TANT 15 - Abhyankar's lemma and local Kronecker-Weber

Hello there! These are notes for the fifteenth 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 defined the ramification filtration of a finite, Galois extension of complete, discretely valued fields. This filtration gives us a way of studying the Galois group of any extension of complete, discretely valued fields. In this lecture we will concentrate on Galois extensions of a local field that are abelian, i.e. the Galois group is commutative. In particular we will see that every (tamely ramified) abelian extension of $\mathbb{Q}_{p}$ is contained in a cyclotomic extension.

In order to do so, we will need a lemma which will enable us to kill the ramification in an extension of valued fields after changing the base field: this is the so-called Abhyankhar's lemma.

Abhyankhar's lemma

Let's start with a preliminary result on the composition of unramified extensions.

 Lemma 1  Consider the diagram of discretely valued fields on the right, where $K \subseteq L$ is finite and unramified. Then $E \subseteq E L$ is also finite and unramified.

 Proof  We can assume that everything is complete, because completing preserves finite extensions (see Exercise 5 of the twelfth lecture) and preserves the value groups when the fields are non-Archimedean (see Exercise 2 of the ninth lecture). Moreover, we can assume that $K = L \cap E$ because the extension $L \cap E \subseteq L$ is also unramified.

Now, since everything is complete we can apply Theorem 1 of the thirteenth lecture to see that there exists a polynomial $f(x) \in K[x]$ such that its reduction $\overline{f}(x) \in \kappa_{K}[x]$ is separable and $L \cong K[x]/(f(x))$. Moreover, since $K = L \cap E$ we have that $EL \cong E[x]/(f(x))$ and this concludes the proof because the polynomial $\overline{f}(x) \in \kappa_{E}[x]$ is still separable. Q.E.D.

 Remark 2  In the proof of the previous lemma we have introduced new notation. Namely, if $(K,\phi)$ is a non-Archimedean valued field sometimes we will omit $\phi$ and we will write $A_K$ for the valuation ring, $\mathfrak{m}_K$ for the maximal ideal and $\kappa_{K}$ for the residue field.

Using the previous lemma we can prove a powerful result on tamely ramified extensions, which goes under the name of Abhyankar's lemma.

 Theorem 2  Let $(K,\phi)$ be a discretely valued field and let $K \subseteq L \subseteq M$ and $K \subseteq E \subseteq M$ be finite field extensions. Suppose that $K \subseteq L$ is tamely ramified and that $e(L \mid K)$ divides $e(E \mid K)$. Then $E \subseteq L E$ is unramified.

 Proof   Again as before, we may assume that everything is complete. Let $L_0 \subseteq L$ be the inertia field of the extension $K \subseteq L$ (see Lemma 7 of the thirteenth lecture). Then $K \subseteq L_0$ is unramified (i.e. $e(L_0 \mid K) = 1$ ), and thus we can apply the previous lemma to see that $E \subseteq L_0 E$ is unramified, i.e. $e(L_0 E \mid E) = 1$. Thus we can use the multiplicativity of the ramification and inertia indexes (see Exercise 3) to see that $E \subseteq L E$ is unramified if and only if $L_0 E \subseteq L E$ is unramified.

Observe now that the extension $L_0 \subseteq L$ is tamely and totally ramified, because $L_0$ is the inertia field of the extension $K \subseteq L$. Thus we can apply Theorem 6 of the thirteenth lecture to see that $L = L_0(\sqrt[e]{\pi_{L_0}})$ for some uniformizer $\pi_{L_0} \in L_0$, where $e = e(L \mid K) = e(L \mid L_0) = [L \colon L_0]$. Let now $\pi_{L_0 E} \in L E$ be any uniformizer and recall that $\pi_{L_0 E}^{e(L_0 E \, \mid \, L_0)}/\pi_{L_0} \in A_{L_0}^{\times}$. We can now use the fact that $e = e(L \mid K)$ divides $e(E \mid K) = e(L_0 E \mid L_0)$ to write $e(L_0 E \mid L_0) = e \, e'$ and get that $$\frac{\pi_{L_0}}{\pi_{L_0 E}^{e(L_0 E \, \mid \, L_0)}} = \left( \frac{\sqrt[e]{\pi_{L_0}}}{\pi_{L_0 E}^{e'}} \right)^e = u \in A_{L_0}^{\times}$$ which implies that $L E = L_0 E(\sqrt[e]{\pi_{L_0}}) = L_0 E(\sqrt[e]{u})$ for $u \in A_{L_0}^{\times}$. Let now $f(x) := x^e - u \in A_{L_0}[x]$ and observe that $\overline{f}(x) \in \kappa_{L_0}[x]$ is separable because $\overline{u} \neq 0$ and the characteristic of $\kappa_{L_0}$ equals the characteristic of $\kappa_{K}$, which is coprime to $e$ by hypothesis. Thus we may invoke again Theorem 1 of the thirteenth lecture to see that $L_0 E \subseteq L E$ is unramified and conclude. Q.E.D.

 Exercise 3  Prove that the ramification and the inertia index are multiplicative. More precisely, if $K \subseteq L \subseteq M$ are extensions of non-Archimedean valued fields, we have that $e(M \mid K) = e(M \mid L) e(L \mid K)$ and $f(M \mid K) = f(M \mid L) f(L \mid K)$.

The (local) theorem of Kronecker and Weber

Abhyankar's lemma allows us to "kill the ramification" of any tamely ramified extension $K \subseteq L$ if we extend the base field $K$ to a field $E$ whose ramification index $e(E \mid K)$ is a multiple of $e(L \mid K)$. This allows us immediately to prove that every tamely ramified abelian extension is cyclotomic, i.e. it is contained in $\mathbb{Q}_p(\zeta_n)$ for some $n \in \mathbb{N}$. Before proving this, let's have a look at the various types of cyclotomic extensions. First of all, we can see how many roots of unity are already contained in $\mathbb{Q}_p$.

 Exercise 4  Prove that the group of roots of unity $\mu_n(\mathbb{Q}_p) \leq \mathbb{Q}_p^{\times}$ is a cyclic group of order equal to $\gcd(n,p-1)$ if $p \geq 3$ and of order equal to $\gcd(n,2)$ if $p = 2$. (Hint: prove a similar statement for $\mu_n(\mathbb{F}_p)$ and use Hensel's lemma when $p \nmid n$. If $p \mid n$ you need to do a little more... See for example Theorem 3.1 of these notes by Keith Conrad).

 Example 5  Let $m \in \mathbb{N}$ and suppose that $p \nmid m$. Then the reduction of the $m$-th cyclotomic polynomial $\Phi_m(x)$ modulo $p$ is separable, and thus we can apply Theorem 1 of the thirteenth lecture to see that the extension $\mathbb{Q}_p \subseteq \mathbb{Q}_p(\zeta_m)$ is unramified. Moreover, if $d = [\mathbb{Q}_p(\zeta_m) \colon \mathbb{Q}_p]$ we know from Corollary 2 of the thirteenth lecture that $\mathbb{Q}_p(\zeta_m) = \mathbb{Q}_p(\zeta_{p^d - 1})$. Finally, since this extension is unramified we know from Theorem 1 of the previous lecture that $\operatorname{Gal}(\mathbb{Q}_p(\zeta_{p^d - 1})/\mathbb{Q}_p) \cong \operatorname{Gal}(\mathbb{F}_p(\zeta_{p^d - 1})/\mathbb{F}_p) \cong \mathbb{Z}/d \mathbb{Z}$.

 Example 6  Consider now the cyclotomic extension $\mathbb{Q}_p \subseteq \mathbb{Q}_p(\zeta_{p^r})$ for some $r \in \mathbb{N}$, and assume that $p \geq 3$. We see from Exercise 4 that this extension has order $p^{r-1} \cdot (p - 1)$. Moreover, the situation is exactly like the one we have over $\mathbb{Q}$, and in particular we have that $$\operatorname{Gal}(\mathbb{Q}_p(\zeta_{p^r})/\mathbb{Q}_p) \cong \left( \frac{\mathbb{Z}}{p^r \mathbb{Z}} \right)^{\times} \cong \frac{\mathbb{Z}}{p^{r - 1} \mathbb{Z}} \times \frac{\mathbb{Z}}{(p - 1) \mathbb{Z}}$$ (see Theorem 2.3 and Theorem 2.5 of these notes by Keith Conrad).
Observe now that $1 - \zeta_{p^r} \in \mathbb{Q}_p(\zeta_{p^r})$ is a uniformizer. Indeed its minimal polynomial is $\Phi_{p^r}(1 - x)$ and we have that $\Phi_{p^r}(x) = \sum_{j = 0}^{p - 1} x^{j p^{r - 1}}$, which implies that $\operatorname{N}_{\mathbb{Q}_p(\zeta_{p^r})/\mathbb{Q}_p}(1 - \zeta_{p^r}) = \Phi_{p^r}(1) = p$. This implies immediately that $1 - \zeta_{p^r}$ is a uniformizer of $\mathbb{Q}_p(\zeta_{p^r})$ by the description of the absolute value on $\mathbb{Q}_p(\zeta_{p^r})$ given by Theorem 1 of the twelfth lecture and the fact that $p \in \mathbb{Q}_p$ is a uniformizer.
Moreover, it is not difficult to see that the polynomial $\Phi_{p^r}(1 - x)$ is Eisenstein, and thus that the extension $\mathbb{Q}_p \subseteq \mathbb{Q}_p(\zeta_{p^r})$ is totally ramified.
Finally, Hensel's lemma shows that $\mathbb{Q}_p(\zeta_{p}) \cong \mathbb{Q}_p[x]/(x^{p-1} + p)$, but the analogous statement is false for $r \geq 2$. Indeed, the extension $\mathbb{Q}_3 \subseteq \mathbb{Q}_3[x]/(x^{6} + 3)$ is Galois, but its Galois group is isomorphic to $S_3$ and not to $\mathbb{Z}/2 \mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z} \cong \mathbb{Z}/6 \mathbb{Z}$ as can be seen here.
Finally, if $p = 2$ then again the situation is analogous to what happens over $\mathbb{Q}$, and we have that $$\operatorname{Gal}(\mathbb{Q}_2(\zeta_{2^r})/\mathbb{Q}_2) \cong (\mathbb{Z}/2^r \mathbb{Z})^{\times} \cong \frac{\mathbb{Z}}{2 \mathbb{Z}} \times \frac{\mathbb{Z}}{2^{r - 2} \mathbb{Z}}$$ if $r \geq 2$, whereas $\mathbb{Q}_2(\zeta_2) = \mathbb{Q}_2$. Again the same argument as before shows that this extension is totally ramified.

 Example 7  In general, if $n \in \mathbb{N}$ we can write $n = m p^r$ with $p \nmid m$ and we have that the extension $\mathbb{Q}_p(\zeta_n) = \mathbb{Q}_p(\zeta_m) \mathbb{Q}_p(\zeta_{p^r})$ where $\mathbb{Q}_p \subseteq \mathbb{Q}_p(\zeta_m) \subseteq \mathbb{Q}_p(\zeta_n)$ is the inertia field. The picture to have in mind is the following:

We are now ready to proceed towards the proof of the local theorem of Kronecker and Weber.

 Lemma 8  Let $\mathbb{Q}_p \subseteq K$ be a finite, Galois extension and suppose that $\operatorname{Gal}(K/\mathbb{Q}_p)$ is cyclic of order $q^r$ where $q \in \mathbb{N}$ is a prime and $q \neq p$. Then there exists $m \in \mathbb{N}$ such that $K \subseteq \mathbb{Q}_p(\zeta_m)$, where $\zeta_m$ is a primitive $m$-th root of unity, i.e. $\mathbb{Q}_p(\zeta_m) \cong \mathbb{Q}_p[x]/(\Phi_m(x))$ where $\Phi_m(x) \in \mathbb{Z}[x]$ is the $m$-th cyclotomic polynomial.

 Proof  First of all, $\mathbb{Q}_p \subseteq K$ is tamely ramified because we know from Theorem 10 of the twelfth lecture that $e(K \mid \mathbb{Q}_p)$ divides $q^n$, and thus $p \nmid e(K \mid \mathbb{Q}_p)$. This implies that $I^w(K \mid \mathbb{Q}_p)$ is trivial and thus we can use Corollary 5 of the fourteenth lecture to see that $I(K \mid \mathbb{Q}_p)$ can be embedded into $\mathbb{F}_p^{\times}$. Thus $\mathbb{Q}_p \subseteq K$ is tamely ramified and $e(K \mid \mathbb{Q}_p)$ divides $p - 1$, which is the ramification index of the extension $\mathbb{Q}_p \subseteq \mathbb{Q}_p(\zeta_p)$ by the previous example. Hence Theorem 2 tells us that the extension $\mathbb{Q}_p(\zeta_p) \subseteq K(\zeta_p)$ is an unramified extension of non-Archimedean local fields. Then Corollary 2 of the thirteenth lecture tells us that $K(\zeta_p) = \mathbb{Q}_p(\zeta_p,\zeta)$ for some root of unity $\zeta$ whose order $n$ is coprime to $p$. Thus if we take $m = n p$ we have that $K \subseteq \mathbb{Q}_p(\zeta_m)$. Q.E.D.

 Theorem 9  Let $\mathbb{Q}_p \subseteq K$ be a finite, Galois extension and suppose that $\operatorname{Gal}(K/\mathbb{Q}_p)$ is an abelian group and that $p \nmid \# \operatorname{Gal}(K/\mathbb{Q}_p)$. Then there exists $n \in \mathbb{N}$ such that $K \subseteq \mathbb{Q}_p(\zeta_n)$.

 Proof  We can use the fundamental theorem of finite abelian groups (see also this paper of Gabriel Navarro) to write $\operatorname{Gal}(K/\mathbb{Q}_p) = C_1 \times \cdots \times C_r$ where the $C_j$'s are cyclic groups of prime power order. Thus we can apply Lemma 8 to the extensions $$\mathbb{Q}_p \subseteq K_j \qquad \text{where} \qquad K_j := K^{C_1 \times \cdots \times \widehat{C_j} \times \cdots \times C_r}$$ whose Galois group is isomorphic to $C_j$. Then we get that $K_j \subseteq \mathbb{Q}_p(\zeta_{m_j}$ and all the $m_j$'s are coprime. Thus $K = K_1 \cdots K_r \subseteq \mathbb{Q}_p(\zeta_n)$ where $n = m_1 \cdots m_r$. Q.E.D.

The previous theorem is valid also without the hypothesis that  $p \nmid \# \operatorname{Gal}(K/\mathbb{Q}_p)$. To prove this it is clearly sufficient to apply again the fundamental theorem of finite abelian groups and the following lemma.

 Lemma 10  Let $\mathbb{Q}_p \subseteq K$ be a finite, Galois extension and suppose that $\operatorname{Gal}(K/\mathbb{Q}_p)$ is cyclic of order $p^r$. Then there exists $m \in \mathbb{N}$ such that $K \subseteq \mathbb{Q}_p(\zeta_m)$.

 Proof  We will prove that we can take $m = (p^{p^r} - 1) p^{r + 1}$ if $p \geq 3$. Indeed suppose by contradiction that $K \not\subseteq \mathbb{Q}_p(\zeta_m)$. Then we have that $\{ 1 \} \neq \operatorname{Gal}(K(\zeta_m)/\mathbb{Q}_p(\zeta_m)) \hookrightarrow \operatorname{Gal}(K/\mathbb{Q}_p)$ and thus $\operatorname{Gal}(K(\zeta_m)/\mathbb{Q}_p(\zeta_m))$ is a cyclic group of order $p^s$ with $0 < s \leq r$. Using this we get that $$\operatorname{Gal}(K(\zeta_m)/\mathbb{Q}_p) \cong \operatorname{Gal}(K(\zeta_m)/\mathbb{Q}_p(\zeta_m)) \times \operatorname{Gal}(\mathbb{Q}_p(\zeta_m)/\mathbb{Q}_p) \cong \frac{\mathbb{Z}}{p^s \mathbb{Z}} \times \frac{\mathbb{Z}}{p^r \mathbb{Z}} \times \frac{\mathbb{Z}}{p^r \mathbb{Z}} \times \frac{\mathbb{Z}}{(p - 1) \mathbb{Z}}$$ because $\operatorname{Gal}(\mathbb{Q}_p(\zeta_{p^{r + 1}})/\mathbb{Q}_p) \cong (\mathbb{Z}/p^{r +1} \mathbb{Z})^{\times}$ as it happens over $\mathbb{Q}$ (see Example 5) and $\operatorname{Gal}(\mathbb{Q}_p(\zeta_{p^{p^r} - 1})/\mathbb{Q}_p) \cong \mathbb{Z}/p^r \mathbb{Z}$ (see Example 6). Thus $\operatorname{Gal}(K(\zeta_m)/\mathbb{Q}_p)$ is an abelian group with a (normal) subgroup isomorphic to $(\mathbb{Z}/p \mathbb{Z})^3$, and this implies that there exists a sub-extension $\mathbb{Q}_p \subseteq L \subseteq K(\zeta_m)$ with $\operatorname{Gal}(L/\mathbb{Q}_p) \cong (\mathbb{Z}/p \mathbb{Z})^3$ which is impossible if $p \geq 3$, as we will prove in the first lecture of the last week using Kummer theory.

This proof would hold also for $p = 2$, but the problem is that $\operatorname{Gal}(\mathbb{Q}_2(\zeta_{24})/\mathbb{Q}_2) \cong (\mathbb{Z}/2 \mathbb{Z})^3$. Thus we need to be a little more careful, and to take $m = (2^{2^r} - 1) 2^{r + 2}$. Supposing again by contradiction that $K \not\subseteq \mathbb{Q}_2(\zeta_m)$ we would have that
$$\operatorname{Gal}(K(\zeta_m)/\mathbb{Q}_2) \cong \begin{cases} \frac{\mathbb{Z}}{2^s \mathbb{Z}} \times \frac{\mathbb{Z}}{2^r \mathbb{Z}} \times \frac{\mathbb{Z}}{2^r \mathbb{Z}} \times \frac{\mathbb{Z}}{2 \mathbb{Z}}, \ \text{for some} \ 0 < s \leq r \\ \text{or} \\ \frac{\mathbb{Z}}{2^s \mathbb{Z}} \times \frac{\mathbb{Z}}{2^r \mathbb{Z}} \times \frac{\mathbb{Z}}{2^r \mathbb{Z}}, \ \text{for some} \ 2 \leq s \leq r \end{cases}$$ and this implies that there exists a sub-extension $\mathbb{Q}_2 \subseteq L \subseteq K(\zeta_m)$ with $\operatorname{Gal}(L/\mathbb{Q}_2) \cong (\mathbb{Z}/2 \mathbb{Z})^4$ (in the first case) or $\operatorname{Gal}(L/\mathbb{Q}_2) \cong (\mathbb{Z}/4 \mathbb{Z})^3$ (in the second case), and we will show again in the first lecture of the last week that this is impossible by means of Kummer theory. Q.E.D. 

Thus, as a corollary, we get the important local theorem of Kronecker and Weber:

 Corollary 11  Every finite abelian extension of $\mathbb{Q}_p$ is contained in a cyclotomic field.

Conclusions and references

In this lecture we managed to:

• prove that unramified extensions remain unramified after changing the base field;
• prove that tamely ramified extensions become unramified if we sufficiently extend the base field;
• prove the local theorem of Kronecker and Weber (up to Kummer theory, which will be the subject of next lecture).

References for this lecture include:

• Section 6 of these notes by Peter Stevenhagen;
• these notes by Andrew Sutherland;
• Section 5.2 of the book "Elementary and Analytic Theory of Algebraic Numbers" by Władysław Narkiewicz;
• Chapter 14 of the book "Introduction to Cyclotomic Fields" by Lawrence C. Washington.