TANT 13 - Unramified and totally ramified extensions

Hello there! These are notes for the thirteenth 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 discussed extensions of absolute values. We have seen in particular that for every valued field $(K,\phi)$ and every finite, separable extension $K \subseteq L$ there exist only a finite number of absolute values $\psi \colon L \to \mathbb{R}_{\geq 0}$ which extend $\phi$. Given such an absolute value $\psi$ we have also defined the ramification index $e(\psi \mid \phi)$ and the inertia index $f(\psi \mid \phi)$ which measure how "close" the two absolute values are, in two different ways.

In this lecture we will concentrate on extensions of complete, discretely valued fields which are either unramified, i.e. $e(\psi \mid \phi) = 1$ or totally ramified, i.e.$f(\psi \mid \phi) = 1$. We will prove that all the extensions of this kind have a specific shape, which will turn out to be very handy in the future.

Unramified extensions

The main theorem about finite, unramified extensions of complete and discretely valued fields is that these extensions correspond bijectively to finite, separable extensions of the residue field. This correspondence extends moreover to field homomorphisms, and thus allows us to compute Galois groups of unramified extensions rather easily.

 Theorem 1  Let $(K,\phi)$ be a complete, discretely valued field. Then the map $$\begin{align} \mathcal{F} \colon \{ (L,\psi) \mid (K,\phi) \subseteq (L,\psi) \ \text{finite and unramified} \} &\to \{ \mathfrak{K} \mid \kappa_{\phi} \subseteq \mathfrak{K} \ \text{finite and separable} \} \\ (L,\psi) &\mapsto \kappa_{\psi} \end{align}$$ is a bijection which preserves inclusions of fields. Moreover, for every couple of unramified extensions $(L_1,\psi_1)$ and $(L_2,\psi_2)$ of $(K,\phi)$ we have that the induced map $\operatorname{Hom}_K(L_1,L_2) \to \operatorname{Hom}_{\kappa_{\phi}}(\kappa_{\psi_1},\kappa_{\psi_2})$ is bijective.

 Proof  We construct first of all an inverse for the map $\mathcal{F}$. Let $\mathfrak{K}$ be a finite, separable extension of $\kappa_{\phi}$. Then, by the primitive element theorem, there exists $\alpha_0 \in \mathfrak{K}$ such that $\mathfrak{K} = \kappa_{\phi}(\alpha_0)$ and the minimal polynomial $f_{\alpha_0}(x) \in \kappa_{\phi}[x]$ is separable. Let $f(x) \in A_{\phi}[x]$ be any monic lift of $f_{\alpha_0}(x)$. Then $f(x)$ is irreducible by Hensel's lemma (see Definition 2 and Theorem 5 of the eleventh lecture) and thus $L := K[x]/(f(x))$ is a field. Since $(K,\phi)$ is complete we have a unique extension of $\psi$ to $L$, which we will call $\psi$ (see Theorem 1 of the previous lecture). Since $f_{\alpha_0}(x)$ is separable and irreducible we have that $f_{\alpha_0}'(\alpha_0) \neq 0$ and thus we can apply again Hensel's lemma to find $\alpha \in A_{\psi}$ such that $f(\alpha) = 0$ and $\overline{\alpha} = \alpha_0$. This implies that $L = K(\alpha)$ and that the residue field of $L$ is equal to $\mathfrak{K}$ and thus that $[L \colon K] = \deg(f) = [\mathfrak{K} \colon \kappa_{\phi}] = f(\psi \colon \phi)$. Since $(K,\phi)$ is complete we have that $[L \colon K] = e(\psi \colon \phi) f(\psi \colon \phi) = f(\psi \colon \phi)$, i.e. the extension $(K,\phi) \hookrightarrow (L,\psi)$ is unramified. Thus we defined a map $\mathcal{G}$ which associates to every finite, separable extension of $\kappa_{\phi}$ a finite and unramified extension of $(K,\phi)$, and we have seen that $\mathcal{F} \circ \mathcal{G}$ is the identity map (the residue field of $\mathcal{G}(\mathfrak{K})$ is indeed $\mathfrak{K}$).

Let now $(K,\phi) \subseteq (E,\xi)$ be another finite, unramified extension and suppose that $\kappa_{\xi} = \mathfrak{K}$. Then we have that $L \subseteq E$ because the polynomial $f(x)$, which is the minimal polynomial of $\alpha$ over $K$, has a simple zero in $\kappa_{\xi}$ and thus this simple zero can be lifted to a zero $\beta \in E$ of $f(x)$. But then we have that $\alpha = \beta$ by the uniqueness property of Hensel's lemma, and this implies indeed that $L = K(\alpha) \subseteq E$. Since by assumption $(K,\phi) \subseteq (E,\xi)$ is unramified we have that $[E \colon K] = [\mathfrak{K} \colon \kappa_{\phi}] = [L \colon K]$ which implies that $L = E$. Thus we have proved that there exists a unique finite, unramified extension of $(K,\phi)$ whose residue field equals $\mathfrak{K}$. This shows in particular that $\mathcal{G} \circ \mathcal{F}$ is the identity map, and thus that every finite, unramified extension of $(K,\phi)$ is separable.

Finally, the bijection on homomorphisms can again be deduced using Hensel's lemma. Indeed let $(L_1,\psi_1)$ and $(L_2,\psi_2)$ be two finite, unramified extensions of $(K,\phi)$ and let $\sigma \colon \kappa_{\psi_1} \to \kappa_{\psi_2}$ be a homeomorphism of fields which fixes $\kappa_{\phi}$. Since we know that $\kappa_{\phi} \subseteq \kappa_{\psi_1}$ is separable we can write $\kappa_{\psi_1} = \kappa_{\phi}(\alpha_0)$ and we have that $\sigma(\alpha_0) =: \beta_0$ is a simple root of the minimal polynomial $f_{\alpha_0}(x) \in \kappa_{\phi}[x]$. Thus we can lift $\alpha_0$ to $\alpha \in L_1$ and we have that $L_1 = K(\alpha)$ and we can lift $\beta_0$ to $\beta \in L_2$. Thus we can define a map of fields $\widetilde{\sigma} \colon L_1 \to L_2$ which fixes $K$ simply by setting $\widetilde{\sigma}(\alpha) = \beta$, and it is clear that this is a lift of $\sigma$. Moreover, this lift is unique by the uniqueness property of Hensel's lemma. Q.E.D.

This implies (by taking $L_1 = L_2 =: L$ in the previous theorem) that for every finite, unramified extension $(K,\phi) \hookrightarrow (L,\psi)$ (with $(K,\phi)$ complete and discretely valued) we have that $\operatorname{Aut}_K(L) \cong \operatorname{Aut}_{\kappa_{\phi}}(\kappa_{\psi})$ and thus $K \subseteq L$ is Galois if and only if $\kappa_{\phi} \subseteq \kappa_{\psi}$ is Galois and they have isomorphic Galois groups. This can be used to prove a nice characterization of finite, unramified extensions of non-Archimedean local fields, i.e. non-Archimedean complete fields whose residue field is finite.

 Corollary 2  Let $(K,\phi)$ be a non-Archimedean, local field. Then for every $n \in \mathbb{N}_{\geq 1}$ there exists a unique finite unramified extension $K \subseteq K_n$ of degree $n$. Moreover $K_n = K(\zeta)$ is a cyclic extension generated by a root of unity of order $q^n - 1$, where $q \in \mathbb{N}$ is the order of $\kappa_{\phi}$.

 Proof  Since $K$ is local we know that $\kappa_{\phi}$ is finite (see Theorem 14 of the ninth lecture) and thus we know that for every $n \in \mathbb{N}$ there exists a unique finite extension $\kappa_{\phi} \subseteq \mathfrak{K}_n$ of degree $n$. Moreover, this extension is Galois and $\operatorname{Gal}(\mathfrak{K}/\kappa_{\phi}) \cong \mathbb{Z}/n \mathbb{Z}$. Finally, if $q := \# \kappa_{\phi}$ we have that $\mathfrak{K} = \kappa_{\phi}(\zeta_0)$ with $\zeta_0^{q^n - 1} = 1$. Thus if $p$ is the characteristic of $\kappa_{\phi}$ we have that $q = p^r$ for some $r \in \mathbb{N}_{\geq 1}$ and the minimal polynomial of $\zeta_0$ is a factor of the reduction of the cyclotomic polynomial $\Phi_{q^n - 1}(x) \in \mathbb{Z}[x]$ modulo $p$, which is separable because $q^n -1$ is coprime to $p$ (see this note of Keith Conrad for a proof of all these results).

Using what we have written in the previous paragraph we know from Theorem 1 that $K$ has a unique finite unramified extension $K \subseteq K_n$ and we have that $\operatorname{Gal}(K_n / K) \cong \mathbb{Z} / n \mathbb{Z}$. Moreover, we can use Hensel's lemma to see that the minimal polynomial of $\zeta_0$ lifts to a polynomial $f(x) \in K[x]$ which is a factor of $\Phi_{q^n - 1}(x) \in \mathbb{Z}[x] \to K[x]$. Finally we know that $K_n = K(\zeta)$, where $\zeta$ is a lift of $\zeta_0$, and thus its an $m$-th root of unity in $K$. Q.E.D.

Totally ramified extensions

We have seen in the previous section that unramified extensions of a complete non-Archimedean field correspond bijectively to separable extensions of its residue field. In this section we will deal with totally ramified extensions, i.e. extensions where the residue field does not change. We will prove in particular that they are all simple, and generated by roots of a special kind of polynomials, called Eisenstein polynomials.

 Definition 3  Let $R$ be a ring and $I \subseteq R$ be an ideal. We say that a polynomial $f(x) = \sum_{j = 0}^n a_j x^j \in A[x]$ is Eisenstein with respect to $I$ if $a_0,\dots,a_{n - 1} \in I$, $a_n \notin I$ and $a_0 \notin I^2$. If $(K,\phi)$ is a non-Archimedean valued field we say that a polynomial $f(x) \in A_{\phi}[x]$ is an Eisenstein polynomial if it is Eisenstein with respect to $\mathfrak{m}_{\phi}$.

Eisenstein polynomials are nice because they give origin to totally ramified extensions, as we will see in Lemma 4. Moreover, every totally ramified extension is generated by a root of an Eisenstein polynomial, as we will see in Theorem 6.

 Lemma 4  Let $(K,\phi)$ be a complete, non-Archimedean field and let $f(x) \in A_{\phi}[x]$ be an Eisenstein polynomial. Then $f(x) \in K[x]$ is irreducible and the field extension $K \subseteq K[x]/(f(x))$ is totally ramified.

 Proof  Let's prove first of all that $f(x)$ is irreducible. Suppose that $f(x) = g(x) h(x)$ for two non-constant polynomials $g(x), h(x) \in K[x]$. Then we can apply Gauss' lemma (see the following Exercise 5) to see that $g(x), h(x) \in A_{\phi}[x]$. Thus we have that $\overline{f}(x) = \overline{g}(x) \overline{h}(x) \in \kappa_{\phi}[x]$ and by assumption we have that $\overline{f}(x) = \overline{a_n} x^n$ if $f(x) = \sum_{j = 0}^n a_j x^j$. Since $\kappa_{\phi}$ is a field this implies that $\overline{g}(x)$ and $\overline{h}(x)$ are monomials in $x$, and thus that the constant coefficients of $g(x)$ and $h(x)$ are in $\mathfrak{m}_{\phi}$. This implies that $a_0 \in \mathfrak{m}_{\phi}^2$ which contradicts the fact that $f(x)$ is Eisenstein.

We can now prove very easily that $L := K[x]/(f(x))$ is a totally ramified extension of $K$. Indeed we can apply Theorem 1 of the previous lecture to define an absolute value $\psi \colon L \to \mathbb{R}_{\geq 0}$ which extends $\phi$. We know now that $\kappa_{\psi} \cong \kappa_{\phi}[x]/(g(x))$ where $g(x)$ is a monic irreducible factor of $\overline{f}(x)$. However, we know also from the previous paragraph that $\overline{f}(x)$ is a monomial in $x$, which implies that $g(x) = x$ and thus that $\kappa_{\psi} = \kappa_{\phi}$ and $f(\psi \mid \phi) = 1$. Hence, the extension $K \subseteq L$ is totally ramified, which is what we wanted to prove. Q.E.D.

 Exercise 5  Let $(K,\phi)$ be a valued field, and let $f(x), g(x) \in K[x]$ be two monic polynomials. Prove that if $f(x) g(x) \in A_{\phi}[x]$ then $f(x) \in A_{\phi}[x]$ and $g(x) \in A_{\phi}[x]$.

 Theorem 6  Let $(K,\phi) \subseteq (L,\psi)$ be a finite, totally ramified extension of non-Archimedean, complete and discretely valued fields. Then $L = K(\pi_L)$, where $\pi_L \in L$ is any uniformizer and the minimal polynomial of $\pi_L$ is an Eisenstein polynomial. Moreover, if this extension is tamely ramified we can find a uniformizer $\pi_K \in K$ such that $L = K(\sqrt[e]{\pi_K})$, where $e = [L \colon K]$.

 Proof  By definition of $\pi_L$ we have that $\phi(L^{\times}) = \phi(\pi_L)^{\mathbb{Z}}$, which implies that $e(K(\pi_L) \mid K) = e(L \mid K) = [L \colon K]$. Moreover, we know that $[K(\pi_L) \colon K] =  e(K(\pi_L) \mid K) f(K(\pi_L) \mid K)$ because everything is complete, and thus we have that $e(K(\pi_L) \mid K) = [L \colon K] = [K(\pi_L) \colon K]$ and $f(K(\pi_L) \mid K) = 1$ which implies that $L = K(\pi_L)$ is a totally ramified extension.

Let now $f(x) \in K[x]$ be the minimal polynomial of $\pi_L$ and let $E \supseteq L \supseteq K$ be the splitting field of $f(x)$. Let moreover $\alpha_1,\dots,\alpha_n \in E$ be the roots of $f(x)$. Then we know that $\alpha_j = \sigma_j(\pi_L)$ for some $\sigma_j \in \operatorname{Gal}(E/K)$. Thus we can apply Exercise 7 to see that $\xi(\alpha) = \xi(\pi_L) < 1$ where $\xi \colon E \to \mathbb{R}_{\geq 0}$ is the unique absolute value which extends $\psi$ (and thus $\phi$). Hence if $e = [L \colon K]$ and $f(x) = \sum_{j = 0}^e a_j x^j$ we can use Viète's formulas $$a_{e - k} = (-1)^k \, a_e \, \sum_{1\le i_1 < i_2 < \cdots < i_k\le e} \alpha_{i_1}\, \alpha_{i_2} \, \cdots \, \alpha_{i_k}$$ to see that $\phi(a_{e - k}) = \xi(a_{e - k}) < 1$ for all $k \in \{ 1,\dots,e \}$. Moreover we have that $a_0 = (-1)^e \operatorname{N}_{L/K}(\pi_L)$ which implies that $\phi(a_0) = \psi(\pi_L)^e$ and thus that $\mathfrak{m}_{\phi} = a_0 \, A_{\phi}$. This implies that $a_0 \notin \mathfrak{m}_{\phi}^2$ and thus that $f(x)$ is an Eisenstein polynomial.

Let now $\pi_K' \in K$ be any uniformizer and observe that $\pi_L^e = u \pi_K'$ for some unit $u \in A_{L}^{\times}$. Since $K \subseteq L$ is totally ramified we have that $\kappa_{\psi} = \kappa_{\phi}$ and thus there exists $v \in A_{K}^{\times}$ with $\overline{v} = \overline{u}$. Let now $\beta := v \pi_K' / \pi_L^e \in A_L$ and consider the polynomial $g(x) := x^e - \beta \in L[x]$. Since $\overline{\beta} = 1 \in \kappa_{\psi}$ we have that $g(x)$ has a root in $\kappa_{\psi}$. Suppose now that $K \subseteq L$ is tamely ramified, i.e. that the characteristic of $\kappa_{\phi}$ does not divide $e = e(L \mid K) = [L \colon K]$. Then the derivative $g'(x) = e x^{e - 1}$ does not vanish in $\overline{\beta} = 1$ and thus we can apply Hensel's lemma (see Theorem 5 of the eleventh lecture) to find $\gamma \in A_{L}^{\times}$ with $g(\gamma) = 0$. Thus we have that $$L = K(\pi_L) = K(\gamma \, \pi_L) = K(\sqrt[e]{v \pi_K'})$$ and thus we can take $\pi_K := v \pi_K'$ to be the new uniformizer such that $L = K(\sqrt[e]{\pi_K})$. Q.E.D.

p-adic fields

We have seen in the previous paragraphs that unramified extensions of a complete, discretely valued field can be classified by the extensions of its residue field, and that totally ramified extensions are obtained by adjoining roots of Eisenstein polynomials.
What about general extensions? Well, it turns out that they can be easily split into unramified and totally ramified extensions!

 Lemma 7  Let $(K,\phi) \hookrightarrow (L,\psi)$ be an extension of complete, discretely valued fields, and suppose that the extension $\kappa_{\phi} \subseteq \kappa_{\psi}$ is separable. Then there exists a unique sub-extension $(K,\phi) \hookrightarrow (T,\xi) \hookrightarrow (L,\psi)$ such that $(K,\phi) \hookrightarrow (T,\xi)$ is unramified and $(T,\xi) \hookrightarrow (L,\psi)$ is totally ramified.

 Proof  We know by Theorem 1 that there exists a finite, unramified extension $(K,\phi) \hookrightarrow (T,\xi)$ which corresponds to the separable extension of residue fields $\kappa_{\phi} \subseteq \kappa_{\psi}$. From the proof of the theorem we see that $T = K(\alpha)$ where $\alpha$ is a root of a lift of the minimal polynomial of the primitive element generating the extension $\kappa_{\phi} \subseteq \kappa_{\psi}$. Thus $(T,\xi) \hookrightarrow (L,\psi)$, and this extension is clearly totally ramified because $\kappa_{\xi} = \kappa_{\psi}$. Q.E.D.

We call the valued field $(T,\xi)$ that we found in the previous lemma the inertia field of an extension $(K,\phi) \hookrightarrow (L,\psi)$.
This has an important corollary about finite extensions of the field $\mathbb{Q}_p$ of $p$-adic numbers, which are called $p$-adic fields.

 Definition 8  A $p$-adic field $K$ is a finite extension of $\mathbb{Q}_p$. We will denote by $\mathcal{O}_K$ and $\mathfrak{p}_K$ respectively the closed and open unit balls relative to the unique extension of the $p$-adic absolute value from $\mathbb{Q}_p$ to $K$.

 Corollary 9  Let $(K,\phi)$ be a non-Archimedean local field and $d \in \mathbb{N}_{\geq 1}$. Then $K$ has only finitely many extensions of degree $d$, up to isomorphism. In particular, there exist only finitely many $p$-adic fields of a given degree.

 Proof  Let $n \in \mathbb{N}_{\geq 1}$ and consider the map $$C := \overbrace{\mathfrak{m}_{\phi} \times \cdots \times \mathfrak{m}_{\phi}}^{(n - 1) \, \text{times}} \times (\mathfrak{m}_{\phi} \setminus \mathfrak{m}_{\phi}^2)$$ which is a compact metric space when considered with any product metric. Then the map $$\begin{align} \varphi \colon C &\to K[x] \\ (a_{n - 1},\dots,a_1,a_0) &\mapsto x^n + \sum_{j = 0}^{n - 1} a_j x^{j} \end{align}$$ is a bijection between $C$ and the subset of Eisenstein polynomials in $K[x]$.

The key fact here is that for every $\mathbf{v} \in C$ there exists $\varepsilon \in \mathbb{R}_{> 0}$ such that if $\mathbf{w} \in B_{\varepsilon}(\mathbf{v})$ then $K[x]/(\varphi(\mathbf{v})(x)) \cong K[x]/(\varphi(\mathbf{w})(x))$. This is a consequence of Krasner's lemma (see Exercise 10 and Exercise 11). Since $C$ is compact we can cover it by a finite number of these balls, and this means that the splitting fields of Eisenstein polynomials with coefficients in $K$ are only a finite number (up to isomorphism).

Let now $L$ be any extension of $K$ of degree $d$. Lemma 7 tells us that there exists $T \subseteq L$ such that $K \subseteq T$ is unramified and $T \subseteq L$ is totally ramified. But then we have only a finite number of possibilities for $T$ by Corollary 2, and (when we fix $T$) we have only a finite number of possibilities for $L$ because it is defined by an Eisenstein polynomial with coefficients in $T$. Thus, we have only a finite number of possibilities for $L$, up to isomorphism. Q.E.D.

 Exercise 10   Let $(K,\phi) \hookrightarrow (L,\psi)$ be a Galois extension of non-Archimedean valued fields, and suppose that $(K,\phi)$ is complete. Let $\alpha, \beta \in L$ and suppose that $\psi(\alpha - \beta) < \psi(\alpha - \sigma(\alpha))$ for all $\sigma \in \operatorname{Gal}(L/K)$ such that $\sigma(\alpha) \neq \alpha$. Prove that:

1.  $\psi(\sigma(\alpha) - \beta) = \psi(\alpha - \beta)$ for all $\sigma \in \operatorname{Gal}(L/K)$ such that $\sigma(\beta) = \beta$. To do so use the fact that $\psi$ and $\psi_{\sigma}$ both define two absolute values on $K(\alpha,\beta)$ which is a finite extension of the complete field $(K,\phi)$;
2. $\psi(\sigma(\alpha) - \alpha) \leq \psi(\alpha - \beta)$ for all $\sigma \in \operatorname{Gal}(L/K)$ such that $\sigma(\beta) = \beta$. To do so use that $\psi$ is non-Archimedean;
3. $K(\alpha) \subseteq K(\beta)$ using the fact that $K \subseteq L$ is a Galois extension.

 Exercise 11   Let $(K,\phi)$ be a non-Archimedean, complete valued field and let $n \in \mathbb{N}$. Consider the map  $$\begin{align} \varphi \colon K^{n} &\to K[x] \\ (a_{n - 1},\dots,a_1,a_0) &\mapsto x^n + \sum_{j = 0}^{n - 1} a_j x^{j} \end{align}$$ and consider $K^n$ as a metric space with any product metric. Let moreover $\mathbf{v} \in K^n$ with $\varphi(\mathbf{v})$ irreducible. Prove that there exists $\varepsilon \in \mathbb{R}_{> 0}$ such that for every $\mathbf{w} \in B_{\varepsilon}(\mathbf{v})$ we have that $\varphi(\mathbf{w})$ is irreducible and $K[x]/(\varphi(\mathbf{v})(x)) \cong K[x]/(\varphi(\mathbf{w})(x))$. (Hint: take $E$ to be the Galois closure of the splitting field of $\varphi(\mathbf{v})(x)$ and $\varphi(\mathbf{w})(x)$ and apply the previous exercise to the extension $K \subseteq E$ ).

Conclusions and references

In this lecture we managed to:

• prove a bijective correspondence between finite, unramified extensions of a complete, discretely valued field and finite, separable extensions of its residue field;
• prove that every finite, totally ramified extension of a complete, discretely valued field is generated by a root of an Eisenstein polynomial;
• prove that every local field has only finitely many extensions of a given degree.

References for this lecture include:

• Section 4 of these notes by Peter Stevenhagen;
• Section II.7 of the book "Algebraic number theory" by Jürgen Neukirch;
• Section 4.6 of the book "Number theory" by Helmut Koch.