# The holonomy Lie algebra of a matroid

Clas Löfwall

Seminar in Stockholm 9 dec 2015

##  1 Introduction

I will soon start from the beginning and define a matroid and its Orlik-
Solomon algebra and holonomy Lie algebra, but first I will give some
background from topology and cohomology. A (central) hyperplane arrangement is
a finite number of subspaces of codimension one in a finite dimensional vector
space over ${\mathbb{R}}$, ${\mathbb{C}}$ or even finite fields. The space $X$
which is the complement of the union of the hyperplanes has interesting
properties. If you consider $\mathbb{R}^{n}$ the cohomology is however rather
trivial, there is just an $H^{0}$, since $X$ is a disjoint union of
contractible sets. But it is not a trivial task to compute $|H^{0}|$. The way
to do it is to consider the intersection lattice of the hyperplanes and
compute its Möbius function. The result is that the number of components is
the sum of the absolut values of the Möbius function (Zaslavsky 1975, [20]).

When we change to $\mathbb{C}^{n}$ we get something quite different. If we
think of $\mathbb{C}^{n}$ as $\mathbb{R}^{2n}$ the $\mathbb{C}$-subspaces of
codimension one now has real codimension two and the simplest case is the real
plane minus the origin, whose first cohomology group is non-zero. In fact, the
general complex case may be treated with deRham cohomology using a closed but
not exact one-form for each hyperplane to get the cohomology algebra (at least
over $\mathbb{R}$). This is called the Orlik-Solomon algebra [12] (Brieskorn
[3] showed that the cohomology has no torsion), which we will come back to.
The algebra is determined by knowing which subsets of the defining forms of
the hyperplanes are independent. It is here matroids come in; the main example
of a matroid is the set of independent subsets of a set of vectors. Goresky-
MacPherson [5], Björner-Ziegler [2] and Longueville and Schultz [9] study
general subspace arrangements in $\mathbb{R}^{2n}$, especially subspaces of
codimension 2 whose intersections have even dimension. This is the natural
first step generalization of the complex case. They prove that everything hold
except that the signs in the differential defining the Orlik-Solomon algebra
may be different from the complex case. This is interesting, since it gives us
more freedom in defining holonomy Lie algebras (and my generalization covers
also this case). I should perhaps end this introduction by defining the
holonomy Lie algebra (it will be done more explicitly later with generators
and relations). If $\mathcal{O}$ is the cohomology algebra of $X$, with
coefficients in some field $k$, of a complex arrangement, we have that
$\mathrm{Ext}_{\mathcal{O}}(k,k)$ is the enveloping algebra of the ”homotopy”
Lie algebra ${\mathfrak{g}}$ of the arrangement. The holonomy Lie algebra
$\eta$ of the arrangement is defined as the Lie subalgebra of ${\mathfrak{g}}$
generated by its elements of degree one. The following result by Kohno [6]
highly increases the interest to study $\eta$ ($k$ has characteristic zero).

$\eta\cong\mathrm{gr}(\pi_{1}(X))\otimes k$

where $\mathrm{gr}(\pi_{1}(X))$ is the graded associative with respect to the
lower central series of $\pi_{1}(X)$.

##  2 Matroids

Let $E$ be a finite set. A subset ${\mathcal{M}}$ of $\mathscr{P}(E)$ is a
matroid if the following two axioms are fulfilled.

$\displaystyle 1)$ $\displaystyle\quad A\in{\mathcal{M}},\ B\subset
A\quad\Rightarrow\quad B\in{\mathcal{M}}$ $\displaystyle 2)$
$\displaystyle\quad A,B\in{\mathcal{M}},\ |A|<|B|\quad\Rightarrow\quad\exists
x\in B\setminus A\text{ such that }A\cup\\{x\\}\in{\mathcal{M}}$

The axioms are obviously fulfilled if $E$ is a set of vectors in a vector
space and ${\mathcal{M}}$ is the set of independent subsets of $E$. A matroid
is called representable if it is given in this way. We will use “$A$
independent” and “$A$ dependent” for the cases $A\in{\mathcal{M}}$ and
$A\notin{\mathcal{M}}$ respectively. We will always assume that a matroid is
”simple”, which means that all subsets of size at most two are independent.
This is true for the case when $E$ is a set of defining forms for the
hyperplanes in a central arrangement. Indeed, two different hyperplanes
through origin have non-parallel forms. A graph defines a matroid on its set
of edges. A set of edges is independent if it contains no cycles. The matroid
is simple if there are no loops and no multiple edges. A graph matroid is
representable. If the vertices are $\\{1,\ldots,n\\}$ we may define vectors in
${\mathbb{R}}^{n}$ by defining for each edge between $i$ and $j$ the vector
$e_{i}-e_{j}$ (corresponding to the hyperplanes $x_{i}-x_{j}=0$). The complete
graph $K_{n}$ on $n$ vertices has an interesting holonomy Lie algebra,
obtained by Kohno [6], which we will come back to. The corresponding
arrangement is called the ”braid” arrangement, since its fundamental group is
known to be the pure braid group.

###  2.1 Rank, bases, flats

Let $T\subseteq E$, ${\mathcal{M}}$ a matroid on $E$. An independent subset
$B$ of $T$ is called a basis for $T$ if $B$ is maximal. All bases for $T$ have
the same size, by the exchange property 2). The size of a basis for $T$ is
called $\mathrm{rank}(T)$. If $B$ is a basis for $T$, then $B\cup\\{x\\}$ is
dependent for all $x\in T\setminus B$. A $k$-$flat$ $F$ is a maximal subset of
$E$ of rank $k$. Then $F=B\cup\\{x\in E;\ B\cup\\{x\\}\text{ dependent}\\}$
where $B$ is a basis for $F$. Moreover if $B$ is independent, then
$B\cup\\{x\in E;\ B\cup\\{x\\}\text{ dependent}\\}$ is a $|B|$-flat.

###### Lemma 2.1

If $F_{1}$ and $F_{2}$ are flats, then $F_{1}\cap F_{2}$ is a flat of rank
less than $\min(\mathrm{rank}(F_{1}),\mathrm{rank}(F_{2}))$.

Proof. Let $B$ be a basis for $F_{1}\cap F_{2}$ and extend $B$ to bases
$B_{1}$, $B_{2}$ for $F_{1}$, $F_{2}$ respectively. If $B\cup\\{x\\}$ is
dependent, then $B_{i}\cup\\{x\\}$ is dependent for $i=1,2$. Hence $x\in
F_{1}\cap F_{2}$. If $x\in F_{1}\cap F_{2}\setminus B$ then $B\cup\\{x\\}$ is
dependent. Hence, by the above, $F_{1}\cap F_{2}$ is a flat. $\Box$

We may now define a closure operator $\bar{T}$ as the least flat that contains
$T$. All flats form a lattice $\mathrm{L}({\mathcal{M}})$ under inclusion,
which is graded by rank. If ${\mathcal{M}}$ is the matroid of a hyperplane
arrangement, then $\mathrm{L}({\mathcal{M}})$ is isomorphic to the lattice of
all intersections of the hyperplanes ordered by reverse inclusion, since

$\bigcap_{i\in T_{1}}H_{i}\subseteq\bigcap_{i\in
T_{2}}H_{i}\Leftrightarrow\overline{\\{n_{i};i\in
T_{1}\\}}\supseteq\overline{\\{n_{i};i\in T_{2}\\}}$

where $n_{i}$ is a defining form of $H_{i}$.

The 2-flats are of special interest for us, since they determine the holonomy
Lie algebra. Since the intersection of two 2-flats is a 1-flat or a 0-flat, it
follows that two different 2-flats have at most one element in common. Also
all pairs are contained in a (unique) 2-flat. This is called a 2-partition of
$E$ in [14], where it is also proven that every 2-partition is the set of
2-flats of a matroid of rank $\leq 3$ (the independent 3-sets are the 3-sets
that are not contained in any of the sets in the 2-partition), se [14], Prop.
2.1.21 or [10].

###### Example 2.2

$E=\\{1,2,3,4,5,6,7\\}$. A 2-partition on $E$ is

$\\{\\{1,2,3,4\\},\\{1,5,6\\},\\{2,5,7\\},\\{3,6,7\\},\\{1,7\\},\\{2,6\\},\\{3,5\\},\\{4,5\\},\\{4,6\\},\\{4,7\\}\\}$

The matroid of rank 3 determined by the 2-partition in this example has the
following dependent 3-sets.

$\\{1,2,3\\},\\{1,2,4\\},\\{2,3,4\\},\\{1,5,6\\},\\{2,5,7\\},\\{3,6,7\\}$

Usually one does not give the 2-sets in the partition, since they are
determined by the rest. We will use the following terminology.

###### Definition 2.3

A set-arrangement on a finite set $E$ is a set ${\mathcal{A}}$ of subsets of
$E$ such that $|A|\geq 3$ for all $A\in{\mathcal{A}}$ and $|A\cap B|\leq 1$
for all $A,B\in{\mathcal{A}}$.

A set-arrangement (which may be empty) on $E$ determines a 2-partition on $E$
and vice versa. Observe that for any matroid ${\mathcal{M}}$ we have that
${\mathcal{M}}_{\leq 3}=\\{A\in{\mathcal{M}};\ \mathrm{rank}(A)\leq 3\\}$ is a
matroid with the same set of 2-flats as ${\mathcal{M}}$.

##  3 The Orlik-Solomon algebra and the holonomy Lie algebra

The Orlik-Solomon algebra on a simple matroid ${\mathcal{M}}$ on
$\\{e_{1},\ldots,e_{n}\\}$ is

$\mathrm{OS}({\mathcal{M}})=E(e_{1},\ldots,e_{n})/\langle\partial
e_{S}\rangle$

where $\\{e_{1},\ldots,e_{n}\\}$ is the exterior algebra, $\partial e_{i}=1$,
$\partial$ is a derivation, $S=\\{e_{i_{1}},\ldots,e_{i_{k}}\\}$ is dependent
and $e_{S}=e_{i_{1}}\cdots e_{i_{k}}$, and $\langle\partial e_{S}\rangle$ is
the ideal generated by $\\{\partial e_{S};\ S\text{ dependent}\\}$. The
algebra is graded by letting the degree of $e_{i}$ be 1 for all $i$. The
Hilbert series $\mathrm{OS}({\mathcal{M}})(z)$, i.e., the generating series
for the dimensions of the different graded pieces is obtained from the Möbius
function of the lattice $\mathrm{L}({\mathcal{M}})$, e.g., if
$\mathrm{rank}({\mathcal{M}})=3$, then
$\mathrm{OS}({\mathcal{M}})(z)=1+nz+dz^{2}+(d+1-n)z^{3}$, where
$d=\sum(|A|-1)$ and the sum is taken over all 2-flats $A$. Observe that
$e_{S}=0$ in $\mathrm{OS}({\mathcal{M}})$ if $S$ is dependent, since
$e_{S}=e_{i_{1}}\partial e_{S}$, whence $\mathrm{OS}({\mathcal{M}})_{i}=0$ if
$i>\mathrm{rank}({\mathcal{M}})$, which also follows from the fact that
$\mathrm{L}({\mathcal{M}})$ vanishes above $\mathrm{rank}({\mathcal{M}})$.
(Also, as was mentioned in the introduction, the number of components of a
hyperplane arrangement in ${\mathbb{R}}^{n}$ can be obtained by evaluating the
Hilbert series $\mathrm{OS}({\mathcal{M}})(z)$ at $z=1$.)

The quadratic relations are $ab-ac+bc$ for all $\\{a,b,c\\}$ which are
dependent, i.e., are contained in a 2-flat. In order to define the holonomy
Lie algebra we want to compute the orthogonal complement of these quadratic
relations. In this process it is enough to consider one 2-flat at a time,
since a quadratic monomial occurs only in one 2-flat. So it is enough to
consider a matroid ${\mathcal{M}}_{2}$ of rank 2 on $\\{1,\ldots,n\\}$, where
$\\{1,\ldots,n\\}$ is the only 2-flat (${\mathcal{M}}_{2}$ is realized as $n$
lines in ${\mathbb{C}}^{2}$). Write $\mathrm{OS}({\mathcal{M}}_{2})$ as a
quotient of a free non-commutative algebra. Here is a (triangular) independent
set of quadratic relations: $e_{1}^{2},\ldots,e_{n}^{2},\
e_{j}e_{i}+e_{i}e_{j}$ for $j>i$, followed by the relations defined by the
triples $123,\ 134,\ 145,\ldots,1(n-1)n,\
234,\ldots,2(n-1)n,\ldots,(n-2)(n-1)n$. In total this gives $n+{n\choose
2}+{n-1\choose 2}=n^{2}-(n-1)$ independent relations. Hence the orthogonal
complement is of dimension at most $n-1$. Let $x_{1},\ldots,x_{n}$ be a dual
basis of $e_{1},\ldots,e_{n}$. An element in the orthogonal complement is a
linear combination of $[x_{i},x_{j}]=x_{i}x_{j}-x_{j}x_{i}$. Here is a set of
solutions:

$\displaystyle[e_{1},e_{1}]+[e_{1},e_{2}]+\cdots+[e_{1},e_{n}]$
$\displaystyle[e_{3},e_{1}]+[e_{3},e_{2}]+\cdots+[e_{3},e_{n}]$
$\displaystyle\vdots$
$\displaystyle[e_{n},e_{1}]+[e_{n},e_{2}]+\cdots+[e_{n},e_{n}]$

In a suitable order, the leading monomials are
$e_{1}e_{2},e_{1}e_{3},\ldots,e_{1}e_{n}$ and hence this gives $n-1$
independent elements in the orthogonal complement, so it is in fact a basis.
Summing the above elements, we get also
$[e_{2},e_{1}]+[e_{2},e_{2}]+\cdots+[e_{2},e_{n}]$. Letting
$c=\sum_{i=1}^{n}e_{i}$ we get that the holonomy Lie algebra of the matroid of
rank 2 on $\\{e_{1},\ldots e_{n}\\}$ is

$\eta={\mathcal{F}}(e_{1},\ldots,e_{n})/\langle[e_{i},c];i=1,\ldots,n\rangle$

where ${\mathcal{F}}(e_{1},\ldots,e_{n})$ is the free Lie algebra on
$e_{1},\ldots,e_{n}$. Hence $\eta/\langle c\rangle$ is free on $n-1$
generators and $\eta$ is the direct product of a free Lie algebra on $n-1$
generator and a free Lie algebra on one generator. The holonomy Lie algebra of
an arbitrary simple matroid on $E=\\{e_{1},\ldots,e_{n}\\}$ (or: of a set-
arrangement on $E$) is the free Lie algebra on $e_{1},\ldots,e_{n}$ modulo
relations as above for each 2-flat with at least three elements (or: for each
set in the set-arrangement). The variables in a 2-flat with two elements
commute (or: the pair of variables which are not contained in any set of the
set-arrangement commute). The question is how these ”local” Lie algebras
defined from the 2-flats are glued together.

##  4 Structure of the holonomy Lie algebra

Much of the theory for the structure of the full holonomy Lie algebra in terms
of the local ones does not depend on how the relations look like. Hence we
assume that we are given a set-arrangement ${\mathcal{A}}$ on
$E=\\{x_{1},\ldots,x_{n}\\}$ with local Lie algebras
$L_{A}={\mathcal{F}}(A)/\langle R_{A}\rangle$, where
$R_{A}\subset{\mathcal{F}}(A)^{\prime}=[{\mathcal{F}}(A),{\mathcal{F}}(A)]$,
for each $A\in{\mathcal{A}}$. Define ${\mathcal{L}}={\mathcal{F}}(E)/\langle
R\rangle$, where

$R=\cup_{A\in{\mathcal{A}}}R_{A}\cup\\{[x_{i},x_{j}];\ \\{x_{i},x_{j}\\}\text{
is not a subset of any }A\in{\mathcal{A}}\\}$

Define a Lie homomorphism $\pi_{A}:\ {\mathcal{F}}(E)\to L_{A}$ by sending
$x\in E\setminus A$ to zero. Since any Lie-monomial in $R\setminus R_{A}$
contains at least one generator $\notin A$, the map factors and defines a
surjection $\pi_{A}:\ {\mathcal{L}}{}\to L_{A}$. There is also a natural map
$s_{A}:\ L_{A}\to{\mathcal{L}}{}$ and $\pi_{A}\circ s_{A}=\mathrm{id}_{L_{A}}$
for each $A\in{\mathcal{A}}$. Hence we have a split exact sequence of Lie
algebras for each $A\in{\mathcal{A}}$,

$0\to I_{A}\to{\mathcal{L}}{}\to L_{A}\to 0$ (1)

Now consider the derived Lie algebras, ${\mathcal{L}}{}^{\prime}$,
$L_{A}^{\prime}$ and the Lie homomorphisms $\pi_{A}^{\prime},s_{A}^{\prime}$.
We have that $\pi_{B}^{\prime}\circ s_{A}^{\prime}=0$ for $B\neq A$. There is
a surjective Lie homomorphism
$\pi=\oplus\pi_{A}^{\prime}:{\mathcal{L}}{}^{\prime}\to\oplus L_{A}^{\prime}$,
which has a section $s=\sum s_{A}^{\prime}$ that is a module homomorphism (not
Lie in general). Hence we have an exact sequence of Lie algebras

$0\to
I\to{\mathcal{L}}{}^{\prime}\to\oplus_{A\in{\mathcal{A}}}L_{A}^{\prime}\to 0$
(2)

and $\sum\mathrm{im}(s_{A}^{\prime})\subset{\mathcal{L}}{}^{\prime}$ is a
direct sum, since $s$ is injective.

###### Example 4.1

In [17] the following set-arrangement is studied

$(124),\ (135),\ (236),\ (167),\ (258),\ (457),\ (468)$

and it is (almost) proved that the ideal $I$ in (2) is one-dimensional in each
degree $\geq 3$.

###### Proposition 4.2

$I=0\ \Leftrightarrow\ [x,\mathrm{im}(s_{A}^{\prime})]=0\text{ for all
}A\text{ and all }x\in E\setminus A$

Proof.

$\displaystyle
I=0\Rightarrow\ker(\oplus\pi_{B}^{\prime})=0\Rightarrow[x,\mathrm{im}(s_{A}^{\prime})]=0,x\notin
A\Rightarrow$ $\displaystyle\sum\mathrm{im}(s_{A}^{\prime})\text{ is an ideal
containing }{\mathcal{L}}{}^{\prime}\Rightarrow$ $\displaystyle s\text{
surjective }\Rightarrow s\text{ iso }\Rightarrow\pi\text{ mono }\Rightarrow
I=0$

$\Box$

###### Definition 4.3

A Lie algebra ${\mathcal{L}}{}$ of a set-arrangement ${\mathcal{A}}$ is said
to be decomposable if $I=0$.

In order to get more precise results, we introduce a condition on the local
Lie algebras which is satisfied for the holonomy Lie algebra.

###### Definition 4.4

A Lie algebra ${\mathcal{L}}{}$ of a set-arrangement ${\mathcal{A}}$ is said
to satisfy replacement if for any $A\in{\mathcal{A}}$ and any $x\in A$ such
that also $x\in B$ for some $B\in{\mathcal{A}}$, $B\neq A$, it holds that
$L_{A}^{\prime}\subseteq\mathrm{subalg}_{L_{A}}(A\setminus\\{x\\})$.

Here $\mathrm{subalg}_{L}(X)$ denotes the Lie subalgebra of $L$ generated by
the set $X\subseteq L$.

The following proposition is easily proved by induction.

###### Proposition 4.5

If ${\mathcal{L}}{}$ satisfies replacement, then

  * •

$I=0\Leftrightarrow[x,[y,z]]$ for all $A\in{\mathcal{A}}$ and all $x\in
E\setminus A$ and all $y,z\in A$

  * •

$I_{A}=\mathrm{subalg}_{\mathcal{L}}(E\setminus A)$ for all
$A\in{\mathcal{A}}$

The first part of the proposition is a result in [15] and is implemented in
Macaulay2.

The formulas (1) and (2) may be generalized in the following way.

###### Definition 4.6

Given a set-arrangement ${\mathcal{A}}$ on $E$, a subset ${\mathcal{B}}$ of
${\mathcal{A}}$ is closed in ${\mathcal{A}}$ if for all
$A\in{\mathcal{A}}\setminus{\mathcal{B}}$ it holds that
$|A\cap\mathrm{supp}({\mathcal{B}})|\leq 1$ where
$\mathrm{supp}({\mathcal{B}})=\cup_{B\in{\mathcal{B}}}B$.

Let ${\mathcal{A}}$ be a set-arrangement on $E$ and ${\mathcal{B}}$ a subset
of ${\mathcal{A}}$ which is closed in ${\mathcal{A}}$. Define
$L_{{\mathcal{B}}}={\mathcal{F}}(\mathrm{supp}({\mathcal{B}}))/\langle
R_{{\mathcal{B}}}\rangle$, where

$R_{{\mathcal{B}}}=\cup_{B\in{\mathcal{B}}}R_{B}\cup\\{[x,y];\
x,y\in\mathrm{supp}({\mathcal{B}})\text{ and }\forall B\in{\mathcal{B}}\
\\{x,y\\}\not\subseteq B\\}$

It follows that $L_{{\mathcal{B}}}=L_{A}$, when ${\mathcal{B}}=\\{A\\}$ and
given the definitions of the local Lie algebras $L_{A}$ for
$A\in{\mathcal{A}}$, $L_{{\mathcal{B}}}$ is the Lie algebra of ${\mathcal{B}}$
viewed as a set-arrangement on $\mathrm{supp}({\mathcal{B}})$.

A map $\pi_{{\mathcal{B}}}:{\mathcal{L}}\to L_{{\mathcal{B}}}$ is defined by
sending the variables in $E\setminus\mathrm{supp}({\mathcal{B}})$ to zero. A
natural map $s_{{\mathcal{B}}}:L_{{\mathcal{B}}}\to{\mathcal{L}}$ may be
defined since two different elements in $\mathrm{supp}({\mathcal{B}})$ is not
contained in a single $A$, where $A\not\in{\mathcal{B}}$. Also,
$\pi_{{\mathcal{B}}}\circ s_{{\mathcal{B}}}=\mathrm{id}_{L_{{\mathcal{B}}}}$.
Hence we have a split exact sequence of Lie algebras

$0\to I_{{\mathcal{B}}}\to{\mathcal{L}}{}\to L_{{\mathcal{B}}}\to 0$ (3)

Now suppose we have disjoint closed subsets ${\mathcal{B}}_{i},\ i=1,\ldots k$
of ${\mathcal{A}}$, such that $\cup_{i=1}^{k}{\mathcal{B}}_{i}={\mathcal{A}}$.
In the same way as above, we may define a surjective Lie homomorphism
$\pi=\oplus\pi_{{\mathcal{B}}_{i}}^{\prime}:{\mathcal{L}}{}^{\prime}\to\oplus
L_{{\mathcal{B}}_{i}}^{\prime}$, which has a section $s=\sum
s_{{\mathcal{B}}_{i}}^{\prime}$ that is a module homomorphism (not Lie in
general). Hence we have an exact sequence of Lie algebras

$0\to
J\to{\mathcal{L}}{}^{\prime}\to\oplus_{i=1}^{k}L_{{\mathcal{B}}_{i}}^{\prime}\to
0$ (4)

and
$\sum\mathrm{im}(s_{{\mathcal{B}}_{i}}^{\prime})\subset{\mathcal{L}}{}^{\prime}$
is a direct sum.

###### Proposition 4.7

We have

  * •

$J$ is generated as an ideal by
$[x,\mathrm{im}(s_{{\mathcal{B}}_{i}}^{\prime})]\text{ for all }i\text{ and
all }x\in E\setminus\mathrm{supp}({\mathcal{B}}_{i})$

  * •

$J=0\ \Leftrightarrow\ [x,\mathrm{im}(s_{{\mathcal{B}}_{i}}^{\prime})]=0\text{
for all }i\text{ and all }x\in E\setminus\mathrm{supp}({\mathcal{B}}_{i})$

The proof is the same as the proof of Proposition 4.2. We also have the
following generalization of Proposition 4.5.

###### Proposition 4.8

If ${\mathcal{L}}{}$ satisfies replacement, then

  * •

$J$ is generated as an ideal by $[x,[y,z]]\text{ for all
}y,z\in\mathrm{supp}({\mathcal{B}}_{i}),\ x\in
E\setminus\mathrm{supp}({\mathcal{B}}_{i}),\ i=1,\ldots,k$

  * •

$J=0\Leftrightarrow[x,[y,z]]=0\text{ for all
}y,z\in\mathrm{supp}({\mathcal{B}}_{i}),\ x\in
E\setminus\mathrm{supp}({\mathcal{B}}_{i}),\ i=1,\ldots,k$

  * •

$I_{{\mathcal{B}}_{i}}=\mathrm{subalg}_{\mathcal{L}}(E\setminus\mathrm{supp}({\mathcal{B}}_{i}))$
for all $i=1,\ldots k$

###  4.1 Application to graphs

Paulo Lima-Filho and Hal Schenk [8] proved a general formula for the series of
the holonomy Lie algebra of graph arrangements. We will now describe their
technique. Let $G$ be a graph and ${\mathcal{A}}_{G}$ the corresponding set-
arrangement on the edge set consisting of 2-flats of size at least three. A
2-flat has at most three elements, since 4 edges always contain 3 edges which
is not a triangle and hence has rank 3. Hence ${\mathcal{A}}_{G}$ consists of
all triangles in the graph. Let $L_{G}$ be the holonomy Lie algebra of
${\mathcal{A}}_{G}$. Hence $L_{G}$ is the free Lie algebra on the set of edges
modulo all $[x,y]$ where $x,y$ is not part of a triangle and also for each
triangle $(a,b,c)$ there are the relations $[a,b]-[b,c]$ and $[a,c]-[c,b]$. A
subgraph of $G$ is closed in the meaning above if for all pairs $a,b$ of edges
in the subgraph such that there is a triangle $(a,b,c)$ in $G$, it holds that
also $c$ is in the subgraph.

Let $A$ be a vertex and $V$ the subgraph of all edges going out from $v$. Let
$K$ be the subgraph of edges between endpoints of the edges in $V$, let
$G_{1}=V\cup K$ and $G_{2}=G\setminus V$. Then

  * •

$G_{1}$ and $G_{2}$ are closed in $G$.

  * •

$K$ is closed in $G_{1}$ and $G_{2}$.

  * •

If $K$ is complete, then $\ker(L_{G}\xrightarrow{\pi}L_{G_{2}})$ is free on
the edges in $V$.

The first two claims are easily checked. We prove the last claim. We know that
$\ker(\pi)=\mathrm{subalg}_{L_{G}}(V)$. Define an action of $L_{G_{2}}$ on
${\mathcal{F}}(V)$ by

$x\circ v=\begin{cases}0&\text{if }x\notin K\\\ [v,w]&\text{if }v,w\in V\text{
and }(x,v,w)\text{ is a triangle}\end{cases}$

We want to prove that the relations in $L_{G_{2}}$ operate as zero. If $[x,y]$
is a relation, then either one of $x,y$ is not in $K$, and then $[x,y]$
operates as zero or $x,y$ have no common vertex and then $x\circ v=0$ or
$y\circ v=0$ for all $v\in V$. Now suppose $(x,y,z)$ is a triangel in $G_{2}$.
If two of $x,y,z$ are not in $K$ then $[x,y],[y,z],[z,x]$ all operate as zero.
If two of $x,y,z$ are in $K$, then also the third is. We can look at the
subgraph consisting of four vertices $A,B,C,D$ and the four edges $v_{1}=AB,\
v_{2}=AC,\ v_{3}=AD$ and $x=BC,\ y=CD,\ z=BD$. A computation shows that
$[x,y]\circ v_{1}=[y,z]\circ v_{1}=[z,x]\circ v_{1}=[v_{1},[v_{3},v_{2}]]$. It
follows that there is a semi-direct product of $L_{G_{2}}$ operating on
${\mathcal{F}}(V)$ which is a quotient of $L_{G}$ and at the same time this
Lie algebra maps onto $L_{G}$. Hence they are isomorphic and the claim
follows.

###### Corollary 4.9

(Kohno) The holonomy Lie algebra of the complete graph $K_{n}$ is a successive
semi-direct product of free Lie algebras on $1,2,\ldots,n-1$ generators and
the series is $\Pi_{i=1}^{n-1}1/(1-it)$.

###### Corollary 4.10

(Papadima-Suciu) The holonomy Lie algebra $L_{G}$ of a graph $G$ is not
decomposable iff $G$ contains $K_{4}$ as a subgraph.

Proof. We know that $L_{G}$ is not decomposable iff there are three edges
$x,y,z$ which is not a triangle such that $[x,[y,z]]\neq 0$ in $L_{G}$.
Suppose $K_{4}$ is a subgraph of $G$. By the last claim above we know that the
three edges $x,y,z$ going from one vertex in $K_{4}$ satisfy $[x,[y,z]]\neq 0$
in $L_{K_{4}}$. We prove by induction that $L_{K_{4}}$ is a Lie subalgebra of
$L_{G}$. Suppose $A$ is a vertex in $G$ which is not in the subgraph $K_{4}$
and consider the subdivision of $G$ in $G_{1},K,G_{2}$ as above. Since $G_{2}$
is closed in $G$, $L_{G_{2}}$ is a Lie subalgebra of $L_{G}$. Since $K_{4}$ is
a subgraph of $G_{2}$ it follows by induction that $L_{K_{4}}$ is a Lie
subalgebra of $L_{G_{2}}$ and hence of $L_{G}$ which proves one direction of
the claim. Suppose there are three edges $x,y,z$ in $G$ which is not a
triangle such that $[x,[y,z]]\neq 0$ in $L_{G}$. Then $[y,z]\neq 0$ and hence
there is an edge $u\neq x$ such that $y,z,u$ is a triangle and
$[x,[y,z]]=[x,[z,u]]=[x,[u,y]]$. Let the vertices of the triangle be $A,B,C$,
where $y=AB,z=AC,u=BC$. The edge $x$ must be connected to the triangle, say
$x=AD$. Then $[x,u]=[x,BC]=0$ and hence $[x,[y,z]]=[[x,z],u]=[u,[x,y]]$ which
implies that $[x,z]\neq 0$ and $[x,y]\neq 0$. Hence there are edges
$v=CD,w=BD$ and it follows that $ABCD$ is a $K_{4}$. $\Box$

Here is an example of a non-representable set-arrangement, where the same
technique as above may be used to obtain the structure of the holonomy Lie
algebra $L$. It is the ”Fano” matroid of the seven non-zero vectors in
$\mathbb{Z}_{2}^{3}$ (and hence representable in characteristic 2). There are
seven triples of dependent vectors, which may be numbered as follows

$(123),\ (145),\ (167),\ (246),\ (257),\ (347),\ (356)$

We have an exact sequence $0\to\mathrm{subalg}_{L}(4567)\to L\to L_{(123)}\to
0$. We may define an action of $L_{(123)}$ onto ${\mathcal{F}}(4567)$ using
the six last triples and checking that the relations operate as zero. This
gives that $\mathrm{subalg}_{L}(4567)$ is free and we get that $L$ is a
successive semi-direct product of free Lie algebras on 4,2,1 generators.
Dividing $L$ by the ideal generated by $[1,2]$ gives the ”non-Fano” holonomy
Lie algebra with set-arrangement as above but with (123) deleted. Its series
seems difficult to determine.

Here is the method by Lima-Filho and Schenk [8] to obtain the series for
$L_{G}$ for an arbitrary graph $G$. Consider the subdivision of $G$ above. We
have a commutative diagram of surjective maps of Lie algebras,

$\begin{CD}L_{G}@>{}>{}>L_{G_{1}}\\\ @V{}V{}V@V{}V{}V\\\
L_{G_{2}}@>{}>{}>L_{K}\end{CD}$

which they proved is a pullback diagram. There is also a pushout diagram of
injective maps of Lie algebras,

$\begin{CD}L_{K}@>{}>{}>L_{G_{1}}\\\ @V{}V{}V@V{}V{}V\\\
L_{G_{2}}@>{}>{}>L_{G}\end{CD}$

yielding $L_{G}\cong(L_{G_{1}}\oplus L_{G_{2}})/L_{K}$ (where $L_{K}$ is
considered as the ”diagonal” in the direct sum). From this they find their
formula by induction. The result is that the series contains the series for
the free Lie algebra on $j$ generators, $1/(1-jt)$, with exponent

$\sum_{s\geq j}(-1)^{s-j}{s\choose j}\kappa_{s}$

where $\kappa_{s}$ is the number of complete subgraphs of $G$ on $s+1$
vertices.

Question 1: Is this exponent always positive?

In [17] there are examples of ”exotic” holonomy Lie algebras, e.g., the LCS
series, i.e., the Hilbert series of the enveloping algebra, can be
transcendental, disproving a natural question, inspired by the result above
for graphs, namely if the holonomy Lie algebra always can be obtained in a
finite number of steps from free Lie algebras by extensions and pullbacks. But
the following question might be open.

Question 2: Is the global dimension finite for the enveloping algebra of a
holonomy Lie algebra $\eta$, i.e., is it true that
$\mathrm{Tor}_{i}^{\mathrm{U}(\eta)}(k,k)$ for big enough $i$?

## References

  * [1] Anders Björner, Subspace arrangements, First European Congress of Mathematics, Progress in Mathematics, vol 3, (1994), 321–370.
  * [2] Anders Björner and Günter Ziegler, Combinatorial stratification of complex arrangements, J. of the Amer. Math. Soc. 5, (1992), 105–149.
  * [3] Egbert Brieskorn, Sur les groupes de tresses, Séminaire Bourbaki 24e année 1971/72, Lecture Notes in Math. 317 (1973), 21–44.
  * [4] Graham Denham and Alexander Suciu, On the homotopy Lie algebra of an arrangement, Michigan Math. J. 54:2 (2006), 319–340.
  * [5] Mark Goresky and Robert MacPherson, Stratified Morse theory, Ergebnisse der Mathematik und ihrer Grenzgebiete, vol 14, Springer Verlag, New York, 1989.
  * [6] Toshitake Kohno, On the holonomy Lie algebra and the nilpotent completion of the fundamental group of the complement of hypersurfaces, Nagoya Math. J. 92 (1983), 21–37.
  * [7] Lionel Levine, Orlik-Solomon algebras of hyperplane arrangements, notes 2004 available at http://www.math.cornell.edu/$\sim$levine/orliksolomonfinal.pdf.
  * [8] Paulo Lima-Filho and Hal Schenck, The holonomy Lie algebra and the LCS formula for subarrangements of $A_{n}$, International Mathematics Research Notices, 8 (2009), 1421-1432.
  * [9] Mark de Longueville and Carsten Schultz, The cohomology rings of complements of subspace arrangements, Math. Annalen 319, (2001), 625–646.
  * [10] Clas Löfwall, Decompositions theorems for a generalization of the holonomy Lie algebra of an arrangement, to appear in Communications in Algebra.
  * [11] Clas Löfwall, “liedim, A Mathematica program for calculating in finitely presented graded Lie superalgebras”, Available on line via the page http://www.math.su.se/forskning/datorprogram-utvecklade-vid-institutionen-1.70986, or direct from http://www2.math.su.se/liedim/.
  * [12] Peter Orlik and Louis Solomon, Combinatorics and topology of complements of hyperplanes, Inventiones Math. 56, (1980), 167–189.
  * [13] James Oxley, What is a matroid? available from https://www.math.lsu.edu/$\sim$oxley/survey4.pdf
  * [14] James Oxley, Matroid theory, Oxford Graduate Texts in Mathematics 21, second edition, (2011).
  * [15] Stefan Papadima and Alexander I. Suciu, When does the associated graded Lie algebra of an arrangement group decompose?, Comment. Math. Helv. 81:4 (2006), 859–875.
  * [16] Irena Peeva, Hyperplane arrangements and linear strands in resolutions, Trans. Amer. Math. Soc. 355 (2003), 609?618.
  * [17] Jan-Erik Roos, The Homotopy Lie Algebra of a Complex Hyperplane Arrangement Is Not Necessarily Finitely Presented, Experimental Math. 17:2 (2006), 129–143.
  * [18] Hal Schenk, Hyperplane Arrangements: Computations and Conjectures, in Advanced Studies in Pure Mathematics, 62, (2012) 323-358.
  * [19] Richard Stanley, An Introduction to Hyperplane Arrangements, Lecture notes, IAS/Park City Mathematics Institute, (2004).
  * [20] Thomas Zaslavsky, Facing up to arrangements: Face-count formulas for partitions of space by hyperplanes Thesis (MIT, 1974) and Memoirs Amer. Math. Soc., No. 154, (1975).


# An Elementary Proof of the Local Kronecker-Weber Theorem

Jochen Koenigsmann, Benedikt Stock

###### Abstract

We will present a novel elementary and self-contained proof of the local
Kronecker-Weber theorem. Apart from discrete valuation theory, it does not
make use of any tools beyond those introduced in a second undergraduate course
on algebra. In particular, we will not make use of results from local class
field theory or Galois cohomology.

_“There are several ways to prove the Kronecker–Weber theorem, even if class
field theory […] seems inevitable.”_ – P. Guillot (Preface, [Guillot18])

##  1 Introduction

The Kronecker-Weber theorem states that every finite abelian extension of the
rational numbers is contained in a cyclotomic extension. A number of proofs of
this result are known. Culler [Culler07] presents two proofs; an elementary
approach based on David Hilbert’s original proof and one that makes use of
(global) class field theory. Another classical approach goes via the local
Kronecker-Weber theorem – the subject of this article – where the role of
$\mathbb{Q}$ is replaced by $\mathbb{Q}_{p}$. Once the local version is
established, it is not hard to deduce the global version via completions,
Minkowski’s bound, and basic properties of discriminants (cf. [Guillot18],
Chapter 14 for an excellent presentation of a complete proof).

Similar to the global version, a few different proofs of the local Kronecker-
Weber theorem are known. Perhaps the most classical one goes via the
fundamental theorem of local class field theory, a result that itself is
usually either proved via the cohomological approach (cf. [Guillot18]) or
Lubin-Tate theory (cf. [Childress08, Hazewinkel75]). Our proof is distinct
from those as it makes no mention of local class field theory at all. Papers
by Rosen [Rosen81] and Kozuka [Kozuka90] have a similar goal in mind. However,
the new approach presented here differs from these in that it is self-
contained and does not even mention Lubin-Tate constructions or Galois
cohomology. Instead, our considerations are merely based on discrete valuation
theory combined with basic results from (infinite) Galois theory and Kummer
theory.

In the next section we will remind the reader of some facts about local fields
before presenting the proof of the local Kronecker-Weber theorem in section 3.

##  2 Some facts about local number fields

Let us first fix some notation. We call finite extensions of $\mathbb{Q}_{p}$
_local number fields_. For a finite local number field $L$, we denote by
$\mathcal{O}_{L}$ its ring of integers, by $\operatorname{\mathcal{M}}_{L}$
the unique maximal ideal of $\mathcal{O}_{L}$, and by
$\overline{L}=\mathcal{O}_{L}/\operatorname{\mathcal{M}}_{L}$ its residue
field. Moreover, we take $v_{L}$ to mean the unique extension of the $p$-adic
valuation $v_{p}$ to $L$. Finally, for an extension $L/F$ of local number
fields of degree $n$, we denote by $e=e(L/F)$ and $f=f(L/F)$ the ramification
index and inertia degree of the extension, respectively.

We remind the reader of the fundamental equality $n=ef$ (cf. [Guillot18],
Theorem 2.38; [Neukirch99], II. 6.8). What is more, there exists an
intermediate field $L_{0}$, called the _inertia subfield_ of $L$, with the
property that for any intermediate field $F\subseteq K\subseteq L$, the
extension $K/F$ is unramified if and only if $K\subseteq L_{0}$ (cf.
[Guillot18], Corollary 2.43).

We shall also remind the reader that we have very explicit descriptions for
the cyclotomic extensions of $\mathbb{Q}_{p}$ (cf. [Guillot18], Propositions
2.46 and 2.48) as summarized in the next lemma. Here and in the remaining
text, we let $C_{n}$, $\mu_{n}$, and $\zeta_{n}$ denote the cyclic group of
order $n$, the group of $n$th roots of unity, and a primitive $n$th root of
unity, respectively.

######  Lemma 2.1.

Let $L=\mathbb{Q}_{p}(\mu_{N})$ be a cyclotomic extension of $\mathbb{Q}_{p}$.

  * •

If $(N,p)=1$, then the extension $L/\mathbb{Q}_{p}$ is unramified, and its
degree $n$ is the smallest natural number such that $p^{n}\equiv 1\
(\mathrm{mod}\ N)$. In particular, $L=L_{0}$ and
$\text{Gal}(L/\mathbb{Q}_{p})\cong C_{n}$.

  * •

If $N=p^{n}$, then $L/\mathbb{Q}_{p}$ is totally ramified of degree
$p^{n-1}(p-1)$. Moreover,
$\text{Gal}(L/\mathbb{Q}_{p})\cong(\mathbb{Z}/p^{n}\mathbb{Z})^{\times}$, and
$1-\zeta_{p^{n}}$ is a uniformizer of $L$ with
$v_{L}(1-\zeta_{p^{n}})=\frac{1}{p^{n-1}(p-1)}$ and field-theoretic norm
$N_{L/\mathbb{Q}_{p}}(1-\zeta_{p^{n}})=p$.

######  Corollary 2.2.

The roots of unity in $\mathbb{Q}_{p}$ are exactly $\mu_{p-1}$ for $p$ odd,
and $\mu_{2}$ for $p=2$.

The following subgroups of the unit groups of local number will play a central
role in our argument.

######  Definition 2.3 (higher unit groups).

Let $L$ be a local number field. The _$n$ -th higher unit group_ is given by
$U^{(0)}_{L}=\mathcal{O}_{L}^{\times}$ for $n=0$ and
$U^{(n)}_{L}=\\{x\in\mathcal{O}_{L}^{\times}:x\equiv 1\ (\mathrm{mod}\
\operatorname{\mathcal{M}}_{L}^{n})\\}$ for $n\geq 1$.

One can easily classify the quotients of consecutive higher unit groups. For a
local number field $L$,

$U_{L}^{(0)}/U_{L}^{(1)}\cong\overline{L}^{\times}$

(consider the map $x\mapsto\overline{x}$ from
$U_{L}^{(0)}=\mathcal{O}_{L}^{\times}$ to $\overline{L}^{\times}$), and

$U_{L}^{(n)}/U_{L}^{(n+1)}\cong\overline{L}$

for any integer $n\geq 1$ (take a uniformizer $\pi\in\mathcal{O}_{L}$ and
consider the map $x\mapsto\overline{~{}\frac{x-1}{\pi^{n}}~{}}$ from
$U^{(n)}_{L}$ to $\overline{L}$).

######  Lemma 2.4.

Let $L$ be a local number field, $\pi$ denote a uniformizer of $L$, and
$f=f(L/\mathbb{Q}_{p})$. Then we may represent every $a\in
U_{L}^{(1)}/U_{L}^{(n+1)}$ uniquely as

$a=1+\sum_{i=1}^{n}a_{i}\pi^{i},$

with coefficients $a_{i}\in\\{0\\}\cup\mu_{p^{f}-1}$.

###### Proof.

From the isomorphisms just presented, we see that the size is correct. To show
uniqueness, assume two sums $1+\sum_{i=1}^{n}a_{i}\pi^{i}$ and
$1+\sum_{i=1}^{n}a_{i}^{\prime}\pi^{i}$ represented the same element. Then
there would be a smallest $i$ with $a_{i}\neq a_{i}^{\prime}$ and we derive a
contradiction by considering both sums modulo $U_{L}^{(i+1)}$. ∎

The following lemma is a neat result that will come in handy multiple times.

######  Lemma 2.5.

Let $F/\mathbb{Q}_{p}$ be a finite extension with $\zeta_{p}\in F$ and
valuation $v$ extending $v_{p}$. Then
$\overline{\frac{p}{(1-\zeta_{p})^{p-1}}}=-1$.

###### Proof.

Write

$g(X)=\sum_{i=0}^{p-1}X^{i}=\prod_{i=1}^{p-1}(X-\zeta_{p}^{i}),$

so

$\frac{p}{(1-\zeta_{p})^{p-1}}=\frac{g(1)}{(1-\zeta_{p})^{p-1}}=\prod_{i=1}^{p-1}\frac{1-\zeta^{i}_{p}}{1-\zeta_{p}}=\prod_{i=1}^{p-1}\sum_{j=0}^{i-1}\zeta_{p}^{j}.$

By lemma 2.1, $\zeta_{p}-1$ is a uniformizer of $\mathbb{Q}_{p}(\zeta_{p})$.
Hence, $\zeta_{p}-1\in\operatorname{\mathcal{M}}_{F}$, i.e., $\zeta_{p}\equiv
1\ (\mathrm{mod}\ \operatorname{\mathcal{M}}_{F})$. This implies
$\zeta_{p}^{j}\equiv 1\ (\mathrm{mod}\ \operatorname{\mathcal{M}}_{F})$ for
all integers $j$, so $\sum_{j=0}^{i-1}\zeta_{p}^{j}\equiv i\ (\mathrm{mod}\
\operatorname{\mathcal{M}}_{F})$. This means that
$\prod_{i=1}^{p-1}\sum_{j=0}^{i-1}\zeta_{p}^{j}\equiv(p-1)!\ (\mathrm{mod}\
\operatorname{\mathcal{M}}_{F})$, so the claim follows from Wilson’s theorem
since $p\in\operatorname{\mathcal{M}}_{F}$. ∎

The next two corollaries and their proofs are slight adaptations of an idea
first observed by Koenigsmann ([Koenigsmann03], Lemma 3.2). Following our
philosophy of a self-contained presentation, we present it here again for the
convenience of the reader.

######  Corollary 2.6.

Let $F/\mathbb{Q}_{p}$ finite with $\zeta_{p}\in F$ and ramification index
$e$. Then
$U^{(\frac{ep}{p-1}+1)}\subseteq\left(U^{(\frac{e}{p-1})}\right)^{p}$.

###### Proof.

First, note that the statement makes sense as $\zeta_{p}\in F$ and
$\mathbb{Q}_{p}(\zeta_{p})/\mathbb{Q}_{p}$ is totally ramified of degree $p-1$
by lemma 2.1, so $p-1|e$.

Now let us denote by $\Pi$ a uniformizer of $F$ and by $v$ the unique
valuation of $F$ extending $v_{p}$. Moreover, define $\pi:=1-\zeta_{p}$, which
is a uniformizer of $\mathbb{Q}_{p}(\zeta_{p})$ with $v(\pi)=\frac{1}{p-1}$ by
lemma 2.1. Now, by (the extension of) Lagrange’s theorem, we see that
$v\left(\Pi^{\frac{e}{p-1}}\right)=v(\pi)$, so
$\left(\Pi^{\frac{e}{p-1}}\right)=(\pi)$ and
$\left(\Pi^{\frac{ep}{p-1}}\right)=(\pi)^{p}$. This implies in particular that
we may rewrite $U^{(\frac{e}{p-1})}=1+(\pi)$ and
$U^{(\frac{ep}{p-1}+1)}=1+\pi^{p}(\Pi)$.

So take $x\in U^{(\frac{ep}{p-1}+1)}$ and write $x=1+\pi^{p}a$ with
$a\in(\Pi)$. We need to show that the polynomial $Y^{p}-(1+\pi^{p}a)=0$ has a
root in $1+(\pi)$, i.e., a solution of the form $Y=1+\pi X$, where
$X\in\mathcal{O}_{F}$. Dividing the polynomial by $\pi^{p}$ and plugging in
$Y=1+\pi X$, this becomes equivalent to finding a root $X\in\mathcal{O}_{F}$
of

$f(X)=\left(\frac{Y}{\pi}\right)^{p}-\left(\frac{1}{\pi^{p}}+a\right)=\left(X+\frac{1}{\pi}\right)^{p}-\left(\frac{1}{\pi^{p}}+a\right).$

Now,

$\displaystyle f(X)$
$\displaystyle=\sum_{i=0}^{p}\binom{p}{i}\left(\frac{1}{\pi}\right)^{p-i}X^{i}-\left(\frac{1}{\pi^{p}}+a\right)$
$\displaystyle=\sum_{i=1}^{p}\binom{p}{i}\left(\frac{1}{\pi}\right)^{p-i}X^{i}-a,$

and we claim that $f(X)\in\mathcal{O}_{F}[X]$. Indeed, as observed above,
$v(\pi)=\frac{1}{p-1}$, so the coefficients of $X^{i}$ ($i>0$) satisfy

$v\left(\binom{p}{i}\left(\frac{1}{\pi}\right)^{p-i}\right)=\underbrace{v\left(\binom{p}{i}\right)}_{\geq
1\text{~{}since~{}}p|\binom{p}{i}}-(p-i)\underbrace{v\left(\pi\right)}_{=\frac{1}{p-1}}\geq
0.$

We can say even more. For $1<i\leq p$, the above inequality is strict while we
get equality for $i=1$. As $a\in(\Pi)$, this means that

$\displaystyle\overline{f}(X)=\overline{\frac{p}{\pi^{p-1}}}X+X^{p}.$ (1)

But by the previous lemma 2.5$,\overline{\frac{p}{\pi^{p-1}}}=-1$, so the
claim follows immediately from Hensel’s lemma since $\overline{f}(X)$ is
separable and splits in $\mathbb{F}_{p}$. ∎

Adapting the proof of the previous lemma slightly even gives rise to a partial
converse.

######  Corollary 2.7.

Let $F/\mathbb{Q}_{p}$ be a finite extension with $\zeta_{p}\in F$ and assume
$1+(1-\zeta_{p})^{\frac{e}{p-1}}a\in F^{p}$ for some $a\in\mathcal{O}_{F}$.
Then there exists $x\in\overline{F}$ such that $x^{p}-x-\overline{a}=0$

###### Proof.

Setting $\pi=1-\zeta_{p}$ as before, $1+\pi^{\frac{e}{p-1}}a\in F^{p}$ means
that the equation $Y^{p}-(1+\pi^{p}a)=0$ has a root in $F$. Now, analogous to
the above, we set $X=\frac{Y-1}{\pi}\Leftrightarrow Y=1+\pi X$, giving rise to
a root of the polynomial $f(X)$ from above. Previously, $a\in(\pi)$ (so
$\overline{a}=0$), while now we have $a\in\mathcal{O}_{F}$. Therefore,
equation (1) now becomes $\overline{f}(X)=X^{p}-X-\overline{a}$, and the claim
follows. ∎

In some cases, the inclusion stated in corollary 2.6 even becomes an equality.

######  Lemma 2.8.

If $K=\mathbb{Q}_{p}(\zeta_{p})$, then
$U_{K}^{(p+1)}=\left(U_{K}^{(1)}\right)^{p}$.

###### Proof.

The inclusion $U^{(p+1)}\subseteq\left(U^{(1)}\right)^{p}$ was shown in
corollary 2.6 (as $e(F/\mathbb{Q}_{p})=p-1$ by lemma 2.1).

For the reversed inclusion, we let $x\in\left(U^{(1)}\right)^{p}\subseteq
U^{(1)}$. Then it suffices to show that $x$ represents a unit in
$U^{(1)}/U^{(p+1)}$. Write $x=(1+a\pi)^{p}$ for $a\in\mathcal{O}_{F}$ and
$\pi=1-\zeta_{p}$. As $\frac{p}{\pi^{p-1}}\in-1+(\pi)$ by lemma 2.5, we see
that $p\in-\pi^{p-1}+(\pi)^{p}$. Then

$x=(1+a\pi)^{p}=1+\underbrace{pa\pi}_{\in-a\pi^{p}+(\pi)^{p+1}}+\underbrace{\sum_{i=2}^{p-1}\binom{p}{i}(a\pi)^{i}}_{\in
p(\pi)^{2}=(\pi)^{p+1}}+a^{p}\pi^{p}\equiv 1+(a^{p}-a)\pi^{p}\ (\mathrm{mod}\
(\pi^{p+1})),$

Hence, by lemma 2.4, $x$ may be represented as $1+(a^{p}-a)\pi^{p}$ in
$U^{(1)}/U^{(p+1)}$. But $\overline{a}^{p}-\overline{a}\equiv 0\
(\mathrm{mod}\ p)$, so $x$ is indeed reduced to 1 modulo $U^{(p+1)}$. ∎

######  Corollary 2.9.

Let $F/\mathbb{Q}_{p}$ be a finite extension of degree $n$ with $\zeta_{p}\in
F$. Then $\dim_{\mathbb{F}_{p}}(F^{\times}/F^{\times p})=n+2$.

###### Proof.

As usual, we let $\pi$ denote a uniformizer of $F$ and let $f$ be the inertia
degree of $F$. It is then easy to see that
$F^{\times}=\langle\pi\rangle\times\mu_{p^{f}-1}\times U_{F}^{(1)}$ (cf.
[Guillot18], Proposition 4.2). Now
$\mu_{p^{f}-1}\cong\overline{F}^{\times}\cong\overline{F}^{\times p}$ by the
properties of finite fields. Hence, if we add $\pi$ to our basis, all that
remains to be shown is that $U^{(1)}/(F^{\times p}\cap
U^{(1)})=U^{(1)}/\left(U^{(1)}\right)^{p}$ has $\mathbb{F}_{p}$-dimension
$n+1$. In other words, we want to show that
$\left|U^{(1)}/\left(U^{(1)}\right)^{p}\right|=p^{n+1}$. To do so, we note
that

$U^{(1)}/\left(U^{(1)}\right)^{p}\equiv\frac{U^{(1)}/U^{(\frac{ep}{p-1}+1)}}{(U^{(1)})^{p}/U^{(\frac{ep}{p-1}+1)}}$

since $U^{(\frac{ep}{p-1}+1)}\subseteq\left(U^{(\frac{e}{p-1})}\right)^{p}$ by
corollary 2.6. Now we already know that

$\left|U^{(1)}/U^{(\frac{ep}{p-1}+1)}\right|=\left(p^{f}\right)^{\frac{ep}{p-1}}=p^{\frac{np}{p-1}}=p^{n+\frac{n}{p-1}}$

by lemma 2.4. Next, we want to calculate the cardinality of
$(U^{(1)})^{p}/U^{(\frac{ep}{p-1}+1)}$. To do so, consider an element of $a\in
U^{(1)}/U^{(\frac{ep}{p-1}+1)}$ and represent it as

$a=1+a_{1}\pi+\dots+a_{\frac{e}{p-1}}\pi^{\frac{e}{p-1}}+b\pi^{\frac{e}{p-1}+1}$

with $a_{i}\in\\{0\\}\cup\mu_{p^{f}-1}$ and $b\in\mathcal{O}$ by lemma 2.4. It
is easy to see that $a^{p}\in U^{(\frac{ep}{p-1}+1)}$ if $a_{i}=0$ for all
$i\in[\frac{e}{p-1}]$, independent of $b$ (note that $p\in(\pi)^{e}$). Hence,
elements of $(U^{(1)})^{p}/U^{(\frac{ep}{p-1}+1)}$ can be just represented as

$a^{p}=(1+a_{1}\pi+\dots+a_{\frac{e}{p-1}}\pi^{\frac{e}{p-1}})^{p}.$

Multiplying this out shows that $a^{p}\in U^{(\frac{ep}{p-1}+1)}$ if and only
if $a_{i}=0$ for all $i\in[\frac{e}{p-1}-1]$ and $a_{\frac{e}{p-1}}$ is a
$p$th power. Hence,
$\left|(U^{(1)})^{p}/U^{(\frac{ep}{p-1}+1)}\right|=(p^{f})^{\frac{e}{p-1}}/p=p^{\frac{n}{p-1}-1}$,
so

$\left|U^{(1)}/\left(U^{(1)}\right)^{p}\right|=\frac{\left|U^{(1)}/U^{(\frac{ep}{p-1}+1)}\right|}{\left|(U^{(1)})^{p}/U^{(\frac{ep}{p-1}+1)}\right|}=\frac{p^{n+\frac{n}{p-1}}}{p^{\frac{n}{p-1}-1}}=p^{n+1}$

as required. ∎

######  Corollary 2.10.

If $K=\mathbb{Q}_{p}(\zeta_{p})$, then the elements of the form
$p^{a_{0}}\left(1+\sum_{i=1}^{p}a_{i}\pi^{i}\right)K^{\times p}$ with
$a_{i}\in\\{0,\dots,p-1\\}$ form a set of representatives of
$K^{\times}/K^{\times p}$.

###### Proof.

By corollary 2.9, we know that $K^{\times}/K^{\times p}$ has
$\mathbb{F}_{p}$-dimension $p+1$, so the size is correct. Hence, it suffices
to show that each element in $K^{\times}/K^{\times p}$ is represented by one
of these elements. So take some element in $K^{\times}/K^{\times p}$ and
represent it as $\pi^{n}u$ for some $n\in\mathbb{Z}$ and
$u\in\mathcal{O}_{K}^{\times}$.

Let us first show that we may represent $\pi^{n}\equiv p^{a_{0}}\
(\mathrm{mod}\ O_{K}^{\times}K^{\times p})$ for some
$a_{0}\in\\{0,\dots,p-1\\}$. Indeed, if we write $p=\pi^{p-1}u^{\prime}$ for
some $u^{\prime}\in\mathcal{O}_{K}^{\times}$, then all that remains to be
shown is that there exists an $a_{0}$ such that
$\pi^{n}\equiv\pi^{(p-1)a_{0}}\ (\mathrm{mod}\ K^{\times p})$. Now just take
$a_{0}$ to be a solution of the linear congruence $(p-1)a_{0}\equiv n\
(\mathrm{mod}\ p)$.

Let us now focus on the representation of $u\in\mathcal{O}_{K}^{\times}$. We
claim that, modulo $K^{\times p}$, we may assume $u\in U_{K}^{(1)}$. Indeed,
it suffices to find $a\in K^{\times}$ such that $ua^{p}\in U_{K}^{(1)}$, and
this can be achieved by Hensels’ lemma. Then, by lemma 2.8,

$\displaystyle\frac{U_{K}^{(1)}K^{\times p}}{K^{\times
p}}\cong\frac{U_{K}^{(1)}}{K^{\times p}\cap
U_{K}^{(1)}}=\frac{U_{K}^{(1)}}{\left(U_{K}^{(1)}\right)^{p}}=\frac{U_{K}^{(1)}}{U_{K}^{(p+1)}},$
(2)

so the result follows from lemma 2.4. ∎

##  3 Proof of the local Kronecker-Weber theorem

We shall prove the following explicit version of the local Kronecker-Weber
theorem.

######  Theorem 3.1 (Local Kronecker-Weber Theorem).

Let $F/\mathbb{Q}_{p}$ be an abelian extension of degree $p^{n}\cdot m$ with
integers $m$ and $n$ such that $(p,m)=1$. Then
$F\subseteq\mathbb{Q}_{p}(\zeta_{p^{n+2}},\zeta_{p^{p^{n}\cdot m}-1})$.

Assume that $F/\mathbb{Q}_{p}$ has ramification index $e=p^{r}e^{\prime}$
(where $(e^{\prime},p)=1$) and inertia degree $f$. We want to split the proof
of the theorem into two steps, considering the cases $e=p^{r}$ and
$e=e^{\prime}$ with $(e,p)=1$ separately. To see why this is possible, note
that $G=\text{Gal}(F/\mathbb{Q}_{p})$ being finite abelian implies that it is
the direct product of its Sylow subgroups. We let $H_{1}$ denote the Sylow
$p$-subgroup of $G$, $H_{2}$ be the subgroup of $G$ generated by all Sylow
$l$-subgroups with $l\neq p$, and denote the corresponding fixed fields by
$L_{1}$ and $L_{2}$, respectively. Then, if we let $s$ be maximal with
$p^{s}|f$, the fundamental equality and the Galois correspondence will give us
that $[L_{1}:\mathbb{Q}_{p}]=\frac{ef}{p^{r+s}}$ and
$[L_{2}:\mathbb{Q}_{p}]=p^{r+s}$. By maximality of $r$ and $s$, this means
that $L_{1}L_{2}=F$. Hence, if we show that $L_{1}$ and $L_{2}$ are contained
in cyclotomic extensions, we get that $F$ is contained in the compositum of
those cyclotomic extensions (which, of course, is itself cyclotomic). Hence,
we are only left with showing that $L_{1}$ and $L_{2}$ are indeed contained in
the cyclotomic extension described in theorem 3.1.

We will treat both cases separately, starting with $L_{1}$. Following standard
terminology, we shall refer to the $L_{1}$-case as the _tame_ case and to the
$L_{2}$-case as the _wild_ case.

###  3.1 The tame case

Note that since $[L_{1}:\mathbb{Q}_{p}]=\frac{ef}{p^{r+s}}$ and by maximality
of $r$ and $s$, $(e(L_{1}/\mathbb{Q}_{p}),p)=(f(L_{1}/\mathbb{Q}_{p}),p)=1$.
Hence, the following lemma (with $F$ taking the role of $L_{1}$) applies in
our setting.

######  Lemma 3.2.

Let $F/\mathbb{Q}_{p}$ be abelian with ramification index $e$ and inertia
degree $f$ such that $(e,p)=1$. Then $e|p-1$.

###### Proof.

We will establish the claim by showing that every prime power dividing $e$
also divides $p-1$. So let us fix some prime $q|e$ (in particular $q\neq p$)
and let $r$ and $s$ be maximal such that $q^{r}|e$ and $q^{s}|f$. Now, let $H$
be the group generated by all Sylow $l$-subgroups of
$G=\text{Gal}(F/\mathbb{Q}_{p})$ with $l\neq q$ and let $L$ denote the fixed
field of $H$. Then $[L:\mathbb{Q}_{p}]=[G:H]$ is the maximal power of $q$ in
$|G|=ef$, so $L/\mathbb{Q}_{p}$ is of degree $q^{r+s}$ by the choice of $r$
and $s$. Moreover, let $L_{0}$ denote the inertia subfield of
$L/\mathbb{Q}_{p}$. Again by the choice of $r$ and $s$, we get
$[L_{0}:\mathbb{Q}_{p}]=q^{s}$. Hence,
$L_{0}=\mathbb{Q}_{p}(\zeta_{p^{q^{s}}-1})$ by lemma 2.1 and the uniqueness of
inertia fields. We also set $E=\mathbb{Q}_{p}(\zeta_{p^{q^{s+r}}-1})$, so
$E/\mathbb{Q}_{p}$ is unramified of degree $q^{s+r}$ by lemma 2.1. In
particular, the tower law implies $[\overline{E}:\overline{L_{0}}]=q^{r}$. The
following diagram gives an overview over all the extensions just described.

${EL}$${F}$${E}$${L}$${L_{0}}$${\mathbb{Q}_{p}}$$\scriptstyle{q^{r}}$$\scriptstyle{q^{s}}$$\scriptstyle{q^{r}}$$\scriptstyle{q^{r+s}}$

Let now $\pi$ be a uniformizer of $L$. As $L$ is ramified of degree $q^{r}$,
we have $v_{L}\left(\frac{\pi^{q^{r}}}{p}\right)=0$ so
$\frac{\pi^{q^{r}}}{p}\in\mathcal{O}_{L}^{\times}$. Then, since
$\overline{L_{0}}=\overline{L}$, we may find
$u\in\mathcal{O}_{L_{0}}^{\times}$ such that
$\overline{u}=\overline{\frac{~{}\pi^{q^{r}}}{p}~{}}$. This means that
$\frac{\pi^{q^{r}}}{pu}\in U_{L}^{(1)}$. Now, applying Hensel’s lemma with
$X^{q^{r}}-l$ for any $l\in U_{L}^{(1)}$, we see that $U_{L}^{(1)}\leq
L^{\times q^{r}}$ since $q\neq p$. Hence,
$\sqrt[q^{r}]{\frac{\pi^{q^{r}}}{pu}}\in L$ and thus $\sqrt[q^{r}]{pu}\in L$.
In particular,
$v_{L}(\sqrt[q^{r}]{pu})=\frac{1}{q^{r}}(v(p)+v(u))=\frac{1}{q^{r}}$, so we
even get $L=L_{0}(\sqrt[q^{r}]{pu})$.

Now, by the properties of finite fields, the polynomial
$X^{q^{r}}-\overline{u}$ will split into distinct linear factors in
$\overline{E}$, so $u$ is a $q^{r}$-th power in $E$ by Hensel’s lemma. In
particular, this implies that $EL=E(\sqrt[q^{r}]{pu})=E(\sqrt[q^{r}]{p})$.
Note that $EL/\mathbb{Q}_{p}$ is abelian as the compositum of two abelian
extensions. In particular, this implies that the intermediate extension
$\mathbb{Q}_{p}(\sqrt[q^{r}]{p})$ must be Galois, so we need
$\zeta_{q^{r}}\in\mathbb{Q}_{p}$. However, by corollary 2.2, this implies that
$q^{r}|p-1$. ∎

With this lemma at hand, we can immediately treat the tame case. Indeed, as
$(e(L_{1}/\mathbb{Q}_{p}),p)=(f(L_{1}/\mathbb{Q}_{p}),p)=1$, we have $n=0$ in
the setting of theorem 3.1, so the following lemma gives rise to subextensions
of the required form.

######  Corollary 3.3.

If $F/\mathbb{Q}_{p}$ is abelian with ramification index $e$ and inertia
degree $f$ such that $(e,p)=1$, then
$F\subseteq\mathbb{Q}_{p}(\zeta_{p},\zeta_{p^{ef}-1})$.

###### Proof.

Let $\pi$ denote a uniformizer of $F$ and define $u=\frac{\pi^{e}}{-p}$. Then
$v_{F}(u)=0$, so $u\in\mathcal{O}_{F}^{\times}$. Also set
$E=\mathbb{Q}_{p}(\zeta_{p^{ef}-1})$, so $EF$ is the unique unramified
extension of $F$ of degree $e$ by lemma 2.1. Therefore,
$\overline{EF}/\overline{F}$ is an extension of finite fields of degree $e$.
Hence, by the properties of finite fields, the separable polynomial
$X^{e}-\overline{u}$ will split in $\overline{EF}$, so $u$ becomes an $e$-th
power in $EF$ by Hensel’s lemma. As $\sqrt[e]{u}=\frac{\pi}{\sqrt[e]{-p}}$,
this means $\sqrt[e]{-p}\in EF$. But
$v(\sqrt[e]{-p})=\frac{1}{e}v(p)=\frac{1}{e}$, so the subextension
$E(\sqrt[e]{-p})$ of $EF/E$ is of degree $e$ which means that we must have
$EF=E(\sqrt[e]{-p})$. By the previous lemma 3.2, $e|p-1$, so
$E(\sqrt[e]{-p})\subseteq E(\sqrt[p-1]{-p})$ as represented in the following
diagram.

${E(\sqrt[p-1]{-p})}$${EF}$${E}$${F}$${\mathbb{Q}_{p}}$$\scriptstyle{e}$

We claim $\mathbb{Q}_{p}(\sqrt[p-1]{-p})=\mathbb{Q}_{p}(\zeta_{p})$ from which
the result will follow since then

$F\subseteq EF=E(\sqrt[e]{-p})\subseteq
E(\sqrt[p-1]{-p})=\mathbb{Q}_{p}(\sqrt[p-1]{-p})(\zeta_{p^{ef}-1})=\mathbb{Q}_{p}(\zeta_{p},\zeta_{p^{ef}-1}).$

To show $\mathbb{Q}_{p}(\sqrt[p-1]{-p})=\mathbb{Q}_{p}(\zeta_{p})$, we first
note that, if $w$ denotes the unique extension of $v_{p}$ to
$\mathbb{Q}_{p}(\sqrt[p-1]{-p})$, then
$w(\sqrt[p-1]{-p})=\frac{1}{p-1}w(-p)=\frac{1}{p-1}$, so
$[\mathbb{Q}_{p}(\sqrt[p-1]{-p}):\mathbb{Q}_{p}]=p-1$. By lemma 2.1, also
$[\mathbb{Q}_{p}(\zeta_{p}):\mathbb{Q}_{p}]=p-1$. Hence, it suffices to show
that $\mathbb{Q}_{p}(\sqrt[p-1]{-p})\subseteq\mathbb{Q}_{p}(\zeta_{p})$. If we
denote

$f(X)=\sum_{i=0}^{p-1}X^{i}=\prod_{i=1}^{p-1}(X-\zeta_{p}^{i}),$

then

$-p=-f(1)=-\prod_{i=1}^{p-1}\left((1-\zeta_{p})\sum_{j=0}^{i-1}\zeta_{p}^{j}\right)=(1-\zeta_{p})^{p-1}(-1)\prod_{i=1}^{p-1}\sum_{j=0}^{i-1}\zeta_{p}^{j}.$

To show that $\sqrt[p-1]{-p}\in Q_{p}(\zeta_{p})$, it hence suffices to show
that $a:=(-1)\prod_{i=1}^{p-1}\sum_{j=0}^{i-1}\zeta_{p}^{j}$ is a $(p-1)$th
power in $\mathbb{Q}_{p}(\zeta_{p})$. As already observed in the proof of
lemma 2.5, $\prod_{i=1}^{p-1}\sum_{j=0}^{i-1}\zeta_{p}^{j}\equiv-1\
(\mathrm{mod}\ (\pi))$, so $a\equiv(-1)(-1)\equiv 1\ (\mathrm{mod}\ (\pi))$.
Therefore, the equation $X^{p-1}-a$ is reduced to the reducible polynomial
$X^{p-1}-1$ in $\overline{\mathbb{Q}_{p}(\zeta_{p})}=\mathbb{F}_{p}$, which
implies that $a$ is indeed a $(p-1)$th power in $\mathbb{Q}_{p}(\zeta_{p})$ by
Hensel’s lemma. ∎

###  3.2 The wild case

The wild case is more involved. The broad idea is to show that the (infinite)
maximal abelian pro-$p$ extension of $\mathbb{Q}_{p}$ is a cyclotomic
extension from which we can read off the relevant cyclotomic extension
containing our given finite abelian extension of $\mathbb{Q}_{p}$. This will
be achieved via infinite Galois theory. In a first step, we find an “upper
bound” for the Galois group of this maximal abelian pro-$p$ extension of
$\mathbb{Q}_{p}$. To do so, we calculate as a base case the compositum of all
$C_{p}$ extensions of $\mathbb{Q}_{p}(\zeta_{p})$ that are abelian over
$\mathbb{Q}_{p}$111We work with $\mathbb{Q}_{p}(\zeta_{p})$ instead of
$\mathbb{Q}_{p}$ to be able to use arguments from Kummer theory; this does
nothing except potentially enlarging our cyclotomic extensions. before
deducing the infinite case. We then proceed to show that the calculated Galois
group actually appears as the Galois group of a cyclotomic extension. The
concrete calculations will then allow us to explicitly describe for each
finite abelian extension a cyclotomic extension containing it.

For ease of notation, we set $K:=\mathbb{Q}_{p}(\zeta_{p})$ and denote by $v$
the extension of $v_{p}$ to $K$. Observe that $K$ is just the trivial
extension for $p=2$. For $p>2$, on the other hand, $K/\mathbb{Q}_{p}$ is
totally ramified of degree $p-1$ with uniformizer $\pi:=\zeta_{p}-1$ and
$\text{Gal}(K/\mathbb{Q}_{p})\cong(\mathbb{Z}/p\mathbb{Z})^{\times}$ by lemma
2.1. The latter group is cyclic as the multiplicative group of the field
$\mathbb{F}_{p}$. We let $k$ be a generator of
$(\mathbb{Z}/p\mathbb{Z})^{\times}$, so
$\text{Gal}(K/\mathbb{Q}_{p})=\langle\sigma\rangle$, where
$\sigma(\zeta_{p})=\zeta_{p}^{k}$. Crucially, note that $k$ is a primitive
root modulo $p$.

Following the structure described above, the first step is the base case,
i.e., classifying the (compositum of the) $C_{p}$-extensions of $K$ that are
abelian over $\mathbb{Q}_{p}$. By Kummer theory, the $C_{p}$-extensions of $K$
are in a one-to-one correspondence with the 1-dimensional
$\mathbb{F}_{p}$-subspaces of $K^{\times}/K^{\times p}$. For $p=2$, all these
subspaces will trivially give rise to an extension that is abelian over
$\mathbb{Q}_{2}$. The case $p>2$ will require some more care. In particular,
it requires a careful consideration of the action of $\sigma$ on the elements
of the higher unit groups of $K$, as summarized in the next lemma.

######  Lemma 3.4.

Let $a\in 1+\sum_{i=j}^{p}a_{i}\pi^{i}+(\pi)^{p+1}$ with $a_{j}\neq 0$ (so
$a\in U_{K}^{(j)}$). Then $\sigma(a)\in 1+k^{j}a_{j}\pi^{j}+(\pi)^{j+1}$.
Moreover, $\sigma(a)\in a^{k}K^{\times p}$ for $j=p$.

###### Proof.

First observe that

$\sigma(\pi)=\sigma(\zeta_{p}-1)=\zeta_{p}^{k}-1=(\zeta_{p}-1)(\zeta_{p}^{k-1}+\zeta_{p}^{k-2}+\dots+1)=\pi(\zeta_{p}^{k-1}+\zeta_{p}^{k-2}+\dots+1).$

Now $\overline{\zeta_{p}^{k-1}+\zeta_{p}^{k-2}+\dots+1}=k$, which means
$\zeta_{p}^{k-1}+\zeta_{p}^{k-2}+\dots+1\in k+(\pi)$, so

$\displaystyle\sigma(\pi)\in k\pi+(\pi)^{2}.$ (3)

Hence,

$\sigma(\pi^{j})=\sigma(\pi)^{j}\in k^{j}\pi^{j}+(\pi)^{j+1},$

and, similarly,

$\sigma(\pi^{i})=\sigma(\pi)^{i}\in
k^{i}\pi^{i}+(\pi)^{i+1}\subseteq(\pi)^{j+1}$

for $i>j$. Since $\sigma$ fixes the integers $a_{i}$, the preceding two
equations yield that

$\sigma(a)=1+a_{j}\sigma(\pi^{j})+\sum_{i=j+1}^{p}a_{i}\sigma(\pi^{i})\in
1+k^{j}a_{j}\pi^{j}+(\pi)^{j+1}.$

In particular, $\sigma(a)\in 1+k^{p}a_{p}\pi^{p}+(\pi)^{p+1}$ if $j=p$. By
lemma 2.4, we then get $\sigma(a)\equiv
1+ka_{p}\pi^{p}\equiv(1+a_{p}\pi^{p})^{k}\equiv\left(a\right)^{k}\
(\mathrm{mod}\ U_{K}^{(p+1)})$, so $\sigma(a)\in a^{k}U_{K}^{(p+1)}$. As
$U_{K}^{(p+1)}=\left(U_{K}^{(1)}\right)^{p}$ by lemma 2.8, the result follows.
∎

We can now show the following characterization.

######  Lemma 3.5.

The compositum of all $C_{p}$-extensions $F/K$ such that $F/\mathbb{Q}_{p}$ is
abelian is

  * (i)

$\mathbb{Q}_{2}(\sqrt{-1},\sqrt{2},\sqrt{5})$ with
$\text{Gal}(\mathbb{Q}_{2}(\sqrt{-1},\sqrt{2},\sqrt{5})/\mathbb{Q}_{2})\cong
C_{2}\times C_{2}\times C_{2}$ for $p=2$, and

  * (ii)

$\mathbb{Q}_{p}(\zeta_{p^{2}},\zeta_{p^{p}-1})$ with
$\text{Gal}(\mathbb{Q}_{p}(\zeta_{p^{2}},\zeta_{p^{p}-1})/\mathbb{Q}_{p})\cong
C_{p}\times C_{p}$ for $p>2$.

###### Proof.

_(i)._ Observe that it is enough to show that $(-1,2,5)$ is a basis of
$\mathbb{Q}_{2}^{\times}/\mathbb{Q}_{2}^{\times 2}$ as an
$\mathbb{F}_{2}$-vector space. Indeed, the spanning property shows that any
$C_{2}$-extensions of $\mathbb{Q}_{2}$ is contained in
$\mathbb{Q}_{2}(\sqrt{-1},\sqrt{2},\sqrt{5})$, while linear independence
implies the claim about the Galois group.

By corollary 2.9, we know that $\mathbb{Q}_{2}^{\times}/\mathbb{Q}_{2}^{\times
2}$ has dimension $3$. Hence, it suffices to show that $(-1,2,5)$ spans
$\mathbb{Q}_{2}^{\times}/\mathbb{Q}_{2}^{\times 2}$. So let
$x\in\mathbb{Q}_{2}^{\times}$ and write $x=2^{n}y$ with $y\in
U_{\mathbb{Q}_{2}}^{(1)}$. Then $x\equiv y\ (\mathrm{mod}\
\mathbb{Q}_{2}^{\times 2})$ if $n$ is even, and $x\equiv 2y\ (\mathrm{mod}\
\mathbb{Q}_{2}^{\times 2})$ if $n$ is odd, so
$\mathbb{Q}_{2}^{\times}/\mathbb{Q}_{2}^{\times 2}$ is spanned by $2$ together
with a basis of $U_{\mathbb{Q}_{2}}^{(1)}/\mathbb{Q}_{2}^{\times}\cap
U_{\mathbb{Q}_{2}}^{(1)}=U_{\mathbb{Q}_{2}}^{(1)}/\left(U_{\mathbb{Q}_{2}}^{(1)}\right)^{2}=U_{\mathbb{Q}_{2}}^{(1)}/U_{\mathbb{Q}_{2}}^{(3)}$
(here we used lemma 2.8). By lemma 2.4,
$U_{\mathbb{Q}_{2}}^{(1)}/U_{\mathbb{Q}_{2}}^{(3)}$ is spanned by elements
represented by $y=1+2a_{1}+4a_{2}$ for $a_{1},a_{2}\in\\{0,1\\}$. So let us
consider the four different cases for $(a_{0},a_{1})$.

  * •

$(a_{1},a_{2})=(0,0)$. Then $y=1=(-1)^{0}\cdot 5^{0}$.

  * •

$(a_{1},a_{2})=(0,1)$. Then $y=5=(-1)^{0}\cdot 5^{1}$.

  * •

$(a_{1},a_{2})=(1,0)$. Then $y=3\equiv-5=(-1)^{1}\cdot 5^{1}\ (\mathrm{mod}\
U_{\mathbb{Q}_{2}}^{(3)})$.

  * •

$(a_{1},a_{2})=(1,1)$. Then $y=7\equiv-1=(-1)^{1}\cdot 5^{0}\ (\mathrm{mod}\
U_{\mathbb{Q}_{2}}^{(3)})$.

Hence $(-1,5)$ indeed spans
$U_{\mathbb{Q}_{2}}^{(1)}/U_{\mathbb{Q}_{2}}^{(3)}$.

_(ii)_ As $\mathbb{Q}_{p}(\zeta_{p^{2}})$ and
$\mathbb{Q}_{p}(\zeta_{p^{p}-1})$ are $C_{p}$-extensions of $K$ abelian over
$\mathbb{Q}_{p}$ by lemma 2.1, it is clear that the compositum contains
$\mathbb{Q}_{p}(\zeta_{p^{2}},\zeta_{p^{p}-1})$. Moreover, lemma 2.1 also
tells us that $K(\zeta_{p^{2}})/K$ is totally ramified while
$K(\zeta_{p^{p}-1})/K$ is unramified. This means that both subextensions are
linearly disjoint over $K$, which implies
$\text{Gal}(K(\zeta_{p^{2}},\zeta_{p^{p}-1})/K)\cong C_{p}\times C_{p}$.
Therefore, it only remains to see that any $C_{p}$-extensions $F/K$ with
$F/\mathbb{Q}_{p}$ abelian is contained in
$\mathbb{Q}_{p}(\zeta_{p^{2}},\zeta_{p^{p}-1})$.

By Kummer theory, we have $F=K(\sqrt[p]{u})$ for some $u\in K^{\times}$; by
corollary 2.10 we may represent
$u=p^{a_{0}}\left(1+\sum_{i=1}^{p}a_{i}\pi\right)K^{\times p}$ with
$a_{i}\in\\{0,\dots,p-1\\}$. As a first step, we show that abelianity of
$F/\mathbb{Q}_{p}$ implies $a_{0}=0$. For the sake of obtaining a
contradiction, let us assume $a_{0}\neq 0$. On the one hand,

$\sigma(u)\in
p^{a_{0}}\sigma\left(1+\sum_{i=1}^{p}a_{i}\pi^{i}\right)K^{\times p}.$

On the other hand, by the uniqueness of splitting fields and the fundamental
theorem of Kummer theory222Extend $\sigma$ linearly to an isomorphism
$K(\sqrt[p]{u})\cong K(\sqrt[p]{\sigma(u)}$). By the uniqueness of splitting
fields, this must be an equality, so the fundamental theorem of Kummer theory
implies $\sigma(u)\in\langle u\rangle K^{\times p}$., we have
$\sigma(u)\in\langle u\rangle K^{\times p}$, say $\sigma(u)\in u^{r}K^{\times
p}$ with $r\in\\{1,...,p-1\\}$. Hence,

$\sigma(u)\in
u^{r}K^{\times}=p^{ra_{0}}\left(1+\sum_{i=1}^{p}a_{i}\pi^{i}\right)^{r}K^{\times
p}.$

As the exponent of $p$ determines the valuation of $\sigma(u)$ modulo $1$, we
get $r=1$, which means

$\sigma(u)\in
uK^{\times}=p^{a_{0}}\left(1+\sum_{i=1}^{p}a_{i}\pi^{i}\right)K^{\times p}.$

We claim that this implies $0=a_{1}=\dots=a_{p-2}$. Indeed, otherwise we could
let $a_{j}$ denote the first non-zero $a_{i}$ ($i\in\\{1,\dots p-2\\}$). Then
lemma 3.4 implies $\sigma(u)\in 1+k^{j}a_{j}\pi^{j}+(\pi)^{j+1}$. By the
previous equation and lemma 2.4, this would imply $k^{j}\equiv 1\
(\mathrm{mod}\ p)$, which is impossible since $k$ is a primitive root modulo
$p$.

Hence, $u=p^{a_{0}}(1+a_{p-1}\pi^{p-1}+a_{p}\pi^{p})$. If we write
$\frac{\pi^{p-1}}{u^{\prime}}=p$ for $u^{\prime}\in\mathcal{O}_{K}$ and let
$l\in\\{0,...,p-1\\}$ be such that $l=\overline{a_{p-1}u^{\prime}}$, we get

$u=p^{a_{0}}(1+a_{p-1}\pi^{p-1}+a_{p}\pi^{p})\in
p^{a_{0}}(1+lp+a_{p}\pi^{p})K^{\times
p}=p^{a_{0}}(1+lp)(1+a_{p}\pi^{p})K^{\times p},$

If we could show that $K(\sqrt[p]{1+a_{p}\pi^{p}})\subseteq
K(\zeta_{p^{p}-1})$, we would get that
$K(\zeta_{p^{p}-1},\sqrt[p]{p^{a_{0}}(1+lp)})/\mathbb{Q}_{p}$ is abelian as
the compositum of the two abelian extensions
$\mathbb{Q}_{p}(\zeta_{p^{p}-1})/\mathbb{Q}_{p}$ and $F/\mathbb{Q}_{p}$. In
particular, the intermediate extension
$\mathbb{Q}_{p}(\sqrt[p]{p^{a_{0}}(1+lp)})/\mathbb{Q}_{p}$ would have to be
Galois. However, as $\mathbb{Q}_{p}$ does not contain the $p$th roots of unity
by corollary 2.2, this is impossible for $a_{0}\neq 0$, contradicting our
assumption.

To finish the proof of $a_{0}=0$, it therefore only remains to show that
$K(\sqrt[p]{1+a_{p}\pi^{p}})\subseteq K(\zeta_{p^{p}-1})$. This is clearly the
case for $a_{p}=0$. Otherwise, let us denote $u^{\prime}=1+a_{p}\pi^{p}\in
U_{K}^{(p)}$. Then $u^{\prime}\notin K^{\times p}$ by equation (2) and lemma
2.4, so the extension $K(\sqrt[p]{u^{\prime}})/K$ is a $C_{p}$-extension by
Kummer theory. Moreover, we claim that it is unramified. Indeed, $u^{\prime}$
is a $p$-th power in $K(\sqrt[p]{u^{\prime}})$, so corollary 2.7 tells us that
we may find $\overline{x}\in\overline{K(\sqrt[p]{u^{\prime}})}$ such that
$\overline{x}^{p}-\overline{x}-\overline{a_{p}}=0$. As this is not the case
for $\overline{a_{p}}\neq 0$ in $\overline{K}=\mathbb{F}_{p}$, we see that
$K(\sqrt[p]{u^{\prime}})/K$ gives rise to a (degree-$p$) extension of residue
fields, so it is unramified. Hence, by lemma 2.1,
$K(\sqrt[p]{u^{\prime}})=K(\zeta_{p^{p}-1})$, completing the proof that
$a_{0}=0$.

It remains to consider the remaining $a_{i}$. If $u\in\langle
u^{\prime}\rangle K^{\times p}$ for some $u^{\prime}=1+a_{p}^{\prime}\pi^{p}$
with $a_{p}^{\prime}\in\\{1,\dots,p-1\\}$, then $\sqrt[p]{u}\in
K(\sqrt[p]{u^{\prime}})$ and, by the last paragraph,
$F=\mathbb{Q}_{p}(\zeta_{p},\sqrt[p]{u^{\prime}})=\mathbb{Q}_{p}(\zeta_{p},\zeta_{p^{p}-1})\subseteq\mathbb{Q}_{p}(\zeta_{p^{2}},\zeta_{p^{p}-1})$
as claimed. So assume $u\notin\langle u^{\prime}\rangle K^{\times p}$. Then,
since $K(\sqrt[p]{uu^{\prime}})/\mathbb{Q}_{p}$ is abelian as a subextension
of the compositum of the two abelian extensions $K/\mathbb{Q}_{p}$ and
$\mathbb{Q}_{p}(\zeta_{p^{p}-1})/\mathbb{Q}_{p}$, we see that
$\sigma(uu^{\prime})\in\langle uu^{\prime}\rangle$ (by the same argument as in
footnote 2). As $\sigma(uu^{\prime})=\sigma(u)\sigma(u^{\prime})$, together
with lemma 3.4, this implies

$\sigma(uu^{\prime})\in\langle uu^{\prime}\rangle K^{\times p}\cap
u^{r}\left(u^{\prime}\right)^{k}K^{\times p},$

which forces $r=k$. Hence

$\sigma(u)K^{\times p}=\left(1+\sum_{i=1}^{p}a_{i}\pi^{i}\right)^{k}K^{\times
p},$

which, comparing coefficients, forces $a_{1}\neq 0$ by lemma 3.4 since $k$ is
a primitive root modulo $p$. On the other hand, if we write
$\sigma(\pi)=k\pi+b\pi^{2}$ for some $b\in\mathcal{O}_{K}$ as in (3), we get

$\sigma(u)K^{\times
p}=\left(1+\sum_{i=1}^{p}a_{i}(k\pi+b\pi^{2})^{i}\right)K^{\times p}.$

Expanding and comparing coefficients with the previous equation, we can
uniquely find $a_{i}$ in terms of $a_{i-1},\dots a_{1}$. Hence, all $a_{i}$
are uniquely determined by $a_{1}$. This means that we can find at most $p-1$
distinct values of $u$ that satisfy $\sigma(u)\in u^{k}K^{\times p}$. On the
other hand, by the very definition of $\sigma$, we notice that
$u=\zeta_{p}^{i}$ for $i\in\\{1,\dots p-1\\}$ satisfies this equation. Thus,
we get that $F\subseteq
K(\sqrt[p]{\zeta_{p}})=\mathbb{Q}_{p}(\zeta_{p^{2}})\subseteq\mathbb{Q}_{p}(\zeta_{p^{2}},\zeta_{p^{p}})$
as claimed. ∎

From this base case, we can now conclude the infinite case. While this is
straight-forward for $p>2$, the case $p=2$ requires the following slightly
technical lemma.

######  Lemma 3.6.

$\mathbb{Q}_{2}(i)$ is not contained in a $C_{4}$-extension of
$\mathbb{Q}_{2}$.

This lemma at hand, we may then deduce.

######  Corollary 3.7.

Let $\tilde{F}$ be the unique maximal pro-$p$ extension of $K$ that is abelian
over $\mathbb{Q}_{p}$. Then $\text{Gal}(\tilde{F}/K)$

  * (i)

is a quotient of $C_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ for $p=2$,
and

  * (ii)

is a quotient of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ for $p>2$.

###### Proof.

_(ii)_ By the previous lemma, the compositum of all $C_{p}$-extensions is
exactly $L:=K(\zeta_{p^{2}},\zeta_{p^{p}-1})$. As $C_{p}$-extensions are the
minimal subextensions of $\tilde{F}/K$, the Galois group
$\text{Gal}(\tilde{F}/L)$ is exactly the intersection of all maximal subgroups
of $\text{Gal}(\tilde{F}/K)$ by the infinite Galois correspondence (cf.
[Szamuely09], Theorem 1.3.11). In other words, $\text{Gal}(\tilde{F}/L)$ is
the Frattini subgroup of $\text{Gal}(\tilde{F}/K)$ and
$\text{Gal}(L/K)\cong\text{Gal}(\tilde{F}/K)/\text{Gal}(\tilde{F}/L)$ is its
Frattini quotient. As $\text{Gal}(L/K)\cong C_{p}\times C_{p}$, we see that,
as pro-$p$ groups,

$\text{rank}(\text{Gal}(\tilde{F}/K))=\text{rank}(\text{Gal}(L/K))=\text{rank}(C_{p}\times
C_{p})=2,$

so $\text{Gal}(\tilde{F}/K)$ must be a quotient of the free abelian pro-$p$
group of rank $2$, which is $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$.

_(i)_ By the same argument as in (ii), we could now conclude that the Galois
group of the maximal abelian pro-$2$ extension of $\mathbb{Q}_{2}$ is a
quotient of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. However,
by lemma 3.6, $\mathbb{Q}_{2}(i)$ is not contained in a $C_{4}$-extension of
$\mathbb{Q}_{2}$, so $\text{Gal}(\tilde{F}/K)$ must actually be a quotient of
$C_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. ∎

To show lemma 3.6, we must examine the quadratic extensions of
$\mathbb{Q}_{2}(i)$. By Kummer theory, these extensions correspond exactly to
the 1-dimensional $\mathbb{F}_{2}$-subspaces of
$\mathbb{Q}_{2}(i)^{\times}/\mathbb{Q}_{2}^{\times 2}(i)$. The next lemma
calculates a basis of $\mathbb{Q}_{2}(i)^{\times}/\mathbb{Q}_{2}^{\times
2}(i)$ as an $\mathbb{F}_{2}$-vector space explicitly.

######  Lemma 3.8.

$(1+i,1+2i,2,5)$ is a basis of
$\mathbb{Q}_{2}(i)^{\times}/\mathbb{Q}_{2}^{\times 2}(i)$ as an
$\mathbb{F}_{2}$-vector space.

###### Proof.

$\mathbb{Q}_{2}(i)^{\times}/\mathbb{Q}_{2}^{\times 2}(i)$ has dimension $4$ by
corollary 2.9. To show that $(1+i,1+2i,2,5)$ is indeed a basis of
$\mathbb{Q}_{2}(i)^{\times}/\mathbb{Q}_{2}^{\times 2}(i)$, it hence suffices
to show that the four elements are linearly independent over
$\mathbb{Q}_{2}^{\times 2}(i)$. So assume
$x=(1+i)^{\alpha}\cdot(1+2i)^{\beta}\cdot 2^{\gamma}\cdot
5^{\delta}\in\mathbb{Q}_{2}^{\times 2}(i)$ for
$\alpha,\beta,\gamma,\delta\in\\{0,1\\}$, say $y^{2}=x$. Then, denoting by $N$
the field-theoretic norm map of the extension
$\mathbb{Q}_{2}(i)/\mathbb{Q}_{2}$,

$2^{\alpha}\cdot 5^{\beta}\cdot 4^{\gamma}\cdot
25^{\delta}=N(x)=N(y^{2})=N(y)^{2}\in\mathbb{Q}_{2}^{\times 2},$

so $\alpha=\beta=0$ by (the proof of) lemma 3.5(i). It remains to exclude the
three cases $(\gamma,\delta)=(1,0),(0,1),(1,1)$.

  * •

$(\gamma,\delta)=(1,0)$ would mean that $2\in\mathbb{Q}_{2}(i)^{\times 2}$, so
$\mathbb{Q}_{2}(i)=\mathbb{Q}_{2}(i,\sqrt{2})$. But
$\zeta_{8}=\frac{1+i}{\sqrt{2}}\in\mathbb{Q}_{2}(i,\sqrt{2})$, and
$\mathbb{Q}_{2}(\zeta_{8})$ totally ramified of degree $4\neq 2$ over
$\mathbb{Q}_{2}$ by lemma 2.1.

  * •

$(\gamma,\delta)=(1,0)$ would mean that
$\mathbb{Q}_{2}(i,\sqrt{5})=\mathbb{Q}_{2}(i)$. Modulo
$U_{\mathbb{Q}_{2}}^{(3)}=\left(U_{\mathbb{Q}_{2}}^{(1)}\right)^{2}\subseteq\mathbb{Q}_{2}(i)^{\times
2}$, $1+1\cdot 4\equiv 1+(-1)\cdot 4=-3$, so
$\mathbb{Q}_{2}(i,\sqrt{5})=\mathbb{Q}_{2}(i,\sqrt{-3})$. Now
$\zeta_{3}=\frac{-1+\sqrt{-3}}{2}\in\mathbb{Q}_{2}(i,\sqrt{-3})$. But
$\mathbb{Q}_{2}(\zeta_{3})$ is the unique unramified extension of
$\mathbb{Q}_{2}$ of degree 2 by lemma 2.1.

  * •

$(\gamma,\delta)=(1,1)$ would mean that
$\mathbb{Q}_{2}(i)=\mathbb{Q}_{2}(i,\sqrt{2\cdot 5}$). However, this would
imply that $\mathbb{Q}_{2}(i,\sqrt{5})=\mathbb{Q}_{2}(i,\sqrt{2})$, which
contradicts what we have shown in the previous two items.

∎

We can now deduce lemma 3.6.

###### Proof (lemma 3.6)..

Let $F/\mathbb{Q}_{2}$ be a Galois extension of degree 4 containing
$\mathbb{Q}_{2}(i)$. Then $[F:\mathbb{Q}_{2}(i)]=2$, so
$F=\mathbb{Q}_{2}(i)(\sqrt{x})$ by Kummer theory. By the previous lemma, we
may assume $x=(1+i)^{\alpha}\cdot(1+2i)^{\beta}\cdot 2^{\gamma}\cdot
5^{\delta}$ for $\alpha,\beta,\gamma,\delta\in\\{0,1\\}$. Our goal is to show
that $\alpha=\beta=0$. In that case, $x$ is one of 2, 5, 10, and either way
$\mathbb{Q}_{2}(\sqrt{x})$ is an intermediate extension of degree 2, so
$F/\mathbb{Q}_{2}$ has two distinct proper non-trivial intermediate fields, so
it cannot be a $C_{4}$-extension.

So let us show $\alpha=\beta=0$. If $\sigma$ denotes the generator of
$\text{Gal}(\mathbb{Q}_{2}(i)/\mathbb{Q}_{2})$ (i.e., $\sigma$ is complex
conjugation), then, by the uniqueness of splitting fields and Kummer theory
(as in footnote 2), we have $\sigma(x)\in\langle
x\rangle\mathbb{Q}_{2}^{\times 2}$, i.e., $x\equiv\sigma(x)\ (\mathrm{mod}\
\mathbb{Q}_{2}^{\times 2})$. Now assume $\alpha,\beta\neq 0$. Then, in
$\mathbb{Q}_{2}(i)^{\times}/\mathbb{Q}_{2}^{\times 2}$, we have

$\displaystyle\begin{array}[]{rrll}&x&\equiv\sigma(x)&\ (\mathrm{mod}\
\mathbb{Q}_{2}^{\times 2})\\\ \Longleftrightarrow&(1+i)\cdot(1+2i)\cdot
2^{\gamma}\cdot 5^{\delta}&\equiv(1-i)\cdot(1-2i)\cdot 2^{\gamma}\cdot
5^{\delta}&\\\ &&\equiv(1-i)\cdot(1-2i)\cdot 2^{\gamma}\cdot
5^{\delta}(1+i)^{2}(1+2i)^{2}&\\\ &&\equiv(1+i)\cdot(1+2i)\cdot
2^{\gamma+1}\cdot 5^{\delta+1}&\ (\mathrm{mod}\ \mathbb{Q}_{2}^{\times 2})\\\
\Longleftrightarrow&0&\equiv 10&\ (\mathrm{mod}\ \mathbb{Q}_{2}^{\times
2}),\end{array}$

contradicting the linear independence proved before. Similarly, we get
contradictions $2,5\equiv 0\ (\mathrm{mod}\ \mathbb{Q}_{2}^{\times 2})$ for
$(\alpha,\beta)=(1,0),(0,1)$, respectively. ∎

For the next step of our argument, we want to show that $\tilde{F}$ from
corollary 3.7 is cyclotomic. To do so, we prove that the groups calculated in
corollary 3.7 actually appear as the Galois groups of cyclotomic extensions of
$K$.

######  Lemma 3.9.

Let $F=\mathbb{Q}_{p}(\zeta_{p^{p^{\infty}}-1},\zeta_{p^{\infty}})$. Then

  * (i)

$\text{Gal}(F/K)=\text{Gal}(F/\mathbb{Q}_{2})\cong
C_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ for $p=2$, and

  * (ii)

$\text{Gal}(F/K)\cong\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ for $p>2$.

###### Proof.

_(ii)._ From elementary infinite Galois theory, we have
$\text{Gal}(F/K)\cong\varprojlim\text{Gal}(E/K)$, where $E$ runs over the
directed partially ordered set $I$ of finite Galois subextensions of $F/K$
(cf. [Szamuely09], Proposition 1.3.5). By definition of $F$, these finite
subextensions are of the form
$E=\mathbb{Q}_{p}(\zeta_{p^{p^{n}}-1},\zeta_{p^{m}})$ for some non-negative
integers $m,n$. Now $K(\zeta_{p^{p^{n}}-1})/K$ is unramified of degree $p^{n}$
with $\text{Gal}(K(\zeta_{p^{p^{n}}-1})/K)\cong C_{p^{n}}$ by lemma 2.1.
$K(\zeta_{p^{m}})/K$, on the other hand, is totally ramified by lemma 2.1, and

$\text{Gal}(K(\zeta_{p^{m}})/\mathbb{Q}_{p})=\text{Gal}(\mathbb{Q}_{p}(\zeta_{p^{m}})/\mathbb{Q}_{p})\cong(\mathbb{Z}/p^{m}\mathbb{Z})^{\times}\cong
C_{p^{m-1}(p-1)}\cong C_{p^{m-1}}\times C_{p-1}.$

Hence, as $(p^{m-1},p-1)=1$,
$\text{Gal}(K(\zeta_{p^{m}})/K)\cong\text{Gal}(\mathbb{Q}_{p}(\zeta_{p^{m}})/\mathbb{Q}_{p})/\text{Gal}(K/\mathbb{Q}_{p})\cong
C_{p^{m-1}}$. Finally, $K(\zeta_{p^{p^{n}}-1})$ and $K(\zeta_{p^{m}})$ are
linearly disjoint as one extension is unramified and the other one totally
ramified. Thus, $\text{Gal}(E/K)=C_{p^{n}}\times C_{p^{m-1}}$, and the claim
follows since $\mathbb{Z}_{p}=\varprojlim_{k}C_{p^{k}}$.

_(i)._ As $\mathbb{Q}_{2}(\zeta_{2^{\infty}})/\mathbb{Q}_{2}$ is totally
ramified and $\mathbb{Q}_{2}(\zeta_{2^{2^{\infty}}-1})/\mathbb{Q}_{2}$ is
unramified, we see that both extensions are linearly disjoint over
$\mathbb{Q}_{2}$, so
$\text{Gal}(F/\mathbb{Q}_{2})\cong\text{Gal}(\mathbb{Q}_{2}(\zeta_{2^{\infty}})/\mathbb{Q}_{2})\times\text{Gal}(\mathbb{Q}_{2}(\zeta_{2^{2^{\infty}}-1})/\mathbb{Q}_{2})$.
By lemma 2.1,

$\text{Gal}(\mathbb{Q}_{2}(\zeta_{2^{2^{\infty}}-1})/\mathbb{Q}_{2})\cong\varprojlim_{k}\text{Gal}(\mathbb{Q}_{2}(\zeta_{2^{2^{k}}-1})/\mathbb{Q}_{2})\cong\varprojlim_{k}C_{2^{k}}=\mathbb{Z}_{2}.$

Hence, it remains to show that
$\text{Gal}(\mathbb{Q}_{2}(\zeta_{2^{\infty}})/\mathbb{Q}_{2})\cong
C_{2}\times\mathbb{Z}_{2}$. To do so, we will set
$a_{k}:=\zeta_{2^{k}}+\zeta_{2^{k}}^{-1}$ and show that all finite
subextensions $\mathbb{Q}_{2}(\zeta_{2^{k}})/\mathbb{Q}_{2}$ ($k\geq 2$) are
given as $\mathbb{Q}_{2}(\zeta_{2^{k}})=\mathbb{Q}_{2}(i,a_{k})$, where
$\text{Gal}(\mathbb{Q}_{2}(i)/\mathbb{Q}_{2})\cong C_{2}$,
$\text{Gal}(\mathbb{Q}_{2}(a_{k})/\mathbb{Q}_{2})\cong C_{2^{k-2}}$, and
$\mathbb{Q}_{2}(i)$ and $\mathbb{Q}_{2}(a_{k})$ are linearly disjoint over
$\mathbb{Q}_{2}$. From elementary infinite Galois theory, we then indeed get

$\text{Gal}(\mathbb{Q}_{2}(\zeta_{2^{\infty}})/\mathbb{Q}_{2})\cong\varprojlim_{k}\text{Gal}(\mathbb{Q}_{2}(\zeta_{2^{k}})/\mathbb{Q}_{2})\cong\varprojlim_{k}\text{Gal}(\mathbb{Q}_{2}(i)/\mathbb{Q}_{2})\times\text{Gal}(\mathbb{Q}_{2}(a_{k})/\mathbb{Q}_{2})\cong\varprojlim_{k}C_{2}\times
C_{2^{k-2}}=C_{2}\times\mathbb{Z}_{2}.$

So let us show all the required properties. We already know that
$\mathbb{Q}_{2}(i)/\mathbb{Q}_{2}$ is Galois with
$\text{Gal}(\mathbb{Q}_{2}(i)/\mathbb{Q}_{2})\cong C_{2}$ from lemma 2.1, so
let us show the more difficult claim
$\text{Gal}(\mathbb{Q}_{2}(a_{k})/\mathbb{Q}_{2})\cong C_{2^{k-2}}$. We first
show that $[\mathbb{Q}_{2}(a_{k}):\mathbb{Q}_{2}]=2^{k-2}$ and then exhibit a
generator of the Galois group of order $2^{k-2}$. As a first step, we note
that $a_{k}=\zeta_{2^{k}}+\zeta_{2^{k}}^{-1}$ is given recursively by
$a_{2}=0$ and $a_{k+1}=\sqrt{2+a_{k}}$. Indeed, inductively we see that
$a_{2}=i+(-i)=0$ and

$a_{k+1}^{2}=(\zeta_{2^{k+1}}+\zeta_{2^{k+1}}^{-1})^{2}=\underbrace{\zeta_{2^{k+1}}^{2}+\left(\zeta_{2^{k+1}}^{-1}\right)^{2}}_{=\zeta_{2^{k}}+\zeta_{2^{k}}^{-1}=a_{k}}+2(\underbrace{\zeta_{2^{k+1}}\zeta_{2^{k+1}}^{-1}}_{=1})=a_{k}+2.$

Similarly, we see that $a_{k}$ is the root of a polynomial of degree
$2^{k-2}$, so $[\mathbb{Q}_{2}(a_{k}):\mathbb{Q}_{2}]\leq 2^{k-2}$.

Next, if $v$ denotes the unique extension of $v_{2}$ to
$\mathbb{Q}_{2}(\zeta_{2^{k}}+\zeta_{2^{k}}^{-1})$ for $k\geq 3$, we claim
that then $v(a_{k})=\frac{1}{2^{k-2}}$. Indeed, for $k=3$, we have
$v(a_{3})=v(\sqrt{2})=\frac{1}{2}v(2)=\frac{1}{2}$. Inductively, it follow
that

$v(a_{k})=v\left(\sqrt{a_{k-1}+2}\right)=\frac{1}{2}v(a_{k-1}+2)=\frac{1}{2^{k-1}}$

as $v(a_{k-1}+2)=\min\\{v(a_{k-1}),v(2)\\}$ since
$v(a_{k})=\frac{1}{2^{k-2}}<1=v(2)$ by the inductive hypothesis. This means
that $\mathbb{Q}_{2}(a_{k})$ is ramified of degree $2^{k-2}$, so
$[\mathbb{Q}_{2}(a_{k}):\mathbb{Q}_{2}]\geq 2^{k-2}$. Together with the
previous paragraph, we see that
$[\mathbb{Q}_{2}(a_{k}):\mathbb{Q}_{2}]=2^{k-2}$.

It remains to show that we can find a generator of order $2^{k-2}$ of
$\text{Gal}(\mathbb{Q}_{2}(a_{k})/\mathbb{Q}_{2})$ for $k\geq 3$. To do so, it
suffices to find $\sigma\in\text{Gal}(\mathbb{Q}_{2}(a_{k},i)/\mathbb{Q}_{2})$
of order $2^{k-2}$ with $\sigma(i)=i$ (then $\sigma|_{\mathbb{Q}_{2}(a_{k})}$
has the required properties). The advantage of exhibiting $\sigma$ is that it
automatically implies $\mathbb{Q}_{2}$-linear disjointness of
$\mathbb{Q}_{2}(i)$ and $\mathbb{Q}_{2}(a_{k})$. Indeed, if we let $L$ denote
the non-trivial fixed field of $\langle\sigma\rangle$, then

$\underbrace{[\mathbb{Q}_{2}(a_{k},i):L]}_{=|\langle\sigma\rangle|=2^{k-2}}\underbrace{[L:\mathbb{Q}_{2}]}_{>1}=[\mathbb{Q}_{2}(a_{k},i):\mathbb{Q}_{2}]=[\mathbb{Q}_{2}(a_{k},i):\mathbb{Q}_{2}(a_{k})]\underbrace{[\mathbb{Q}_{2}(a_{k}):\mathbb{Q}_{2}]}_{=2^{k-2}},$

so $[\mathbb{Q}_{2}(a_{k},i):\mathbb{Q}_{2}(a_{k})]>1$, which implies
$i\notin\mathbb{Q}_{2}(a_{k})$ and hence that $\mathbb{Q}_{2}(i)$ and
$\mathbb{Q}_{2}(a_{k})$ are linearly disjoint over $\mathbb{Q}_{2}$

To find $\sigma$, observe that over $\mathbb{Q}$, we can easily see that
$\mathbb{Q}(i,a_{k})=\mathbb{Q}(\zeta_{2^{k}})$. Indeed, clearly
$\mathbb{Q}(i,a_{k})\subseteq\mathbb{Q}(\zeta_{2^{k}})$, and the degree of the
minimal polynomial of $a_{k}$ over $\mathbb{Q}$ is at least the degree over
$\mathbb{Q}_{2}$ so $[\mathbb{Q}(a_{k}):\mathbb{Q}]\geq 2^{k-2}$. Now, by the
recursive formula, the extension $\mathbb{Q}(a_{k})/\mathbb{Q}$ is totally
real. Hence, $i\notin\mathbb{Q}(a_{k})$ and so
$[\mathbb{Q}(a_{k},i):\mathbb{Q}]\geq
2^{k-1}=[\mathbb{Q}(\zeta_{2^{k}}):\mathbb{Q}]$, which shows
$\mathbb{Q}(i,a_{k})=\mathbb{Q}(\zeta_{2^{k}})$. The argument even shows that
$\mathbb{Q}(a_{k})$ and $\mathbb{Q}(i)$ are $\mathbb{Q}$-linearly independent,
so
$\text{Gal}(\mathbb{Q}(\zeta_{2^{k}})/\mathbb{Q})\cong\text{Gal}(\mathbb{Q}(i)/\mathbb{Q})\times\text{Gal}(\mathbb{Q}(a_{k})/\mathbb{Q})\cong
C_{2}\times\text{Gal}(\mathbb{Q}(a_{k})/\mathbb{Q})$. On the other hand, by
Galois theory, we know that
$\text{Gal}(\mathbb{Q}(\zeta_{2^{k}})/\mathbb{Q})\cong\text{Aut}(\mu_{2^{k}})\cong(\mathbb{Z}/2^{k}\mathbb{Z})^{\times}\cong
C_{2}\times C_{2^{k-2}}$. Therefore, necessarily
$\text{Gal}(\mathbb{Q}(a_{k})/\mathbb{Q})\cong C_{2^{k-2}}$. In particular, we
can find an element
$\sigma^{\prime}\in\text{Gal}(\mathbb{Q}(\zeta_{2^{k}})/\mathbb{Q})$ of order
$2^{k-2}$ with $\sigma^{\prime}(i)=i$. Now recall that we can embed
$\text{Gal}(\mathbb{Q}_{2}(\zeta_{2^{k}})/\mathbb{Q}_{2})$ in
$\text{Gal}(\mathbb{Q}(\zeta_{2^{k}})/\mathbb{Q})$ via $\iota^{*}$ with
$\iota^{*}(\tau)=\iota^{-1}\circ\tau\circ\iota$, where
$\iota:\mathbb{Q}(a_{k},i)\hookrightarrow\mathbb{Q}_{2}(a_{k},i)$ is a
completion. By lemma 2.1,
$|\text{Gal}(\mathbb{Q}_{2}(\zeta_{2^{k}})/\mathbb{Q}_{2})|=2^{k-1}=|\text{Gal}(\mathbb{Q}(\zeta_{2^{k}})/\mathbb{Q})|$,
so $\iota^{*}$ is an isomorphism. In particular, it is surjective so we can
find $\sigma\in\text{Gal}(\mathbb{Q}_{2}(\zeta_{2^{k}})/\mathbb{Q}_{2})$ with
$\iota^{*}(\sigma)=\sigma^{\prime}$. By definition of $\iota^{*}$, $\sigma$
will then be of order $2^{k-1}$ and satisfy $\sigma(i)=i$. ∎

By lemma 2.1, $F$ from the previous lemma is a pro-$p$ extension of $K$ that
is abelian over $\mathbb{Q}_{p}$. In particular, $F\subseteq\tilde{F}$. Hence,
together with corollary 3.7 and the previous lemma, this gives rise to maps

$C_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}\twoheadrightarrow\text{Gal}(\tilde{F}/K)\stackrel{{\scriptstyle|_{F}}}{{\twoheadrightarrow}}\text{Gal}(F/K)\cong
C_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$

for $p=2$, and

$\mathbb{Z}_{p}\times\mathbb{Z}_{p}\twoheadrightarrow\text{Gal}(\tilde{F}/K)\stackrel{{\scriptstyle|_{F}}}{{\twoheadrightarrow}}\text{Gal}(F/K)\cong\mathbb{Z}_{p}\times\mathbb{Z}_{p}$

for $p>2$. In either case, all epimorphisms are necessarily isomorphisms.
Therefore, $F=\tilde{F}$ by the infinite Galois correspondence.

We can now readily deduce the wild case. Indeed, given an extension
$L/\mathbb{Q}_{p}$ of degree $p^{n}$, we can see that the maximal abelian
$p$-extension of $K$ of exponent $p^{n}$ that is also abelian over
$\mathbb{Q}_{p}$ is contained in
$\mathbb{Q}_{2}(\zeta_{4},\zeta_{2^{n+2}},\zeta_{2^{2^{n}}-1})=\mathbb{Q}_{2}(\zeta_{2^{n+2}},\zeta_{2^{2^{n}}-1})$
for $p=2$ and in $\mathbb{Q}_{p}(\zeta_{p^{n}},\zeta_{p^{p^{n}}-1})$ for
$p>2$. Either way,
$L\subseteq\mathbb{Q}_{p}(\zeta_{p^{n+2}},\zeta_{p^{p^{n}}-1})$.