TITLE: Factoring Cartesian bitwise join of bit vectors
QUESTION [3 upvotes]: (This question has been substantially revised in an attempt to word it clearly.)
I am wondering if anyone has seen this problem. Let $[n] = \{1,\ldots,n\}$ for an integer $n$. Consider two finite multisets of bitstrings, $$ A=\{a_i\}_{i\in[r]} \quad\text{and} \quad B=\{b_j\}_{j\in[s]} \;,$$ where each $a_i \in \{0, 1\}^n$ and each $b_j \in \{0, 1\}^n$. We may compute from this a "Cartesian bitwise join" $C$, which is also a multiset of bitstrings $C= \{c_{i,j}\}_{(i,j) \in [r] \times [s]}$ for size $|C| = |A| \cdot |B| = rs$, where
$$ c_{i,j} = a_i \vee b_j$$ where "$\vee$" is bitwise-OR.
Question. If you are given such a multi-set $C$, without any particular indexing, can you find multi-sets $A$ and $B$ of which $C$ is the Cartesian bit-wise join? Equivalently: given a sequence $C = (c_k)_{k\in[m]}$ consisting of bitstrings in $\{0,1\}^n$, can you find sequences $A = (a_i)_{i\in[r]}$ and $B = (b_j)_{j\in[s]}$ such that $m = rs$, and a bijection $\varphi: [r] \times [s] \to [m]$, such that $$c_{\varphi(i,j)} = a_i \vee b_j$$ for all $(i,j) \in [r] \times [s]$?

This problem is equivalent to the following problem. Consider the ring $R$ whose elements are strings in $\{0,1\}^n$, where addition is done bit-wise by the operation $x + y = x \oplus y \oplus 1$ (where $\oplus$ is parity) with $1^n$ being the identity, and multiplication is the bit-wise OR with $0^n$ being the identity. For some particular ordering of the elements of $A$, we may present $A$ as an element of $R^r$, and similarly present $B$ as an element of $R^s$.
The Cartesian bitwise join corresponds to the outer product of $A$ and $B$,
$$ C \;=\; A B^\top \;=\; \begin{bmatrix} a_1 \vee b_1 & a_1 \vee b_2 &\cdots& a_1 \vee b_s \\
a_2 \vee b_1 & a_2 \vee b_2 & \cdots & a_2 \vee b_s \\
\vdots & \vdots & & \vdots \\
a_r \vee b_1 & a_r \vee b_2 & \cdots & a_r \vee b_s
\end{bmatrix}\;,$$
except interpreted as a multiset rather than a matrix (that is, the elements of $C$ are not assigned to particular coefficients of an $r \times s$ matrix). Given a multiset $C$ with composite cardinality $m \in \mathbb N$, can its elements be mapped to coefficients of an $r \times s$ array (such that $m = rs$) which is an outer product of the above form?
(By interchanging 0s and 1s in all of the bitstrings, we may replace the bit-wise OR "$\vee$" with the bit-wise AND "$\wedge$", in which case we may take $R$ to be the ring whose elements are strings in $\{0,1\}^n$, where addition is as vectors over $\mathbb Z_2$, and multiplication is the bit-wise AND. )

One may consider additional constraints on this problem, such as minimum Hamming weight of the bitstrings in $A$ and $B$ (in the Cartesian join formulation). We certainly require that $A,B \ne \{0^n\}$ to prevent a trivial solution.

This problem arises in an EE & complexity theory context. I am looking for other cases or analysis of it.
The problem is somewhat analogous to factoring but does not seem to have a unique factorization.
Note that bitvectors can be naturally modelled as hypergraphs so maybe something from that theory is applicable. I did research hypergraph products but that theory does not seem to apply directly. The problem translates to finding two hypergraphs $H'$ and $H''$ given a hypergraph $H$, such that the $H$ can be formed by taking all pairwise unions of edges in $H'$ and $H''$.

REPLY [5 votes]: Passing to the dual hypergraphs to look for products.
As you mention in the comments, we may interpret the bitstrings contained in $A$, $B$, and $C$ as edges in hypergraphs.

Vertices are the elements of $[n]$;
An edge $x \in \{0,1\}^n$ contains a vertex $v \in [n]$ if $x_v = 1$.

(Because these are multi-sets, they're actually multi-hyper-graphs. We may consider the edges to have some generic labels to distinguish them.) We may consider the dual hypergraph $H$ in which the vertices and the edges are interchanged; that is, where

The (labelled) bitstrings $x \in \{0,1\}^n$ now distinct vertices, and
A bitstring $x \in \{0,1\}^n$ belongs to an edge represented (that is, labelled) by $e \in [n]$ if and only if $x_e =1$;

— that is, a bitstring $x \in \{0,1\}^n$ describes all of the edges which are incident to it; of which there are always $n$ in this graphical model, albeit some of them may be the "empty edge".
This puts us in a much better position to describe the hypergraph as a product in some conventional manner. The hypergraph $H_C$ obtained from a multiset $C$ now has a composite number of vertices $m$:  we would like hypergraphs $H_A$ and $H_B$ such that $V(H_C)$ has some particular bijective correspondance with $V(H_A) \times V(H_B)$.
A diagonally restricted tensor product of hypergraphs.
Consider the alternative formulation of the problem, replacing bit-wise OR with bit-wise AND, alluded to in the recent revision of the problem. (To do this, we take the bitwise complement of all of the bitstrings.)  The Cartesian bit-wise meet condition (rather than "join", as we're using "$\wedge$" now in place of "$\vee$") corresponds to the existence of a particular bijection $$\varphi: V(H_A) \times V(H_B) \to V(H_C)$$ such that the edge labelled $e \in [n]$ in $H_C$ contains only the vertices $\varphi(a,b) \in V(H_C)$ such that both $a \in V(H_A)$ and $b \in V(H_B)$ are contained in the respective edges labelled by $e$ in $H_A$ and $H_B$. This seems somewhat different from most graph products, in that we do not take combinations of arbitrary edges, but only those with consistent labels.

If we suspended this label consistency condition, the join construction would correspond to a hypergraph tensor product (generalizing the tensor product of graphs, also commonly known as the direct product in the literature), in which we define edge-pairs $(e,f) \in E(H_A) \times E(H_B)$ which contain all vertices $(a,b) \in V(H_A) \times V(H_B)$ such that $a \in e$ and $b \in f$. This glosses over some issues to do with edges which contain $(a,b)$ and $(a,b')$ for a fixed co-ordinate $a$; but the construction we consider is appropriate to hypergraphs, taking the tensor product of the incidence matrices (characteristic functions of the edges, ranging over the vertices) to form the incidence matrix of $H_A \otimes H_B$.
To re-obtain the Cartesian bit-wise join of $H_A$ and $H_B$, we must
restrict to the diagonal edges $(e,e) \in E(H_A) \times E(H_B) =
   [n] \times [n]$, which we re-label by $(e,e) \mapsto e$. Note that $H_A$ and $H_B$ are edge-labelled hypergraphs, as their edges are non-negotiably labelled by the elements of $[n]$ from the formulation of the vertices as bitstrings with a fixed order. (Of course, when we are searching for hypergraphs $H_A$ and $H_B$, re-labelling the edges simply corresponds to considering a different hypergraph.) Selecting the diagonal is a plausible restriction, if somewhat unnatural in the context of graph products.

The resulting reformualtion.
From this, asking "does $C$ arise from the Cartesian bitwise join of $A$ and $B$?" is equivalent to asking:

Is an edge-labelled hypergraph $H_C$ with $m$ vertices and at msot $n$ edges, with distinct edge-labels selected from $[n]$ isomorphic to the diagonal subgraph (i.e. taking only the diagonally labelled edges) of the tensor product $H_A \otimes H_B$ of two labelled hypergraphs $H_A$ and $H_B$, which also have at most $n$ edges each and also have edges labels over $[n]$?

I'm not sure if you're going to come closer to a natural graph product formulation of than this. I rather suspect that if this does play a deep role in computational complexity that someone has studied something like this; but without knowing what areas of complexity theory it is pertinent to, I would not be able to tell you where to look, beyond perhaps investigating tensor products of hypergraphs.


TITLE: Show that $\alpha^n+\beta^n+\gamma^n=2^{2n+1}\,cos\,\frac{n\pi}{3}+\left(-\frac12\right)^n$
QUESTION [0 upvotes]: $$z^n=r^n(\cos{n}\theta+i\sin{n}\theta)$$
$$\alpha=2+2\sqrt3\,i$$
$${\Rightarrow}\,(2+2\sqrt3\,i)^n=\left(\sqrt{2^2+(2\sqrt{3})^2}\right)^n\left(\cos\frac{n\pi}{3}+i\sin\frac{n\pi}{3}\right)$$
$${\Rightarrow}\,\alpha^n=4^n\left(\cos\frac{n\pi}{3}+i\sin\frac{n\pi}{3}\right)$$
$$\beta=2-2\sqrt3\,i$$
$${\Rightarrow}\,(2-2\sqrt3\,i)^n=\left(\sqrt{2^2+(2\sqrt{3})^2}\right)^n\left(\cos(-\frac{n\pi}{3})+i\sin(-\frac{n\pi}{3})\right)$$
$${\Rightarrow}\,\beta^n=4^n\left(\cos(-\frac{n\pi}{3})+i\sin(-\frac{n\pi}{3})\right)$$
$$\gamma=-\frac12$$
$${\Rightarrow}\,\gamma^n=\left(-\frac12\right)^n$$
$$\alpha^n+\beta^n+\gamma^n=$$
$$\left(4^n\left(\cos\frac{n\pi}{3}+i\sin\frac{n\pi}{3}\right)\right)+\left(4^n\left(\cos(-\frac{n\pi}{3})+i\sin(-\frac{n\pi}{3})\right)\right)+\left(-\frac12\right)^n$$
What have I done wrong here? The Mark Scheme has $$\,\beta^n=4^n\left(\cos(\frac{n\pi}{3})-i\sin(\frac{n\pi}{3})\right)$$
Rewriting the equation with this in mind
$$\alpha^n+\beta^n+\gamma^n=$$
$$\left(4^n\left(\cos\frac{n\pi}{3}+i\sin\frac{n\pi}{3}\right)\right)+\left(4^n\left(\cos(\frac{n\pi}{3})-i\sin(\frac{n\pi}{3})\right)\right)+\left(-\frac12\right)^n$$
$${\Rightarrow}\,2\left(4^{n}\cos\frac{n\pi}{3}\right)+\left(-\frac12\right)^n$$
$${\Rightarrow}\,2\left(2^{2n}\cos\frac{n\pi}{3}\right)+\left(-\frac12\right)^n$$
Now I'm stuck, where does the $^{n+1}$ come from? And what happens to the $2$?
$$2^{2n+1}\,cos\,\frac{n\pi}{3}+\left(-\frac12\right)^n$$

REPLY [0 votes]: If you look at the sine and cosine graphs you will see that:
$$cos{\theta}=cos(-\theta)$$
$$sin(-\theta)=-sin\theta$$
Therefore:
$$\beta^n=\left(4^n\left(\cos(-\frac{n\pi}{3})+i\sin(-\frac{n\pi}{3})\right)\right)=\left(4^n\left(\cos(\frac{n\pi}{3})-i\sin(\frac{n\pi}{3})\right)\right)$$
Using the exponent rule:
$$(x^m)(x^n)=x^{m+n}$$
$$2(2^{2n})=(2^1)(2^{2n})=2^{2n+1}$$


TITLE: $\lim_{n \to \infty}(1 - \frac{1}{n^{p}})^{n}$ for $p< 1$ and $p > 1$
QUESTION [0 upvotes]: This
answers $\lim_{n \to \infty}(1 + \frac{1}{n^{p}})^{n}$ for $p< 1$ and $p > 1$
Is there a similar approach for the one below, or an approach that adapts what the link shows?
$\lim_{n \to \infty}(1 - \frac{1}{n^{p}})^{n}$ for $p< 1$ and $p > 1$

REPLY [2 votes]: Note that for $p,q>0$, as $n$ goes to $+\infty$,
$$\left(1 - \frac{1}{n^{p}}\right)^{n^q}=\exp\left(n^q\ln(1 - \frac{1}{n^{p}})\right)\sim\exp(-n^{q-p})$$
(because $\ln(1+x)\sim x$ when $x$ goes to $0$).
Hence
$$\lim_{n\to +\infty} \left(1 - \frac{1}{n^{p}}\right)^{n^q}=
\left\{\begin{array}{lr}
        0, & \text{if $q>p$, } \\
        1, & \text{if $q<p$, }\\
        e^{-1}, & \text{if $q=p$. }
        \end{array}\right.
$$


TITLE: Topological , Homeomorphic version of $|S \times S|=|S| $
QUESTION [1 upvotes]: Give example of a subset $A$ of $\mathbb R$ such that with respect to some topology , $ A$ is homeomorphic to $A\times A$ .
In set theory ZF it is known to be equivalent to A.C. that for any infinite set $S$ , there is a bijection between $S$ and $S \times S$ , what would be a suitable homeomorphic(topology) version of this theorem ?

REPLY [4 votes]: The trivial examples are a one-point subspace and an infinite discrete subspace of $\Bbb R$. The simplest non-trivial examples for $A$ are the rationals, the Cantor set, the irrationals, and the Cantor set minus any one of its points: each is homeomorphic to its square. In none of these cases is this entirely trivial to prove.

The Cantor set is homeomorphic to $\{0,1\}^{\Bbb N}$, where $\{0,1\}$ has the discrete topology; this is a fairly common exercise and not enormously difficult to prove. It’s easy to show that $\{0,1\}^{\Bbb N}$ is homeomorphic to its square.
The irrationals are homeomorphic to $\Bbb N^{\Bbb N}$; this is a bit harder to prove than the fact that the Cantor set is homeomorphic to $\{0,1\}^{\Bbb N}$. However, it’s easy to show that $\Bbb N^{\Bbb N}$ is homeomorphic to its square.
Every countable metric space without isolated points is homeomorphic to $\Bbb Q$, and $\Bbb Q^2$ is a countable metric space without isolated points, so $\Bbb Q$ is homeomorphic to its square. The characterization theorem isn’t exactly hard, but it’s certainly not trivial, either.
The Cantor set minus a point is homeomorphic to $\Bbb N\times C$, where $C$ is the Cantor set. Thus, its square is homeomorphic to $(\Bbb N\times C)\times(\Bbb N\times C)$, which in turn is homeomorphic to $\Bbb N^2\times C^2$ and hence to $\Bbb N\times C$.

For the general question, note that if $X$ is an infinite space with the discrete topology, then $X\times X$ is homeomorphic to $X$ if and only if $|X\times X|=|X|$: any homeomorphism between $X\times X$ and $X$ is a bijection, and conversely.


TITLE: Commutator of covariant derivative for rank 2 tensor
QUESTION [2 upvotes]: I am a newbie at tensor notation and I have been told to prove the identity

$$ (\nabla_a\nabla_b - \nabla_b\nabla_a)X^a_{\ \ \ b}=- R^e_{\ \ \ bcd}X^a_{\ \ \ e}+ R^a_{\ \ \ ecd}X^e_{\ \ \ b} $$

I am aware of the definition of the Riemman tensor
$(\nabla_a\nabla_b - \nabla_b\nabla_a)X_c= R_{abc}^{\quad d}X_{d}$ and using the metric connection $\nabla_a g_{bc}=0$ I have already shown
$(\nabla_a\nabla_b - \nabla_b\nabla_a)X^c= -R_{abd}^{\quad c}X^{d}$
but still I cannot figure out how to obtain the first one... :(
I am working torsion-free and I am aware of the formula that relates the Riemman tensor with the Christoffel symbols.
PS:  My guess is to go with $X^a_{\ \ \ b} = v^a w_b$ but not sure if that is the right way to go.

REPLY [1 votes]: PS: My guess is to go with $X^a_{\ \ \ b} = v^a w_b$ but not sure if
that is the right way to go.

It is. Schematically:
$$
R\sim\left[\nabla,\nabla\right]
$$
Any derivative worth that name satisfies:
$$
\nabla\left(uv\right)=\left(\nabla u\right)v+u\nabla v
$$
Covariant derivatives act like,
$$
\nabla\sim\partial+\Gamma
$$
for contravariant vectors.
$$
\nabla\sim\partial-\Gamma
$$
for covariant vectors.
Finally, the tangent space at a point is a vector space. Is that enough of a clue?


text:
\begin{document}

\begin{abstract}

We give a construction of the genus field for Kummer $\ell^n$-cyclic extensions of rational congruence function fields, where $\ell$ is a prime number. First, we compute the genus field of a field contained in a cyclotomic function field, and then for the general case.
This generalizes the result obtained by Peng for a Kummer $\ell$-cyclic extension. Finally, we study the extension $(K_1K_2)_{\eu{ge}}/(K_1)_{\eu{ge}}(K_2)_{\eu{ge}}$, for $K_1$, $K_2$ abelian extensions of $k$.

\end{abstract}

\maketitle

\section{Introduction}\label{S1}

The origin of genus fields dates back to C. F. Gauss \cite{Gau1801} in his work about binary quadratic forms. For a finite field extension $K$ of $\Q$, the rational number field, the genus field $\g K$ is defined as the maximum unramified field extension of $K$ such that it is the composite $Kk^{\ast}$, where $k^{\ast}$ is an abelian field extension over $\Q$. This definition is due to A. Fr\"ohlich \cite{Fro83}. We have $K\subseteq K_{{\eu {ge}}}\subseteq K_H$, where $K_H$ denotes the Hilbert class field over $K$. Originally the concept of genus fields was given for quadratic field extensions of ${\Q}$.

Using Dirichlet characters, H. W. Leopoldt \cite{Leo53} determined the narrow genus field $K_{{\eu {gex}}}$ of a finite abelian field extension $K$ over ${\Q}$, i.e, $K_{{\eu {gex}}}$ is the maximum abelian field extension of $\Q$ such that $K_{{\eu {gex}}}/K$ is unramified at any finite prime of $K$. This generalizes the results of H. Hasse \cite{Has51} who introduced genus theory for quadratic extensions of number fields.

M. Ishida described the narrow genus field $K_{{\eu {gex}}}$ of any finite extension of ${\Q}$ \cite{Ish76}. X. Zhang \cite{Xia85} gave a simple expression of $K_{{\eu {ge}}}$ of any finite abelian extension $K$ of $\Q$ using Hilbert ramification theory.

For function fields, the notion of Hilbert class field as the maximum unramified abelian extension of a congruence function field $K/{\ma F}_q$ is not suitable since it contains all the constant field extensions $K_m:=K{\ma F}_{q^m}$ for every natural number $m$. Therefore the maximum unramified abelian extension of $K$ is of infinite degree over $K$.

M. Rosen \cite{Ros87} gave a definition of an analogue of the Hilbert class field of $K$ for a fixed finite nonempty set $S$ of prime divisors of $K$. Given a finite nonempty set $S$ of places of a global function field $K$, the Hilbert class field (relative to $S$) $K_{H,S}$ of $K$ is defined as the maximum unramified abelian extension of  $K$ such that the places in $S$ are completely descomposed in $K_{H,S}$. Using Rosen's definition of Hilbert class field, it is possible to give a proper concept of genus fields along the lines of number fields.

R. Clement \cite{Cle92} found a narrow genus field of a cyclic extension of $k:={\ma F}_q(T)$ of prime degree $\ell$ dividing $q-1$. She found the genus field using class field theory and defining the Hilbert class field following the ideas of H. Hasse \cite{Has51}. Later, S. Bae and J. K. Koo \cite{BaeKoo96} were able to generalize the results of Clement with the methods developed by Fr\"ohlich \cite{Fro83}.

G. Peng \cite{Pen2003} explicitly described the genus theory for Kummer function fields extensions of prime degree. Later S. Hu and Y. Li \cite{HuLi2010} described explicitly the ambiguous ideal  classes and the genus field of an Artin--Schreier extension of a  rational congruence function field. In \cite{RockM}, it was studied the particular case of abelian finite $p$-extensions and the explicit description of their genus field was given. That article also studied finite abelian extensions of global rational functions fields, and the term {\it conductor of constants} was introduced for those extensions and it was computed in terms of other invariants of the field.

In \cite{MaRzVi2013, MaRzVi2015} it was developed a theory of genus fields of congruence function fields using Rosen's definition of Hilbert class field. Similarly to the case of numer fields, it was defined the genus field $\g K$ of $K$, as the maximum extension of $K$ such that $K\subseteq\g K\subseteq K_{H,S}$ with $\g K=K k^{\ast}$ and $k^{\ast}/k$ an abelian extension.
When $K/k$ is an abelian extension, $\g K$ is the maximum unramified extension over $K$ such that the primes in $S$ are fully decomposed in $\g K/K$. The methods used there were based on Leopoldt's ideas using Dirichlet characters and gave a general description of $\g K$ in terms of the Dirichlet characters associated to the field $K$. The genus field $\g K$ was obtained for an abelian extension $K$ of $k$ and $S$ the set of the infinite primes.
The method was used to give $\g K$ explicitly when $K/k$ is a cyclic extension of prime degree $\ell\smid q-1$ (Kummer) or cyclic of degree $p$ where $p$ is the characteristic (Artin--Schreier) and also when $K/k$ is a $p$--cyclic extension (Witt). Then, the method was used in \cite{BaRzVi2013} to describe $\g K$ explicitly when $K/k$ is an $\ell^n$-cyclic extension, where $\ell$ is a prime number such that $\ell^n\mid q-1$, $K/k$ is a cyclotomic extension and under a strong restriction.

In this paper, we will use the results obtained in \cite{MaRzVi2013} to describe explicitly the genus field of a cyclic extension of degree $\ell^n$ where $\ell^n\smid q-1$. In this way, we complete what was developed in \cite{BaRzVi2013}. Here we consider a cyclic Kummer extension of degree $\ell^n$, not necessarily cyclotomic and without any restrictions. In case $n=1$ this is the result of Peng. Our method is based on Leopoldt's ideas and therefore they differ from the methods used by Peng which are based on the global function fields analog to the exact hexagon of P. E. Conner and J. Hurrelbrink \cite{Hexagon}. In \cite{MaRzVi2013} the case $n=1$ is described in a little bit different way from how it was originally presented by Peng. Here we will show that using our methods it is possible to give the same description as in the original article.

Finally, given two finite abelian extensions $K_i/k$, $i=1,2$, we have $(K_1)_{\eu{ge}}(K_2)_{\eu{ge}}$ $\subseteq (K_1K_2)_{\eu{ge}}$ but in general there is no equality. This fact is a big obstruction to study the genus field of a finite abelian extension $K/k$ since we cannot just study the components say $K=K_1\dots K_s$. In Section \ref{S5} we study the extension $(K_1K_2)_{\eu{ge}}/(K_1)_{\eu{ge}}(K_2)_{\eu{ge}}$. The main result is that the degree of this extension divides $(q-1)^2$. It is worth to mention that the factor $(q-1)^2$ is due to two different facts. One factor comes from the cyclotomic case where in a composition of two fields such that $\p$ ramifies in both components, in the composition some ramification of $\p$ shifts to descomposition since we have tame ramification. The other factor comes from the various descomposition groups given in the main theorem of the structure of the genus field $\g K=E_\eu{ge}^HK$. As a consequence we obtain that for extensions of degree relatively prime to $q-1$, we have $(K_1)_{\eu{ge}}(K_2)_{\eu{ge}}= (K_1K_2)_{\eu{ge}}$. This was used in \cite{RockM}

\section{Notation}\label{S2}

The notation we will use throughout the paper is the following.

\begin{itemize}
    \item[]$\gcd\quad$ denotes the greatest common divisor;
    \item[]${\rm lcm}\quad$ denotes the least common multiple;
    \item[]$\nu_\ell\quad$ denotes the valuation respect to $\ell$, a prime number;
    \item[]$k\quad$ denotes the rational function field ${\mathbb F}_q(T)$;
    \item[]$R_T\quad$ denotes the polynomial ring ${\mathbb F}_q[T]$;
    \item[]$R_T^+\quad$ denotes the set of monic and irreducible polynomials in $R_T$;
    \item[]$K\quad$ denotes a cyclic finite Kummer extension of $k$;
    \item[]$\p\quad$ denotes the infinite prime in $k$;
    \item[]$S_\infty(K)\quad$ denotes the set of primes in $K$ over $\p$;
    \item[]$\e{E/F}\quad$ denotes the ramification index of $\p$ in the field extension $E/F$;
    \item[]$\f{E/F}\quad$ denotes the inertia degree of $\p$ in the field extension $E/F$;
    \item[]$\Lambda_N\quad$ denotes the $N$-torsion of the Carlitz's module for $N\in R_T\setminus\{0\}$;
    \item[]$\g K\quad$ denotes the genus field of $K$;
    \item[]$K_{{\eu {gex}}}\quad$ denotes the extended or narrow genus field of $K$;
    \item[]$\lam N\quad$ denotes the cyclotomic function field over $k$ determined by $N$;
    \item[]$\lam N^+\quad$ denotes the maximum real subfield of $\lam N$;
    \item[]$L_n\quad$ denotes the subfield of ${\lam{{T^{-n-1}}}}$ fixed by ${\mathbb F}^*_q$;
    \item[]$K_m\quad$ denotes the field $K{\mathbb F}_{q^m}$;
    \item[]${_n}K\quad$ denotes the field $KL_n$;
    \item[]${_n}K_m\quad$ denotes the field $K{\mathbb F}_{q^m}L_n$.
    \item[]
\end{itemize}

Let $\ell$ be a prime number such that $\ell^n\smid q-1$.
Let $D\in R_T$, $D=P_1^{\alpha_1}\cdots P_r^{\alpha_r}$ with $P_1,\dots,P_r\in R_T^+$ different polynomials, $r\geq 1$, $\alpha_1,\dots,\alpha_r\in\N$ and $1\leq \alpha_j\leq \ell^n-1$, $ 1\leq j\leq r$. Let $E:=k(\sqrt[\ell^n]{(-1)^{\deg D}D})$. We have $E\subseteq \lam{D}$\cite[Corolario 9.5.12]{VilCC}. We denote $D^*:=(-1)^{\deg D}D$, that is $k(\sqrt[\ell^n]{(-1)^{\deg D}D})=k(\sqrt[\ell^n]{D^*})$. Note that if $\ell^n\smid \deg D$, then $k=k(\sqrt[\ell^n]{(-1)^{\deg D}})$ and thus $k(\sqrt[\ell^n]{D^*})=k(\sqrt[\ell^n]{D})$.

Let $\alpha_j=b_j\ell^{a_j}$ with $\gcd(b_j,\ell)=1$ and $\deg P_j=c_j\ell^{d_j}$ with $\gcd(c_j,\ell)=1$, $1\leq j\leq r$. For each $1\leq j\leq r$, we define $E_j:=k(\sqrt[\ell^n]{(P_j^{\alpha_j})^*})$. We have $E_j\subseteq k(\Lambda_{P_j})$, $1\leq j\leq r$. Note that
$\sqrt[\ell^n]{(-1)^{\deg P_j^{\alpha_j}}P^{\alpha_j}}=\sqrt[\ell^n]{(-1)^{b_j\ell^{a_j} \deg P_j}P_j^{b_j\ell^{a_j}}}=\sqrt[\ell^{n-a_j}]{(-1)^{\deg P_j^{b_j}}P_j^{b_j}}$. Further $\sqrt[\ell^{n-a_j}]{(-1)^{\deg P_j^{b_j}}P_j^{b_j}}=(\sqrt[\ell^{n-a_j}]{(-1)^{\deg P_j}P_j})^{b_j}$ $\in k(\sqrt[\ell^{n-a_j}]{(-1)^{\deg P_j}P_j})$. We have that $P_j^{b_j}$ is $\ell$-power free and thus $P_j$ is fully ramified in $k(\sqrt[\ell^{n-a_j}]{(-1)^{\deg P_j^{b_j}}P_j^{b_j}})/k$. Therefore
\begin{align*}
[k(\sqrt[\ell^{n-a_j}]{(-1)^{b_j\deg P_j}P_j^{b_j}}):k]&=\deg(X^{\ell^{n-a_j}}-(-1)^{b_j\deg P_j}P_j^{b_j})\\
&=\deg(X^{\ell^{n-a_j}}-(-1)^{\deg P_j}P_j)\\
&=[k(\sqrt[\ell^{n-a_j}]{(-1)^{\deg P_j}P_j}):k].
\end{align*}

It follows that $E_j=k(\sqrt[\ell^{n-a_j}]{P_j^*})$.

Define $M:=E_1\cdots E_r$. We have that $M/k$ is the maximum extension of $E$ unramified on every finite prime and contained in a cyclotomic function field \cite[Proposition 3.3]{MaRzVi2013}. In particular $\g E\subseteq M$. We have that $\p$ is fully ramified in $M/\g E$.
The field $M$ is known as the {\it narrow genus field} of $E$ and it is denoted by $E_{\eu {gex}}$.
From Abhyankar's Lemma \cite[Theorem 12.4.4]{Vil2006}, we have
\[
e_{\infty}(E_{\eu {gex}}/k)={\rm lcm}[e_{\infty}(E_j/k)\smid 1\leq j\leq r]=:\ell^m.
\]

Since $\p$ is unramified in the extension $\g E/E$, i.e, $\e{\g E/E}=1$, we have in general $$[E_{\eu {gex}}:\g E]=\e{E_{\eu {gex}}/\g E}\e{\g E/E}=\e{E_{\eu {gex}}/E}.$$

Let $P\in R_T$ and let $F:=k(\sqrt[\ell^n]{(P^{\alpha})^*})$ with $\alpha=b\ell^a$, $\gcd(b,\ell)=1$. We denote by $e_{P}(F/k)$ the  ramification index of the polynomial $P$ in the extension $k(\sqrt[\ell^n]{(P^{\alpha})^*})/k$. Let $e_{\infty}(E/k)$ denote the ramification index of the infinite prime $\p$, in the extension $k(\sqrt[\ell^n]{(P^{\alpha})^*})/k$.

We have the following proposition.

\begin{prop}\label{p1} Let $P\in R_T$ and $k(\sqrt[\ell^n]{(P^{\alpha})^*})=k(\sqrt[\ell^{n-a}]{P^*})$. Then
\begin{gather*}
    e_{P}(k(\sqrt[\ell^n]{(P^{\alpha})^*})/k)=\frac{\ell^n}{\gcd(\alpha,\ell^n)}=\ell^{n-a}
    \quad\textit{ and }\\
    e_{\infty}(k(\sqrt[\ell^n]{(P^{\alpha})^*})/k)
    =\frac{\ell^n}{\gcd(\ell^n,\deg P^\alpha)}
    =\frac{\ell^{n-a}}{\gcd(\ell^{n-a},\deg P)}.
\end{gather*}
\end{prop}
\begin{proof}
See \cite[Subsection 5.2]{MaRzVi2016} or \cite[Teorema 10.3.1]{VilCC}.
\end{proof}

Let $\deg P=c\ell^d$ with $\gcd(c,\ell)=1$. Then $\gcd(\ell^{n-a},\deg P)=\ell^{{\rm min}\{n-a,d\}}$. From Proposition \ref{p1} it follows that
    \begin{gather*}
        e_{\infty}\left( k\left(\sqrt[\ell^{n-a}]{P^*}\right)/k\right)
        =\ell^{n-a-{\rm min}\{n-a,d\}}.
    \end{gather*}

Thus
\begin{gather}\label{divRaM}
    e_{\infty}(E_j/k)=\ell^{n-a_j-{\rm min}\{n-a_j,d_j\}}\smid \ell^m=e_{\infty}(E_{\eu {gex}}/k),\,\, 1\leq j\leq r.
\end{gather}

Let $K/\F$ be a finite abelian field extension of $k$, where $\p$ is tamely ramified. Let $N\in R_T$ and $u\in{\ma N}$ be such that $\g K{}\subseteq {\lam N}_{u}$.
Let $E_{{\eu {ge}}}$ be the genus field of $E:=\lam N\cap
K_{u}$ and let $E_{{\eu {ge}},u}=E_{{\eu {ge}}}{\ma F}_{q^u}$.
Let $H$ be the decomposition group of $\S K$, the set of primes in $K$ over $\p$, in $\g E{}K/K$. Let $H'
=H|_{\g E{}}$.
Then the genus field of $K$ is
$
K_{{\eu {ge}}}= (\g E{}K)^H=\g E^{H'}K
$
\cite[Theorem 4.2]{MaRzVi2015}.
We have the following diagram
\begin{tiny}
\begin{gather}\label{diag}
\xymatrix{
&k(\Lambda_N)\ar@{-}[rrrrr]\ar@{-}[d]&&&&&k(\Lambda_N)_u\ar@{-}[d]\\
&E_{{\eu {ge}}}\ar@{-}[dl]_{H'=H|_{\g E{}}}\ar@{-}[ddr]\ar@{-}[rrr]&&&
E_{{\eu {ge}}}K\ar@{-}[ddr]\ar@{-}[dl]_H\ar@{-}[rr]\ar@/_2pc/@{-}[ddd]
|!{[dl];[d]}\hole|!{[ddll];[ddr]}\hole
&&K_{{\eu {ge}},u}=E_{{\eu {ge}},u}\ar@{-}[dd]\\
\g E^{H'}\ar@{-}[ddr]\ar@{-}[rrr]|!{[ur];[drr]}\hole
&&&(\g E{}K)^H\ar@{-}[ddr]\ar@{--}[r]&\g K{}
\ar@{-}[dd]|!{[dll];[dr]}\hole\\
&&E\ar@{-}[rrr]|!{[ru];[rrd]}\hole \ar@{-}[dl]&&&EK\ar@{-}[r]\ar@{-}[dl]
&E_u=K_u\ar@{-}[dd]\ar@{-}[dll]\\
&E\cap K\ar@{-}[rrr]\ar@{-}[d]&&&K\\
&k\ar@{-}[rrrrr]&&&&&k_u}
\end{gather}
\end{tiny}

\section{Cyclotomic $\ell^n$ case}\label{S3}

We start with the following proposition that bounds the type of ramification in Kummer $\ell^n$-extensions.

\begin{prop}\label{imenosj} Let $P,Q\in R_T^+$ and let $J:=k(\sqrt[\ell^n]{(P^\alpha)^*})$ and $F:=k(\sqrt[\ell^n]{(Q^\beta)^*})$. Suppose that $e_P(J/k)\leq e_Q(F/k)$ and $e_{\infty}(F/k)\leq e_{\infty}(J/k)$ with $1<e_{\infty}(J/k)$. Then $\nu_\ell(\deg P)\leq\nu_\ell(\deg Q)$.
\end{prop}
\begin{proof}
Since $e_P(J/k)\leq e_Q(F/k)$, we have $1\leq\frac{e_Q(F/k)}{e_P(J/k)}$. On the other hand, from $e_{\infty}(F/k)\leq e_{\infty}(J/k)$ we obtain
\begin{gather*}
    \frac{e_Q(F/k)}{\gcd(\deg Q, e_Q(F/k))}=e_{\infty}(F/k)\leq
    e_{\infty}(J/k)=\frac{e_P(J/k)}{\gcd(\deg P, e_P(J/k))}.
\end{gather*}
Since $e_{\infty}(J/k)\neq 1$, it follows that $\nu_\ell(\deg P)< e_P(J/k)$, and $\gcd(\deg P,e_P(J/k))=\ell^{\nu_\ell(\deg P)}$. Also, since $\gcd(\deg Q, e_Q(F/k))\smid \deg Q$, we obtain
\begin{gather*}
    \frac{e_Q(F/k)}{\ell^{\nu_\ell(\deg Q)}}\leq\frac{e_Q(F/k)}{\gcd(\deg Q, e_Q(F/k))}\leq \frac{e_P(J/k)}{\gcd(\deg P, e_P(J/k))}=\frac{e_P(J/k)}{\ell^{\nu_\ell(\deg P)}}.
\end{gather*}

Hence
$$1\leq \frac{e_Q(F/k)}{e_P(J/k)}\leq \ell^{\nu_\ell(\deg Q)-\nu_\ell(\deg P)}.$$ Therefore $0\leq \nu_\ell(\deg Q)-\nu_\ell(\deg P)$, i.e, $\nu_\ell(\deg P)\leq\nu_\ell(\deg Q)$.
\end{proof}

The main result for the Kummer cyclotomic $\ell^n$-cyclic case is the next theorem.

\begin{teo}\label{Pengln} Let $E =k(\sqrt[\ell^n]{D^*})$, with $D=P_1^{\alpha_1}\cdots P_r^{\alpha_r}$, $1\leq \alpha_j\leq \ell^n-1$, $\alpha_j=b_j\ell^{a_j}$ with $\gcd(b_j,\ell)=1$, $1\leq j \leq r$, $P_1,\dots, P_r\in R_T^+$ different with $\deg P_j=c_j\ell^{d_j}$, ${\rm gcd}(c_j,\ell)=1$, $1\leq j\leq r$. We order the polynomials $P_1,\dots,P_r$ such that $0=a_1\leq\cdots\leq a_r\leq n-1$.

Let $E_{\eu {gex}}:=E_1\cdots E_r$ with $E_j=k(\sqrt[\ell^{n-a_j}]{P_j^*})$, $1\leq j\leq r$. Let
\begin{align*}
    e_\infty(E/k)=\ell^t\;\textit{with }\;t&=n-{\rm min}\{n,\nu_\ell(\deg D)\},\\
    &\\
    e_\infty(E_{\eu {gex}}/k)=\ell^m\;\textit{with }\;m&=\underset{1\leq j\leq r}{\rm max}\nu_\ell(e_\infty(E_j/k))\\
    &={\rm max}\{n-a_j-{\rm min}\{n-a_j,d_j\} \smid 1\leq j\leq r\}.
\end{align*}

Let $i_0$, $1\leq i_0\leq r$, be such that $n-a_{i_0}-{\rm min}\{n-a_{i_0},d_{i_0}\}=m$ and $n-a_j-d_j<m$ for $j>{i_0}$.
For $m>0$ we have $\gcd(\deg P_{i_0},\ell^n)=\ell^{d_{i_0}}$, and therefore there exist $a, b\in\Z$ such that $a\,\deg P_{i_0}+b\ell^n=\ell^{d_{i_0}}$. For $j<{i_0}$, we have $d_{i_0}\leq d_j$. Let $z_j:=-ac_j\ell^{d_j-d_{i_0}}$. For $j>{i_0}$, let $y_j\equiv -c_jc_{i_0}^{-1}\mod \ell^{n}\in\Z$.

Then $$\g E=F_1\cdots F_r,$$ where $F_j=E_j$ with $1\leq j\leq r$ if $m=t$, i.e, $\g E=E_{\eu {gex}}$, and if $m>t\geq 0$, then
\begin{gather}\label{knorr}
    F_j:=
    \begin{cases}
    k\left(\sqrt[\ell^{n-a_j}]{P_jP_{i_0}^{z_j}}\right)&\text{if $j<{i_0}$,}\\
    k\left(\sqrt[\ell^{d_{i_0}+t}]{P_{i_0}^*}\right)&\text{if $j={i_0}$,}\\
    k\left(\sqrt[\ell^{n-a_j}]{P_jP_{i_0}^{y_j\ell^{d_j-d_{i_0}}}}\right)&\text{if $j>{i_0}$ and $d_j\geq d_{i_0}$},\\
    k\left(\sqrt[\ell^{n-a_j+d_{i_0}-d_j}]{P_j^{\ell^{d_{i_0}-d_j}}P_{i_0}^{y_j}}\right)&\text{if $j>{i_0}$ and $d_{i_0}> d_j$.}
    \end{cases}
\end{gather}
\end{teo}
\begin{proof}
First suppose that $m=t$. Then $$[E_{\eu {gex}}:\g E]=\frac{\e{E_{\eu {gex}}/k}}{\e{E/k}}=\ell^{m-t}=1.$$ It follows that $$E_{\eu {gex}}=\g E.$$

Now, suppose that $m>t$. Let $i_0$ be as before. From (\ref{divRaM}) we have
$$e_{\infty}\left( k\left(\sqrt[\ell^{n-a_{i_0}}]{P_{i_0}^*}\right)/k\right)=\eu{\ell^{n-a_{i_0}}}{\gcd(\deg P_{i_0}, \ell^{n-a_{i_0}})}=\ell^m.$$

Therefore, $\gcd(\deg P_{i_0}, \ell^{n-a_{i_0}})=\ell^{d_{i_0}}$ and $m=n-a_{i_0}-{\rm min}\{n-a_{i_0}, d_{i_0}\}=n-a_{i_0}-d_{i_0}>t\geq0$.

Because $\gcd(\deg P_{i_0},\ell^n)=\ell^{d_{i_0}}$, there exist $a,b\in\Z$, such that
\begin{equation}\label{combilin}
        a\deg P_{i_0}+b\ell^n=\ell^{d_{i_0}}.
\end{equation}

In particular $ac_{i_0}+b\ell^{n-d_{i_0}}=1$ and therefore $\gcd(a,\ell)=1$.
From Proposition \ref{imenosj} we have $\ell^{d_{i_0}}\smid \deg P_j$ for $1\leq j\leq {i_0}-1$, so that from (\ref{combilin}) we obtain

$$\deg P_j+\left(-a\frac{\deg P_j}{\ell^{d_{i_0}}}\right)\deg P_{i_0}=(b\,\deg P_j)\ell^{n-d_{i_0}}.$$

We have $-a\frac{\deg P_j}{\ell^{d_{i_0}}}\in \Z$ with $\gcd(a,\ell)=1$. Let $z_j:=-a\frac{\deg P_j}{\ell^{d_{i_0}}}=-ac_j\ell^{d_j-d_{i_0}}$ and $Q_j:=P_jP_{i_0}^{z_j}$, $1\leq j\leq {i_0}-1$. By construction we have that $\ell^n\smid \deg Q_j$, since $d_j\geq d_{i_0}$. Therefore, $\p$ is unramified at $k\left(\sqrt[\ell^{n-a_j}]{Q_j}\right)/k$, i.e, $e_{\p}\left(k\left(\sqrt[\ell^{n-a_j}]{Q_j}\right)/k\right)=1$, (Proposition \ref{p1}).
Also, for $j<{i_0}$, we have
\begin{align}
        e_{P_j}\left(k\left(\sqrt[\ell^{n-a_j}]{Q_j}\right)/k\right)&=\ell^{n-a_j},\label{I}\\
        e_{P_{i_0}}\left(k\left(\sqrt[\ell^{n-a_j}]{Q_j}\right)/k\right)&=\frac{\ell^{n-a_j}}{\gcd(z_j,\ell^{n-a_j})}
        = \ell^{n-a_j-{\rm min}\{n-a_j, d_j-d_{i_0}\}}.\nonumber
\end{align}

 Now $d_j-d_{i_0}\geq {\rm min}\{n-a_j,d_j-d_{i_0}\}$ so that $n-a_j-{\rm min}\{n-a_j,d_j-d_{i_0}\}\leq n-a_j-d_j+d_{i_0}\leq n-a_{i_0}-d_{i_0}+d_{i_0}=n-a_{i_0}$ since $e_\infty(E_j/k)=\ell^{n-a_j-d_j}\smid e_\infty(E_{i_0}/k)=\ell^{n-a_{i_0}-d_{i_0}}$. Therefore
 \begin{equation}
        e_{P_{i_0}}\left(k\left(\sqrt[\ell^{n-a_j}]{Q_j}\right)/k\right)\leq \ell^{n-a_{i_0}}.\label{II}
\end{equation}

Now consider $j>{i_0}$. We have two possibilities: $d_{i_0}\leq d_j$ or $d_{i_0}>d_j$.

First suppose that $d_{i_0}\leq d_j$. Consider the polynomial $Q_j:=P_jP_{i_0}^{x_j}$, with $x_j:=y_j\ell^{d_j-d_{i_0}}$. Hence
\begin{align*}
        \deg Q_j&=\deg P_j+x_j\deg P_{i_0}
        =c_j\ell^{d_j}+y_jc_{i_0}\ell^{d_j}.
\end{align*}

Then $\deg Q_j=\ell^{d_j}(c_j+y_jc_{i_0})=A\ell^B$ with $\gcd(A,\ell)=1$. Note that we may choose $B\geq n+d_j$, since this is equivalent to $c_j+y_jc_{i_0}\equiv 0\mod \ell^{n}$, and this is possible because $\gcd(c_{i_0}c_j,\ell)=1$.

Finally, since $\gcd(c_{i_0}c_j,\ell)=1$, we select $y_j\in\Z$ such that $y_j\equiv -c_jc_{i_0}^{-1}\mod \ell^{n}$. Note that in particular, ${\rm gcd}(y_j,\ell)=1$. Thus, define $F_j:=k\left(\sqrt[\ell^{n-a_j}]{Q_j}\right)$. We have $\deg Q_j=A\ell^B$ with $B\geq n$. Note that
\begin{gather}
        e_{P_j}(F_j/k)=e_{P_j}(E_j/k)=\ell^{n-a_j},\label{III}\\
\begin{align}        e_{P_{i_0}}(F_j/k)&=\frac{\ell^{n-a_j}}{\gcd(y_j\ell^{d_j-d_{i_0}},\ell^{n-a_j})}=\ell^{n-a_j-{\rm min}\{n-a_j,d_j-d_{i_0}\}}\nonumber\\
&\leq \ell^{n-a_{i_0}}=e_{P_{i_0}}(E_{i_0}/k)
\end{align}\label{IV}\\
{\rm{and}}\quad e_{\infty}\left(F_j/k\right)=\frac{\ell^{n-a_j}}{\gcd(\deg Q_j,\ell^{n-a_j})}=\ell^{n-a_j-n+a_j}=1.\label{V}
\end{gather}

Now, suppose that $d_{i_0}>d_j$.
Let $Q'_j:=P_j^{\ell^{d_{i_0}-d_j}}P_{i_0}^{y_j}$, with $y_j$ as before, that is, $y_j\equiv -c_jc_{i_0}^{-1}\mod \ell^{n}$, $Q'_j\in R_T$. Define $$F_j:=k\left(\sqrt[\ell^{n-a_j+d_{i_0}-d_j}]{P_j^{\ell^{d_{i_0}-d_j}}P_{i_0}^{y_j}}\right).$$

We have
\begin{align}\label{JGi}
       e_{P_j}(F_j/k)&=\frac{\ell^{n-a_j+d_{i_0}-d_j}}{\gcd(\ell^{d_{i_0}-d_j},\ell^{n-a_j+d_{i_0}-d_j})}
       =\ell^{n-a_j+d_{i_0}-d_j-d_{i_0}+d_j}\nonumber\\
       &=\ell^{n-a_j}=e_{P_j}(E_j/k),
\end{align}
and
\begin{gather*}
       e_{P_{i_0}}(F_j/k)=\frac{\ell^{n-a_j+d_{i_0}-d_j}}{\gcd(y_j,\ell^{n-a_j+d_{i_0}-d_j})}
       =\ell^{n-a_j+d_{i_0}-d_j}.
\end{gather*}

Since $n-a_j-d_j<m=n-a_{i_0}-d_{i_0}$, then $n-a_j-d_j+d_{i_0}<n-a_{i_0}$. It follows that
\begin{equation}
        e_{P_{i_0}}(F_j/k)<\ell^{n-a_{i_0}}=e_{P_{i_0}}(E_{i_0}/k).\label{VI}
\end{equation}

Finally
\begin{align*}
       \deg Q'_j&=\deg P_j^{\ell^{d_{i_0}-d_j}}P_{i_0}^{y_j}=\ell^{d_{i_0}-d_j}\deg P_j+y_j\deg P_{i_0}\\
       &=\ell^{d_{i_0}-d_j}c_j\ell^{d_j}+y_jc_{i_0}\ell^{d_{i_0}}=\ell^{d_{i_0}}(c_j+y_jc_{i_0})=\ell^{d_{i_0}}(A\ell^{n-d_j})
       =A\ell^{n+d_{i_0}-d_j}.
\end{align*}

We have $A\ell^{n+d_{i_0}-d_j}=A\ell^{n-d_j+d_{i_0}}$, with $\nu_\ell(A)=0$ and $n-d_j+d_{i_0}>n$. It follows that
\begin{equation}
        e_{\infty}(F_j/k)=\frac{\ell^{n-a_j+d_{i_0}-d_j}}{\gcd(A\ell^{n+d_{i_0}-d_j},\ell^{n-a_j+d_{i_0}-d_j})}=\ell^{n-a_j+d_{i_0}-d_j-n+a_j-d_{i_0}+d_j}=1.\label{VII}
\end{equation}

Let $L:=F_1\cdots F_{{i_0}-1}F_{{i_0}+1}\cdots F_r$. From (\ref{I}) -- (\ref{VII}) we have $L\subseteq \g E$. We will prove that $\g E=LF_{i_0}$, with $F_{i_0}$ as in (\ref{knorr}). Consider the sets
\[
    \mathcal J=\{j\in\{1,2,\dots,r\}\smid j>{i_0},\;n-a_j-d_j>t\;\textit{ and }\;d_{i_0}> d_j\}
\]
and $\I=\{1,2,\dots,r\}\setminus (\J\cup \{{i_0}\})$. We order the elements of $\J=\{j_1,\dots,j_s\}$ so that $d_{i_0}-d_{j_1}\leq d_{i_0}-d_{j_2}\leq\cdots\leq d_{i_0}-d_{j_s}$. For $j_u\in\J$ we have $F_{j_u}=k(\sqrt[\ell^{n-a_{j_u}+d_{i_0}-d_{j_u}}]{P_{j_u}^{\ell^{d_{i_0}-d_{j_u}}}P_{i_0}^{y_{j_u}}})$.

Let $I_{j}$ be the inertia group of the prime $P_{j}$ in $F_{j}/k$ for $1\leq j\leq r$. We have $|I_{j}|=e_{P_j}(F_j/k)=\ell^{n-a_j}$, $j\neq i_0$. Let $F'_{j}:=F_{j}^{I_{j}}$. If $j_u\in\J$ we have $F'_{j_u}=k(\sqrt[\ell^{d_{i_0}-d_{j_u}}]{P_{i_0}^*})$, $1\leq u\leq s$, and $F'_{j_1}\subseteq F'_{j_2}\subseteq \cdots\subseteq F'_{j_s}$. If $j\in\I$ we have $F'_j=k$.

For $F_i$ and $F_j$ given as in (\ref{knorr}) with $i\neq j$, we obtain $$\Gal{(F_iF_j/F_i\cap F_j)}\cong \Gal{(F_i/F_i\cap F_j)}\times\Gal{(F_j/F_i\cap F_j)}.$$

In addition, we have
\begin{gather*}
        |\Gal{(F_i/F'_i)}|=|\Gal{(F_i/F_i^{I_i})}|=|I_i|=e_{P_i}(F_i/k)=\ell^{n-a_i},\textit{ and}\\
        |\Gal{(F_j/F'_j)}|=|\Gal{(F_j/F_j^{I_j})}|=|I_j|=e_{P_j}(F_j/k)=\ell^{n-a_j}.
\end{gather*}

Hence $\Gal{(F_iF_j/F_i\cap F_j)}\cong I_i\times I_j.$

Now, from Abhyankar's Lemma \cite[Theorem 12.4.4]{Vil2006}, we have
\[
    e_{P_i}(F_iF_j/k)={\rm lcm}[e_{P_i}(F_i/k),e_{P_i}(F_j/k)]={\rm lcm}[e_{P_i}(F_i/k),1]=e_{P_i}(F_i/k)=|I_i|.
\]

Similarly we have $e_{P_j}(F_iF_j/k)=e_{P_i}(F_j/k)=|I_j|$. Let $I'_i$ and $I'_j$ be the inertia groups of $P_i$ and $P_j$ in $F_iF_j/k$ respectively. Then $I_i\times\{e\}<I'_i$ and $\{e\}\times I_j<I'_j$ are such that $|I_i\times\{e\}|=|I'_i|$ and $|\{e\}\times I_j|=|I'_j|$. Therefore, the maximum unramified field subextension at $P_i$ and $P_j$ of $F_iF_j/k$ is $(F_iF_j)^{I_iI_j}=(F_iF_j)^{I_i\times I_j}$.

Also note that for  $i,j\in\J$, $i\neq j$, we have that $P_i$ and $P_j$ are unramified in $F_i\cap F_j$.
Hence $F_i\cap F_j\subseteq F'_i$. Similarly $F_i\cap F_j\subseteq F'_j$. Thus $F_i\cap F_j\subseteq F'_i\cap F'_j\subseteq F_i\cap F_j$. Therefore $F_i\cap F_j= F'_i\cap F'_j$. If we assume that $i<j$, then $F_i\cap F_j=F'_i$. Therefore
\begin{align*}
    [F_iF_j:k]&=[F_iF_j:F_i\cap F_j][F_i\cap F_j:k]=[F_iF_j:F'_i][F'_i:k]\\
    &=[F_iF_j:F'_j][F'_j:F'_i][F'_i:k]=[F_iF_j:F_j][F_j:F'_j][F'_j:k]\\
    &=[F_i:F'_i][F_j:F'_j][F'_j:k].
\end{align*}

Now, say that $\I=\{i_1,\dots,i_{s'}\}$. Using induction it follows that the fields $F_{i_w}$, with $i_w\in\I$, for $1\leq w\leq s'$ satisfy
\begin{itemize}
        \item[a.1.-] $F_{i_1}\cdots F_{i_{w-1}}\cap F_{i_w}=k$,
        \item[b.1.-] $I_{i_1}\cdots I_{i_w}\cong I_{i_1}\times\cdots\times I_{i_w}$,
        \item[c.1.-] $[F_{i_1}\cdots F_{i_w}:k]=\prod\limits_{j=1}^w[F_{i_j}:k]=\prod\limits_{j=1}^w \ell^{n-a_{i_j}}$,
        \item[d.1.-] $(F_{i_1}\cdots F_{i_w})^{I_{i_1}\cdots I_{i_w}}=k$.
\end{itemize}

Similarly, the fields $F_{j_w}$, $1\leq w\leq s$, with $j_w\in\J$, satisfy
\begin{itemize}
    \item[a.2.-] $F_{j_1}\cdots F_{j_{w-1}}\cap F_{j_w}=F'_{j_{w-1}}$,
    \item[b.2.-] $I_{j_1}\cdots I_{j_w}\cong I_{j_1}\times\cdots\times I_{j_w}$,
    \item[c.2.-] $[F_{j_1}\cdots F_{j_w}:k]=(\prod\limits_{n=1}^{w-1}[F_{j_n}:F'_{j_n}])[F'_w:k]=(\prod\limits_{n=1}^{w-1} \ell^{n-a_{j_n}})\ell^{d_i-d_w}$,
    \item[d.2.-] $(F_{j_1}\cdots F_{j_w})^{I_{j_1}\cdots I_{j_w}}=F'_{j_w}$.
\end{itemize}

Let $L=\prod\limits_{i\in\I} F_i \prod\limits_{j\in\J} F_j$. We will see that $(\prod\limits_{i\in\I}F_i)\cap (\prod\limits_{j\in\J} F_j)=k$. Otherwise, let $A\neq k$ be a proper subfield of $\prod\limits_{i\in\I}F_i$. Then at least one $P_i$ with $i\in \I$ should ramify in $A$, since otherwise we would have from (b.1) and (d.1) that $A\subseteq (\prod\limits_{i\in\I}F_i)^{\prod\limits_{i\in\I}I_i}=k$. Therefore, in every nontrivial proper subfield of $\prod\limits_{i\in\I}F_i$ at least one $P_i$ with $i\in\I$ is ramified. Now, in every subfield of $\prod\limits_{j\in\J} F_j$ some $P_j, j\in \mathcal J$ or $P_{i_0}$ is ramified but none of these is ramified in $\prod\limits_{i\in\I} F_i$. It follows that $(\prod\limits_{i\in\I}F_i)\cap (\prod\limits_{j\in\J} F_j)$ does not have proper subfields distinct of $k$. As consequence of (c.1) and (c.2) we obtain
\begin{equation}\label{nueve}
\begin{split}
    [L:k]&=[F_{i_1}\cdots F_{i_{s'}}:k][F_{j_1}\cdots F_{j_s}:k]\\&=\prod\limits_{j=1}^{s'} \ell^{n-a_{i_j}}(\prod\limits_{n=1}^{s-1} \ell^{n-a_{j_n}})\ell^{d_{i_0}-d_s}=(\prod\limits_{\stackrel{j=1}{j\neq {i_0}}}^{r} \ell^{n-a_{j}})\ell^{d_{i_0}-d_s}.
\end{split}
\end{equation}

Next we will see that $L\cap F_{i_0}=F'_s$. Let $C:=L\cap F_{i_0}$. We have $F'_s\subseteq C$. Since $C\subseteq F_{i_0}$, every prime $P_j$ with $1\leq j\leq r$, $j\neq {i_0}$, is unramified in $C$. Let $I=\prod\limits_{\stackrel{j=1}{j\neq {i_0}}}^r I_j$. From (b.1) and (b.2) we obtain $C\subseteq L^I$. From (d.1) and (d.2) it follows
\[
    L^I=(F_{i_1}\cdots F_{i_{s'}})^{I_{i_1}\cdots I_{i_{s'}}}(F_{j_1}\cdots F_{j_s})^{I_{j_1}\cdots I_{j_s}}=F'_s.
\]

Therefore $C=L^I=F'_s$.

Finally, note that $[M:\g E]=[M:LF_{i_0}]$. By the Galois correspondence we have $[LF_{i_0}:L]=[F_{i_0}:L\cap F_{i_0}]$. Thus
\begin{align*}
    [F_{i_0}:L\cap F_{i_0}]&=[k(\sqrt[\ell^{d_{i_0}+t}]{P_{i_0}}):k(\sqrt[\ell^{d_{i_0}-d_s}]{P_{i_0}^*})]\\
    &=\frac{[k(\sqrt[\ell^{d_{i_0}+t}]{P_{i_0}}):k]}{[k(\sqrt[\ell^{d_{i_0}-d_s}]{P_{i_0}^*}):k]}=\ell^{d_{i_0}+t-d_{i_0}+d_s}.
\end{align*}

It follows that $[LF_{i_0}:L]=[F_{i_0}:L\cap F_{i_0}]=[F_{i_0}:F'_{i_0}]=\ell^{d_{i_0}+t-d_{i_0}+d_s}=\ell^{d_s+t}$. From equation (\ref{nueve}) we obtain
\begin{align*}
    [M:LF_{i_0}]&=\frac{[M:k]}{[LF_{i_0}:k]}=\frac{[M:k]}{[LF_{i_0}:L][L:k]}=\frac{\prod\limits_{j=1}^{r} \ell^{n-a_{j}}}{\ell^{d_s+t}(\prod\limits_{\stackrel{j=1}{j\neq {i_0}}}^{r} \ell^{n-a_{j}})\ell^{d_{i_0}-d_s}}
    \\
    &=\frac{\ell^{n-a_{i_0}}}{\ell^{d_{i_0}+t}}=\ell^{n-a_{i_0}-d_{i_0}-t}.
\end{align*}

On the other hand, we have $m=n-a_{i_0}-d_{i_0}$. Therefore $[M:LF_{i_0}]=\ell^{m-t}=[M:\g E]$ and by construction we have $L(\sqrt[\ell^{d_{i_0}+t}]{P_{i_0}^*})\subseteq \g E$. It follows that $\g E=L(\sqrt[\ell^{d_{i_0}+t}]{P_{i_0}^*})$.
\end{proof}

\begin{no} {\rm{In the definition of $F_j$ given in \eqref{knorr}, for $j<i_0$ if $1\leq n-a_j-{\rm min}\{n-a_j, d_j\}\leq t$, it can be defined simply $F_j=E_j$.
}}
\end{no}

\section{General $\ell^n$ case}\label{S4}

Recall that $H$ is the decomposition group of $\p$ in $\g EK/K$ (see diagram (\ref{diag})).

\begin{teo}\label{Tprin} Let $K =k(\sqrt[\ell^n]{\gamma D})\subseteq \lam D_u$, with $\gamma\in \F^*$, $D=P_1^{\alpha_1}\cdots P_r^{\alpha_r}$, $1\leq \alpha_j\leq \ell^n-1$, $\alpha_j=b_j\ell^{a_j}$ with $\gcd(b_j,\ell)=1$, $1\leq j \leq r$, $P_1,\dots, P_r\in R_T^+$ different polynomials. We order the polynomials $P_1,\dots,P_r$ so that $0=a_1\leq\cdots\leq a_r\leq n-1$. Let $E=K_u\cap\lam D$, $t$ as in Theorem \ref{Pengln} and $\alpha=\nu_{\ell}(|H|)$. Let $H':=H\smid_{\g E}$. Then $E_{\eu{ge}}^{H'}=F_1\cdots F_{{i_0}-1}F_{{i_0}+1}\cdots F_r (\sqrt[\ell^{d_{i_0}+(t-\alpha)}]{P_{i_0}^*})$ where $F_j$ are given in (\ref{knorr}) for all $j$. Thus
\[
 \g K=E_{\eu{ge}}^{H'}K=\prod_{\stackrel{i=1}{i\neq i_0}}^r F_i K(\sqrt[\ell^{d_{i_0}+(t-\alpha)}]{P_{i_0}^*}).
\]

Further, if $d={\rm{min}}\{n,\nu_\ell(\deg D)\}$, we have
\[
|H|=\ell^\alpha=[\F(\sqrt[\ell^n]{(-1)^{\deg D}\gamma}):\F(\sqrt[\ell^d]{(-1)^{\deg D}\gamma})].
\]
\end{teo}
\begin{proof}
From Theorem \ref{Pengln} we have $\e{F_j/k}=1$ for $j\neq i_0$ (i.e, $\e{L/k}=1$). Therefore $e_{\infty}(E_{\eu {ge}}/k)={\rm lcm}[e_{\infty}(F_j/k)\smid 1\leq j\leq r]=e_{\infty}(F_{i_0}/k)=\ell^t.$ That is, the ramification of $\p$ in $\g E/k$ depends only on the ramification of $\p$ in the extension $F_{i_0}/k$. Since $\g E/\g E^+$ is a cyclic extension and $[\g E:\g E^{H'}]=[\g E:F_1\cdots F_{{i_0}-1}F_{{i_0}+1}\cdots F_r (\sqrt[\ell^{d_{i_0}+(t-\alpha)}]{P_{i_0}^*})]$ it follows that
\[
\g E^{H'}=F_1\cdots F_{{i_0}-1}F_{{i_0}+1}\cdots F_r (\sqrt[\ell^{d_{i_0}+(t-\alpha)}]{P_{i_0}^*}).
\]

We have $EK=K(\sqrt[\ell^n]{(-1)^{\deg D}\gamma})$ and $EK/K$ is unramified, in fact, $EK/K$ is a constant extension \cite[Subsection 5.3]{MaRzVi2016}. From \cite[Theorem 6.2.1]{Vil2006}, we have $$\f{EK/k}=[\F(\sqrt[\ell^n]{(-1)^{\deg D}\gamma}):\F]$$ and from \cite[Proposition 2.8]{BaRzVi2013} we have $$\f{K/k}=[\F(\sqrt[\ell^{d}]{(-1)^{\deg D}\gamma}):\F] \;\;{\rm with}\;\; d={\rm{min}}\{n,\nu_\ell(\deg D)\}.$$ Then $\f{EK/k}=\f{EK/K}\f{K/k}$ where $\f{EK/K}=|H|=\ell^\alpha$.
$$\xymatrix{
&EK\ar@{-}[dd]^{|H|=\ell^\alpha}& \\
&&\\
& K \ar@{-}[dl]^{\f{K/k}} &&\g K\ar@{-}[ll]_{f_\infty=1} \\
k\ar@/^2pc/@{-}[uuur]^{\f{EK/k}}&&&
}$$

Hence
\begin{align*}
    |H|&=\ell^\alpha=\frac{\f{EK/k}}{\f{K/k}}=\frac{[\F(\sqrt[\ell^n]{(-1)^{\deg D}\gamma}):\F]}{[\F(\sqrt[\ell^d]{(-1)^{\deg D}\gamma}):\F]}\\
&=[\F(\sqrt[\ell^n]{(-1)^{\deg D}\gamma}):\F(\sqrt[\ell^d]{(-1)^{\deg D}\gamma})].
\end{align*}
\end{proof}

\begin{cor}[Case $n=1$, G. Peng  \cite{MaRzVi2013}]\label{T5.1.6}
Let $K:=k(\sqrt[\ell]{\gamma D})$ with $\gamma\in\F^*$, where $D=P_1^{\alpha_1}\cdots P_r^{\alpha_r}\in R_T$ is a monic $\ell$--power free polynomial, $P_i\in R_T^+$ and $d_j=\nu_\ell(\deg P_j)$, $1\leq \alpha_j\leq \ell-1$, $1\leq j\leq r$. Let $m={\rm max}\{1-{\rm min}\{1,d_j\} \smid 1\leq j\leq r\}$ and $d={\rm{min}}\{1,\nu_\ell(\deg D)\}$. We assume that $m=1-{\rm {min}}\{1,d_r\}$.
Then
\begin{gather}\label{pengnew}
    K_{{\eu {ge}}}=
    \begin{cases}
    E_1\cdots E_rK&\text{if $m=1-d$, $E=K$ or $E\neq K$ and $d=1$}\\
    F_1\cdots F_{r-1}K&\text{if $m>1-d$, $E=K$ or $E\neq K$ and $d=0$ or $1$}\\
    \end{cases},
\end{gather}
where $E_j=k(\sqrt[\ell]{P_j^*})$, $1\leq j\leq r$, $F_j=k(\sqrt[\ell]{P_jP_{r}^{z_j}})$ $1\leq j\leq r-1$, $z_j=-a\deg P_j$, with $a\deg P_r + b\ell=1$, when $d_r=0$ for some $b$.
\end{cor}
\begin{proof}
Writing $\alpha_j=b_j\ell^{a_j}$, $1\leq j\leq r$, we have $0=a_1=\cdots=a_r$.
Let $i_0$ be as in Theorem \ref{Tprin}, that is, $i_0$, $1\leq i_0\leq r$, is such that $n-a_{i_0}-{\rm min}\{n-a_{i_0},d_{i_0}\}=m$. Then $i_0=r$.

On the other hand we have two cases. If we write $\epsilon:=(-1)^{\deg D}\gamma$, then
\begin{itemize}
    \item[$a)$\,] $\epsilon\in (\F^*)^\ell$,
    \item[$b)$\,] $\epsilon\notin (\F^*)^\ell$.
\end{itemize}

{\large $a)$} If $K=E$ (see \cite[Theorem 4.2]{MaRzVi2015}). Therefore $\g K=\g E$.
If $m=t$, then $\g E=E_1\cdots E_r$, with $E_j=k(\sqrt[\ell]{P_j^*})$, $1\leq j\leq r$, so that $\g K=E_1\cdots E_r$. If $m>t$, we have $m=1$ and $t=0$ and from Theorem $\ref{Pengln}$ we have $\g E=F_1\cdots F_r$, where if $z_j:=-ac_j\ell^{d_j-d_{r}}$ then
\begin{gather}\label{knorPeng}
    F_j:=
    \begin{cases}
    k\left(\sqrt[\ell]{P_jP_{r}^{z_j}}\right)&\text{if $j<{r}$,}\\
    k\left(\sqrt[\ell^{d_{r}+t}]{(-1)^{\deg P_r}P_{r}}\right)&\text{if $j={r}$}.
    \end{cases}
\end{gather}

Note that $d_r=0$ and since $t=0$, we obtain $F_r=k$. Hence $\g K=F_1\cdots F_{r-1}$.

{\large $b)$} We have that $K\neq E$ and that $\g K=E_{\eu{ge}}^{H'}K$ (where $H$ is as in Theorem \ref{Tprin}), with $\g E$ as in case $a)$. We also have $|H'|=\ell^\alpha=[\F(\sqrt[\ell]{(-1)^{\deg D}\gamma}):\F(\sqrt[\ell^d]{(-1)^{\deg D}\gamma})]$, where $\alpha= 0$ or $1$, that is $|H'|=1$ or $\ell$.

If $d=1$, then $|H'|=1$, that is $\alpha=0$. Thus $\g K=\g EK$, with $\g E$ as in $a)$. Therefore $\g K=E_1\cdots E_rK$ if $m=t$ and $\g K=F_1\cdots F_{r-1}K$ if $m>t$ with $F_j$ as in (\ref{knorPeng}).

If $d=0$, we have $|H'|=\ell$, that is $\alpha=1$. Since $d=1-t$, we have $1=t\leq m\leq 1$. Therefore $\ell\nmid \deg P_r$, that is $d_r=0$. From Theorem \ref{Tprin} and from $d_r+t-\alpha=0$, we obtain
\[
\g E^{H'}=F_1\cdots F_{r-1} (\sqrt[\ell^{d_{i_0}+(t-\alpha)}]{P_{r}^*})=F_1\cdots F_{r-1}.
\]

Therefore $\g K=F_1\cdots F_{r-1}K$.
\end{proof}

\begin{ejm} {\rm{Let $K=\F(\sqrt[\ell^n]{\gamma D})$ with $\gamma=5$, $\ell = 3$, $n=10$, $q=472393$ and $D=P_1^{\alpha_1}P_2^{\alpha_2}P_3^{\alpha_3}P_4^{\alpha_4}$ $P_5^{\alpha_5}P_6^{\alpha_6}P_7^{\alpha_7}P_8^{\alpha_8}$, where $\alpha_j=b_j\ell^{a_j}$, $\deg P_j=c_j\ell^{b_j}$, $1\leq j\leq 8$. Let
$a_1=0$, $a_2=1$, $a_3=3$, $a_4=3$, $a_5=4$, $a_6=7$, $a_7=8$ and $a_8=9$. Let $d_1=5$, $d_2=7$, $d_3=2$, $d_4=3$, $d_5=2$, $d_6=0$, $d_7=10$ and $d_8=0$. Since ${\rm gcd}(b_jc_j,\ell)=1$, $1\leq j\leq 8$, we can select $c_1=2$, $b_1=b_2=b_4=b_5=b_6=b_7=b_8=c_2=c_4=c_6=c_7=c_8=1$ and $b_3=c_3=c_5=5$. We have $m={\rm max}\{n-a_j-{\rm min}\{n-a_j,d_j\} \smid 1\leq j\leq r\}={\rm max}\{5,2,5,4,4,3,0,1\}=5$ and $t=10-{\rm min}\{10,\nu_\ell(\deg D)\}$ where $\deg D=\ell^5(\ell^3(b_2c_2+b_7c_7\ell^{10}+b_8c_8\ell)+b_1c_1+b_3c_3+\ell(b_4c_4+b_5c_5+b_6c_6\ell))$, and $\nu_\ell(b_1c_1+b_3c_3+\ell(b_4c_4+b_5c_5+b_6c_6\ell))=3$. Thus $\nu_\ell(\deg D)=8$, $\deg D=387459855$ and $t=2$. We have $i_0=3$, $F_3=k\left(\sqrt[\ell^{4}]{P_3^*}\right)$ and
\begin{align*}
F_1&=k\left(\sqrt[\ell^{10}]{P_1P_3^{z_1}}\right),\; F_2=k\left(\sqrt[\ell^{9}]{P_2P_3^{z_2}}\right),\; F_4=k\left(\sqrt[\ell^{7}]{P_4P_{3}^{y_4\ell}}\right),\;\\ F_5&=k\left(\sqrt[\ell^{6}]{P_5P_{3}^{y_5\ell^0}}\right),\; F_6=k\left(\sqrt[\ell^{5}]{P_6^{\ell^{2}}P_{3}^{y_6}}\right),\; F_7=k\left(\sqrt[\ell^{2}]{P_7P_{3}^{y_7\ell^{8}}}\right),\\ F_8&=k\left(\sqrt[\ell^{3}]{P_8^{\ell^{2}}P_{3}^{y_8}}\right),
\end{align*}
with $z_1=-2a\ell^3$, $z_2=-a\ell^5$ and $y_j\equiv -c_j5^{-1}\mod \ell^{10}=-c_j11810\mod 3^{10}$, $4\leq j\leq 8$, where $a5+b\ell^{8}=1$. We may choose $y_4=y_6=y_7=y_8=47239$, $y_5=59048$, $a=-1312$ and $b=1$.
Therefore
\begin{align*}
 \g K=K(\sqrt[3^{10}]{P_1P_3^{70848}}, &\sqrt[3^{9}]{P_2P_3^{318816}},\sqrt[3^{7}]{P_4P_{3}^{141717}},\sqrt[3^{6}]{P_5P_{3}^{59048}},\\&\sqrt[3^{5}]{P_6^{9}P_{3}^{47239}},\sqrt[3^{2}]{P_7P_{3}^{309935079}},\sqrt[3^{3}]{P_8^{9}P_{3}^{47239}}, \sqrt[3^{4-\alpha}]{P_{3}^*}).
\end{align*}
with
\[
|H'|=3^\alpha=[\F(\sqrt[3^{10}]{-\gamma}):\F(\sqrt[3^8]{-\gamma})].
\]

From \cite[Theorem 4 (1)]{ExtEulerCrit}, $5\notin(\F^*)^\ell$ and $[\F(\sqrt[3^{10}]{-\gamma}):\F(\sqrt[3^8]{-\gamma})]=3^2$. Therefore $\alpha=2$ and $\sqrt[3^{4-\alpha}]{P_{3}^*}=\sqrt[3^{2}]{P_{3}^*}$.
}}
\end{ejm}

\section{On the field extension  $(K_1K_2)_{\eu{ge}}/(K_1)_{\eu{ge}}(K_2)_{\eu{ge}}$}\label{S5}

Let $K_1$ and $K_2$ be function fields over $k=\F(T)$. Let $K:=K_1K_2$ and suppose that $K\subseteq {_n}{\lam N}_m$. Let $E:=E_1E_2$, with $E_i:={_n}(K_i)_m\cap \lam N$, $i=1,2$. We have $E={_n}K_m\cap \lam N$.

\begin{prop}\label{prop5} Let $L_1,L_2\subseteq\lam N$, $L:=L_1L_2$, $L^+:=L\cap\lam N^+$ and $L_i^+:=L_i\cap\lam N^+$, $i=1,2$. Then $[L^+:(L_1)^+(L_2)^+]\smid q-1$.
\end{prop}

\begin{proof}
Let $G:=\Gal(\lam N/E)$, $G_1:=\Gal(\lam N/L_1)$ and $G_2:=\Gal(\lam N/L_2)$. We have $G=G_1\cap G_2$. Let $L^+=L\cap\lam N^+=\lam N^{G\F^*}$, $(L_1)^+=L_1\cap\lam N^+=\lam N^{G_1\F^*}$ and $(L_2)^+=L_2\cap\lam N^+=\lam N^{G_2\F^*}$. Since $G\F^*<G_1\F^*\cap G_2\F^*$, we have
\[
L_1^+L_2^+= \lam N^{G_1\F^*}\lam N^{G_2\F^*}= \lam N^{G_1\F^*\cap G_2\F^*}\subseteq \lam N^{G\F^*}=L^+.
\]

$$\xymatrix{
\lam N^+\ar@{-}[r]^{\F^*}&\lam N\ar@{-}[d]^{G}\ar@/^1pc/@{--}[ddr]^{G_2}&\\
&L=L_1L_2\ar@{-}[dl]\ar@{-}[dr]& \\
L_1\ar@/^1pc/@{--}[uur]^{G_1} && L_2
}$$

Thus
\begin{align*}
    [L^+:L_1^+L_2^+]&=|G_1\F^*\cap G_2\F^*:G\F^*|
    =\frac{|G_1\F^*\cap G_2\F^*|}{|G\F^*|}\\
    &=\frac{\frac{|G_1\F^*||G_2\F^*|}{|G_1G_2\F^*|}}{\frac{|G||\F^*|}{|G\cap\F^*|}}
    =\frac{|G_1\F^*||G_2\F^*||G\cap\F^*|}{(q-1)|G||G_1G_2\F^*|}\\
    &=\frac{|G_1||\F^*||G_2||\F^*||G\cap\F^*|}{(q-1)|G||G_1\cap\F^*||G_2\cap\F^*||G_1G_2\F^*|}\\
    &=(q-1)\frac{|G_1||G_2|}{|G|}\frac{|G\cap \F^*|}{|G_1\cap \F^*||G_2\cap \F^*|}\frac{1}{|G_1G_2\F^*|}\\
    &=(q-1)\frac{|G_1||G_2|}{|G_1\cap G_2|}\frac{1}{|G_1G_2\F^*|}\frac{|G\cap \F^*|}{|G_1\cap \F^*||G_2\cap \F^*|}\\
    &=(q-1)\frac{|G_1 G_2|}{|G_1G_2\F^*|}\frac{|G\cap \F^*|}{|G_1\cap \F^*||G_2\cap \F^*|}\\
    &=(q-1)\frac{1}{|(G_1\cap \F^*)(G_2\cap \F^*)|}\frac{1}{[L_1\cap L_2:L_1^+\cap L_2^+]}.
\end{align*}

Let $\alpha:=|(G_1\cap \F^*)(G_2\cap \F^*)|[L_1\cap L_2:L_1^+\cap L_2^+]$. Since $[L^+:L_1^+L_2^+]\in\Z$, we have $\alpha\smid q-1$, and $[L^+:L_1^+L_2^+]\smid q-1$.
\end{proof}

\begin{prop}\label{prop6}
Let $L\subseteq\lam N$. If $\lam N^+\subseteq L\subseteq\lam N$, then $\g L=L$.
\end{prop}
\begin{proof}
We have $L_{\eu{gex}}=\lam N$ and from \cite[Theorem 2.1]{RockM} it follows that $\g L=L_{\eu{gex}}^+L=\lam N^+L=L$.
\end{proof}

Let $E=E_1E_2\subseteq \lam N$. Let $Y_1$ and $Y_2$ be the groups of Dirichlet characters associated with $(E_1)_{\eu {gex}}$ and $(E_2)_{\eu {gex}}$ respectively. Then, from Leopoldt's Theorem \cite[Proposition 14.4.1]{VilCC}, we have that $Y=Y_1Y_2$ is the group of characters associated with $E_{\eu {gex}}$, where $Y_1Y_2$ is the group of Dirichlet characters associated with the field $(E_1)_{\eu {gex}}(E_2)_{\eu {gex}}$ \cite[Proposition 9.4.33]{VilCC}, i.e, $E_{\eu {gex}}=(E_1)_{\eu {gex}}(E_2)_{\eu {gex}}$. From Proposition \ref{prop5} we obtain that $[E_{\eu {gex}}^+:(E_1)_{\eu {gex}}^+(E_2)_{\eu {gex}}^+]\smid q-1$.

We have the following result.

\begin{prop} Let $K_1,K_2\subseteq {_n}{\lam N}_m$, $K=K_1K_2$ and $E_1$, $E_2$ as before. If $\g K=(K_1)_{\eu {ge}}(K_2)_{\eu {ge}}$, then $\g E=(E_1)_{\eu {ge}}(E_2)_{\eu {ge}}$.
\end{prop}
\begin{proof}
Let $K=K_1K_2\subseteq {_n\lam N_m}$ and $E:=E_1E_2 \subseteq \lam N$. Consider the fields $(\g K)_m$ and $(\g E)_m$. Now, from  \cite[Theorem 2.2]{RockM} we have $(\g E)_m=(\g K)_m$ and $((K_i)_{\eu {ge}})_m=((E_i)_{\eu {ge}})_m$, $i=1,2$. Therefore
\begin{align*}
(\g E)_m=(\g K)_m&=((K_1)_{\eu {ge}}(K_2)_{\eu {ge}})_m=((K_1)_{\eu {ge}})_m((K_2)_{\eu {ge}})_m\\
&=((E_1)_{\eu {ge}})_m((E_2)_{\eu {ge}})_m
=((E_1)_{\eu {ge}}(E_2)_{\eu {ge}})_m.
\end{align*}

Finally, by the Galois correspondence, it follows
\begin{align*}
    \g E=(\g E)_m\cap\lam N=((E_1)_{\eu {ge}}(E_2)_{\eu {ge}})_m\cap\lam N=(E_1)_{\eu {ge}}(E_2)_{\eu {ge}}.
\end{align*}
\end{proof}

The converse of Proposition \ref{prop6} does not hold in general, that is, if $\g E=(E_1)_{\eu{ge}}(E_2)_{\eu{ge}}$, then the equality $\g K=(K_1)_{\eu{ge}}(K_2)_{\eu{ge}}$ may fail.

\begin{ejm} {\rm{Let $K_1=k(\sqrt[\ell]{P_1})$, $K_2=k(\sqrt[\ell]{\gamma P_2})$ and $K=K_1K_2$ be such that $P_1, P_2\in R_T^+$ are different polynomials with $\deg P_1=a$, $1\leq a <\ell$, $\deg P_2=\ell-a$ and $\gamma\notin(\F^*)^\ell$. Then we have $E_i=k(\sqrt[\ell]{(P_i)^*})$, $i=1,2$ and $E=E_1E_2$. If $\chi_{P_i}$ is the Dirichlet character associated with the field $E_i$, $i=1,2$, by Leopoldt's Theorem (\cite[Proposition 14.4.1]{VilCC}), we have $E_i=(E_i)_\eu{ge}$, $i=1,2$ and $E_{\eu {ge}}=E_1E_2=E$. Therefore $\g E=E=E_1E_2=(E_1)_\eu{ge}(E_2)_\eu{ge}$.

Now, since $\p$ is ramified in $K_i/k$, $i=1,2$, we have that $\p$ is of degree $1$ in $K_i$ and $(K_i)_\eu{ge}=K_i$.
On the other hand, from Abhyankar's Lemma, the remification of $\p$ in $K/k$ is equal to ${\rm lcm}[e_\infty(K_1/k),e_\infty(K_2/k)]=\ell$. Since $k(\sqrt[\ell]{\gamma P_1P_2})\subseteq K$, with $\deg P_1P_2=\deg P_1+\deg P_2=\ell$, thus, $\ell\smid \deg P_1P_2$ and $\gamma\notin(\F^*)^\ell$, we obtain $f_\infty(K/k)=\ell$. Also we have $[K:k]=\ell^2=f_\infty(K/k) e_\infty(K/k)$. It follows that $h_\infty(K/k)=1$ and $\deg(S_\infty(K))=f_\infty(K/k)=\ell$. Thus $[K{\ma F}_{q^\ell}:K]=\ell$ and $K\subsetneq K{\ma F}_{q^\ell}\subseteq \g K$, i.e, $\g K\neq K=K_1K_2=(K_1)_\eu{ge}(K_2)_\eu{ge}$.
}}
\end{ejm}

The main result of this section is the following theorem.

\begin{teo} Let $K_1,K_2/k$ be abelian finite field extensions. Then
$$[(K_1K_2)_{\eu{ge}}:(K_1)_{\eu{ge}}(K_2)_{\eu{ge}}]\smid (q-1)^2.$$
\end{teo}
\begin{proof}
We have $K_1,K_2\subseteq {_n\lam N_m}$ for some $m\in\N$, $n\in\N\cup \{0\}$ and $N\in R_T$. Let $K=K_1K_2\subseteq {_n\lam N_m}$. Let $E_i={_n}(K_i)_m\cap \lam N$, $i=1,2$ y $E={_n}K_m\cap\lam N$. Then $E=E_1E_2$. Now, since $E_{\eu{gex}}^+\cap\g E=E_{\eu{gex}}\cap\lam N^+\cap\g E=E_{\eu{gex}}\cap\g E^+=\g E^+$, we have the following Galois square.
\[
\xymatrix{
E_{\eu{gex}}^+\ar@{-}[d]\ar@{-}[r]&E_{\eu{gex}}^+\g E\ar@{-}[d] \\
E_{\eu{ge}}^+\ar@{-}[r]&\g E
}
\]

From \cite[Theorem 2.1]{RockM} we have $\g E=E_{\eu{gex}}^+ E$. Then from the Galois correspondence we obtain $\g E^+=E_{\eu{gex}}^+$. Similarly $(E_i)_{\eu{ge}}^+=(E_i)_{\eu{gex}}^+$, $i=1,2$. We have $[E_{\eu {gex}}^+:(E_1)_{\eu {gex}}^+(E_2)_{\eu {gex}}^+]\smid q-1$. In particular $[E_{\eu {ge}}^+:(E_1)_{\eu {ge}}^+(E_2)_{\eu {ge}}^+]=[E_{\eu {gex}}^+:(E_1)_{\eu {gex}}^+(E_2)_{\eu {gex}}^+]\smid q-1$. Finally, we obtain the following diagram.
\[
\xymatrix{
E_{\eu{ge}}^+\ar@{-}[ddd]\ar@{-}[r]&E_{\eu{ge}}\ar@{-}[dd]\ar@{-}[r]&\g EK\ar@{-}[d]\\
&&(E_1)_{\eu{ge}}(E_2)_{\eu{ge}}K\\
&(E_1)_{\eu{ge}}(E_2)_{\eu{ge}}\ar@{-}[ru]&\\
(E_1)_{\eu{ge}}^+(E_2)_{\eu{ge}}^+\ar@{-}[ru]&&
}
\]

It follows that
\[
[\g EK:(E_1)_{\eu {ge}}(E_1)_{\eu {ge}}K]\smid [E_{\eu {ge}}:(E_1)_{\eu {ge}}(E_2)_{\eu {ge}}]\smid [E_{\eu {gex}}^+:(E_1)_{\eu {ge}}^+(E_2)_{\eu {ge}}^+].
\]

Therefore $[\g EK:(E_1)_{\eu {ge}}(E_1)_{\eu {ge}}K]\smid q-1$.

On the other hand, from \cite[Theorem 2.2]{RockM} we have that the following extensions $(E_i)_{\eu {ge}}K_i/(K_i)_{\eu {ge}}$, $i=1,2$, are constant extension fields of order $|H_i|\smid q-1$, $i=1,2$, were $H_i$ is the decomposition group of $S_\infty(K_i)$ in $(E_i)_{\eu {ge}}K_i$, $i=1,2$. Also, if $t_i:=\deg S_\infty(K_i)$, then $(E_i)_{\eu {ge}}K_i=(K_i)_{\eu {ge}}{\ma F}_{q^{t_i|H_i|}}$, $i=1,2$. It follows that
\begin{align*}
    (E_1)_{\eu {ge}}K_1(E_2)_{\eu {ge}}K_2&=(K_1)_{\eu {ge}}{\ma F}_{q^{t_1|H_1|}}(K_2)_{\eu {ge}}{\ma F}_{q^{t_2|H_2|}}\\
    &=(K_1)_{\eu {ge}}(K_2)_{\eu {ge}}{\ma F}_{q^{{\rm lcm}[t_1|H_1|,t_2|H_2|]}}\\
    &\subseteq (K_1)_{\eu {ge}}(K_2)_{\eu {ge}}{\ma F}_{q^{{\rm lcm}[t_1(q-1),t_2(q-1)]}}\\
    &= (K_1)_{\eu {ge}}(K_2)_{\eu {ge}}{\ma F}_{q^{{\rm lcm}[t_1,t_2](q-1)}}.
\end{align*}

We have ${\ma F}_{q^{{\rm lcm}[t_1,t_2]}}\subseteq (K_1)_{\eu {ge}}(K_2)_{\eu {ge}}$, i.e, the field $(K_1)_{\eu {ge}}(K_2)_{\eu {ge}}{\ma F}_{q^{{\rm lcm}[t_1,t_2](q-1)}}$ is an extension of constants of $(K_1)_{\eu {ge}}(K_2)_{\eu {ge}}$ of degree at most $q-1$. We have
\[
(K_1)_{\eu {ge}}(K_2)_{\eu {ge}}\subseteq (E_1)_{\eu {ge}}(E_2)_{\eu {ge}}K\subseteq (K_1)_{\eu {ge}}(K_2)_{\eu {ge}}{\ma F}_{q^{{\rm lcm}[t_1,t_2](q-1)}}.
\]

Therefore, $(E_1)_{\eu {ge}}(E_2)_{\eu {ge}}K/(K_1)_{\eu {ge}}(K_2)_{\eu {ge}}$ is an extension  of constants of degree at most $(q-1)$. Finally, we have $[\g EK:(E_1)_{\eu {ge}}(E_2)_{\eu {ge}}K]\smid [\g E:(E_1)_{\eu {ge}}(E_2)_{\eu {ge}}]\smid q-1$. Thus
\begin{align*}
[\g K:(K_1)_\eu{ge}(K_2)_\eu{ge}]&\smid[\g EK:(E_1)_{\eu {ge}}(E_2)_{\eu {ge}}K][(E_1)_{\eu {ge}}(E_2)_{\eu {ge}}K:(K_1)_{\eu {ge}}(K_2)_{\eu {ge}}]\\
&\smid (q-1)^2.
\end{align*}
\end{proof}

Let $H$, $H_1$ and $H_2$ be the decomposition groups of $\p$ in
$\g EK/K$, $(E_1)_{\eu {ge}}K_1/K_1$ and $(E_2)_{\eu {ge}}K_2/K_2$ respectively and let $H':=H|_{\g E}$, $H'_i:=H_i|_{(E_i)_{\eu {ge}}}$, $i=1,2$.

\begin{prop} With the previous notation, we have
\[
(E_1)_{\eu {ge}}^{H'_1}(E_2)_{\eu {ge}}^{H'_2}\subseteq E_{\eu {ge}}^{H'}.
\]
\end{prop}
\begin{proof}
Since $K=K_1K_2$, then $(K_i)_\eu{ge}\subseteq \g K$, $i=1,2$. Because $\g K=E_\eu{ge}^{H'}K$ and $(K_i)_\eu{ge}=(E_i)_\eu{ge}^{H'_i}K_i$, $i=1,2$, it follows that $(E_i)_\eu{ge}^{H'_i}K\subseteq E_\eu{ge}^{H'}K$, $i=1,2$. Now, since $E_i\subseteq (E_i)_{\eu{ge}}$, from the Galois correspondence it follows that $E_i^{H'_i}\subseteq (E_i)_{\eu{ge}}^{H'_i}$, $i=1,2$. Hence $E_i^{H'_i}K\subseteq (E_i)_\eu{ge}^{H'_i}K\subseteq E_\eu{ge}^{H'}K$. Thus, we have the following Galois square.
\[
\xymatrix{
\g E\ar@{-}[d]_{H'}\ar@{-}[r]&\g EK\ar@{-}[d]^H\\
\g E^{H'}\ar@{-}[d]\ar@{--}[r]&\g E^{H'}K=\g K\ar@{-}[d]\\
\g E\cap (E_i)_{\eu {ge}}^{H'_i}K\ar@{-}[d]\ar@{--}[r]&(E_i)_{\eu {ge}}^{H'_i}K\ar@{-}[d]\\
\g E\cap (E_i)^{H'_i}K_i\ar@{-}[d]\ar@{--}[r]&(E_i)^{H'_i}K\ar@{-}[d]\\
\g E\cap K\ar@{-}[r]&K
}
\]

From the Galois correspondence, we have $\g E\cap (E_i)^{H'_i}K\subseteq \g E\cap (E_i)_{\eu {ge}}^{H'_i}K\subseteq \g E\cap \g K=\g E^{H'}$, where $(E_i)^{H'_i}\subseteq \g E\cap (E_i)^{H'_i}K$ and $(E_i)_{\eu {ge}}^{H'_i}\subseteq \g E\cap (E_i)_{\eu {ge}}^{H'_i}K$. Therefore $\;(E_i)_{\eu {ge}}^{H'_i}\subseteq\g E^{H'}$, $i=1,2$.
\end{proof}

\begin{teo} With the previous notation, we have $|H|\smid {\rm lcm}[|H'_1|,|H'_2|]$.
\end{teo}
\begin{proof}
We have that $\p$ is totally ramified in the extension $\g E/E_\eu{ge}^{H'}$ and that $\p$ is totally decomposed in $\g E/(E_1)_\eu{ge}(E_2)_\eu{ge}$ since
\[
E=E_1E_2\subseteq (E_1)_\eu{ge}(E_2)_\eu{ge}\subseteq \g E.
\]

It follows that $\g E=E_\eu{ge}^{H'}(E_1)_\eu{ge}(E_2)_\eu{ge}$, since if $L:=E_\eu{ge}^{H'}(E_1)_\eu{ge}(E_2)_\eu{ge}$, in the extension $\g E/L$, $\p$ is totally ramified and totally decomposed.

Now, $(E_1)_{\eu{ge}}/(E_1)_\eu{ge}^{H'_1}$ is totally ramified in $\p$ with ramification index $|H'_1|$, so that
$$\e{(E_1)_\eu{ge}(E_2)_\eu{ge}/(E_1)_\eu{ge}^{H'_1}(E_2)_\eu{ge}}\smid |H'_1|$$
\[
\xymatrix{
(E_1)_\eu{ge}\ar@{-}[r]\ar@{-}[d]_{|H'_1|}&(E_1)_\eu{ge}(E_2)_\eu{ge}^{H'_2}\ar@{-}[d]^{e_{\infty,1}\smid |H'_1|}\\
(E_1)_\eu{ge}^{H'_1}\ar@{-}[r]&(E_1)_\eu{ge}^{H'_1}(E_2)_\eu{ge}^{H'_2}
}
\]

Similarly $\e{(E_1)_\eu{ge}(E_2)_\eu{ge}/(E_1)_\eu{ge}(E_2)_\eu{ge}^{H'_2}}\smid |H'_2|\smid \e{(E_2)_{\eu{ge}}/(E_2)_\eu{ge}^{H'_2}}=[(E_2)_{\eu ge}:(E_2)_\eu{ge}^{H'_2}]$
\[
\xymatrix{
(E_2)_\eu{ge}\ar@{-}[r]\ar@{-}[d]_{|H'_2|}&(E_1)_\eu{ge}^{H'_1}(E_2)_\eu{ge}\ar@{-}[d]^{e_{\infty,2}\smid |H'_2|}\\
(E_2)_\eu{ge}^{H'_2}\ar@{-}[r]&(E_1)_\eu{ge}^{H'_1}(E_2)_\eu{ge}^{H'_2}
}
\]

We have $(E_1)_\eu{ge}(E_2)_\eu{ge}=((E_1)_\eu{ge}(E_2)_\eu{ge}^{H'_2})((E_1)_\eu{ge}^{H'_1}(E_2)_\eu{ge})$ and the following diagram
\[
\xymatrix{
&(E_1)_\eu{ge}(E_2)_\eu{ge}\ar@{-}[dr]&\\
(E_1)_\eu{ge}(E_2)_\eu{ge}^{H'_2}\ar@{-}[ur]&&(E_1)_\eu{ge}^{H'_1}(E_2)_\eu{ge}\ar@{-}[dl]^{e_{\infty,2}\smid |H'_2|}\\
&(E_1)_\eu{ge}^{H'_1}(E_2)_\eu{ge}^{H'_2}\ar@{-}[ul]^{e_{\infty,1}\smid |H'_1|}&\\
}
\]

From Abhyankar's Lemma
\[
\e{(E_1)_\eu{ge}(E_2)_\eu{ge}/(E_1)_\eu{ge}^{H'_1}(E_2)_\eu{ge}^{H'_2}}=e_0:={\rm lcm}[e_{\infty,1},e_{\infty,2}]\smid {\rm lcm}[|H'_1|,|H'_2|].
\]

Finally, we have $\p$ is totally ramified in $\g E/\g E^{H'}$ and totally decomposed in $\g E/(E_1)_\eu{ge}(E_2)_\eu{ge}$.
\[
\xymatrix{
E_\eu{ge}^{H'}\ar@{-}[r]^{|H'|}\ar@{-}[d]&E_\eu{ge}\ar@{-}[d]^{\substack{\p\\
\text{totally decomposed}}}\\
(E_1)_\eu{ge}^{H'_1}(E_2)_\eu{ge}^{H'_2}\ar@{-}[r]_{e_0}&(E_1)_\eu{ge}(E_2)_\eu{ge}
}
\]

Thus $|H'|\smid [\g E:(E_1)_\eu{ge}(E_2)_\eu{ge}]\, e_0$. Thus $|H'|\smid e_0$ and $e_0\smid {\rm lcm}[|H'_1|,|H'_2|]$. Therefore $$|H'|\smid {\rm lcm}[|H'_1|,|H'_2|].$$
\end{proof}

\begin{teo} Let $K_1,K_2\subseteq {_n}{\lam N}_m$, $K=K_1K_2$, $F:={_n}K\cap {\lam N}_m$ and $F_i:={_n}K_i\cap {\lam N}_m$, $i=1,2$. Then
\[
\g K=(K_1)_\eu{ge}(K_2)_\eu{ge}\textit{ if and only if } \g F=(F_1)_\eu{ge}(F_2)_\eu{ge}.
\]
\end{teo}
\begin{proof}
Since $K=K_1K_2$ we have ${_n}K={_n}{(K_1K_2)}={_n}{(K_1)}{_n}{(K_2)}$. From the Galois correspondence it follows that $F=F_1F_2$.

From \cite[Theorem 4.4]{MaRzVi2013} we have that $\g K=M\g F$ where $M=K\lam N_m\cap L_n$, $\g F$ is the genus field of $F$. We have $M\cap \g F\subseteq L_n\cap{\lam N}_m =k$. Then $\g K=M\g F$ and $(K_i)_\eu{ge}=M_i(F_i)_\eu{ge}$, where $M_i=K_i{\lam N}_m\cap L_n$, $i=1,2$. If $V$ is the first ramification group of $\p$ in $\g K/k$, then $\g F=K_{\eu{ge}}^V$, and we have that $[\g K:\g F]=[M:k]=|V|$.
\[
\xymatrix{
F_{{\eu {ge}}}\ar@{-}[rr]^V\ar@{-}[d]&&\g K=M\g F\ar@{-}[d]\\
F\ar@{-}[rr]\ar@{-}[dd]_{\substack{\p \text{\ is tamely}\\
\text{ramified}}}&&FM\ar@{-}[dd]^{
\substack{\S M\text{\ is tamely}\\ \text{ramified}}}\\
&K\ar@{-}[ru]\ar@{-}[dl]\\ k\ar@{-}[rr]^V&&M
}
\]

Then, by the Galois correspondence we have $\g F=\g K\cap \lam N_m$ and $[\g K:M]=[\g F:k]$.

Now, first suppose that $\g K=(K_1)_\eu{ge}(K_2)_\eu{ge}$. Since
\[
M\g F=\g K=(K_1)_\eu{ge}(K_2)_\eu{ge}=M_1(F_1)_\eu{ge}M_2(F_2)_\eu{ge}=M_1M_2((F_1)_\eu{ge}(F_2)_\eu{ge}).
\]

Note that $M{\lam N}_m=M_1{\lam N}_m M_2{\lam N}_m=M_1M_2{\lam N}_m$. Therefore, by the Galois correspondence $M=M_1M_2$. Thus $M\g F=M_1M_2((F_1)_\eu{ge}(F_2)_\eu{ge})=M((F_1)_\eu{ge}(F_2)_\eu{ge})$. It follows that $\g F=(F_1)_\eu{ge}(F_2)_\eu{ge}$.

Conversely, assume that $\g F=(F_1)_\eu{ge}(F_2)_\eu{ge}$. Then
\[
\g K=M\g F=M_1(F_1)_\eu{ge}M_2(F_2)_\eu{ge}=(K_1)_\eu{ge}(K_2)_\eu{ge}.
\]
\end{proof}

config: arxiv


TITLE: Can the mass of an orbiting object and the object being orbited be determined by the distance and orbit velocity alone?
QUESTION [2 upvotes]: This question is the inverse of: "Could an object orbit while moving at twice the speed, but at the same distance, if it had half the mass?"
I'm curious about the nature of orbits, but am not well enough versed in mathematics to understand Kepler's laws well. I have been wondering if the mass of a planet and a star it orbits could be determined based solely on the distance and speed of the orbit, or if the ability to orbit at a given speed/distance was based relatively on the mass of both objects (i.e., we could determine the ratio of the mass of the two objects, but not the actual mass).

REPLY [2 votes]: No, if you only know the distance between the objects and the relative orbital velocity of the planet, you cannot determine its mass. In fact, if you only know the distance and velocity at one particular moment, you don't have enough information to determine the orbit.
Suppose we know the distance $r$ and the relative orbital velocity $\vec{v}=(v_r,v_T)$ of a planet at a given moment. Here, $v_r=\dot{r}$ is the radial velocity component, and $v_T$ the tangential component. The orbit of the planet has two constants of motion: the
specific orbital energy $E$ and the specific relative angular momentum $h$:
$$
\begin{align}
E &= \frac{1}{2}v_{r}^2 + \frac{1}{2}v_{T}^2 - \frac{\mu}{r}= -\frac{\mu}{2a},\\
h^2 &= r^2\,v^2_{T} = \mu a(1-e^2),
\end{align}
$$
where $a$ is the semi-major axis of the orbit, $e$ is the orbital eccentricity, and $\mu = G(M_\text{p} + M_\text{s})$, with $M_\text{p}$ the mass of the planet and $M_\text{s}$ the mass of the star. So we have two equations and three unknowns $(\mu,a,e)$, in other words we need additional information to solve them.
For instance, if we assume that the orbit is circular, then $e=0$, and we can solve for $\mu$ and $a$. Another possibility is that we know the distance and velocity at two instances $t_1$ and $t_2$, then
$$
\frac{1}{2}v_{r}^2(t_1) + \frac{1}{2}v_{T}^2(t_1) - \frac{\mu}{r(t_1)}=\frac{1}{2}v_{r}^2(t_2) + \frac{1}{2}v_{T}^2(t_2) - \frac{\mu}{r(t_2)},
$$
and we can solve for $\mu$, and, from the first two equations, we know $a$ and $e$. A third possibility is that we know the orbital period $T$ of the planet. In that case, we can use Kepler's Third Law:
$$
T^2 = (2\pi)^2\frac{a^3}{\mu},
$$
which, in combination with the first two equations, yields $(\mu,a,e)$. In any case though, we can only derive $\mu = G(M_\text{p} + M_\text{s})$, i.e. we only know the sum of the masses (which will be dominated by the mass of the star).
If you want to derive the mass of the planet, you need to know the motion of the planet and the star with respect to their common centre of mass. If $(r_\text{p},v_{r,\text{p}},v_{T,\text{p}})$ and $(r_\text{s},v_{r,\text{s}},v_{T,\text{s}})$ are the position and velocity of the planet and the star with respect to their common centre of mass, then
$$
\begin{align}
E_\text{p} &= \frac{1}{2}v_{r,\text{p}}^2 + \frac{1}{2}v_{T,\text{p}}^2 - \frac{\mu_\text{p}}{r_\text{p}}= -\frac{\mu_\text{p}}{2a_\text{p}},\\
h^2_\text{p} &= r^2_\text{p}\,v^2_{T,\text{p}} = \mu_\text{p} a_\text{p}(1-e^2),\\
E_\text{s} &= \frac{1}{2}v_{r,\text{s}}^2 + \frac{1}{2}v_{T,\text{s}}^2 - \frac{\mu_\text{s}}{r_\text{s}}= -\frac{\mu_\text{s}}{2a_\text{s}},\\
h^2_\text{s} &= r^2_\text{s}\,v^2_{T,\text{s}} = \mu_\text{s} a_\text{s}(1-e^2),
\end{align}
$$
and in addition, if we know $T$,
$$
T^2 = (2\pi)^2\frac{a^3}{\mu} = (2\pi)^2\frac{a^3_\text{p}}{\mu_\text{p}} = (2\pi)^2\frac{a^3_\text{s}}{\mu_\text{s}}.
$$
In principle, these are five equations with five unknowns $(\mu_\text{p},\mu_\text{s},a_\text{p},a_\text{s},e)$ (actually, these are six equations, but they're not independent; also note that the eccentricities of the orbits are the same). Also
$$
\begin{align}
r &= r_\text{p} + r_\text{s},\\
a &= a_\text{p} + a_\text{s},\\
M_\text{p}r_\text{p} &= M_\text{s}r_\text{s},\\
M_\text{p}a_\text{p} &= M_\text{s}a_\text{s},\\
\mu_\text{p} &= \frac{GM_\text{s}^3}{(M_\text{p} + M_\text{s})^2},\\
\mu_\text{s} &= \frac{GM_\text{p}^3}{(M_\text{p} + M_\text{s})^2}.
\end{align}
$$
Once the equations are solved, you can derive $a$ and $\mu$ (using Kepler's Third Law again). So you have $M_\text{p}/M_\text{s}$ and $M_\text{p} + M_\text{s}$, so that you can derive $M_\text{p}$ and $M_\text{s}$ separately.
In practice, it is usually too difficult to measure $r_\text{p}$ and $r_\text{s}$, because the centre of mass will be very close to the centre of the star. But by measuring the velocities at two instances $t_1$ and $t_2$, we can treat $r_\text{p}$ and $r_\text{s}$ as additional unknowns and calculate them as well.


TITLE: Problem with Cosets in the proving of $K/H \trianglelefteq G/H$
QUESTION [1 upvotes]: We have $H \trianglelefteq G$ and $H \subseteq K \leq G$.
we can prove if $K \trianglelefteq G$ then $K/H \trianglelefteq G/H$ like this:
We already showed that $K/H \leq G/H$. If we show that $\forall Hg \in G/H$ and $\forall Hx \in K/H$ : $(Hg)^{-1}(Hx)(Hg) \in K/H$ then the proof is complete. we have that $Hg \in G/H \Rightarrow g \in G$ and $Hx \in K/H = \{Ht | t \in K\} \Rightarrow \exists t \in K$ such that $Hx = Ht$.
$\begin{align*}
g \in G, t \in K & \ \ \Longrightarrow  \ \ g^{-1}tg \in K \\
& \ \ \Longrightarrow  \ \  H(g^{-1}tg) \in K/H \\
& \ \ \Longrightarrow  \ \ (Hg^{-1}) (Ht)(Hg) \in K/H\\
& \ \ \Longrightarrow  \ \ (Hg)^{-1} (Ht)(Hg) \in K/H\\
\end{align*}$
I can't undestand this part of proof

$Hx \in K/H = \{Ht | t \in K\} \Rightarrow \exists t \in K$ such that
  $Hx = Ht$

why we use $t$ ? can't we continue with $x$ and say $Hx \in K/H \Rightarrow x \in K$ ?

REPLY [2 votes]: There is a typo. $x\in K/H$ so it should say $\exists t\in K\;(x=H t)$.


TITLE: Normalize ratio metric
QUESTION [0 upvotes]: I have two values: one raw value number $r$ and one predicted value number $p$.
Now I want to give a metric how similar the two are on a scale from 0 to 1.
In the best case the $r=p$ which I would give a score of 1.
I first thought of defining a metric by $\frac{p}{r}$. This has the problem that if $p$ is larger then $r$ the metric generates values larger than 1. However I am a bit clueless how to normalize this.

REPLY [1 votes]: It is fairly standard practice to begin such a process by defining the error $\epsilon$ between $p$ and $r$ as the absolute value of their difference: $\epsilon = |r-p|$.
The next step is to consider whether a relative error is perhaps more appropriate than the absolute error. For example, when $p$ becomes a factor $10$ larger, it is reasonable to assume that the same is true for $r$ and $\epsilon$. A safe and sensible choice for the relative error might be $\epsilon = |r-p|/p$. But the denominator could also be $r$ or $(p+r)/2$ or $\sqrt{pr}$ or something similar.
If you are happy with your (absolute or relative) error, you can convert into a metric of your choice. You say that would like $m=1$ when $r=p$ and around zero if $r$ and $p$ are very dissimilar. Okay, then $m = 1 - \epsilon$ might work for you. You may also consider $m = exp(-\epsilon)$.


TITLE: exponential Radioactive decay
QUESTION [0 upvotes]: A doctor injects you with 500 milligrams of a radioactive compound for medical imaging. It has a half-life of 6 hours. How long will it take for 99.9% of it to decay? Round to the nearest hour.
Answer
99.9*500 = 500e^kt
99.9 = e^kt
99.9 = e^6k
ln99.9 = lne^6k
4.6041 = 6k
k = 0.76736
I actually made some progress. Now I'm stuck here. How do i proceed?

REPLY [1 votes]: Always divide percentages by $100$ to get proportions in contexts like this. If $99.9\%$ (a proportion of $0.999$) has decayed, you're left with $0.001$ times the original total. Let's work in hours. Since $e^{-6k}=\frac12$, $e^{-kt}=0.001\implies t=6\frac{\ln 1000}{\ln 2}\approx60$.


TITLE: Proof of trig identity using t-formulae
QUESTION [2 upvotes]: Show that $$2\arctan x = \arccos\frac{1-x^2}{1+x^2}$$
if $ x > 0$, and
$$2\arctan x = -\arccos\frac{1-x^2}{1+x^2}$$ if $x <0.$
I have been able to equate the expressions by letting $x=\tan(y/2)$, but I do not know how to show the different signs of the equation for $x>0$ and $x<0$.

REPLY [2 votes]: The simplest way to go is probably to calculate $\;\cos(2\arctan x)$.
Set $\theta=\arctan x$ and use the duplication formula:
\begin{align}
\cos 2\theta&=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-x^2}{1+x^2},\\
\text{so }\qquad
2\arctan x&\equiv \pm\arccos\frac{1-x^2}{1+x^2}\mod 2\pi.
\end{align}
Now either $\;0\le \theta<\frac\pi2,\;$  so $\;0\le 2\theta<\pi$, which is (included in, strictly speaking) the range of  arccos. In this case we conclude that
$$ 2\arctan x=\arccos\frac{1-x^2}{1+x^2},.$$
or $\;-\frac\pi 2<\theta<0$, so $\;-\pi<2\theta<0$, and for the same reason, we conclude in this case that
$$ 2\arctan x=-\arccos\frac{1-x^2}{1+x^2}.$$