Last time on the seminar, we used the Kronecker-Weber Theorem to discuss number theory in $\mathbb{F}_1$. Due to my almost nonexistent knowledge of number theory, I thought it would be a great exercise to work this theorem out. It is anything but trivial and it requires a lot of lemmas and fundamental theorems, which I will write about this time.

The theorem and notations

Theorem (Global Kronecker-Weber Theorem) If $K/\mathbb{Q}$ is a finite abelian Galois extension, then there exists an $n$ such that $K \subseteq \mathbb{Q}(\zeta_n)$, with $\zeta_n$ a primitive root of $1$.

There is also a local version of this theorem, given by replacing $\mathbb{Q}$ by $\mathbb{Q}_p$ for some prime $p$. This theorem has been proved in various ways, the “easiest” one by using class field theory. I will give another proof, based on the book “Introduction to cyclotomic fields” by Lawrence C. Washington. It will use the Hasse principle, which we will prove in the first step of the proof.

Theorem If the local Kronecker-Weber Theorem is true for every prime $p$, then so is the global Kronecker-Weber Theorem.

For the remainder of this series, $p$ is always a prime and $\mathbb{Q}_p$ are the $p$-adic numbers with ring of integers $\mathbb{Z}_p$. For an arbitrary number or local field $K$, one can define $K_\mathfrak{p}$ as the completion of $K$ with respect to a prime ideal $\mathfrak{p}$. For 2 extensions $L_1$ and $L_2$ of a number or local field $K$, we denote the compositum as $L_1L_2$.

Preliminary work

Next proposition is probably one of the most important theorems in number theory and is worth a post itself, but one has to choose a subject and stick to it.

Proposition Let $L/K$ be a finite Galois extension, where $K$ may be a number field or a local field. Let $\mathfrak{p}$ be a prime of $K$. Then we have the following equality $$\mathfrak{p}=\mathfrak{b}_1^e \mathfrak{b}_2^e\ldots \mathfrak{b}_g^e,$$ with $\mathfrak{b}_i$ primes in $L$ lying above $K$. $e$ is called the ramification index. Let $f$ be the degree of the extension $\mathcal{O}_L \bmod \mathfrak{b}_i / \mathcal{O}_K \bmod \mathfrak{p}$. Then we have $$ \left[L:K\right] = efg.$$ $\mathfrak{p}$ is said to be totally ramified in $L$ if $e =n $ and to be unramified if $e=1$. Moreover, for all $1 \leq i<j \leq g$, there exists a $\sigma \in Gal(L/K)$ such that $\sigma(\mathfrak{b}_i) = \mathfrak{b}_j$.

In local fields, it is easy to see that if $K \subset L \subset M$ with ramification indices $e_{LK},e_{ML},e_{MK}$ and with extension degrees of the residue fields $f_{LK},f_{ML},f_{MK}$, then we have $e_{LK}e_{ML} = e_{MK}$ and $f_{LK}f_{ML}=f_{MK}$. In general, one sees that if $K \subset L \subset M$ and a prime $\mathfrak{p}$ of $K$ ramifies in $L$, then it ramifies in $M$.

In case of cyclotomic fields, next theorem says something about ramification.

Theorem A prime $\mathfrak{p}$ of $K$ ramifies in a cyclotomic extension $K(\zeta_n)$ if and only if $p|n$.

One may hope that there are extensions of $\mathbb{Q}$ that aren’t ramified at any prime. Unfortunately, there aren’t any: if $K/\mathbb{Q}$ is unramified at all primes, $K = \mathbb{Q}$. This is easily proved using a theorem of Minkowski (for a proof, see here) and the fact that if a prime divides the discriminant of a number field $K$, it always ramifies (the converse is also true).

A useful theorem is the following, where $\mathbb{Q}$ can be replaced by $\mathbb{Q}_p$ in both this theorem and its corollary:

Theorem Let $K$ and $L$ be finite Galois extensions of $\mathbb{Q}$. Then $Gal(KL/\mathbb{Q}) \cong H$, with $H$ a subgroup of $Gal(K/\mathbb{Q}) \times Gal(L/\mathbb{Q})$ consisting of all elements $(\phi,\psi)$ such that $\phi|_{K \cap L} = \psi|_{K \cap L}$.

Corollary Let $L/\mathbb{Q}$ be an abelian Galois extension, with $$Gal(L/\mathbb{Q}) = \prod_{i=1}^m G_i.$$ Then $$L = \prod_{i=1}^m L^{G_i}.$$

Since all abelian groups are products of cyclic groups of prime order, we can reduce Kronecker-Weber to Galois extensions with Galois group a cyclic group of prime order, since previous corollary gives the general case.

Definition Let $L/K$ be a Galois extension, with $\mathfrak{p}$ a prime of $K$ and $\mathfrak{b}$ a prime lying above $\mathfrak{p}$. Define $$D_\mathfrak{b}=\left\{\sigma \in Gal(L/K)|\sigma(\mathfrak{b}) = \mathfrak{b}\right\}$$ as the decomposition group of $\mathfrak{b}$ with inertia group $$I_\mathfrak{b} = \left\{\sigma \in D_\mathfrak{b}|\sigma(x) \equiv x \bmod \mathfrak{b} \text{ for all } \alpha \in \mathcal{O}_L\right\}.$$

In general, 2 decomposition groups $D_{\mathfrak{b}_1}$ and $D_{\mathfrak{b}_2}$ with $\mathfrak{b}_1$ and $\mathfrak{b}_2$ primes lying above $\mathfrak{p}$, will be conjugated. Moreover, we have: $$ \sigma D_{\mathfrak{b}_1} \sigma^{-1} = D_{\mathfrak{b}_2} \Rightarrow \sigma I_{\mathfrak{b}_1} \sigma^{-1} = I_{\mathfrak{b}_2}.$$
Knowing this, it’s easy to see that if you work in abelian extensions, there is only one decomposition group and one inertia group. In this case, it’s easier to write $I_{\mathfrak{p}}$ instead of $I_{\mathfrak{b}}$, since the inertia group doesn’t depend on $\mathfrak{b}$ anymore.
If you work in a local field $K$ which is an extension of $\mathbb{Q}_p$, you can discard the index $\mathfrak{b}$ and write $I_K$ instead.

The order of $D_\mathfrak{b}$ is $ef$, since the number of elements in the orbit of $\mathfrak{b}$ is $g$.  The order of $I_\mathfrak{b}$ will be $e$ and since $$\{id_L\}\subseteq I_\mathfrak{b} \subseteq D_\mathfrak{b} \subseteq Gal(L/K),$$ we have a tower of extensions $$L\supseteq L^{I_\mathfrak{b}} \supseteq L^{D_\mathfrak{b}} \supseteq K,$$ on which we can use the index calculation $$n = \left[L:K\right] = \left[L:L^{I_\mathfrak{b}}\right]\left[L^{I_\mathfrak{b}}:L^{D_\mathfrak{b}}\right]\left[L^{D_\mathfrak{b}}:K\right]=efg.$$
Furthermore, $\mathfrak{p}$ will be unramified in $L^{I_\mathfrak{b}}$. In case of a unique inertia group, $L^{I_\mathfrak{b}}$ is the largest field between $L$ and $K$ that has this property. Also, $I_\mathfrak{b}$ is always a normal subgroup of $D_\mathfrak{b}$ (not only in case of abelian extensions).

Next four theorems will prove vital for proving the ‘local implies global’ Kronecker-Weber Theorem.

Theorem Let $L/K$ be a Galois extension of number fields, with $\mathfrak{p}$ a prime of $K$ with a prime $\mathfrak{b}$ lying over. Then we have $$Gal(L_\mathfrak{b}/K_\mathfrak{p}) \cong D_\mathfrak{b}$$ and the inertia groups of $\mathfrak{b}$ are isomorphic in both extensions.

Theorem Let $L/K/\mathbb{Q}_p$ be a tower of finite Galois extensions. Then there exists a surjective homomorphism $f:I_L \rightarrow I_K$.

Theorem The inertia group of $\mathbb{Q}_p(\zeta_n)/\mathbb{Q}_p$ is $(\mathbb{Z}/p^e\mathbb{Z})^*$, where $e$ is determined by $p^e|n, p^{e+1} \nmid n.$

Theorem The Galois group of an abelian extension $K/\mathbb{Q}$ is generated by the inertia groups $I_{p}$’s, with $p$ running over all primes of $\mathbb{Q}$.

Local implies global

All these theorem imply the first step in the proof of Kronecker-Weber.

Theorem If the local Kronecker-Weber Theorem is true for every prime $p$, then so is the global Kronecker-Weber Theorem.

Proof Let $K/\mathbb{Q}$ be an abelian extension and $p$ a rational prime that ramifies, with $\mathfrak{b}$ a prime lying above $p$. Take completions $K_\mathfrak{b}$ and $\mathbb{Q}_p$. We know that $Gal(K_\mathfrak{b}/\mathbb{Q}_p) = D_\mathfrak{b}$, so it is necessarily abelian, since it is a subset of $Gal(K/\mathbb{Q})$. Assuming local Kronecker-Weber, we know that $K_\mathfrak{b} \subset \mathbb{Q}_p(\zeta_{n_p})$ for some $n_p$. Now, let $p^{e_p}$ be the exact power of $p$ that divides $n_p$ and let $$n = \prod_{p \text{ ramifies in } K} p^{e_p}.$$ We will now prove that $K \subset \mathbb{Q}(\zeta_n)$.
Take $L = K(\zeta_n)$. The main objective will be to prove that $\left[L:\mathbb{Q}\right] \leq \phi(n)$, since $\left[\mathbb{Q}(\zeta_n):\mathbb{Q}\right] = \phi(n)$ and $\zeta_n \in L$ implies that $\mathbb{Q}(\zeta_n) = L$ and so $K \subset \mathbb{Q}(\zeta_n)$.
We know that a prime $p$ is unramified in $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ if $p$ doesn’t divide $n$, so the primes that ramify in $K$ and $\mathbb{Q}(\zeta_n)$ are the same. So $L$ and $K$ have the same ramifying primes. Take a prime $p$ of $\mathbb{Q}$ that ramifies in $L$, with a prime $\mathfrak{b}$ lying over it. By one of the above theorems, we can calculate $I_{\mathfrak{b}}$ locally. Since we are working in abelian extensions and consequently there is only one inertia group, we denote $I_p$ instead of $I_{\mathfrak{b}}$. Taking completions and using that $K_\mathfrak{b} \subset \mathbb{Q}_p(\zeta_{n_p})$, we get $$K_\mathfrak{b}(\zeta_n) \subset \mathbb{Q}_p(\zeta_{n_p},\zeta_n) = \mathbb{Q}_p(\zeta_m),$$ with $m = lcm(n_p,n)$. Since the highest power of $p$ that divides both $m$ and $n$ is $e_p$, we have that the inertia groups of both $\mathbb{Q}_p(\zeta_m)$ and $\mathbb{Q}_p(\zeta_n)$ are $(\mathbb{Z}/p^{e_p}\mathbb{Z})^*$. Since $\mathbb{Q}_p(\zeta_n) \subset K_\mathfrak{b}(\zeta_n) \subset \mathbb{Q}_p(\zeta_m)$, we have that $I_p = (\mathbb{Z}/p^{e_p}\mathbb{Z})^*$ and so $|I_p| = \phi(p^{e_p})$. Since all the $I_p$ generate $Gal(L/\mathbb{Q})$ and we work in abelian extensions, we get $$|Gal(L/\mathbb{Q})|\leq \prod_{p \text{ ramifies}} |I_p|  = \phi(n).$$ So $\left[L:\mathbb{Q}\right] \leq \phi(n)$ and we are done !

Next time, I’ll post (some of) the proof for the local Kronecker-Weber Theorem.

About Kevin De Laet

Interested in mathematics, specifically Group & Representation theory, currently studying in Antwerp (and sometimes Brussels)

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Last time, I proved that the Hasse principle is valid for Kronecker-Weber Theorem. So in order to prove Kronecker-Weber, the local version still has to be proved. As noted before, you only need to prove it for Galois extensions $K/\mathbb{Q}_p$ with Galois group $\mathbb{Z}/q^m\mathbb{Z}$, where $q$ is a prime number that can be equal to $p$. There will be 3 cases to prove:

• The extension is unramified.
• The extension is ramified, but the ramification degree $e$ is not divisible by $p$ (also known as tamely ramified).
• The extension is ramified with ramification degree $e$ divisible by $p$ (wildly ramified).

In all these cases, we will have need for Hensel’s Lemma:

Hensel’s Lemma Let $L$ be a local field, with ring of integers $\mathcal{O}_L$ and maximal ideal $\mathfrak{b}$. Let $l$ be it’s residue field. Suppose $f \in \mathcal{O}_L [x]$, monic with restriction $\bar{f}$ in $l[x]$. Suppose$\bar{f}$ has a root $\alpha$ in $l$, such that $\bar{f}’(\alpha) \neq 0$. Then there exists a root $\beta \in \mathcal{O}_L$ of $f$, with $\beta \equiv \alpha \bmod \mathfrak{b}$.

Let us start the unramified case, which will be the easiest. We will use the notation from the previous post.

$K/\mathbb{Q}_p$ is unramified

There is a much stronger result for this, which I will prove here. Taking $K = \mathbb{Q}_p$ will give what we want.

Theorem Suppose $L/K$ is an unramified, finite Galois extension, where $L$ and $K$ are finite extensions of $\mathbb{Q}_p$ (not necessarily Galois over $\mathbb{Q}_p$). Then there exists an $n \in \mathbb{N},p \nmid n$ so that $L = K(\zeta_n)$. Furthermore, $Gal(L/K)$ will be cyclic.

Proof Take $L/K$ like in the theorem, with prime $\mathfrak{p}$ in both extensions, since $e=1$. Let $l$ and $k$ be the residue fields of respectively $L$ and $K$. Since $e = 1$, $Gal(L/K) = Gal(l/k)$. The right hand side is a Galois group of an extension in finite fields, hence it is cyclic. Since $l/k$ is a Galois extension, there exists a primitive element $\alpha$ so that $l = k[\alpha]$. Consequently, there exists an $n$, with $\gcd(n,p)=1$, such that $X^n-1$ has a root in $l$, namely $\alpha$. Using Hensel’s Lemma, we find a $\beta \in \mathcal{O}_L$ as a root of $X^n-1$ and $\beta \equiv \alpha \bmod \mathfrak{p}$. Since $\beta$ is a root of $1$ and, due to being equivalent to $\alpha$, has order at least $n$, we have $$\left[K(\beta):K\right] \geq \left[k[\alpha]:k\right] = \left[l:k\right] = \left[L:K\right].$$ But $\beta \in L$, so we get $L = K(\beta)$ and $\beta = \zeta_n$.

This was the easy one, the next two cases are harder.

$K/\mathbb{Q}_p$ is tamely ramified.

This case follows from 3 lemmas. I will give the proofs depending on how difficult or technical it is.

Lemma Let $p\neq 2$ be a prime. Then $\mathbb{Q}_p$ contains the $p-1$-th roots of unity, but none other.

Proof Let $\mu_p$ be the set of roots of unity in $\mathbb{Q}_p$. Thanks to Hensel’s Lemma, we have a surjective morphism $\phi:\mu_p \rightarrow \mathbb{F}_p^*$, so we need to prove that it is injective. So suppose we have a root of unity of $(1+tp)^n$. Using Newton’s binomial formula, we get $$\sum_{i = 1}^n \binom{n}{i} (tp)^{i}=0.$$ Now, there are 2 cases: $t = 0$ and we are done, or $$ \sum_{i = 1}^n \binom{n}{i} (tp)^{i-1}=0,$$ but then, since $p|0$, $p|\binom{n}{1}=n.$ Since $\mathbb{Q}_p$ contains the $n$-th root of $1$ and $p|n$, $\mu_p$ must contain the $p$-th roots of unity. So we are reduced to $(1+tp)^p = 1$. But then we have $$\sum_{i = 1}^p \binom{p}{i} (tp)^{i}=0.$$ If $t \neq 0$, we have $$\sum_{i = 1}^p \binom{p}{i} (tp)^{i-1}=0.$$But every term is divisible by $p^2$ except for the first one, which is only divisible by $p$, so we have a contradiction. This concludes the proof.

In case of $p=2$, $\mathbb{Q}_2$ only has $-1,1$ as roots of unity. This is proved by noticing that $(\mathbb{Z}/8\mathbb{Z})^* = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

Lemma Let $L/K/\mathbb{Q}_p$ be a tower of finite extensions, with $L/K$ Galois. Let $\mathfrak{p}_K$ be the unique maximal ideal of $K$ and suppose that that $L/K$ is totally ramified of degree $e$. Then there exists a $\pi \in K$ a generator of $\mathfrak{p}_K$ and a root $\alpha$ of the polynomial $f(X) = X^e – \pi$, such that $L = K(\alpha)$.

Proof Use of the non-archimedean absolute value, choosing a uniformizing parameter and using the fact that elements of the Galois group preserve absolute values, this is a long but not very difficult proof and will be omitted.

Lemma $\mathbb{Q}_p((-p)^{\frac{1}{p-1}}) = \mathbb{Q}_p(\zeta_p)$

Proof It can be proved that the maximal ideal of $\mathbb{Q}_p(\zeta_p)$ is given by $(1-\zeta_p)$ and that $(p) = (1-\zeta_p)^{p-1}$. Consequently, $(1-\zeta_p)^p = (p(1-\zeta_p))$. Take the polynomial $$g(X) = \frac{(X+1)^p-1}{X}=\sum_{i=1}^{p} \binom{p}{i}X^{i-1},$$ which is the minimal polynomial of $\zeta_p-1$, so we get $$0 = g(\zeta_p) \equiv (\zeta_p-1)^{p-1}+p \bmod (\zeta_p-1)^p.$$ This can be modified to $$u = \frac{(\zeta_p-1)^{p-1}}{-p} \equiv 1 \bmod (\zeta_p-1).$$ Now look at the polynomial $f(X) = X^{p-1}- u$. Then $f(1) \equiv 0 \bmod (\zeta_p – 1)$ and $(\zeta_p – 1)\nmid f’(1)$. Hensel’s Lemma (again) implies that there exists an $\alpha \in \mathbb{Q}_p(\zeta_p)$ as a root of $f$. But $(-p)^{\frac{1}{p-1}} = \frac{\zeta_p-1}{\alpha}$, so $(-p)^{\frac{1}{p-1}} \in \mathbb{Q}_p(\zeta_p)$. Now, the minimal polynomial of $(-p)^{\frac{1}{1-p}}$ is $X^{p-1}+p$, since it is irreducible due to Eisenstein’s criterion, so $(-p)^{\frac{1}{1-p}}$ and $\zeta_p$ have the same degree over $\mathbb{Q}_p$. This gives $\mathbb{Q}_p((-p)^{\frac{1}{1-p}}) = \mathbb{Q}_p(\zeta_{p})$.

Now, for the proof of the tamely ramified case. Let $L/\mathbb{Q}_p$ be a tamely ramified abelian extension of ramification degree $e, p\nmid e$, with $K/\mathbb{Q}_p$ the maximal unramified subextension. We already know that $K \subset \mathbb{Q}_p(\zeta_n)$ for some $n$. We also have $L/K$ is totally ramified with degree $e$. From one of the above lemmas, we know that $L = K(\pi^{\frac{1}{e}})$ with $\pi$ a generator of the unique maximal ideal $\mathfrak{p}_K$ of $\mathcal{O}_K$. $K/\mathbb{Q}_p$ is unramified, so $p$ is also a generator of $\mathfrak{p}_K$. This gives $\pi = -up$ for some unit $u$ of $\mathcal{O}_K$. $u$ is a unit, $p \nmid e$, so the discriminant of $f(X) = X^e-u$ is not divisible by $p$. This implies that $K(u^{\frac{1}{e}})/K$ is unramified. $K \subset \mathbb{Q}_p(\zeta_n)$, so $$K(u^{\frac{1}{e}}) \subset K(\zeta_M) \subset \mathbb{Q}_p(\zeta_{Mn}),$$ with $M$ an integer. Let $T$ be the compositum of $\mathbb{Q}_p(\zeta_{Mn})$ and $L$. $T$ is an abelian extension of $\mathbb{Q}_p$ and $u^{\frac{1}{e}},\pi^{\frac{1}{e}} \in T$, which gives $(-p)^{\frac{1}{e}} \in T$. Since $\mathbb{Q}_p((-p)^{\frac{1}{e}}) \subset T$ and $T/\mathbb{Q}_p$ is abelian, it is an Galois extension over $\mathbb{Q}_p$. Since all the roots of $X^e +p$ are $\zeta_e^k (-p)^{\frac{1}{e}}, k=0 \ldots e-1$, we know that $\zeta_e \in \mathbb{Q}_p((-p)^{\frac{1}{e}})$. $\mathbb{Q}_p((-p)^{\frac{1}{e}})$ is totally ramified and since every subfield of a totally ramified subfield is totally ramified in case of local fields, $\mathbb{Q}_p(\zeta_e)$ is also totally ramified. But $p \nmid e$, so it can’t ramify, except when it’s the trivial extension, so $\zeta_e \in \mathbb{Q}_p$ and $e | p-1$. We have a tower of extensions $$\mathbb{Q}_p((-p)^{\frac{1}{e}}) \subset \mathbb{Q}_p((-p)^{\frac{1}{p-1}}) \subset \mathbb{Q}_p(\zeta_p).$$ Consequently, $$L = K(\pi^{\frac{1}{e}}) \subset K(u^{\frac{1}{e}},(-p)^{\frac{1}{e}}) \subset \mathbb{Q}_p(\zeta_{Mnp}),$$ which concludes this case.

$K/\mathbb{Q}_p$ is widely ramified

This is the hardest case, so I won’t give the entire proof, but I will describe the way the proof works. We know that we just have to work with extensions with Galois group $\mathbb{Z}/q^{m}\mathbb{Z}$, with $q$ a prime. Since we work in the widely ramified case, this breaks down to $q = p$. There are 2 cases to consider: $p = q = 2$ or $p = q \neq 2$ (2 is always a bad prime). Let us first proof the last case.

Take an abelian widely ramified extension $L/\mathbb{Q}_p$ of degree $p^m$, with cyclic Galois group. We first construct 2 Galois extensions of $\mathbb{Q}_p$ of degree $p^m$, one totally ramified, the other unramified:

• Consider the extension $\mathbb{Q}_p(\zeta_{p^{m+1}})$ with Galois group $\mathbb{Z}/p^{m+1}\mathbb{Z}$ and let $K_r$ be the fixed field by the subgroup of order $p-1$. This will be a totally ramified extension of degree $p^m$ with cyclic Galois group.
• Take an irreducible polynomial in $\mathbb{F}_p[X]$ of degree $p^m$, lift it to an irreducible polynomial in $\mathbb{Z}_p[X]$ and construct the corresponding Galois extension. This will give an unramified extension $K_u$ with Galois group $\mathbb{Z}/p^m\mathbb{Z}$.

As we know from the unramified case, $K_u \subset \mathbb{Q}_p(\zeta_n)$ for some $n$. Since one extension is unramified and the other is totally ramified, we necessarily have $K_u \cap K_v = \mathbb{Q}_p$ and consequently, $Gal(K_uK_v/\mathbb{Q}_p) = (\mathbb{Z}/p^m\mathbb{Z}^2)$. Suppose that $L \not\subset \mathbb{Q}_p(\zeta_n,\zeta_{p^{m+1}})$, then we should have $$Gal(L(\zeta_n,\zeta_{p^{m+1}})) = (\mathbb{Z}/p^m\mathbb{Z})^2 \times\mathbb{Z}/p^k\mathbb{Z},$$ with $k > 0$. This implies that there is a Galois extension of $\mathbb{Q}_p$ with Galois group $(\mathbb{Z}/p\mathbb{Z})^3$. The next part is to prove that there doesn’t exist such an extension, but this is very hard and requires knowledge of Kummer Theory..

The case $p=q=2$ is similar. One concludes that there should exist a Galois extension of $\mathbb{Q}_2$ with Galois group $(\mathbb{Z}/2\mathbb{Z})^4$ or $(\mathbb{Z}/4\mathbb{Z})^3$ and once again it is proved that this is impossible.

About Kevin De Laet

Interested in mathematics, specifically Group & Representation theory, currently studying in Antwerp (and sometimes Brussels)

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