TITLE: Find least $n$ such that $\sum_{2\leq k \leq n} \frac{1}{k} \geq 5$
QUESTION [1 upvotes]: Let $n$ be the least positive integer such that $\sum_{2\leq k \leq n} \frac{1}{k} \geq 5$. Choose the correct option:
(a) $n \leq 32$
(b) $32<n\leq96$
(c) $96<n\leq729$
(d) $729<n$

I know the series $\sum \frac{1}{n}$ diverges. But I am finding it difficult on how to find the solution to the above problem.
Any help is appreciated. Thanks in Advance.

REPLY [1 votes]: One can have even more precise bracketings.
We can use for example the following Euler-Maclaurin expansion (due to Ramanujan) with first terms :
$$\sum_{k=1}^n \dfrac{1}{k} =\dfrac12\ln(2p)+\gamma+\dfrac{1}{12p}-\dfrac{1}{120 p^2}+\dfrac{1}{630p^3}-\dfrac{1}{1680p^4}+\cdots \tag{1}$$
Where $p:=\dfrac{n(n+1)}{2}$ and $\gamma\approx 0.57721566490153286060...$ is the Euler-Mascheroni constant.
giving for $n=227$ for the result with
(This the third formula in "Asymptotic expansions" here)
In particular, (1) gives very precise bracketings like this one:
$$\dfrac12\ln(2p)+\gamma+\dfrac{1}{12p}-\dfrac{1}{120 p^2}<\sum_{k=1}^n \dfrac{1}{k}< \dfrac12(1/2)\ln(2p)+\gamma+\dfrac{1}{12p}\tag{2}$$
according to the fact that the error done in truncating an alternate series  whose general term is decreasing is less (in absolute value) than the first omitted term.
Here is the answer : $\color{red}{n=227}$ because (2) gives :
$$6.004366708345565 < \sum_{k=1}^{\color{red}{227}}\dfrac{1}{k} < 6.004366708358010$$
whereas :
$$5.999961422001953  < \sum_{k=1}^{226} \dfrac{1}{k} <   5.999961422014619$$
(please note the very narrow bracketting : 11 significant digits !)
Therfore, the good answer was answer c).


TITLE: Do side-rational triangles of the same area admit side-rational dissections?
QUESTION [25 upvotes]: Call a polygon side-rational if the lengths of all its sides are rational.  Call a dissection of a polygon side-rational if all of the polygons within the dissection are side-rational.  Then my question is as in the title:
Do any two rational triangles of the same area have a common side-rational dissection?
Of course it's a classical result that any two polygons of the same area have a common dissection; a standard proof goes through triangulating the polygons, using a triangle-to-rectangle dissection, partitioning the rectangles for both polygons into a set of common-area rectangles, and then using a rectangle-to-rectangle dissection.  But many of the steps in this process will involve non-rational sides (for instance, the rectangle corresponding to a triangle of rational sides but irrational area certainly can't have rational sides), and it's not clear that such a dissection could be patched up.
This question was prompted by rediscovering my common dissection of two very special rational triangles, the $4-5-6$ triangle and the $3-8-10$ triangle (both of area $\frac{15}{4}\sqrt{7}$):

All the lengths here can be shown straightforwardly with the law of sines and law of cosines; these triangles are a somewhat special case because so many of the angles involved are multiples of each other, but it seems likely that any rational triangle will inherently be special in somewhat similar ways.
EDIT: the original version of this question asked about arbitrary polygons, but having thought about it more I would be amazed if that result is true.  The area of a (side-)rational triangle is always the square root of a rational number (by Heron's formula, for instance), but already rational-sided quadrilaterals can have arbitrary area: just imagine 'flexing' a rhombus with all sides of unit length, from a square down to a trivial, collapsed shape: it will continuously take on all areas from $1$ to $0$.  This implies, for instance, that there's a rhombus with all sides of unit length and area $\frac\pi4$; such a rhombus can't be dissected into rational triangles, and it's hard to imagine any shared dissection it would have with, e.g., the rhombus with all sides of length $2$ and area $\frac\pi4$.  (Which admittedly isn't a proof that one doesn't exist!)
A fair bit of rummaging around the web has turned up nothing on the topic; if anyone has pointers to similar matters, they'd be very much appreciated.

REPLY [17 votes]: Let $T,T'$ be any side-rational triangles of the same area.
We construct a side-rational dissection between them.
It will be convenient to scale $T,T'$ by a common factor so that
each has an integral side.  Position $T$ and $T'$ so that
their chosen integral sides are on the same line $\ell$,
and $T,T'$ are on the same side of $\ell$.
We first dissect $T,T'$ into side-rational parallelograms $P_1,P'_1$
each of which has a side of length $1$.  If $T$ has vertices $A,B,C$ with
$|BC|$ integral, let $M_B,M_C$ be the midpoints of the sides opposite $B,C$,
cut $T$ into the half-size triangle $A M_B M_C$ and a residual trapezoid,
and combine them to form a parallelogram $P$ with sides $B C$ and $B M_C$.
Then divide $BC$ and the opposite side of $P$ into unit segments,
connect corresponding points to cut $P$ into $|BC|$ congruent parallelograms
with one side of length $1$ and the other parallel to $B M_C$,
and join them by the unit-length sides to form a parallelogram $P_1$.
Apply the same procedure to $T'$ to dissect it into $P'_1$.
It is easy to see that all the polygons involved in each dissection
have rational sides, each parallel to a side of $T$ or $T'$.
Since the unit sides of $P_1,P'_1$ are on $\ell$
and $P_1,P'_1$ lie on the same side of $\ell$,
and these parallelograms have the same area,
there is a line $m$ parallel to $\ell$ that contains
the opposite sides of both $P_1$ and $P'_1$,
Tile the strip betweem $\ell$ and $m$ with parallel copies of $P_1$.
A finite number of these copies intersect $P'_1$.
These give a dissection of $P'_1$ into $P_1$, and thus of $T'_1$ into $T_1$.
The fact that this dissection is side-rational then follows from
repeated application of the following lemma:
If triangle $ABC$ has rational sides then the altitude from any vertex
divides the opposite side into rational segments.  (One of the two segments
may have zero or negative length if the triangle is right or obtuse.)
Proof: let the sides opposite $A,B,C$ have lengths $a,b,c$.
Suppose the altitude $h$ from $A$ divides $BC$ into segments of lengths
$\beta,\gamma$.  Then $h^2 = b^2 - \beta^2 = c^2 - \gamma^2$, so
$\beta^2 - \gamma^2 = b^2 - c^2 \in \bf Q$.  But
$\beta^2 - \gamma^2 = (\beta+\gamma) (\beta-\gamma) = a (\beta-\gamma)$,
and $a \in \bf Q$, so $\beta - \gamma \in \bf Q$.
Since $\beta\pm\gamma$ are both rational, so are $\beta$ and $\gamma$. $\Box$
Now in our dissection each edge is parallel to an edge of $T$ or $T'$, and by the lemma each vertex
projects to a point on $\ell$ at rational distances from
the common vertices of $P_1$ and $P'_1$.  Hence each side is
a rational fraction of some side of $T$ or $T'$ and is thus rational.  $\Box\Box$


text:
\begin{document}
\selectlanguage{english}

\begin{abstract}
  Given a number field $K$, a finite abelian group $G$ and finitely many elements $\alpha_1,\ldots,\alpha_t\in K$, we construct abelian extensions $L/K$ with Galois group $G$ that realise all of the elements $\alpha_1,\ldots,\alpha_t$ as norms of elements in $L$. In particular, this shows existence of such extensions for any given parameters.

  Our approach relies on class field theory and a recent formulation of Tate's characterisation of the Hasse norm principle, a local-global principle for norms. The constructions are sufficiently explicit to be implemented on a computer, and we illustrate them with concrete examples.
\end{abstract}

\maketitle
\tableofcontents
\section{Introduction}
Attached to each extension $L/K$ of number fields comes the field-theoretic norm map $\norm_{L/K}:L^\times \to K^\times$. Given an extension $L/K$, a classical problem is to study which elements of $K$ are in the image of this norm map. We consider the inverse problem: given elements $\alpha_1,\ldots,\alpha_t\in K^\times$, we construct an extension $L/K$ that realises all of them as norms.

Of course the trivial extension does the trick, so we prescribe moreover a given degree and even a given Galois group $G$. It is easy to prove the existence of a degree-$n$-extension $L/K$ whose normal closure has full Galois group $S_n$, such that a given $\alpha\in K^\times$ is a norm: one may adjoin to $K$ a root of a polynomial $X^n+a_{n-1}X^{n-1}+\cdots+a_1X+(-1)^n\alpha$, with coefficients $a_1,\ldots,a_{n-1}$ sufficiently generic in~$K$ so that Hilbert's irreducibility theorem applies.

On the opposite end, one can look at Abelian extensions. In this situation,  D.~Loughran, R.~Newton and the first-named author have recently proved the following result.

\begin{theorem}{\cite[Theorem 1.1]{Frei}}\label{thm:existence}
  Let $K$ be a number field, $G$ a finite abelian group and $\alpha_1,\ldots,\alpha_t\in K^\times$. Then there is a normal extension $L/K$ with $\Gal(L/K)\simeq G$ and  $\{\alpha_1,\ldots,\alpha_t\}\subset\norm_{L/K}(L^\times)$.
\end{theorem}

In \cite{Frei}, this follows from an analytic result regarding the density of such extensions. The appendix to \cite{Frei} by Y.~Harpaz and O.~Wittenberg provides in addition an algebro-geometric proof of Theorem \ref{thm:existence}, using descent, a version of the fibration method developed in their work \cite{HW18}, and a version of Hilbert's irreducibility theorem.
\smallskip

Both of these proofs are not constructive, and the main achievement of our work is the direct construction of an extension $L/K$ that satisfies the conclusion of Theorem \ref{thm:existence}. Our construction is based on explicit class field theory, the availability of places of $K$ that satisfy certain Chebotarev-type conditions, and a version of Tate's criterion for the Hasse norm principle (see Theorem \ref{thm:HNP} in \S \ref{sec:HNP}), which was formulated in \cite{HasseNormPrinciple,Frei} and is of central importance for the quantitative arguments in these papers. Some of our constructions are reminiscent of arguments appearing in \cite{Jehne}.
\smallskip

A stronger version of Theorem \ref{thm:existence} is proved in \cite[Corollary 4.11]{Frei}. There, one can even specify finitely many places at which the extension $L/K$ is required to have a prescribed admissible local structure. Our constructions are sufficiently general to permit the same restrictions, thus also recovering \cite[Corollary 4.11]{Frei}. See \S \ref{sec:local} for precise statements.
\smallskip

Our constructions are explicit enough to be implemented on a computer. In \S \ref{sec:computation}, we provide some remarks on how this can be done, though one should note that \emph{efficiency} is not a central focus of this work. Nevertheless, we illustrate our constructions in \S \ref{sec:illustrations} with concrete examples obtained by hand and aided by computer.

\subsection*{Outline of the construction}
By class field theory, we can specify the extension $L/K$ through a
continuous epimorphism~$\rho$ from the id\`ele class group~$I_K/K^\times$ of $K$ to $G$.
We construct the epimorphism~$\rho$ in such a way that $\alpha_1,\ldots,\alpha_t$ are local norms at all places, and moreover $L/K$ satisfies the \emph{Hasse norm principle}, a local global principle for norms reviewed in \S \ref{sec:HNP}.
We rely
on
a criterion for the Hasse norm principle to hold,
in terms of the decomposition groups; see Theorem \ref{thm:HNP}.
This is especially amenable to class field theory and allows us to reformulate the desired conditions, precisely~\eqref{eq:cond_local_norms}--\eqref{eq:cond_psi}, in terms of conditions on the local components~$\rho_v$ of the epimorphism~$\rho$, namely~\eqref{eq:rho_local_norms}--\eqref{eq:rho_triv_S}.

The epimorphism~$\rho$
will be
induced by a simpler object, which we call a \emph{characteristic morphism}, introduced in \S \ref{sec:char_morph}. The former is defined on classes of id\`eles, and the latter on actual integral $S$-id\`eles, for a conveniently chosen set of places~$S$.
Lemmas~\ref{lemma} and~\ref{eq:rho_local_norms} explain explicitly how the characteristic morphism inducing $\rho$ determines the corresponding decomposition subgroups $G_v=\rho_{v}({K_{v}}^\times)\leq G$ of the extension~$L/K$.

We actually study in~\S\ref{sec:local_norms} a very particular class of characteristic morphisms, of the form~$\rho^T_S$ of~\eqref{characteristic morphisms type T}, which are are explicitly determined by the data of a finite set~$T$ of places~$v_i$ of~$K$. To guarantee that $\alpha_1,\ldots,\alpha_t$ are local norms at all places, we give elementary sufficient conditions~\eqref{eq:local_norm_elementary}, to be checked only at these places~$v_i$. This amounts to
Chebotarev-type conditions on the places~$v_i$, ensuring an unlimited supply of places to choose from.

In order to also satisfy the Hasse norm principle, we use
the
mentioned criterion,
Theorem \ref{thm:HNP},
which
requires us
to produce sufficiently many large decomposition groups, and a ``spanning type'' property of these. At this point it will help to work with auxiliary characteristic morphisms~$\rho'_S$ into an auxiliary group~$G'\simeq(\Lambda^2 G)^2$, related to~$G$ by an explicit map~$\Psi:G'\to G$ given in~\eqref{eq:def_Psi}. In~\S\ref{sec:forcing_HNP} we devise a concrete condition~\eqref{eq:basis_condition} on this auxiliary characteristic morphism~$\rho'_S$
to ensure the criterion of Theorem \ref{thm:HNP}
for the characteristic morphism~$\Psi\circ\rho'_S$. It involves places~$v_i$ and
auxiliary
places~$w_i$. The
relevant
decomposition groups
are those
at
the places~$v_i$, and the auxiliary place~$w_i$ is chosen to ensure
that
the decomposition group~${G'}_{v_i}\leq G'$ will be large enough.
The condition~\eqref{eq:basis_condition} has then to do with relative position of these decomposition groups~$G'_{v_i}$, guaranteeing the ``spanning type'' property
for
~$\Psi\circ\rho'_S$.

We finally explain in \S\ref{sec:choosing_places} how one can find a set of places satisfying~\eqref{eq:basis_condition}
through an incremental construction. Here, the auxiliary places $w_i$ have to satisfy Chebotarev-type conditions of a form studied earlier in \S\ref{sec:supply}.
If~$G$ can be generated by~$\knew$ elements, then we can stop as soon as we
have produced
~$\kprimenew={\knew \choose 2}$ places~$v_i$, and as many of the corresponding~$w_i$.

Folding up, the set~$T=\{v_1;w_1;...;v_{\kprimenew};w_{\kprimenew}\}$ determines an auxiliary characteristic morphism~\eqref{characteristic morphisms type T} into~$G'$ which satisfies condition~\eqref{eq:basis_condition}, see Theorem~\ref{thm:choose_places}. Composing with~$\Psi$
yields
a characteristic morphism into~$G$ and an induced epimorphism~$\rho:I_K/K^\times\to G$,
which
corresponds to an
extension~$L/K$. By Theorem~\ref{thm:main_construction} this~$\rho$ will satisfy~\eqref{eq:rho_local_norms}--\eqref{eq:rho_triv_S}, and hence
we are done.

\smallskip
Our constructions rely on reduction maps modulo well-chosen places of $K$. We require these places to satisfy certain Chebotarev conditions, and their abundance is secured by the Chebotarev density theorem. In practice, we find suitable places through exhaustive search.

\subsection*{Acknowledgements}
This collaboration was initiated when C.F.~presented the results of \cite{Frei} at the ninth WOMBL one-day meeting at University of Cambridge in 2019, and continued during a visit of R.R.~to give a seminar talk at The University of Manchester. We thank the organisers and hosting institutions for their hospitality. C.F.~was supported by EPSRC grant EP/T01170X/1. R.R.~was supported by ERC grant GeTeMo 617129, and Leverhulme Research Project Grant ``Diophantine problems related to Shimura varieties''.

\section{Constructive proof of Theorem \ref{thm:existence}}
\subsection{Notation}
We fix the number field $K$, the elements $\alpha_1,\ldots,\alpha_t\in
K^\times$ and the finite abelian group $G$ henceforth. We write $\places$ for
the set of all places of $K$. By $S$, we will always denote a finite subset of $\places$ that contains all archimedean places. For a non-archimedean place $v$, we write $K_v$ for the completion of $K$ at $v$, $\OO_v$ for the subring of $v$-adic integers, and $\OO_v^\times$ for the multiplicative group of $v$-adic units. The ring of $S$-integers of $K$ is defined as $\OO_S=K\cap\bigcap_{v\notin S}\OO_v$, and its unit group, the group of $S$-units, is $\OO_S^\times=K\cap\bigcap_{v\notin S}\OO_v^\times$. Here and in similar situations, the index $v\notin S$ is understood to run over $\places\smallsetminus S$. The symbol $v$ will also be used for various objects defining the place $v$, including the corresponding prime ideals in $\OO_S$ and $\OO_v$, and the exponential valuation at $v$. We write, for example $\alpha\bmod v$ for the image of $\alpha\in \OO_v$ in the residue field $\F_v$ at $v$. For any field $F$ and $e\in\N$, we will write  $F^{\times e}=(F^\times)^e$ for the group of non-zero $e$-th powers in $F$.

We will always assume that $S$ is large enough, in particular we will require that
\begin{equation}
  \label{eq:S}
  \begin{aligned}
    &S \text{ contains all archimedean places, }\\
    &\OO_S\text{ is a principal ideal domain, and}\\
    &\{\alpha_1,\ldots,\alpha_t\}\subset\OO_S^\times.
  \end{aligned}
\end{equation}

\subsection{Class field theory}
By global class field theory, extensions $L/K$ with an isomorphism $\Gal(L/K)\to G$ are parameterised
by epimorphisms $\rho : I_K/K^\times\to G$. Here, $I_K$ is the id\`ele group of $K$, in which $K^\times$ is embedded diagonally, and a morphism between topological groups is a continuous group homomorphism. We endow the finite group $G$ with the discrete topology. The extension given by such an epimorphism $\rho$ will be denoted by $L_\rho/K$. By the interplay between local and global class field theory, the local behaviour of the extension $L_\rho$ corresponding to $\rho$ at $v$ is described by the restriction $\rho_v$ of $\rho$ to the subgroup $K_v^\times\subset I_K/K^\times$, embedded by sending $\alpha\in K_v^\times$ to the class of the id\`ele $(1,\ldots,1,\alpha,1,\ldots)$ that has component $\alpha$ at place $v$ and $1$ at every other place.

\subsection{Hasse norm principle}\label{sec:HNP}
Our goal is to construct an epimorphism $\rho$ whose corresponding extension $L=L_\rho$ satisfies $\{\alpha_1,\ldots,\alpha_t\}\subset\norm_{L/K}(L^\times)$. Since it is hard to detect global norms directly, we rely on local norms and a suitable local-global principle. Recall that the norm $\norm_{L/K}:L^\times\to K^\times$ extends to a homomorphism $\norm_{L/K}:I_{L}\to I_K$. By definition, the \emph{Hasse norm principle} holds for the extension $L/K$, if
\begin{equation*}
\norm_{L/K}(L^\times)= K^\times\cap\norm_{L/K}(I_L).
\end{equation*}
We call the elements of the  right-hand set \emph{local norms at all places} of $K$, so the Hasse norm principle, if valid for $L/K$, allows us to detect global norms as local norms at all places. The Hasse norm theorem \cite{Hasse} asserts the validity of the Hasse norm principle when $\Gal(L/K)$ is \emph{cyclic}. However, it may fail already in the case of biquadratic extensions, with the first counter-example of $\Q(\sqrt{-3},\sqrt{13})/\Q$ also due to Hasse.

We rely on the following criterion for the Hasse norm principle, which is based
on Tate's cohomological description of the knot group $(K^\times\cap\norm_{L/K}(I_L))/\norm_{L/K}(L^\times)$ (e.g.~\cite[\S 11.4]{Tate}).

\begin{theorem}[{\cite[Lemma 4.2]{Frei}}, {\cite[\S6]{HasseNormPrinciple}}]\label{thm:HNP} Let $G$ be a finite abelian group and $\rho:I_K/K^\times\to G$ a surjective homomorphism. Then the Hasse norm principle holds for the extension $L_\rho/K$ if and only if the natural map
\begin{equation}\label{LG criterion}
\bigoplus_{v\in\places}{\bigwedge}^2 \rho_v(K_v^\times) \to {\bigwedge}^2 G
\end{equation}
is surjective.
\end{theorem}

\subsection{Local conditions}\label{sec:local} For additional flexibility, we allow ourselves to impose arbitrary local conditions at the finite set of places $S$, as long as these conditions are realised by some sub-$G$-extension that admits $\alpha_1,\ldots,\alpha_t$ as local norms at all places. More precisely, assume we are given a morphism $\psi : I_K/K^\times \to G$, not necessarily surjective, such that $\{\alpha_1,\ldots,\alpha_t\}\subset \norm_{L_\psi/K}(I_{L_\psi})$. We will construct an epimorphism $\varphi : I_K/K^\times \to G$ with the following properties:
\begin{align}
&\text{$\{\alpha_1,\ldots,\alpha_t\}\subset\norm_{L_\varphi/K}(I_{L_\varphi})$,}\label{eq:cond_local_norms}\\
&\text{the extension $L_\varphi/K$ satisfies the Hasse norm principle, and}\label{eq:cond_HNP}\\
& \text{the local restriction $\varphi_v$ agrees with $\psi_v$ on $K_v^\times$ for all $v\in S$.}\label{eq:cond_psi}
\end{align}
In particular, by \eqref{eq:cond_local_norms} and \eqref{eq:cond_HNP} the $\alpha_i$ are global norms from $L_\varphi$, and thus we obtain a  constructive proof of Theorem \ref{thm:existence} and \cite[Corollary 4.11]{Frei}.

Possibly enlarging $S$, we will assume that it contains all places at which $\psi$ is ramified. Let $\spl_S(\psi)$ be the set of all places $v\notin S$ that are split completely in $L_\psi$, i.e.~$\psi_v(K_v^\times)=\{e_G\}$.

In the rest of this section, we will explain how to construct an epimorphism $\rho:I_K/K^\times\to G$ that satisfies the following:
\begin{align}
  &\text{for all $v\in\places$, }\{\alpha_1,\ldots,\alpha_t\}\subset\ker(\rho_v:K_v^\times\to I_K/K^\times\to G),\label{eq:rho_local_norms}\\
  &\text{the natural map }\bigoplus_{v\in\spl_S(\psi)}{\bigwedge}^2 \rho_v(K_v^\times) \to {\bigwedge}^2 G \text{ is surjective,}\label{eq:rho_HNP}\\
  &G=\sum_{v\in\spl_S(\psi)}\rho_v(K_v^\times)\text{, and}\label{eq:rho_surj}\\
  &\text{for all $v\in S$, the local morphism $\rho_v:K_v^\times\to G$ is trivial.}\label{eq:rho_triv_S}
\end{align}

Once we have constructed such an epimorphism $\rho$, we can take $\varphi:I_K/K^\times \to G$ to be the product of $\psi$ and $\rho$. The following lemma justifies this.

\begin{lemma}\label{lem:twist}
If $\rho:I_K/K^\times\to G$ is an epimorphism with the properties \eqref{eq:rho_local_norms}--\eqref{eq:rho_triv_S}, then the product $\varphi = \psi\rho : I_K/K^\times\to G$ is an epimorphism with the properties \eqref{eq:cond_local_norms}--\eqref{eq:cond_psi}.
\end{lemma}

\begin{proof}
For the local morphisms at all places $v$, we have $\varphi_v = \psi_v\rho_v : K_v^\times\to G$. For $v\in\spl_S(\psi)$, the morphism $\psi_v$ is trivial, and hence $\varphi_v=\rho_v$. This shows that
\begin{equation*}
 G=\sum_{v\in\spl_S(\psi)}\varphi_v(K_v^\times)\quad\text{ and that }\quad \bigoplus_{v\in\spl_S(\psi)}{\bigwedge}^2 \varphi_v(K_v^\times) \to {\bigwedge}^2 G\text{ is surjective.}
\end{equation*}
In particular, $\varphi$ is surjective and  \eqref{eq:cond_HNP} holds by Theorem \ref{thm:HNP}. For $v\in S$, the local morphism $\rho_v$ is trivial and thus $\varphi_v = \psi_v$, as desired in \eqref{eq:cond_psi}. For each $\alpha_i$ and $v\in S$, we see by local class field theory that $\alpha_i\in\ker(\psi_v)=\ker(\varphi_v)$. Let $v\notin S$, then $\psi$ is unramified at $v$ and thus $\psi_v$ vanishes on $\OO_{v}^\times$, which contains all of the $\alpha_i$ by \eqref{eq:S}. Hence, $\varphi_v(\alpha_i)=\rho_v(\alpha_i)=e_G$. We have shown that all the local morphisms $\varphi_v$ vanish on all $\alpha_i$, hence \eqref{eq:cond_local_norms} holds by local class field theory.
\end{proof}

\subsection{Characteristic morphisms}\label{sec:char_morph}  Here we describe a hands-on way of constructing an epimorphism $\rho:I_K/K^\times\to G$ through the specification of a simpler object which we call a \emph{characteristic morphism}.

Consider the $S$-id\`eles $I_S = K_S^\times\times\widehat{\OO}_S^\times\subset I_K$, where
\begin{equation*}
K_S^\times = \prod_{v\in S}K_v^\times\quad \text{ and }\quad \widehat{\OO}_S^\times=\prod_{v\notin S}\OO_{v}^\times.
\end{equation*}
Since the $S$-class group $\Pic(\OO_S)$ is trivial by \eqref{eq:S}, we have $I_SK^\times = I_K$. Inclusion of the open subgroup $I_S$ into $I_K$ thus induces an open epimorphism $I_S \to I_K/K^\times$ with kernel $I_S\cap K^\times = \OO_S^\times$, and hence an isomorphism
\begin{equation}\label{eq:isom}
I_S/\OO_S^\times \cong I_K/K^\times.
\end{equation}
With $p: I_S \to \widehat{\OO}_S^\times$ the natural projection, we define the following.
\begin{definition}
  A \emph{characteristic morphism} for $G$ is a continuous epimorphism $\rho_S : \widehat{\OO}_S^\times\to G$ that vanishes on $p(\OO_S^\times)$.
\end{definition}
Since $\rho_S$ is by definition continuous, it vanishes on the closure of $p(\OO_S^\times)$. Note that, by the Dirichlet $S$-unit theorem (e.g. \cite[Chapter VI, Prop.~1.1]{Neukirch}), $\OO_S^\times$ is finitely generated of $\Q$-rank $r=|S|-1$, so it has generators $\gamma_0,\ldots,\gamma_r$, with $\gamma_0$ a root of unity. For $\rho_S$ to vanish on $p(\OO_S^\times)$, it is thus enough that it vanishes on the images of $\gamma_0,\ldots,\gamma_r$ under the diagonal embedding $\OO_S^\times\to\widehat{\OO}_S^\times$.

A characteristic morphism $\rho_S:\widehat{\OO}_S^\times \to G$ induces an epimorphism $\rho: I_K/K^\times\to G$ as follows: as the epimorphism $\rho_S\circ p : I_S \to G$ vanishes on $\OO_S^\times$, it induces an epimorphism $I_S/\OO_S^\times \to G$, which by \eqref{eq:isom} gives the desired $\rho$.

We can also describe the restriction $\rho_v$ of $\rho$ to $K_v^\times\subset I_K/K^\times$ in terms of the characteristic morphism $\rho_S$. To do so, we need the following simple lemma.

\begin{lemma}\label{lemma} Every place $v\in\places\smallsetminus S$ admits a uniformiser $\pi_v\in \OO_{S\cup\{v\}}^\times$, and $\pi_v$ is unique up to multiplication by $S$-units.
\end{lemma}
\begin{proof}
The admissible choices for $\pi_v$ are the generators of the prime ideal $(\OO_v\smallsetminus\OO_v^\times)\cap\OO_S$ of $\OO_S$, which is principal due to \eqref{eq:S}.
\end{proof}

For $v\notin S$, we embed $K_v^\times$ into $\widehat{\OO}_S^\times$ as follows: we identify $\OO_{v}^\times$ with the $v$-component of $\widehat{\OO}_S^\times$. In addition,
we fix a uniformiser $\pi_v\in \OO_{S\cup\{v\}}^\times$ at $v$ as in Lemma \ref{lemma}, which we map to the element of $\widehat{\OO}_S^\times$ that is $1$ at its $v$-component and $\pi_v^{-1}\in \OO_{S\cup\{v\}}^\times\subset \OO_{w}^\times$ at the component indexed by $w\notin S\cup\{v\}$.

\begin{lemma}\label{lem:embedding}
  Let $\rho_S:\widehat{\OO}_S^\times\to G$ be a characteristic morphism and $\rho$ the induced epimorphism $I_K/K^\times \to G$. Then $\rho$ vanishes identically on $K_v^\times$ for $v\in S$. For $v\notin S$, let the embedding $K_v^\times\to \widehat{\OO}_S^\times$ be as specified above. Then the following diagram commutes:
\begin{center}
  \begin{tikzpicture}[>=triangle 60]
\matrix[matrix of math nodes,column sep={60pt,between origins},row
sep={60pt,between origins},nodes={asymmetrical rectangle}] (s)
{
&|[name=a]| K_v^\times &|[name=b]| I_K/K^\times \\

&|[name=c]| \widehat{\OO}_S^\times &|[name=d]| G  \\

};
\draw[->] (a) edge (b)
          (a) edge (c)
          (b) edge node[left] {$\rho$} (d)
          (c) edge node[above] {$\rho_S$} (d);
        \end{tikzpicture}
\end{center}
\end{lemma}

\begin{proof}
The key case is the image of $\pi_v$ for $v\notin S$. In this case,
\begin{equation*}
  (1,\ldots,1,\pi_v,1,\ldots)K^\times = (\pi_v^{-1},\ldots,\pi_v^{-1},1,\pi_v^{-1},\ldots)K^\times,
\end{equation*}
and the latter id\`ele is in $I_S$ by our choice of $\pi_v$. Hence,
\begin{equation*}
  \rho(\pi_v) = (\rho_S\circ p)((\pi_v^{-1},\ldots,\pi_v^{-1},1,\pi_v^{-1},\ldots)) = \rho_S(\pi_v),
\end{equation*}
by definition of the embedding of $K_v^\times$ in $\widehat{\OO}_S^\times$.
\end{proof}

\subsection{Local norms}\label{sec:local_norms}
Here we construct characteristic morphisms whose induced epimorphisms $I_K/K^\times\to G$ have the properties \eqref{eq:rho_local_norms}, \eqref{eq:rho_surj} and \eqref{eq:rho_triv_S}. In particular, $\alpha_1,\ldots,\alpha_t$ are local norms from the corresponding extensions at all places. When $G$ is cyclic, then ${\bigwedge}^2G=0$, so \eqref{eq:rho_HNP} holds automatically and we are done. We will return to \eqref{eq:rho_HNP} for non-cyclic $G$ in \S \ref{sec:forcing_HNP}.

For a non-archimedean place $v\in\places$, we write $\F_v$ for the residue field at $v$ and $q_v=|\F_v|$. Given $e\in\N$ and a finite set $\{x_0,\ldots, x_n\}\subseteq K^\times$, we denote by $T(S;e;\psi;\{x_0, \ldots, x_n\})$ the set of places $v$ of $K$ that satisfy the following:
\begin{equation}
  \label{eq:local_norm_elementary}
  \begin{aligned}
	& v\in\spl_S(\psi),\\
	& q_v\equiv 1\bmod e,\\
	& \{x_0,\ldots,x_n\}\subset \OO_v^\times,\\
	& x_0,\ldots,x_n\text{ are $e$-th powers in the residue field }\F_v^\times.
 \end{aligned}
\end{equation}
Note that $T(S;e;\psi;\{x_0,\ldots,x_n\})$ contains precisely those places in $\places\smallsetminus S$ that split completely in the normal extension $L_\psi(\zeta_e,\sqrt[e]{x_0},\ldots,\sqrt[e]{x_n})/K$, where $\zeta_e$ is a primitive $e$-th root of unity in $\bar{K}$. Hence, by the Chebotarev density theorem, $T(S;e;\psi;\{x_0,\ldots,x_n\})$ is infinite and of density $1/[L_\psi(\zeta_e,\sqrt[e]{x_0},\ldots,\sqrt[e]{x_n}):K]$ in $\places$. Every $v\in T(S;e;\psi;\{x_0,\ldots,x_n\})$ satisfies in particular that
\begin{equation}\label{eq:F_v_isom}
\F_v^\times\otimes\Z/e\Z \simeq \Z/e\Z,
\end{equation}
as $e\mid |\F_v^\times|$. Now let $e$ be a multiple of the
exponent of $G$ and $k$ large enough, so we can fix an epimorphism
$\Phi:(\Z/e\Z)^k\to G$.  Recall that $\{\gamma_0,\ldots,\gamma_r\}$ is a set
of generators of $\OO_S^\times$. For any subset $T=\{v_1,\ldots,v_k\}\subset T(S;e;\psi; \{\gamma_0,\ldots,\gamma_r\})$ of $k$ places, we define the epimorphism
\begin{equation}\label{characteristic morphisms type T}
\rho_S^T:
\widehat{\OO_S}^\times\to \F_{v_1}^\times\times\cdots\times \F_{v_k}^\times\to  (\F_{v_1}^\times\times\ldots\times \F_{v_k}^\times)\otimes \Z/e\Z\xrightarrow{\simeq} (\Z/e\Z)^k\xrightarrow{\Phi} G,
\end{equation}
where the first two maps are the natural epimorphisms, and the third one comes from a choice of the isomorphisms \eqref{eq:F_v_isom}.

\begin{lemma}\label{lem:rho_S_T}
  The morphism $\rho_S^T$ defined in \eqref{characteristic morphisms type T} is a characteristic morphism, and the induced epimorphism $\rho^T:I_K/K^\times\to G$ satisfies \eqref{eq:rho_local_norms}, \eqref{eq:rho_surj} and \eqref{eq:rho_triv_S}.
\end{lemma}

\begin{proof}
  For $\gamma\in K$ and $v\notin S$, write $\gamma_v$ for the local embedding $\gamma\in K_v^\times\subset\widehat{\OO}_S^\times$. We start by showing that $\gamma_v\in\ker\rho_S^T$ for all $v\notin S$ and $\gamma\in\{\gamma_0,\ldots,\gamma_r\}$. If $v\notin T$, this is obviously true by the construction of $\rho_S^T$. For $v\in T$, our choice of $T$ ensures that $\gamma$ is an $e$-th power in $\F_v^\times$, and hence vanishes in $\F_v^\times\otimes\Z/e\Z$.

  Since $\gamma_0,\ldots,\gamma_r$ generate $\OO_S^\times$, we have shown that $\rho_S^T$ vanishes at all local components $\gamma_v$, for $v\notin S$ and $\gamma\in\OO_S^\times$. This implies in particular that $p(\OO_S^\times)\subset \ker\rho_S^T$, so $\rho_S^T$ is a characteristic morphism. Moreover, as $\{\alpha_1,\ldots,\alpha_t\}\subset\OO_S^\times$ by \eqref{eq:S} and since the induced epimorphism $\rho^T$ vanishes on all of $K_v^\times$ for $v\in S$, we immediately get \eqref{eq:rho_local_norms} and \eqref{eq:rho_triv_S}. By the construction of $\rho_S^T$, it is also obvious that the condition \eqref{eq:rho_surj} is satisfied.
\end{proof}

\subsection{Forcing validity of the Hasse norm principle}\label{sec:forcing_HNP} Here we discuss how to choose the morphism $\rho_S^T$ from the previous section in order to guarantee that the induced epimorphism $\rho^T$  satisfies \eqref{eq:rho_HNP}. We will assume that $G$ is not cyclic, as otherwise there is nothing to do.

Fix an epimorphism $\Phinew:(\Z/e\Z)^{\knew}\to G$, with some $\knew\geq 2$. Write $\kprimenew={\knew\choose 2}$ and take a set of $2\kprimenew$ distinct places $T=\{v_1,w_1,v_2,w_2,\ldots,v_{\kprimenew},w_{\kprimenew}\}$ such that
\begin{equation*}
  \{v_1,\ldots,v_{\kprimenew}\}\subset T(S;e;\psi;\{\gamma_0,\ldots,\gamma_r\})\text{ and }  \{w_1,\ldots,w_{\kprimenew}\}\subset T(S;e;1;\{\gamma_0,\ldots,\gamma_r\}).
\end{equation*}
We write
\begin{equation*}
  G'=(\F_{v_1}^\times\times \F_{w_1}^\times\times\cdots\times\F_{v_{\kprimenew}}^\times\times\F_{w_{\kprimenew}}^\times)\otimes \Z/e\Z\simeq ((\Z/e\Z)^2)^{\kprimenew}
\end{equation*}
and consider the auxiliary morphism
\begin{equation}\label{eq:rho_prime}
  \rho_S':\widehat{\OO}_S^\times\to \F_{v_1}^\times\times \F_{w_1}^\times\times\cdots\times\F_{v_{\kprimenew}}^\times\times\F_{w_{\kprimenew}}^\times\to G'
\end{equation}
as in \eqref{characteristic morphisms type T}. (Formally, to fit the definition of $\rho_S^T$ in \eqref{characteristic morphisms type T}, take $(\Z/e\Z)^{2\kprimenew}\to G'$ to be the inverse of the isomorphism coming from \eqref{eq:F_v_isom}.) By Lemma \ref{lem:rho_S_T} (with $\psi=1$), this is a characteristic morphism for $G'$, so it induces an epimorphism $\rho':I_K/K^\times\to G'$. We will show later in \S \ref{sec:choosing_places} that the places in $T$ can be chosen in such a way that the following holds:
\begin{equation}\label{eq:basis_condition}
  \parbox{0.8\textwidth}{There is a $\Z/e\Z$-basis $e_1,e_1',\ldots,e_{\kprimenew},e_{\kprimenew}'$ of $G'$, such that $e_i,e_i'$ is a basis of the decomposition group $G_{v_i}'=\rho'(K_{v_i}^\times)$ for all $1\leq i\leq \kprimenew$.}
\end{equation}
Let $e_1,e_1',\ldots,e_{\kprimenew},e_{\kprimenew}'$ be a basis of $G'$ as in \eqref{eq:basis_condition}, and choose any basis $f_1,\ldots,f_{\knew}$ of $(\Z/e\Z)^{\knew}$, and any enumeration $(f_{m_i},f_{n_i})_{1\leq i\leq \kprimenew}$ of the $\kprimenew={\knew\choose 2}$ pairs $(f_m,f_n)$ with $1\leq m<n\leq \knew$. We define the epimorphism of $\Z/e\Z$-modules $\Psi:G'\to (\Z/e\Z)^{\knew}$ by
\begin{equation}\label{eq:def_Psi}
  \Psi(e_i)=f_{m_i} \text{ and }\Psi(e_i')=f_{n_i} \text{ for }1\leq i\leq \kprimenew,
\end{equation}
and correspondingly the epimorphism $\Phiprimenew = \Phinew\circ\Psi : G'\to G$. With this $\Phiprimenew$ and the set $T$ from above, we define the morphism
\begin{equation*}
  \rho_S^T = \Phiprimenew\circ\rho_S':\widehat{\OO_S}^\times\xrightarrow{\rho_S'} G' \xrightarrow{\Phiprimenew} G.
\end{equation*}

\begin{theorem}\label{thm:main_construction}
  Assume that \eqref{eq:basis_condition} holds. Then the morphism $\rho_S^T:\widehat{\OO}_S^\times\to G$ defined above is a characteristic morphism. The induced epimorphism $\rho=\rho^T:I_K/K^\times\to G$ satisfies \eqref{eq:rho_local_norms}--\eqref{eq:rho_triv_S}. In particular, the extension $L_{\rho}/K$ has Galois group $G$ and $\{\alpha_1,\ldots,\alpha_t\}\subset\norm_{L_{\rho}/K}(L_{\rho}^\times)$.
\end{theorem}

\begin{proof}
  Since $\rho_S^T$ is of the form \eqref{characteristic morphisms type T}, Lemma \ref{lem:rho_S_T} (with $\psi=1$) shows that $\rho_S^T$ is a characteristic morhism and that $\rho$ satisfies \eqref{eq:rho_local_norms} and \eqref{eq:rho_triv_S}. For the induced morphisms, note that $\rho=\Phiprimenew\circ\rho'$. Since $v_1,\ldots,v_{\kprimenew}\in\spl_S(\psi)$, condition \eqref{eq:rho_surj} follows immediately from our assumption \eqref{eq:basis_condition} and the fact that $\Phiprimenew$ is surjective.

  It remains to verify \eqref{eq:rho_HNP}. Since $f_1\ldots,f_{\knew}$ generate $(\Z/e\Z)^{\knew}$ and $\Phinew:(\Z/e\Z)^{\knew} \to G$ is an epimorphism, the group ${\bigwedge}^2G$ is generated by the elements $\Phinew(f_{m_i})\wedge \Phinew(f_{n_i})$ for $1\leq i\leq \kprimenew$. By our hypothesis \eqref{eq:basis_condition}  and the definition of $\Psi$ in \eqref{eq:def_Psi}, we see that $\{\Phinew(f_{m_i}),\Phinew(f_{n_i})\}\subset\rho_v(K_{v_i}^\times)$. Hence, $\Phinew(f_{m_i})\wedge\Phinew(f_{n_i})$ is in the image of the natural map ${\bigwedge}^2\rho_{v_i}(K_{v_i}^\times)\to{\bigwedge}^2G$ for all $1\leq i\leq \kprimenew$.
This is enough to show that
\begin{equation*}
  \bigoplus_{1\leq i\leq \kprimenew}{\bigwedge}^2 \rho_{v_i}(K_{v_i}^\times) \to {\bigwedge}^2G
\end{equation*}
is surjective. As $v_1,\ldots,v_{\kprimenew}\in\spl_S(\psi)$, we have proved \eqref{eq:rho_HNP}.
\end{proof}

\subsection{Supply of places}\label{sec:supply} We still need to specify how to choose the set $\{v_1,w_1\ldots,v_{\kprimenew},w_{\kprimenew}\}$ of places from \S \ref{sec:forcing_HNP} in order to guarantee that $\rho'$ satisfies \eqref{eq:basis_condition}.

For $e\in\N$, a finite set $\{x_0,\ldots,x_n\}\subset K^\times$ and $y\in K^\times$, we define $T(S;e;\{x_0,\ldots,x_n\}; y)$ as the set of all places $w\in T(S;e;1;\{x_0,\ldots,x_n\})$ such that $y\in\OO_w^\times$ and the polynomial $X^e-y$ modulo $w$ is irreducible over the residue field $\F_w$.

Let us show that these sets provide an ample supply of places in those cases that will be of interest to us.

\begin{lemma}\label{lem:cheb}
  Suppose there is a place $v\in T(S;e;1;\{x_0,\ldots,x_n\})$ such that $y$ is a uniformiser at $v$. Then the set $T(S;e;\{x_0,\ldots,x_n\}; y)$ is infinite and of natural density
  \begin{equation*}
\frac{\phi(e)}{e[K(\zeta_e,\sqrt[e]{x_1},\ldots,\sqrt[e]{x_n}):K]}>0
\end{equation*}
  in the set of all places $v$ of $K$ when ordered by $q_v$.
\end{lemma}

\begin{proof}
  Write $F = K(\zeta_e,\sqrt[e]{x_1},\ldots,\sqrt[e]{x_n})$. Since $v$ splits completely in $F$, the element $y$ is a uniformiser for ever place $\tilde{v}$ of $F$ above $v$. Therefore, $f=X^e-y$ is irreducible over $F$ by Eisenstein's criterion in $F_{\tilde{v}}$, and hence the extension $F(\sqrt[e]{y})/F$ is cyclic of degree $e$.

The set $T(S;e;\{x_0,\ldots,x_n\}; y)$ is precisely the set of places $w\in \spl_S(F)$, such that all places $\tilde{w}$ of $F$ above $w$ are inert in $F(\sqrt[e]{y})$.

These are precisely the places $w\notin S$ that do not divide $y$ and whose Frobenius class in $\Gal(F(\sqrt[e]{y})/K)$ consists of generators of the normal subgroup $\Gal(F(\sqrt[e]{y})/F)\simeq \Z/e\Z$. The desired result follows from the Chebotarev density theorem.
\end{proof}

\begin{lemma}\label{lem:irreducibility_enough}
  If $w\in T(S;e;\{x_0,\ldots,x_n\};y)$, then $(y\bmod w)\otimes 1$ generates $\F_w^\times\otimes\Z/e\Z$.
\end{lemma}

\begin{proof}
  Let $a\in \F_w^\times$. Then, $a\otimes 1$ generates $\F_w^\times\otimes\Z/e\Z$ if and only if none of the elements $a,a^2,\ldots,a^{e-1}$ are in $\F_w^{\times e}$. But $a^j\in\F_w^{\times e}$ if and only if $a\in\F_w^{\times e/\gcd(j,e)}$, and thus $a\otimes 1$ generates $\F_w^{\times e}$ if and only if $a\notin \F_w^{\times p}$ for all primes $p\mid e$. The latter condition is clearly satisfied if $X^e-a$ is irreducible over $\F_w$.
\end{proof}

\subsection{Choosing the right places}\label{sec:choosing_places}
With our supply of places guaranteed, we choose the the places $v_1,w_1,\ldots,v_{\kprimenew},w_{\kprimenew}$ as follows. Take $v_1,\ldots,v_{\kprimenew}$ to be any $\kprimenew$ distinct elements of $T(S;e;\psi;\{\gamma_0,\ldots,\gamma_r\})$. We fix a uniformiser $\pi_i\in\OO_{S\cup\{v_i\}}^\times$ for each $v_i$, which exists by Lemma \ref{lemma}.

Moreover, we choose each place $w_i\in T(S;e;\{\gamma_0,\ldots,\gamma_r\}; u_i\pi_i)\smallsetminus\{v_1,\ldots,v_{\kprimenew}\}$, with certain local units $u_i\in K\cap\OO_{v_i}^\times$, to be determined inductively. As $u_i\pi_i$ is a uniformiser at $v_i$, the hypothesis of Lemma \ref{lem:cheb} is satisfied.

Recall from the discussion leading up to Lemma \ref{lem:embedding} that each $K_{v_i}^\times$ is embedded into $\widehat{\OO}_S^\times$ by sending $\OO_{v_i}^\times$ to its factor in $\widehat{\OO}_S^\times$, and $\pi_i$ to $(\pi_i^{-1},\ldots,\pi_i^{-1},1,\pi_i^{-1},\ldots)$.
We will now determine the $u_i$ and $w_i$ inductively, such that the following property holds for all $i$:
\begin{equation}\label{eq:induction}
  \parbox{0.7\textwidth}{The images of $\pi_1,\ldots,\pi_i\in\widehat{\OO}_S^\times$ in $(\F_{w_1}^\times\times\cdots\times\F_{w_i}^\times)\otimes\Z/e\Z$ generate $(\F_{w_1}^\times\times\cdots\times\F_{w_i}^\times)\otimes\Z/e\Z$.}
\end{equation}

For $i=1$, we take $u_1=1$ and choose $w_1\in T(S;e;\{\gamma_0,\ldots,\gamma_r\}; \pi_1)\smallsetminus\{v_1,\ldots,v_{\kprimenew}\}$. Then the image of $\pi_1$ under
\begin{equation*}
\OO_{S\cup\{v_1\}}^\times\to \OO_{w_1}^\times\to\F_{w_1}^\times\to\F_{w_1}^\times\otimes\Z/e\Z
\end{equation*}
generates $\F_{w_1}^\times\otimes\Z/e\Z$, as $X^e-\pi_1$ is irreducible over $\F_{w_1}$. Therefore, we see that $\F_{w_1}^\times\otimes\Z/e\Z$ is also spanned by the image $(\pi_1^{-1}\bmod w_1)\otimes 1$ of $\pi_1\in K_{v_1}\subset\widehat{\OO}_S^\times$ under $\widehat{\OO}_S^\times \to \F_{w_1}^\times\otimes\Z/e\Z$.

For $j>1$, assume that we have determined $u_1,\ldots,u_{j-1}$ and chosen $w_1,\ldots,w_{j-1}$ such that \eqref{eq:induction} holds for $i=j-1$. Then there are $c_1,\ldots,c_{j-1}\in\{0,\ldots,e-1\}$ such that the elements $\pi_j$ and $\pi_1^{c_1}\cdots \pi_{j-1}^{c_{j-1}}$ of $\widehat{\OO}_S^\times$ have the same image in $(\F_{w_1}^\times\times\cdots\times\F_{w_{j-1}}^\times)\otimes\Z/e\Z$.

We set $u_j=\pi_1^{-c_1}\cdots \pi_{j-1}^{-c_{j-1}}\in K\cap\OO_{v_j}^{\times}$ and choose $w_j\in T(S;e;\{\gamma_0,\ldots,\gamma_r\}; u_j\pi_j)\smallsetminus\{v_1,\ldots,v_{\kprimenew},w_1,\ldots,w_{j-1}\}$. Consider the projection
\begin{equation*}
   P : (\F_{w_1}^\times\times\cdots\times\F_{w_{j}}^\times)\otimes\Z/e\Z \to (\F_{w_1}^\times\times\cdots\times\F_{w_{j-1}}^\times)\otimes\Z/e\Z.
\end{equation*}
Then the images of $\pi_1,\ldots,\pi_{j-1}\in\widehat{\OO}_S^\times$ generate the range of $P$ by our inductive hypothesis. By construction, the image of $\pi_1^{-c_1}\cdots\pi_{j-1}^{-c_{j-1}}\pi_j\in\widehat{\OO}_S^\times$ is in $\ker P$. It even generates $\ker P$, since $X^e-u_j\pi_j$ is irreducible over $\F_{w_j}$, so $((u_j\pi_j)^{-1}\bmod w_j)\otimes 1$  generates $\F_{w_j}^\times\otimes\Z/e\Z$. This shows that \eqref{eq:induction} holds for $i=j$, as desired.

Finally, let us show that the places chosen above achieve what we want to satisfy the hypothesis of Theorem \ref{thm:main_construction}.

\begin{theorem}\label{thm:choose_places}
  Let the set $T=\{v_1,w_1,\ldots,v_{\kprimenew},w_{\kprimenew}\}\subset T(S;e;1;\{\gamma_0,\ldots,\gamma_r\})$ be chosen as described above. Then the epimorphism $\rho':I_K/K^\times\to G'$, induced by the auxiliary morphism $\rho_S'$ defined in \eqref{eq:rho_prime}, satisfies \eqref{eq:basis_condition}.
\end{theorem}

\begin{proof}
  For each $1\leq i\leq \kprimenew$, we have $K_{v_i}^\times=\OO_{v_i}^\times\oplus \langle\pi_i\rangle$. Using Lemma \ref{lem:embedding}, we see that $\rho'(\OO_{v_i}^\times)$ is the factor $\F_{v_i}^\times\otimes\Z/e\Z\simeq\Z/e\Z$ of $G'$. Let $e_i$ be a generator of this factor.

  Moreover, take $e_i'$ to be the image of $\pi_i$ in $G'$. Then $e_i,e_i'$ is a basis of $\rho'(K_{v_i}^\times)$. Consider the projection $P : G' \to (\F_{w_1}^\times\times\cdots\times\F_{w_{\kprimenew}}^\times)\otimes\Z/e\Z$. Then clearly $e_1,\ldots,e_{\kprimenew}$ is a basis of $\ker P\cong (\F_{v_1}^\times\times\cdots\times\F_{v_{\kprimenew}})\otimes \Z/e\Z$. Moreover, due to \eqref{eq:induction}, the images $P(e_1'),\ldots,P(e_{\kprimenew}')$ form a basis of the image of $P$.
\end{proof}

\section{Computations with characteristic morphisms}\label{sec:computation}
Let us discuss how to compute a characteristic morphism of the form $\rho_S^T$ as in \eqref{eq:rho_prime} and the map $\Phiprimenew:G'\to G$ in practice, and how to infer from this data some information about the $G$-extension defined by it. We will focus on the case where $\psi=1$, in order to avoid having to specify how $\psi$ is represented. We assume to be given the set $S$ of places satisfying \eqref{eq:S}, the  epimorphism $\Phinew:(\Z/e\Z)^{\knew}\to G$, and $\kprimenew={\knew\choose 2}$.

We need to find places $v_1,w_1,\ldots,v_{\kprimenew},w_{\kprimenew}$ such that \eqref{eq:induction} is satisfied for all $1\leq i\leq \kprimenew$.

\subsection{Finding the places $v_i$}\label{sec:finding_vi}
First, we need a set of generators $\{\gamma_0,\ldots,\gamma_r\}$ for the $S$-units $\OO_S^\times$, which can be computed by well-known algorithms implemented in many computer algebra systems.

As $\psi=1$, the places $v_1,\ldots,v_{\kprimenew}$ have to be members of the set $T(S; e; 1; \{\gamma_0,\ldots,\gamma_r\})$, so they need to be in $\places\smallsetminus S$ and such that
\begin{equation}\label{eq:search_cond_v}
  q_v\equiv 1\bmod e \text{, and the residues of }\gamma_0,\ldots,\gamma_r\text{ are $e$-th powers in }\F_v^\times.
\end{equation}
These places have positive density among all places of $K$ by the Chebotarev density theorem, so we can produce enough of them by an exhaustive search. Effective lower bound versions of Chebotarev's theorem (e.g.~\cite[(3.2)]{Thorner}) guarantee that we will not be unlucky for too long. Note that the condition of $\gamma$ being an $e$-th power in $\F_v^\times$, with $q_v\equiv 1\bmod e$, can be easily checked using Euler's criterion: it holds if and only if $\gamma^{(q_v-1)/e}\equiv 1\bmod v$.

Once $v_i$ has been specified, we fix a residue class $b_i\in\F_{v_i}^\times$ that generates $\F_{v_i}^\times/\F_{v_i}^{\times e}$,
for example a primitive root in $\F_{v_i}^\times$.

We also need a uniformiser $\pi_i\in\OO_{S\cup \{v_i\}}^\times$ for each of the places $v_i$, whose existence is guaranteed by Lemma \ref{lemma}. It can be found, for example, by computing generators $\{g_0,\ldots,g_{r+1}\}$ of $\OO_{S\cup \{v_i\}}^\times$ and solving the linear Diophantine equation $c_0v_i(g_0)+\cdots+c_{r+1}v_i(g_{r+1})=1$ for the exponents in $\pi_i=g_0^{c_0}\cdots g_{r+1}^{c_{r+1}}$, with $v_i(\cdot)$ the exponential valuation at $v_i$.

\subsection{Finding the places $w_i$} The place $w_i$ has to be in $T(S;e;\{\gamma_0,\ldots,\gamma_r\},u_i\pi_i)$. The element $u_i\in \OO_{S\cup\{v_i\}}$ is specified by $u_1=1$ and inductively for $i>1$, as explained in \S \ref{sec:choosing_places}. We can determine $u_i$ by linear algebra with the images of $\pi_1,\ldots,\pi_i\in \widehat{\OO}_S^\times$ in
\begin{equation*}
  (\F_{w_1}^\times\times\cdots\times\F_{w_{i-1}}^\times)\otimes\Z/e\Z
\simeq (\Z/e\Z)^{i-1}.
\end{equation*}
To specify the above isomorphism, fix elements $b_j'\in\F_{w_j}^\times$,
each generating $\F_{w_j}^\times/\F_{w_j}^{\times e}$,
and identify
\begin{equation*}
  (b_1'^{a_1},\ldots,b_{i-1}'^{a_{i-1}})\otimes 1\quad\text{ with }\quad(a_1,\ldots,a_{i-1})\in(\Z/e\Z)^{i-1}.
\end{equation*}
Let $l_{m,j}'\in\Z/e\Z$ be such that $\pi_j^{-1}\bmod w_m$ and $(b_m')^{l_{m,j}'}$ differ only by an $e$-th power in $\F_{w_m}^\times$. For example, if $b_m'$ is a primitive root of $\F_{w_m}^\times$, then $l_{m,j}'$ can be taken as the discrete logarithm of $\pi_j^{-1}$  with respect to the generator $b_m'$ of $\F_{w_m}^\times$, modulo $e$. Recalling how the uniformiser $\pi_j\in K_{v_j}^\times$ is embedded into $\widehat{\OO}_S^\times$, we see that the image of $\pi_j$ in $(\Z/e\Z)^{i-1}$ is then $(l_{1,j}',\ldots,l_{i-1,j}')$. Hence, the exponents in $u_i=\pi_1^{-c_1}\cdots\pi_{i-1}^{-c_{i-1}}$ can be determined by solving the $\Z/e\Z$-linear system
\begin{equation}\label{eq:ui_equations}
  \begin{pmatrix}
    l_{1,1}' & \cdots & l_{1,i-1}'\\
    \vdots & & \vdots\\
    l_{i-1,1}' & \cdots & l_{i-1,i-1}'
  \end{pmatrix}\cdot
  \begin{pmatrix}
    c_1\\\vdots\\c_{i-1}
  \end{pmatrix}=
  \begin{pmatrix}
    l_{1,i}'\\\vdots\\l_{i-1,i}'
  \end{pmatrix}.
\end{equation}
Once $u_i$ has been computed, we can again find $w_i\in T(S;e;\{\gamma_0,\ldots,\gamma_r\},u_i\pi_i)$ by an exhaustive search. In addition to \eqref{eq:search_cond_v}, it has to satisfy that the reduction of  $X^e-u_i\pi_i$ is irreducible over $\F_{w_j}$. Actually, we only need the potentially weaker (if $4\mid e$) condition that $(u_i\pi_i\mod w_i)\otimes 1$ generates $\F_{w_i}^\times\otimes \Z/e\Z$. As explained in the proof of Lemma \ref{lem:irreducibility_enough}, it is enough to verify that
\begin{equation*}
u_i\pi_i\bmod w_i \notin \F_{w_i}^{\times p}\quad\text{ for all primes $p\mid e$},
\end{equation*}
which can again be checked by Euler's criterion.

As the conditions for the $w_i$ are more restrictive than for the $v_i$, we can potentially get smaller places by searching for $w_j$ as soon as possible, i.e.~as soon as $v_{j}$ has been determined, instead of computing first all the $v_i$ and then the $w_i$.

\subsection{Computing $\Phiprimenew$} In addition to the $l_{m,j}'$  already computed, let $l_{m,j}\in\Z/e\Z$ be such that $\pi_j^{-1}\bmod v_m$ differs from $b_m^{l_{m,j}}$ only by an $e$-th power in $\F_{v_m}^\times$, and $l_{m,j}=0$ if $j=m$.

Let us identify our chosen roots $b_i\in \F_{v_i}^\times$ with the basis elements $(1,\ldots,1,b_i,1\ldots,1)\otimes 1 \in G'$, and similarly for the $b_i'$. We call the resulting basis $b_1,b_1',\ldots,b_{\kprimenew},b_{\kprimenew}'$ the \emph{standard basis} of $G'$. Let us compare this basis to the basis $e_1,e_1',\ldots,e_{\kprimenew},e_{\kprimenew}'$ of $G'$, satisfying \eqref{eq:basis_condition}, that is constructed in the proof of Theorem \ref{thm:choose_places}. The matrix describing the change of basis from the standard basis to the $e_i,e_i'$ is
\begin{equation}\label{eq:change_of_basis}
  A =
  \begin{pmatrix}
    1       & l_{1,1}  & 0     & l_{1,2}  & \cdots & 0       & l_{1,\kprimenew}\\
    0       & l_{1,1}' & 0     & l_{1,2}' & \cdots & 0       & l_{1,\kprimenew}' \\
    0       & l_{2,1}  & 1     & l_{2,2}  & \cdots & 0       & l_{2,\kprimenew} \\
    0       & l_{2,1}' & 0     & l_{2,2}' & \cdots & 0       & l_{2,\kprimenew}' \\
    \vdots  & \vdots  & \vdots & \vdots  & \ddots & \vdots  & \vdots  \\
    0       & l_{\kprimenew,1}  & 0      & l_{\kprimenew,2} & \cdots & 1       & l_{\kprimenew,\kprimenew}  \\
    0       & l_{\kprimenew,1}' & 0      & l_{\kprimenew,2}'& \cdots & 0       & l_{\kprimenew,\kprimenew}'
 \end{pmatrix}.
\end{equation}
Let $B$ be the matrix representing the epimorphism $\Psi:G'\to(\Z/e\Z)^{\knew}$ from \eqref{eq:def_Psi} with respect to the bases $e_i,e_i'$ and $f_1,\ldots,f_{\knew}$, and $C$ a matrix representing $\Phinew:(\Z/e\Z)^{\knew}\to G$ with respect to $f_1,\ldots,f_{\knew}$ and any system $g_1,\ldots,g_{\knew}$ of generators for $G$. Then the epimorphism $\Phiprimenew:G'\to G$ is represented with respect to the standard basis of $G'$ and the generators $g_1,\ldots,g_{\knew}$ of $G$ by the matrix $R=CBA^{-1}$.

With the places $v_1,w_1,\ldots,v_{\kprimenew},w_{\kprimenew}$ defining $G'$, the standard basis $b_1,b_1',\ldots,b_{\kprimenew},b_{\kprimenew}'$ of $G$ and the matrix $R$ describing $\Phiprimenew$, we have now computed all the data required to describe the characteristic morphism $\rho_S^T:\widehat{\OO}_S^\times \to G$ from Theorem \ref{thm:main_construction}, and thus the induced epimorphism $\rho:I_K/K^\times\to G$ in a convenient way.

\subsection{Conductor, Artin map and  equations}\label{sec:prime_data}
The epimorphism $\rho:I_K/K^\times\to G$ induced by $\rho_S^T$ defines an extension $L_\rho/K$ together with an isomorphism $\Gal(L_\rho/K)\to G$. Let us see how to easily infer some information about $L_\rho$ from our description of $\rho_S^T$.

First of all, the places $v$ of $K$ that ramify in $L_\rho$ are those, for which the inertia group $\rho(\OO_v^\times)$ does not vanish in $G$. By our construction of $\rho_S'$ in \eqref{eq:rho_prime}, only places in $\{v_1,w_1,\ldots,v_{\kprimenew},w_{\kprimenew}\}$ can ramify. More precisely, the ramified places are all those $v_i,w_i$ for which the corresponding standard basis element $b_i$ or $b_i'$ is not in the kernel of $\Phiprimenew$. If we write $(r_1,\ldots,r_{\knew})^T$ for the corresponding column of $R$ and write $G$ additively, this means that $r_1g_1+\cdots+r_{\knew}g_{\knew}\neq 0$ in $G$. All these places are tamely ramified, so the conductor ideal of $L_\rho$ is just the product of the prime ideals corresponding to these places.

Every place $v\in S$ is split completely, as then $\rho(K_v^\times)$ vanishes in $G$. For places $v\notin S\cup\{v_1,w_1,\ldots,v_{\kprimenew},w_{\kprimenew}\}$, we can compute the Artin symbol $(L_\rho/K, v)\in G$ as follows: compute a uniformiser $\pi_v\in\OO_{S\cup\{v\}}^\times$ at $v$ and $l_{v,i}, l'_{v,i}\in\Z/e\Z$ such that $\pi_v^{-1}=b_i^{l_{v,i}}$ in $\F_{v_i}\otimes\Z/e\Z$ and $\pi_v^{-1}=(b_i')^{l_{v,i}'}$ in $\F_{w_i}\otimes\Z/e\Z$. Then the coordinates of $\rho_S'(\pi_v)\in G'$ with respect to the standard basis are given by $l_v = (l_{v,1},l_{v,1}',\ldots,l_{v,\kprimenew},l_{v,\kprimenew}')^T$. Hence,  the Artin symbol is given by $(L_\rho/K, v) = \rho(\pi_v)= (g_1,\ldots,g_{\knew})\cdot R\cdot l_v$. If $v\in\{v_1,w_1,\ldots,v_{\kprimenew},w_{\kprimenew}\}$ is unramified, we can compute $(L_\rho/K, v)$ in the same way as above, except that we take $l_{v,i}=0$ if $v=v_i$ and $l_{v,i}'=0$ if $v=w_i$.

Given the conductor $\mathfrak f$ and a method to evaluate the Artin symbol $(L_\rho/K,v)$, we have enough information to specify the congruence subgroup of the ray class group modulo $\mathfrak f$ corresponding to $L_\rho$ through an exhaustive search for generating splitting primes. From this data, one can compute polynomials defining $L_\rho$ using algorithms of computational class field theory, see, e.g. \cite{Cohen1,CohenBook,Fieker}. Such algorithms are implemented, for example, in the computer algebra system Magma.

\subsection{Subextensions} If $\Pi:G\to H$ is an epimorphism, then $\Pi\circ\rho$ describes a subextension $L_\Pi$ of $L_\rho$ together with an isomorphism $\Gal(L_\Pi/K)\to H$. The epimorphism $\Pi\circ\rho$ is induced by the characteristic morphism $\Pi\circ\rho_S^T:\widehat{\OO}_S^\times\to H$, and, in a similar way as described in \S \ref{sec:prime_data}, we can compute the ramified primes, splitting primes and equations for $H$.

Given a presentation $G\simeq\Z/n_1\Z\times\cdots\times\Z/n_{r}\Z$, we can
apply this in particular to the projections $\Pi_i:G\to H_i=\Z/n_i\Z$. Then
$L_\rho$ is the compositum of the fields $L_{\Pi_i}$, which we can exploit, for
example, to get equations for $L_\rho$ with lower computational effort.

\subsection{Elements with prescribed norms}
Having determined concrete polynomials defining the number field $L=L_\rho$,
one may want to find elements $\beta_1,\ldots,\beta_t\in L$ such that
$\N_{L/K}(\beta_i)=\alpha_i$ for $1\leq i\leq t$. By our construction, such
$\beta_i$ will exist, and one could find them through an exhaustive search. It
is much faster to use more sophisticated algorithms, see,
e.g. \cite{Simon}, \cite[\S 7.5]{CohenBook}, which solve norm equations
via $S$-units and are implemented in Magma.

\section{Illustrations}\label{sec:illustrations}

\subsection{A biquadratic extension of $\Q$ with norm $37/16$} In this first example, we take $G=(\Z/2\Z)^2$ and find an epimorphism $\rho:I_\Q/\Q^\times\to G$ with $\alpha=37/16\in\norm_{L_\rho/\Q}(L_\rho^\times)$.

Due to the explicit nature of biquadratic fields, arguments simpler than ours would suffice here. A construction yielding distinct primes $p,q$ such that a given rational number $c$ is the norm of an element of $\Q(\sqrt{p},\sqrt{q})$ was communicated to D.~Loughran and the first-named author by J.-L.~Colliot-Th\'el\`ene, in response to a question raised at the workshop \og Rational Points 2017\fg{} in Schloss Schney (Germany). Similarly to our approach here, the primes $p,q$ are chosen in such a way as to guarantee that the extension $\Q(\sqrt{p},\sqrt{q})/\Q$ satisfies the Hasse norm principle and $c$ is a local norm at every place.
\smallskip

Since this example should serve as a first illustration of our approach, we will nevertheless use our setup relying on characteristic morphisms. The set $S=\{\infty,2,37\}\subset\Omega_\Q$ satisfies \eqref{eq:S}, and $\OO_S^\times=\Z[1/2,1/37]^\times=\langle -1,2,37\rangle$.
We can take $e=\knew=2$ and $\Phinew:G\to G$ the identity. With $\kprimenew={2\choose 2}=1$, we are aiming at a set $T$ consisting of a pair $v,w$ of places. The places in $T(S;e;1;\{-1,2,37\})$ are those primes $p$ that satisfy
\begin{equation*}
  \left(\frac{-1}{p}\right)=\left(\frac{2}{p}\right)=\left(\frac{37}{p}\right)=1,
\end{equation*}
with $\left(\frac{\cdot}{\cdot}\right)$ the Legendre symbol. We take $v=41$, the smallest such prime. In addition to the above, the place $w$ then has to satisfy
\begin{equation*}
  \left(\frac{41}{p}\right)=-1,
\end{equation*}
and the smallest choice is $w=137$. Hence, we have found our auxiliary morphism
\begin{equation*}
  \rho_S':\widehat{\Z}_S^\times\to(\F_{41}^\times\times\F_{137}^\times)\otimes \Z/2\Z = G' \simeq(\Z/2\Z)^2=G.
\end{equation*}
By our construction, we get $\rho_S'(\Q_{41}^\times)=G'$, so we can take $\rho_S=\Phiprimenew\circ\rho_S'$ for, e.g., the isomorphism $\Phiprimenew:G'\to G$, that identifies $\F_{41}^\times\otimes\Z/2\Z$ with the first $\Z/2\Z$-factor of $G$ and $\F_{137}^\times\otimes\Z/2\Z$ with the second. We have thus achieved that ${\bigwedge}^2\rho(\Q_{41}^\times)={\bigwedge}^2G$, which confirms the validity of the Hasse norm principle for $\rho$. All elements of $\Z[1/2,1/37]^\times$, including in particular our $\alpha=37/16$, are local norms at all places, and therefore in $\norm_{L_\rho/\Q}(L_\rho^\times)$.
\smallskip

Which biquadratic extension does $\rho$ describe? Via the projections $\Pi_1, \Pi_2: G\to\Z/2\Z$ to the two $\Z/2\Z$-factors of $G$, we can describe $L_\rho$ as the compositum of two quadratic subfields with conductors  $41$ and $137$. Hence, $\rho$ defines the number field $L_\rho=\Q(\sqrt{41},\sqrt{137})$.

We know now that $37/16$ is a norm from this field. In other words, using the norm form coming from the basis $1,\sqrt{41},\sqrt{137},\sqrt{5617}$ of $L_\rho$, the equation
\begin{align*}
  x_1^4 - 82x_1^2x_2^2 + 1681x_2^4 - 274x_1^2x_3^2 - 11234x_2^2x_3^2 + 18769x_3^4 + 44936 x_1x_2x_3x_4&\\ - 11234x_1^2x_4^2 - 460594 x_2^2x_4^2 - 1539058 x_3^2x_4^2 + 31550689 x_4^4 &=37/16
\end{align*}
has rational solutions $(x_1,x_2,x_3,x_4)\in\Q^4$. Indeed, Magma's norm form equation functionality instantly finds the solution
\begin{equation*}
  \left(4449545, -\frac{1389743}{2}, \frac{760267}{2}, -\frac{118739}{2}\right).
\end{equation*}

\subsection{A $\Z/6\Z \times (\Z/3\Z)^3$-extension of $\Q(\sqrt{-47})$ from which $2+3\sqrt{-47}$ is a norm.}
Here we consider an example that is sufficiently generic to demonstrate all the features of our construction. Our computations are assisted by the computer algebra system Sage\footnote{\url{https://www.sagemath.org/}}. Towards the end, we will compute concrete polynomials giving our extension $L$. For this, we use the Magma Calculator\footnote{\url{http://magma.maths.usyd.edu.au/calc/}}. None of the computations took longer than a few seconds.

Take the number field $K=\Q(\sqrt{-47})$ with class number $5$. We construct an
extension $L/K$ with Galois group $G=\Z/6\Z\times(\Z/3\Z)^3$, such that
$2+3\sqrt{-47}\in\norm_{L/K}(L^\times)$.

The principal ideal $(2+3\sqrt{-47})$ factors as the product of two prime ideals of $\OO_K$, namely
\begin{equation*}
  (2+3\sqrt{-47})=\left(7, \frac{3+\sqrt{-47}}{2}\right)\left(61, \frac{21+\sqrt{-47}}{2}\right).
\end{equation*}
We take $S$ to consist of the archimedean place of $K$ and these two prime ideals. Then $\OO_S$ has class number one, so \eqref{eq:S} is satisfied. A quick computation reveals the following set of generators $\{\gamma_0,\gamma_1,\gamma_2\}$ for the $S$-units $\OO_S^\times$: $\gamma_0=-1$, $\gamma_1=2+3\sqrt{-47}$ and $\gamma_2=128+3\sqrt{-47}$.

Take $\Phinew:(\Z/6\Z)^4\to G$ to be the epimorphism mapping each factor $\Z/6\Z$ to the corresponding factor $\Z/6\Z$ or $\Z/3\Z$ of $G$ in the obvious way. With $\knew=4$, we are looking for $\kprimenew={4\choose 2}=6$ pairs of places $v_1,w_1,\ldots,v_6,w_6$ of $K$ to define our auxiliary morphism $\rho_S'$ from \eqref{eq:rho_prime}.

Our places need to be chosen from $T(S;6;1;\{\gamma_0,\gamma_1,\gamma_2\})$, so they must be in $\places\smallsetminus S$ and satisfy the conditions that
\begin{equation}\label{eq:example_cond_v}
  -1,\ 2+3\sqrt{-47}\ \text{ and }\ 128+2\sqrt{-47}\ \text{ are $6$-th power residues in $\F_v^\times$}.
\end{equation}
Here is a list of the (prime ideals corresponding to) the first few places satisfying these conditions, in order of the rational primes lying below:
\begin{align*}
 &(97, (27+\sqrt{-47})/2),\
 (569),\
 (809),\
 (1033),\
 (1381, (1445+\sqrt{-47})/2),\\
 &(1913),\
 (2281, (619+\sqrt{-47})/2),\
 (2377, (3677+\sqrt{-47})/2),\\
 &(2887),\
 (4621),\
 (4789, (2537+\sqrt{-47})/2),\
   (5227),\
   (6101).
\end{align*}
Hence, we take $v_1$ as the first of these places, choose $b_1$ to be a primitive root in $\F_{v_1}^\times$ and compute a uniformiser $\pi_1\in\OO_{S\cup \{v_1\}}^\times$ at $v_1$ as in \S \ref{sec:finding_vi}. The precise values are recorded in Table \ref{tab:v}.

\begin{table}[H]\centering
\begin{tabular}{|c|c|c|c|}
  \hline
  $i$ & $v_i$                         & $b_i$           & $\pi_i$               \\
  \hline
  1   & $(97, (27+\sqrt{-47})/2)$     & $5$             & $-353 + 48\sqrt{-47}$ \\
  2   & $(809)$                       & $1+\sqrt{-47}$  & $809$\\
  3   & $(1033)$                      & $20+\sqrt{-47}$ & $1033$\\
  4   & $(1913)$                      & $7+\sqrt{-47}$  & $1913$\\
  5   & $(2377, (3677+\sqrt{-47})/2)$ & $5$             & $712+81\sqrt{-47}$\\
  6   & $(2887)$                      & $1+\sqrt{-47}$  & $2887$\\
  \hline
\end{tabular}
\caption{The places $v_i$ with primitive roots $b_i$ and uniformisers $\pi_i$.}
\label{tab:v}
\end{table}

\begin{table}[H]\centering
\begin{tabular}{|c|c|c|c|}
  \hline
  $i$ &  $u_i$                 & $w_i$                           & $b_i'$          \\
  \hline
  1   &  $1$                  & $(569)$                          & $2+\sqrt{-47}$ \\
  2   &  $1$                  & $(1381, (1445 + \sqrt{-47})/2)$  & $2$ \\
  3   &  $\pi_2^{-3}$          & $(2281, (619 + \sqrt{-47})/2)$   & $7$ \\
  4   &  $\pi_2^{-5}\pi_3^{-4}$ & $(4789, (2537 + \sqrt{-47})/2)$ & $2$ \\
  5   &  $\pi_1^{-4}\pi_3^{-2}$ & $(4621)$                      & $5+\sqrt{-47}$\\
  6   & $\pi_1^{-4}\pi_2^{-5}\pi_3^{-5}\pi_4^{-2}\pi_5^{-2}$ & $(65 + 12\sqrt{-47})$ & $7$\\
  \hline
\end{tabular}
\caption{The places $w_i$ with factors $u_i$ and primitive roots $b_i$.}
\label{tab:w}
\end{table}

As $u_1=1$, the place $w_1$ has to satisfy in addition to \eqref{eq:example_cond_v} the condition that $X^6-\pi_1$ is irreducible over $\F_{w_1}^\times$, so $\pi_1$ may not be a quadratic or cubic residue in $\F_{w_1}^\times$. We choose $w_1$ as the first place satisfying these conditions and $b_1'$ as a primitive root for $\F_{w_1}^\times$. See Table \ref{tab:w} for the values.

Next, we pick $v_2$ as the next available place from our list and choose a corresponding primitive root $b_2$ and uniformiser $\pi_2\in\OO_{S\cup\{v_2\}}$. As $v_2$ is generated by an inert rational prime, the choice of uniformiser is simple.

Before choosing $w_2$, we need to find $u_2$. To this end, we compute the discrete logarithms $l_{1,1}'=5$ and $l_{1,2}'=0$ (modulo $6$) of $\pi_1^{-1}$ and $\pi_2^{-1}$ in $\F_{w_1}^\times$, with respect to the chosen primitive root $b_1'$. Hence, the images of $\pi_1$ and $\pi_2$ in $\F_{w_1}^\times\otimes\Z/6\Z$ have coordinates $5$ and $0$ with respect to the basis $b_1'\otimes 1$, and thus $\pi_2$ has the same image as $\pi_1^0$. We take $u_2=(\pi_1^0)^{-1}=1$. The condition that $w_2$ needs to satisfy, in addition to \eqref{eq:example_cond_v}, is that $X^6-\pi_2$ is irreducible over $\F_{w_2}$. We take $w_2$ as the first place that satisfies this condition and compute a primitive root $b_2'$ of $\F_{w_2}^\times$.

We choose $v_3$ as the next available place in our list, together with a primitive root $b_3$ and uniformiser $\pi_3$.

For $u_3$, the $\Z/6\Z$-linear system \eqref{eq:ui_equations} takes the form
\begin{equation*}
  \begin{pmatrix}
   5 &0 \\
   4 &5
  \end{pmatrix}\cdot
  \begin{pmatrix}
    c_1\\c_2
  \end{pmatrix}=
  \begin{pmatrix}
    0 \\ 3
  \end{pmatrix},
\end{equation*}
with solution $(c_1,c_2)=(0,3)$. Hence, we take $u_3=\pi_2^{-3}$, and choose $w_3$ as the smallest place that satisfies \eqref{eq:example_cond_v} and moreover that $X^6-u_3\pi_3$ is irreducible over $\F_{w_3}$.

We continue in the same fashion to compute the remaining places $v_4,w_4,v_5,w_5,v_6,w_6$ as well as the corresponding primitive roots and uniformisers (for the $v_i$). The data is collected in Tables \ref{tab:v} and \ref{tab:w}.

This gives us everything that we need to define the auxiliary morphism
\begin{equation*}
  \rho_S' : \widehat{\OO}_S^\times\to G'= (\F_{v_1}^\times\times\F_{w_1}^\times\times\cdots\times\F_{v_6}^\times\times\F_{w_6}^\times)\otimes\Z/6\Z
\end{equation*}
and the standard basis $b_1\otimes 1, b_1'\otimes 1,\ldots,b_6\otimes 1,b_6'\otimes 1$. The change of basis from the standard basis to the $e_i,e_i'$ satisfying \eqref{eq:basis_condition} is described by the matrix \eqref{eq:change_of_basis}, which takes in our case the form

\setcounter{MaxMatrixCols}{20}
\begin{equation*}
  A=\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 3 & 0 & 0 & 0 & 3 & 0 & 1 \\
0 & 5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 \\
0 & 4 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 4 & 0 & 5 & 0 & 3 & 0 & 1 & 0 & 4 & 0 & 0 \\
0 & 1 & 0 & 2 & 1 & 0 & 0 & 0 & 0 & 4 & 0 & 2 \\
0 & 0 & 0 & 2 & 0 & 1 & 0 & 2 & 0 & 2 & 0 & 5 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 4 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 5 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 4 & 0 & 2 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 4 & 0 & 5 & 0 & 4 \\
0 & 3 & 0 & 4 & 0 & 4 & 0 & 0 & 0 & 4 & 1 & 0 \\
0 & 4 & 0 & 1 & 0 & 5 & 0 & 5 & 0 & 0 & 0 & 3
  \end{pmatrix}.
\end{equation*}
An epimorphism $\Psi:G'\to(\Z/6\Z)^4$ as in \eqref{eq:def_Psi} is given, with respect to the basis $e_i,e_i'$ of $G'$ and the standard basis $f_1,f_2,f_3,f_3$ of $(\Z/6\Z)^4$ by the matrix
\begin{equation*}
B=\begin{pmatrix}
  1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1
\end{pmatrix}.
\end{equation*}
The epimorphism $\Phinew:(\Z/6\Z)^4\to G$ just reduces the last three coordinates modulo $3$, so it is represented with respect to the $f_i$ and the standard  generators $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, $(0,0,0,1)$ of $G$ by the identity matrix. In total, our map $\Phiprimenew=\Phinew\circ\Psi:G'\to G$ is given, with respect to the standard basis of $G'$ and the standard generators of $G$, by the matrix
\begin{equation*}
  R=BA^{-1}=\begin{pmatrix}
1 & 5 & 1 & 4 & 1 & 5 & 0 & 4 & 0 & 1 & 0 & 2 \\
0 & 2 & 0 & 2 & 0 & 2 & 1 & 0 & 1 & 2 & 0 & 2 \\
0 & 1 & 0 & 4 & 0 & 3 & 0 & 2 & 0 & 4 & 1 & 5 \\
0 & 4 & 0 & 4 & 0 & 1 & 0 & 4 & 0 & 1 & 0 & 0
            \end{pmatrix}.
          \end{equation*}
          We have now fully specified the characteristic morphism $\rho_S=\Phiprimenew\circ\rho_S':\widehat{\OO}_S^\times\to G$ that induces our epimorphism $\rho:I_K^\times/K^\times\to G$ giving the extension $L_\rho=L$.

          Let us find concrete polynomials defining $L$. To this end, we note that $L$ is the compositum of four subextensions $L_1,L_2,L_3,L_4$ corresponding to the projections $\Pi_j$ from $G$ to its cyclic factors $H_j=\Z/6\Z$ (for $j=1$) and $H_j=\Z/3\Z$ (for $j=2,3,4$).

          The extension $L_j$ is thus induced by the characteristic morphism $\rho_{S,j} = \Phiprimenew_j\circ\rho_S$, where the map $\Phiprimenew_j=\Pi_j\circ\Phiprimenew$ is represented by the $j$-th row $R_j$ of the matrix $R$. We can read off the places of $K$ ramified in $L_j$ as those $v_i,w_i$ whose entry in the $j$-th row of $R$ is not equal to $0$ (modulo $3$ for $j=2,3,4$). Hence, the conductor ideal of $L_j$ is $\mathfrak{f}_j$, with
          \begin{align*}
            \mathfrak{f}_1 &= v_1w_1v_2w_2v_3w_3w_4w_5w_6, \\
            \mathfrak{f}_2 &= w_1w_2w_3v_4v_5w_5w_6 ,      \\
            \mathfrak{f}_3 &= w_1w_2w_4w_5v_6w_6,   \\
            \mathfrak{f}_4 &= w_1w_2w_3w_4w_5.
          \end{align*}
          All places in $S$ split completely in all of the $L_j$. For a place $v\notin S$, we compute a uniformiser $\pi_v\in\OO_{S\cup\{v\}}^\times$ as in \S \ref{sec:finding_vi} and the vector $l_v=(l_{1,v},l_{1,v}',\ldots,l_{6,v},l_{6,v}')$, with $l_{i,v}$ (or $l_{i,v}'$) the discrete logarithm of $\pi_v^{-1}$ in $\F_{v_i}^\times$ (or $\F_{w_i}^\times$) with respect to $b_i$ (or $b_i'$), taken modulo $6$. Then $l_v$ consists of the coordinates of $\rho_S'(\pi_v)$ in the standard basis of $G'$, so $\rho_{S,j}(\pi_v)=0$ if and only if $R_jl_v=0$ (modulo $3$ for $j=2,3,4$). These $v$ are exactly the places not in $S$ that split completely in $L_j$.

In each of the extensions $L_j$, after an exhaustive search for places satisfying these conditions, we present the first few places of $K$ that split completely: for $L_1$, these are the prime ideals
\begin{align*}
  &(7, (3+\sqrt{-47})/2),
 (7,  (11+\sqrt{-47})/2),
 (53, (71+\sqrt{-47})/2),
 (59, (37+\sqrt{-47})/2),\\
 &(61, (21+\sqrt{-47})/2),
 (67),
 (97, (167+\sqrt{-47})/2),
 (103, (57+\sqrt{-47})/2),\\
 &(131, (79+\sqrt{-47})/2),
 (149, (129+\sqrt{-47})/2),
 (157, (253+\sqrt{-47})/2).
\end{align*}
For $L_2$, we get the prime ideals
\begin{align*}
  &(2, (1+\sqrt{-47})/2),
 (5),
 (7, (3+\sqrt{-47})/2),
 (11),
 (41),
 (43),\\
 &(53, (71+\sqrt{-47})/2),
 (59, (37+\sqrt{-47})/2),
 (61, (21+\sqrt{-47})/2),\\
 &(61, (101+\sqrt{-47})/2),
 (67),
 (73),
 (79, (43+\sqrt{-47})/2).
\end{align*}
For $L_3$, the first few splitting places are
\begin{align*}
 &(2, (-1+\sqrt{-47})/2),
 (2, (1+\sqrt{-47})/2),
 (3, (1+\sqrt{-47})/2),\\
 &(7, (3+\sqrt{-47})/2),
 (17, (19+\sqrt{-47})/2),
 (29),
 (43),
 (53, (35+\sqrt{-47})/2),\\
 &(61, (21+\sqrt{-47})/2),
 (71, (109+\sqrt{-47})/2),
 (73),
 (79, (43+\sqrt{-47})/2).
\end{align*}
Finally, the first few splitting primes for $L_4$ are
\begin{align*}
 &(2, (-1+\sqrt{-47})/2),
 (2, (1+\sqrt{-47})/2),
 (5),
 (7, (3+\sqrt{-47})/2),\\
 &(17, (19+\sqrt{-47})/2),
 (19),
 (23),
 (29),
 (37, (45+\sqrt{-47})/2).
\end{align*}
In each case, the presented list of splitting primes is barely enough to generate the correct congruence subgroup of the ray class group modulo $\mathfrak{f}_j$. From this information, Magma obtains the following polynomials defining the $L_j$:

The field $L_1$ is generated by roots of the polynomials
\begin{align*}
  &X^2 + 1625554186831677234132\sqrt{-47} + 1811860086730297979035
\end{align*}
and
\begin{align*}
  X^3 - &3748167037906625162481707496423982188817923353276910868587 X\\
  + & 20932667006810986711572641003097069272613523\\
    &9089132839261455735058433435300040785407916\\
  - &4892889242489320858461993956906597337627420\\
    &962014982904896214316984337417374211250173\sqrt{-47}.
\end{align*}
The field $L_2$ is generated by a root of
\begin{align*}
  X^3 - &514252500054134587252291124378250069242533908387 X\\
     +\frac{1}{2} \Big(- &298646938917972463681140245645712648\\
        &301029855802019621285424376904094635\\
  + &1662861756262325040866728524283288871452\\
    &494124982673546015261665625225\sqrt{-47} \Big).
\end{align*}
For $L_3$, we get the generating polynomial
\begin{align*}
  X^3 - &913736091749824689643505040786706198418283 X\\
      - & 233073217500727285419128281222253866235433000542376262114304064\\
      + & 22955375656018471299613470304327792587793595929519410170152870\sqrt{-47},
\end{align*}
and for $L_4$ the polynomial
\begin{align*}
  X^3 - &4720036166349902544210196434745323 X\\
  - &142867906955019411285258390809644752015601994134080\\
  - &6327448763258896081623032545540797957509248167898\sqrt{-47}.
\end{align*}
Hence, the compositum $L$ is generated over $K$ by roots of all of these polynomials.

\end{document}

config: arxiv


TITLE: $a^{n^2}x^n$ which convergence test?
QUESTION [0 upvotes]: I've been trying to test $a^{n^2}x^n$ for convergence with the ratio, root, leibniz and direct comparison tests but don't seem to get loose of the $n$. any hints?

REPLY [0 votes]: I think you can use the root test here.
$$r=\lim\sup_{n\rightarrow \infty}\sqrt[n]{|a^{n^2}x^n|}=\lim\sup_{n\rightarrow \infty}{|a|^n |x|}=|x|\lim\sup_{n\rightarrow \infty}{|a|^n}$$
Therefore by the root test the series is convergent, divergent and inconclusive (may converge or diverge) when, $x<\frac{1}{\lim\sup_{n\rightarrow \infty}{|a|^n}},\, x>\frac{1}{\lim\sup_{n\rightarrow \infty}{|a|^n}}$ and $x=\frac{1}{\lim\sup_{n\rightarrow \infty}{|a|^n}}$ respectively.


TITLE: How do I calculate how long water will take to warm from 22°C to 35°C?
QUESTION [5 upvotes]: I'm a microbiologist and my physics is not up to par for figuring this out. I've requested our engineering department to perform a temperature mapping of the vessel, but they can't do it for me.  I have a cylinder shaped stainless steel tank. I don't know it's thickness, but estimate it to be 1/2 cm to 1 cm if that is important. It will be filled with $1600$ L of fermentation fluid (I'd guess water or beer would be similar fluid to use). I estimate the tank to have a holding volume of $2000$ L ($2\times 10^6$ cm$^3$). The "water" will be at ~$22°$C and will sit in an incubator (air heating) set to $35°$C.  I need an estimate on how long it will take these fluids to reach at least $34°$C in the air bath incubator.
I tried to figure this out on my own and ran across the formula $A/t=KA(t_h-t_c)/d$.  But solving it is a different ball game. I don't know how to figure out what each letter means to solve this puzzle.
Are there other variables or equations I need to figure this out?

REPLY [1 votes]: Best case all the energy from the incubator goes into the water. Divide the energy needed to raise the temperature of the water your target amount by the incubator's power requirement/power draw (in Watts) for a best-case estimate.
Water's heat capacity is $4.18\ J/g°C$ and since its density is $1000\ g/L$ then the heat capacity is $4180\ J/L°C$ or 4180 Joules per [Liter * temperature difference in Celcius]. You have 1600 L of water that you want to raise 12 degrees Celcius so:
$$4180\ J/LK\ \times 1600\ L\ \times 12\ °C\ =\ 80\ MJ$$
As a worked example, I found this manual for the Thermo Scientific Precision Model 815 Refrigerated Incubator - Microprocessor Controlled. It has a capacity of 566 L:
$$4180\ J/LK\ \times 566\ L\ \times 12\ K\ =\ 28\ MJ$$
and max energy usage of  600 W:
$$\frac{28\ MJ}{600\ W} =\ 13\ hrs$$


TITLE: Are there any slick way of computing the matrix of rotating about a given line passing origin by $\pi$?
QUESTION [2 upvotes]: Let's just consider the lines in $\mathbb{R}^3$. For any line $l$, suppose it is determined by $(x_1:x_2:x_3)$. For any point $(a,b,c)$, after the rotation it is mapped to(as we can compute) $(a',b',c')=(2\lambda x_1-a,2\lambda x_2-b,2\lambda x_3-c)$, where $\lambda=\frac{ax_1+bx_2+cx_3}{x_1^2+x_2^2+x_3^2}$. I computed by using the condition that their middle point is on the line $(x_1:x_2:x_3)$, and $\overline{xx'}$ is orthogonal to the line $l$.
And we can of course compute the matrix of this transformation. But I have trouble checking it indeed satisfies the condition for rotation matrix $\det(A)=1, A^T=A^{-1}$.
I am wondering are their any more algebraic way of finding the matrix? Also, I wish someone could verify whether my answer is correct.
Thanks!

REPLY [2 votes]: Consider a vector $v=(a,b,c)$ and associate to it the skew matrix
$A$ given by\begin{pmatrix}
0 & -c & b \\
c & 0 & -a \\
-b & a &0
\end{pmatrix}
Then $R=e^A$ is rotation about the direction determined by $v$ and angle determined by $\theta=\lVert v\rVert$. Because the powers of $A$ can be computed in terms of $A$, you can find a simple expression for $R$. See here.


TITLE: How to show that $n$th prime is less than $2^n$?
QUESTION [3 upvotes]: I know that there are better lower bounds for number of primes less than n but I don't know how to solve this problem using fundamental proof. I tried using sieve of Eratosthenes and using induction on n to solve the problem but those techniques didn't work. Can someone give me a hint?

REPLY [2 votes]: The inductive step needs Bertrand's postulate. Every number theorist should read its proof once.
Alternatively, if one gets an upper bound on the relative error in $p_n\sim n\ln n$ (a consequence of the prime number theorem), one can use $n\ln n\in o(2^n)$ to prove the result once a suitable base case is checked.