{n + 1}(L/K) \subseteq G_n(L/K) \subseteq \dots \subseteq G_0(L/K) \subseteq G_{-1}(L/K) = \operatorname{Gal}(L/K)$$ by setting $G_n(L/K) := \{ \sigma \in \operatorname{Gal}(L/K) \mid \psi(x - \sigma(x)) \leq \psi(\pi_L)^{n + 1} \ \text{for all} \ x \in A_{\psi} \}$. We call $I(L/K) := G_0(L/K)$ the inertia group of the extension $K \subseteq L$ and $I^w(L/K) := G_1(L/K)$ the wild inertia group of $K \subseteq L$.

 Theorem 2  Let $(K,\phi) \hookrightarrow (L,\psi)$ be a finite, Galois extension of complete, discretely valued fields and suppose that $\kappa_{\phi} \subseteq \kappa_{\psi}$ is separable. Then $\kappa_{\phi} \subseteq \kappa_{\psi}$ is also Galois and there is a surjective map $\rho \colon \operatorname{Gal}(L/K) \twoheadrightarrow \operatorname{Gal}(\kappa_{\psi}/\kappa_{\phi})$. Moreover, $\ker(\rho) = I(L/K)$ and $L^{I(L/K)}$ is the inertia field of the extension $(K,\phi) \hookrightarrow (L,\psi)$.

 Proof  Let $\sigma \in \operatorname{Gal}(L/K)$ and observe that $\psi(\sigma(\alpha)) = \psi(\alpha)$ for every $\alpha \in L$ (see Exercise 3). Thus $\sigma(A_{\psi}) \subseteq A_{\psi}$ and $\sigma(\mathfrak{m}_{\psi}) \subseteq \mathfrak{m}_{\psi}$, which implies that $\sigma$ induces an automorphism $\overline{\sigma} \in \operatorname{Aut}_{\kappa_{\phi}}(\kappa_{\psi})$. Thus if we set $\rho(\sigma) := \overline{\sigma}$ we get a map $\rho \colon  \operatorname{Gal}(L/K) \to \operatorname{Aut}_{\kappa_{\phi}}(\kappa_{\psi})$. Observe that $\rho(\sigma) = \operatorname{Id}_{\kappa_{\psi}}$ if and only if $\sigma(\alpha) \in \mathfrak{m}_{\psi}$ for every $\alpha \in A_{\psi}$, and this is true if and only if $\sigma \in I(L/K)$. Thus, $\ker(\rho) = I(L/K)$.

We can prove at once that $\rho$ is surjective and that $\kappa_{\phi} \subseteq \kappa_{\psi}$ is Galois if we prove that $\kappa_{\phi}$ is the fixed field of $\rho(\operatorname{Gal}(L/K)) \subseteq \operatorname{Aut}_{\kappa_{\phi}}(\kappa_{\psi})$. Clearly $\kappa_{\phi} \subseteq \kappa_{\psi}^{\rho(\operatorname{Gal}(L/K))}$.

Let now $\alpha \in A_{\psi}$ be such that $\overline{\alpha} \in \kappa_{\psi}^{\rho(\operatorname{Gal}(L/K))}$. Suppose first of all that $\kappa_{\phi}$ (and thus $\kappa_{\psi}$ ) have characteristic zero, and let $$\beta := \frac{1}{[L \colon K]} \sum_{\sigma \in \operatorname{Gal}(L/K)} \sigma(\alpha) \in L^{\operatorname{Gal}(L/K)} = K$$ be the Galois-invariant element obtained by averaging over all the elements of the Galois group. Since $\kappa_{\phi}$ has characteristic zero we have that $\beta \in A_{\psi} \cap K = A_{\phi}$ and that $\overline{\beta} = \overline{\alpha} \in \kappa_{\phi}$ because by hypothesis $\overline{\alpha} \in \kappa_{\psi}^{\rho(\operatorname{Gal}(L/K))}$. Thus $\kappa_{\psi}^{\rho(\operatorname{Gal}(L/K))} = \kappa_{\phi}$ as we wanted to prove.

If the residue field $\kappa_{\phi}$ has positive characteristic $p > 0$ the proof proceeds along the same lines, but we need to be slightly more careful. In particular, let $S \subseteq \operatorname{Gal}(L/K)$ be a $p$-Sylow subgroup, and let $\Gamma \subseteq G$ be a set of left coset representatives for $S$. Then we can define $$\beta := \frac{1}{\# \Gamma} \sum_{\sigma \in \Gamma} \sigma\left( \prod_{\tau \in S} \tau(\alpha) \right) \in L^{\operatorname{Gal}(L/K)} = K$$ in a way similar to what we did in the previous paragraph. Observe now that since $p \nmid [\operatorname{Gal}(L/K) \colon S] = \# \Gamma$ we have that $\beta \in A_{\psi} \cap K = A_{\phi}$ because $\mathbb{Z} \setminus p \mathbb{Z} \subseteq A_{\phi}^{\times}$ as the characteristic of $\kappa_{\phi}$ is $p$. Observe now that $$\overline{\beta} := \frac{1}{\# \Gamma} \sum_{\overline{\sigma} \in \rho(\Gamma)} \overline{\sigma}\left( \prod_{\overline{\tau} \in \rho(S)} \overline{\tau}(\overline{\alpha}) \right) = \frac{1}{\# \Gamma} \sum_{\overline{\sigma} \in \rho(\Gamma)} \overline{\sigma}(\overline{\alpha}^{\# S}) = \frac{1}{\# \Gamma} \sum_{\overline{\sigma} \in \rho(\Gamma)} \overline{\alpha}^{\# S} = \overline{\alpha}^{\# S} \in \kappa_{\phi}$$ and since the extension $\kappa_{\phi} \subseteq \kappa_{\psi}$ is separable this implies that $\overline{\alpha} \in \kappa_{\phi}$ because $\# S = p^r$ is a power of $p$ and the polynomial $x^{p^r} - \overline{\beta} \in \kappa_{\phi}[x]$ is not separable. This shows that $\kappa_{\psi}^{\rho(\operatorname{Gal}(L/K))} = \kappa_{\phi}$ and concludes the proof of the fact that $\rho$ is surjective and $\kappa_{\phi} \subseteq \kappa_{\psi}$ is Galois.

To conclude let $T := L^{\ker(\rho)} = L^{I(L/K)}$ and let $\xi$ be the unique absolute value induced on it. What we have proved shows that $$[T \colon K] = [\operatorname{Gal}(L/K) \colon I(L/K)] = \# \operatorname{Gal}(\kappa_{\psi}/\kappa_{\phi}) = [\kappa_{\psi} \colon \kappa_{\phi}]$$ hence $[T \colon K] = [\kappa_{\xi} \colon \kappa_{\phi}]$. Thus the extension $K \subseteq T$ is unramified, and given the uniqueness proved in Lemma 7 of the previous lecture we have that $T$ is the inertia field of the extension $(K,\phi) \hookrightarrow (L,\psi)$.  Q.E.D.

 Exercise 3  Let $(K,\phi)$ be a valued field, and let $\sigma \colon K \to K$ be an automorphism of fields. Prove that the map $\phi_{\sigma} \colon K \to \mathbb{R}_{\geq 0}$ defined by $\phi_{\sigma}(x) := \psi(\sigma(x))$ is an absolute value. Prove moreover that $\sigma$ is continuous with respect to the topology induced by $\phi$ if and only if $\phi_{\sigma}$ and $\phi$ are equivalent absolute values. Use this to prove that for every finite Galois extension of valued fields $(K,\phi) \hookrightarrow (L,\psi)$ where $(K,\phi)$ is complete and discretely valued all the automorphisms $\sigma \in \operatorname{Gal}(L/K)$ are continuous. To prove this last part use Schmidt's theorem on multi-complete fields (see Theorem 1.19 of these notes by Pete L. Clark) and the theorem of Artin and Schreier (see Theorem 3.1 of these notes by Keith Conrad).

We also have more information on the higher ramification groups, and in particular on the wild inertia subgroup $I^w(L/K)$.

 Theorem 4  Let $(K,\phi) \hookrightarrow (L,\psi)$ be a finite, Galois extension of complete, discretely valued fields and let $\pi_L \in L$ be a uniformizer. Define moreover $U_L^{(n)} := \{ x \in A_{\psi}^{\times} \mid x \equiv 1 \, \text{mod} \, \mathfrak{m}_{\psi}^{n} \}$. Then for all $n \in \mathbb{N}_{\geq 1}$ the map $$\begin{align} \chi_n \colon G_{n}(L/K) &\to \frac{U_L^{(n)}}{U_L^{(n + 1)}} \\[1ex] \sigma &\mapsto \frac{\sigma(\pi_L)}{\pi_L} \end{align}$$ is a group homomorphism with kernel $G_{n + 1}(L/K)$. Moreover, $\chi_n$ does not depend on the choice of $\pi_L$.

 Proof  Proving that $\chi_n$ does not depend on the choice of $\pi_L$ is easy. Indeed if $u \in A_{\psi}^{\times}$ and $\sigma \in G_n(L/K)$ we have that $$\psi\left( \frac{\sigma(u)}{u} - 1 \right) = \psi(\sigma(u) - u) \leq \psi(\pi_L)^{n + 1}$$ and thus $\sigma(u)/u \in U_L^{(n+1)}$. Now if $\pi_L' \in L$ is another uniformizer, $\sigma \in G_n(L/K)$ we have that $$\frac{\sigma(\pi_L')}{\pi_L'} = \frac{\sigma(u)}{u} \frac{\sigma(\pi_L)}{\pi_L} = \frac{\sigma(\pi_L)}{\pi_L} \in \frac{U_L^{(n)}}{U_L^{(n+1)}}$$ where $u = \pi_L' / \pi_L \in A_{\psi}^{\times}$.
Recall now that for every $\sigma \in \operatorname{Gal}(L/K)$ we have that $\psi \circ \sigma = \psi$ and thus in particular for every uniformizer $\pi_L \in \mathfrak{m}_{\psi}$ we have that $\sigma(\pi_L)$ is another uniformizer. This implies that for every $\sigma, \tau \in G_{n}(L/K)$ we have that $$\chi_n(\tau \circ \sigma) = \frac{\tau(\sigma(\pi_L))}{\pi_L} = \frac{\tau(\sigma(\pi_L))}{\sigma(\pi_L)} \frac{\sigma(\pi_L)}{\pi_L} = \chi_{n}(\tau) \chi_n(\sigma)$$ which proves that $\chi_n$ is indeed a group homomorphism.

To conclude let $\sigma \in G_n(L/K)$ and observe that $\sigma \in \ker(\chi_n)$ if and only if $\sigma(\pi_L)/\pi_L \in U_L^{(n+1)}$, which is true if and only if $\sigma(\pi_L) - \pi_L \in \mathfrak{m}_{\psi}^{n+2}$, and clearly this is true if $\sigma \in G_{n + 1}(L/K)$, so $G_{n + 1}(L/K) \subseteq \ker(\chi_n)$. To prove the converse suppose that $\sigma(\pi_L) - \pi_L \in \mathfrak{m}_{\psi}^{n+2}$ and observe that $A_{\psi} = A_{\xi}[\pi_L]$ where $(T,\xi)$ is the inertia field of the extension $(K,\phi) \hookrightarrow (L,\psi)$ (combine Theorem 6 of the previous lecture and the proof of Theorem 10 of the twelfth lecture). Thus we can write every $x \in A_{\psi}$ as $x = \sum_{k = 0}^n a_k \pi_L^k$ with $a_k \in A_{\xi}$. Since $\sigma \in G_0(L/K)$ we have that $\sigma$ is trivial on $T = L^{G_0(L/K)}$ and thus $$\sigma(x) - x = \sum_{k = 0}^n a_k (\sigma(\pi_L)^k - x) \in \mathfrak{m}_{\psi}^{n + 2}$$   which implies that $\sigma \in G_{n + 1}(L/K)$. Q.E.D.

Observe that the map $U_L^{(n)} \to \kappa_{\psi}$ defined by $1 + x \pi_L^n \mapsto \overline{x}$ induces isomorphisms $$\left( \dagger \right) \frac{U_L^{(n)}}{U_L^{(n+1)}} \cong \begin{cases} (\kappa_{\psi}^{\times},\cdot), \ \text{if} \ n = 0 \\ (\kappa_{\psi},+), \ \text{if} \ n \geq 1 \end{cases}$$ and these isomorphisms allow us to prove something about the wild inertia group $I^w(L/K)$.

 Corollary 5  Let $(K,\phi) \hookrightarrow (L,\psi)$ be a finite, Galois extension of complete, discretely valued fields. Then $I(L/K) / I^w(L/K)$ is a cyclic group of order coprime to the characteristic of $\kappa_{\phi}$ and if $\operatorname{Gal}(L/K)$ is abelian we have an embedding $I(L/K) / I^w(L/K) \subseteq \kappa_{\phi}^{\times}$. Moreover, $K \subseteq L^{I^w(L/K)} \subseteq L$ is the largest sub-extension of $L$ which is tamely ramified over $K$.

 Proof  Using the isomorphisms $\left( \dagger \right)$ and the maps $\chi_n$ we get embeddings $$\frac{I(L/K)}{I^w(L/K)} \hookrightarrow \kappa_{\psi}^{\times} \qquad \text{and} \qquad \frac{G_n(L/K)}{G_{n+1}(L/K)} \hookrightarrow \kappa_{\psi}$$ which implies that $I(L/K) / I^w(L/K)$ is a cyclic group, because it is a finite subgroup of the group of invertible elements of a field (see this question on Math Stackexchange).
Suppose now that the characteristic of $\kappa_{\phi}$ is $p > 0$. Then the group of $p$-th roots of unity $\mu_p(\kappa_{\phi})$ is trivial, because if  $x \in \kappa_{\phi}$ is such that $x^p = 1$. Then $x^p - 1 = (x - 1)^p = 0$ then $x = 1$. Thus, the order of $I(L/K)/I^w(L/K)$ is coprime to the characteristic of $\kappa_{\psi}$ because if this was not true we would have a non-trivial $p$-th root of unity in $\kappa_{\psi}$.
Suppose now that $\operatorname{Gal}(L/K)$ is abelian. Then for every $\sigma \in \operatorname{Gal}(L/K)$ and $\tau \in I(L/K)$ we have that $$\sigma(\chi_0(\tau)) = \sigma\left( \frac{\tau(\pi_L)}{\pi_L} \right) = \frac{\sigma(\tau(\pi_L))}{\sigma(\pi_L)} = \frac{\tau(\sigma(\pi_L))}{\sigma(\pi_L)} = \chi_0(\tau)$$ and thus $\chi_0$ induces an embedding $I(L/K) / I^w(L/K) \subseteq (\kappa_{\psi}^{\times})^{\operatorname{Gal}(\kappa_{\psi}/\kappa_{\phi})} = \kappa_{\phi}^{\times}$.

Suppose now that $\kappa_{\phi}$ (and thus $\kappa_{\psi}$ ) have characteristic zero. Then the embeddings $G_{n}(L/K)/G_{n+1}(L/K) \hookrightarrow \kappa_{\psi}$ imply that $G_n(L/K) = G_{n+1}(L/K)$ for all $n \geq 1$ because the additive group $\kappa_{\psi}$ does not contain any element of finite order. Then we can use Exercise 6 to see that actually $G_n(L/K)$ is trivial for all $n \geq 1$. Thus $L^{I^w(L/K)} = L$, and this is indeed the largest sub-extension of $L$ which is tamely ramified over $K$, because in this case every extension is tamely ramified.
Suppose instead that the characteristic of $\kappa_{\phi}$ is a prime $p > 0$. Then every element in $G_{n}(L/K)/G_{n+1}(L/K) \hookrightarrow \kappa_{\psi}$ has order $p$, which implies that $G_{n}(L/K)$ is a $p$-group for all $n \geq 1$ (use again Exercise 6). Thus we have that the extension $L^{I^w(L/K)} \subseteq L$ is totally ramified of degree equal to $\# I^w(L/K)$, which is a power of $p$, and the extension $L^{I(L/K)} \subseteq L^{I^w(L/K)}$ is totally ramified of degree equal to $\# I(L/K)/I^w(L/K)$ which, by the previous paragraph, is coprime to $p$. Thus $L^{I^w(L/K)}$ is indeed the maximal tamely ramified sub-extension of $K \subseteq L$. Q.E.D.

 Exercise 6  Let $(K,\phi) \hookrightarrow (L,\psi)$ be a finite, Galois extension of complete, discretely valued fields. Prove that $G_n(L/K)$ is trivial for $n$ big enough. (Hint: prove first of all by contradiction that the ramification filtration stabilizes, i.e. $G_n(L/K) = G_{n+1}(L/K)$ for $n$ big enough).

Conclusions and references

In this lecture, we managed to:

• define the ramification filtration of a finite Galois extension of complete, discretely valued fields;
• relate the first group in this filtration (which is called "inertia group") to the Galois extension of residue fields;
• relate the second group in this filtration to the wild ramification of the extension.

References for this lecture include:

• Section 5 of these notes by Peter Stevenhagen;
• Pages 37-41 of these notes by Peter Bruin and Arno Kret;
• Section II.10 of the book "Algebraic number theory" by Jürgen Neukirch.