text:
\chapter{Preliminaries}

\noindent
In this chapter we summarize basic facts about valued fields, especially Hahn fields, and valued differential fields, as needed
in later chapters. We try to be self contained as to the various algebraic notions involved, and thus we include also some preliminaries on valuation theory and differential algebra.
Almost all of this can be found in more detail and with proofs in \cite[Chapters 2, 3, 4]{ADH}.

\medskip\noindent
We let $m,n$ range over $\mathbb{N}=\{0,1,2,\ldots\}$, the set of natural numbers.  Rings will have a multiplicative identity $1$. We identify $\mathbb{Z}$ in the usual (unique) way with a subring of any ring of characteristic zero. We let $K^\times$ denote the multiplicative group of a field $K$. For a field $K$ of characteristic zero, we identify $\Q$ with a subfield of $K$ in the usual way.

\section{Ordered Sets and Groups}

\subsection*{Ordered sets}
An \textbf{ordered set} $S$ is a set equipped with a distinguished {\em total\/} ordering $\leq$.
If $S$ is an ordered set with ordering $\leq$ we let $\geq, <,$ and $>$ have the usual meaning.
Subsets of $S$ will be considered as ordered sets with the ordering induced by $\leq$.

\medskip \noindent
For an ordered set $S$ and a subset $B$ of $S$ we set $S^{>B}:= \{s\in S: s>b \text{ for all } b \in B\}$, and $S^{>s}:= S^{>\{s\}}$ for an element $s\in S$. Similarly we define $S^{<B}, S^{\leq B}, S^{\geq B}$.
We set $S_\infty := S\cup \{\infty\}$ and extend the ordering of $S$ to an ordering on $S_\infty$ by $s< \infty$ for every $s\in S$.

\medskip\noindent
Let $S$ be an ordered set. Then $S$ is  said to be \textbf{well-ordered} if every non-empty subset of $S$ has a minimum element with respect to $\leq$.
A subset $B$ of $S$ is said to be \textbf{convex in } $S$ if for all $a,b,c\in S$ with $a<b<c$ and $a,c\in B$ we have $b\in B$. The \textbf{convex hull} of $B\subseteq S$ is the smallest subset of $S$, under inclusion, that is convex and contains $B$.

\subsection*{Ordered abelian groups} Let $\Gamma$ be an additively written \textbf{ordered abelian group}, that is, an abelian group with an ordering such that for all $\alpha,\beta,\gamma\in \Gamma$
\[\alpha<\beta\ \Rightarrow\ \alpha+\gamma<\beta+\gamma\]

\medskip\noindent
If $\Gamma$ is an ordered abelian group we write $\Gamma^>$ instead of $\Gamma^{>0}$; similarly with $\Gamma^{<}, \Gamma^{\leq}, \Gamma^{\geq}$. Additionally we set $\Gamma^{\neq} = \Gamma \setminus \{0\}$
Given $\gamma \in \Gamma$ we set $|\gamma| = \begin{cases} \gamma \text{ if } \gamma \geq 0\\ -\gamma \text{ if } \gamma<0. \end{cases}$\\
Ordered abelian groups will be considered as $\Z$-modules in the usual way.

\medskip\noindent
We consider the archimedean equivalence relation on an ordered abelian group $\Gamma$ where the archimedean class of an element $\gamma_0 \in \Gamma$ is given by
\[ [\gamma_0] \ : =\ \{\gamma \in \Gamma \ :\ |\gamma_0| \leq n|\gamma| \text{ and } |\gamma|\leq n|\gamma_0| \text{ for some } n\ >\ 0\}.\]
We will consider the set of archimedean classes as an ordered set where
\[[\gamma]\ \leq\ [\delta] \text{ iff } |\gamma|\ \leq\ n |\delta| \text{ for some } n  .\]
Thus $[0]=\{0\}$, and $[0] \leq [\gamma]$ for all $\gamma \in \Gamma$, with equality iff $\gamma = 0$; moreover
\[[\gamma]\ <\ [\delta] \text{ iff } n|\gamma|\ <\  |\delta| \text{ for all } n  .\]
The \textbf{rank} of $\Gamma$ is the number of archimedean classes of $\Gamma^{\neq}$;  it equals the number of nontrivial convex subgroups of $\Gamma$ if the rank is finite. An ordered abelian group of rank 1 is also said to be \textbf{archimedean}.

\begin{lemma}\label{fGfN1} If $\Delta, \Theta \subseteq \Gamma$ are well-ordered, then $\Delta +\Theta$ is well-ordered, and for every $\gamma\in \Gamma$ there are only finitely many $(\delta,\theta)\in \Delta\times\Theta$ with $\delta+\theta=\gamma$.
\end{lemma}

\begin{lemma}\label{fGstar1}
Let $G\subseteq \Gamma^>$ be a well ordered set.
Then
\[G^*\ :=\ \sum G\ :=\ \{g_1+\cdots +g_n\ :\ g_1,\ldots,g_n\in G\}\]
is well ordered and for every $\gamma\in G^*$ there are finitely many tuples $(n,g_1,\ldots,g_n)$, with $g_1,\ldots, g_n$ elements in $G$ such that $\sum_{i=1}^n g_i = \gamma$.
\end{lemma}

\subsection*{Valuations on abelian groups}\label{VAG}
A valuation on an (additively written) abelian group $\Gamma$ is a surjective map $v:\Gamma \rightarrow S_\infty$, where $S$ is a linearly ordered set, such that for $\alpha, \beta \in \Gamma$
\begin{itemize}
\item[(VG1)] $v(\alpha)= \infty \iff \alpha =0$,
\item[(VG2)]$v(\alpha) = v(-\alpha)$,
\item[(VG3)] $v(\alpha + \beta) \geq \min\{v(\alpha),v(\beta)\}$.
\end{itemize}

\noindent
A \textbf{valued abelian group} is a triple $(\Gamma, S, v)$ where $\Gamma$ is an ordered abelian group, $S$ is an ordered set and $v:\Gamma \rightarrow S_\infty$. It is easy to show that if $v(\alpha)<v(\beta)$ then  $v(\alpha + \beta) = v(\alpha)$; indeed
\[v(\alpha)\ =\ v(\alpha+\beta-\beta)\ \geq\ \min(v(\alpha+\beta),v(\beta))\ \geq\ v(\alpha).\]
We note that given any element $s \in S$, the subsets
\[B(s):=\{\gamma\ :\ v(\gamma) > s \} \text{ and } \overline{B}(s):=\{\gamma:v(\gamma) \geq s\}\] of $\Gamma$ form subgroups with the group operation inherited from $\Gamma$. We denote their quotient by $G(s):= \overline{B}(s)/B(s)$.

\subsection*{Hahn products}

Let $(G_s)_s\in S$ be a family of nontrivial ordered abelian groups indexed by a totally ordered set $S$. Consider the product group $\prod_{s\in S} G_s$ whose elements are the sequences $g=(g_s)_{s\in S}$ such that $g_s\in G_s$ for each $s\in S$. For such $g$, we define the support of $g$ as the set $\{s\in S: g_s \neq 0\}$.

\medskip\noindent
We define the \textbf{Hahn product} of the family to be the subgroup $H[(G_s)_s]$ of $\prod_{s\in S} G_s$ with well ordered support. We can construe $G:= H[(G_s)_s]$ as an ordered valued abelian group, by equipping it with the \textbf{Hahn valuation} $v:G\rightarrow S_\infty$ given by $v(g) = \min(\supp(g))$ for $g\in G\neq$ and $v(0) = \infty$, and the order given by $0<g$ iff $0<g_{s_0}$ for $s_0 = v(g)$.

\section{Hahn Series}\label{HS}

\noindent
By a {\bf monomial set\/} we mean an ordered set $\fM$ whose
elements are thought of as monomials and whose ordering is
denoted by $\preceq$ (or $\preceq_{\fM}$ if we need to
indicate the dependence on $\fM$); the corresponding strict ordering, reverse ordering, and strict reverse ordering are then
denoted by $\prec$, $\succeq$, and $\succ$, respectively. A subset of the monomial set $\fM$ is said to be {\bf well-based\/} if it is well-ordered with respect to the  reverse ordering $\succeq$, equivalently, there is no infinite sequence $\fm_0 \prec \fm_1 \prec \fm_2 \prec \dots$ in the subset.

 Let $\fM$ be a monomial set, let $\fm$ and $\fn$ range over $\fM$, and let $C$ be an (additively written) abelian group. By a
{\bf Hahn series with coefficients in $C$ and monomials in $\fM$\/}
 we mean a function $f: \fM \to C$ whose support
 \[\supp f\ :=\ \{\fm\in \fM:\ f(\fm)\ne 0\}\]
 is well-based as a subset of $\fM$.
We often write such $f$ as a series $\sum_{\fm} f_{\fm}\fm$,
where $f_{\fm}:=f(\fm)$.
We define the {\bf Hahn space\/} $C[[\fM]]$ to be the (additive) abelian group whose elements are
the Hahn series with coefficients in $C$ and monomials in $\fM$,
with addition as suggested by the series notation.

Let $(f_i)_{i\in I}\in C[[\fM]]^I$. We say that $(f_i)$
is \textbf{summable} (or $\sum_i f_i$ exists) if
\begin{itemize}
	\item $\bigcup_i \supp(f_i)$ is well-based;
    \item for each $\fm\in \fM$ there are only finitely many $i$ with $\fm\in \supp(f_i)$.
\end{itemize}
If $(f_i)_{i\in I}$ is summable, we define its sum $f=\sum_{i\in I} f_i\in  C[[\fM]]$ by $f_{\fm}=\sum_{i}f_{i,\fm}$. If $I$ is finite, then $(f_i)$ is summable and $\sum_i f_i$ is the usual finite sum.
If $(f_i)$ and $(g_i)\in C[[\fM]]^I$ are summable, then so is $(f_i+g_i)$, and $\sum_i f_i+g_i=\sum_i f_i + \sum_i g_i$. If $I$ and
$J$ are disjoint sets and $(f_i)\in C[[\fM]]^I$ and $(f_j)\in C[[\fM]]^J$ are summable, then $(f_k)_{k\in I\cup J}$ is summable
and $\sum_{k\in I\cup J}f_k=\sum_{i\in I} f_i + \sum_{j\in J} f_j$.
Finally, if the
family $(f_{i,j})_{(i,j)\in I\times J}$ of elements $f_{i,j}\in C[[\fM]]$ is summable, then
$\sum_j f_{i,j}$ exists for every $i$, and $\sum_i(\sum_j f_{i,j})$ exists, and
likewise $\sum_j(\sum_i f_{i,j})$ exists, and
\[\sum_{i,j} f_{i,j}\ =\ \sum_i\big(\sum_j f_{i,j}\big)\ =\ \sum_j\big(\sum_i f_{i,j}\big).\]

\medskip\noindent
It is convenient to augment $\fM$ by an element $0\notin \fM$ and
extend $\preceq$ to an ordering of $\fM\cup \{0\}$, also denoted by $\preceq$, by declaring
$0\preceq \fm$ (and thus $0\prec \fm$) for all $\fm$.
For $f\in C[[\fM]]$ we define its \textbf{leading monomial} $\fd(f)\in \fM\cup\{0\}$ by
\[\fd(f):= \max \supp f\ \text{ if }f\ne 0, \qquad
\fd(f)=0\ \text{ if }f=0.\]
This allows us to introduce a transitive and reflexive binary relation $\preceq$ on $C[[\fM]]$ by $f\preceq g$ if and only if $\fd(f)\preceq \fd(g)$; replacing in this equivalence $\preceq$ by $\prec$, $\succeq$, and $\succ$, respectively, we introduce likewise transitive
binary relations $\prec$, $\succeq$, and $\succ$ on $C[[\fM]]$.
We also define the equivalence relation $\asymp$ on $C[[\fM]]$ by
$f\asymp g:\Leftrightarrow\ \fd(f)=\fd(g)$.

\subsection*{Hahn rings} By a {\bf monomial group\/} we mean a multiplicative ordered abelian group whose elements are
thought of as monomials. Let $\fM$ be a monomial group; we use the same notations as for monomial sets; in particular, the
ordering is denoted by $\preceq$, and we let $\fm$ and $\fn$
range over $\fM$.
We recall the following basic fact about well-based subsets of $\fM$:

\begin{lemma}\label{fGfN} If $\fG, \fH\subseteq \fM$ are well-based, then $\fG\fH$ is well-based, and for every $\fm$ there are only finitely many $(\fg,\fh)\in \fG\times\fH$ with $\fg\fh=\fm$.
\end{lemma}

\noindent
Next, let $C$ be
a commutative ring with $1\ne 0$. Lemma~\ref{fGfN} allows us to define
for $f,g\in C[[\fM]]$ their product $fg\in C[[\fM]]$ by
$(fg)_{\fh}:=\sum_{\fm\fn=\fh}f_{\fm}g_{\fn}$, so
\[\supp fg\ \subseteq\ (\supp f)(\supp g).\] With this product operation $C[[\fM]]$ is a commutative ring with subrings $C[[\fM^{\preceq 1}]]$ and $C[[\fM^{\succeq 1}]]$. We have the ring embedding of $C$ into $C[[\fM]]$ sending $c\in C$ to $f\in C[[\fM]]$ with $f_1=c$ and $f_{\fm}=0$
for $\fm\ne 1$; we identify $C$ with a subring
of $C[[\fM]]$ via this embedding. We also have the group embedding of
$\fM$ into the
multiplicative group of units of $C[[\fM]]$ sending $\fm$ to
$f\in C[[\fM]]$ with $f_{\fm}=1$, and $f_{\fn}=0$ for
$\fn\ne \fm$; we identify $\fM$ with a subgroup of this multiplicative group of units via this embedding, and identify the element $0$ of $\fM\cup \{0\}$ with the zero element of the ring $C[[\fM]]$. This has the effect that restricting the binary relation
$\preceq$ on $C[[\fM]]$ to $\fM\cup \{0\}$ gives back the originally given ordering $\preceq$ on $\fM\cup\{0\}$, and
likewise with $\prec$, $\succeq$, and $\succ$. Note that $C[[\fM]]^{\prec 1}=C[[\fM^{\prec 1}]]$.

Here is another key fact about well-based subsets of $\fM$:

\begin{lemma}\label{fGstar} Let $\fG$ be a well-based subset of $\fM^{\prec 1}$. Then:
\begin{enumerate}
	\item[\rm{(i)}] $\fG^*:=\bigcup_n \fG^n$ is well-based;
	\item[\rm{(ii)}] for every $\fm$ there are only finitely many
tuples $(n,\fg_1,\dots, \fg_n)$ such that $\fg_1,\dots, \fg_n\in \fG$ and $\fm=\fg_1\cdots \fg_n$.
\end{enumerate}
\end{lemma}

\noindent
For $\varepsilon\in C[[\fM]]^{\prec 1}$ the family $(\varepsilon^n)$ is summable and $(1-\varepsilon)\sum_n\varepsilon^n=1$, by Lemma~\ref{fGstar}. More generally,
let $t=(t_1,\dots, t_n)$ be a tuple of distinct variables and let
\[F\ =\ F(t)\ =\ \sum_{\nu}c_{\nu}t^{\nu}\in C[[t]]\ :=\ C[[t_1,\dots, t_n]]\]
be a formal power series over $C$; here the sum ranges over all multiindices $\nu=(\nu_1,\dots, \nu_n)\in \N^n$, and
$c_{\nu}\in C$, $t^{\nu}:= t_1^{\nu_1}\cdots t_n^{\nu_n}$. For any tuple $\varepsilon=(\varepsilon_1,\dots, \varepsilon_n)$ of
elements of $C[[\fM]]^{\prec 1}$ the family $(c_{\nu}\epsilon^{\nu})$ is summable, where $\varepsilon^{\nu}:= \varepsilon_1^{\nu_1}\cdots \varepsilon_n^{\nu_n}$. Put
\[F(\varepsilon)\ :=\ \ \sum_{\nu}c_{\nu}\varepsilon^{\nu}\in C[[\fM]]^{\preceq 1}\ =\ C[[\fM^{\preceq 1}]].\]
Fixing $\varepsilon$ and varying $F$ we obtain a $C$-algebra morphism
\[F \mapsto F(\varepsilon)\ :\ C[[t]] \to C[[\fM]].\]
Assume that $C$ contains $\mathbb{Q}$ as a subring.
Then we have the formal power series
\[ \exp(t)\ :=\ \sum_{i=0}^\infty t^i/i!\in \mathbb{Q}[[t]], \qquad \log (1+t)\ :=\ \sum_{j=1}^\infty (-1)^{j-1}t^j/j\in \mathbb{Q}[[t]]\]
with $\exp(t_1+t_2)=\exp(t_1)\exp(t_2)$ and $\log(1+t_1+t_2+t_1t_2) =\log(1+t_1)+\log(1+t_2)$
in  $\mathbb{Q}[[t_1,t_2]]\subseteq C[[t_1,t_2]]$. Also
$\log\big(\exp(t)\big) =t$ and
$\exp\big(\log(1+t))= 1+t$
in $\mathbb{Q}[[t]]\subseteq C[[t]]$. Substituting elements of $C[[\fM]]^{\prec 1}$ in these identities yields that
\[\delta\ \mapsto\ \exp(\delta)\ =\ \sum_{i=0}^\infty \delta^i/i!\ :\ C[[\fM]]^{\prec 1} \to 1+C[[\fM]]^{\prec 1},\]
is an isomorphism of the additive subgroup $C[[\fM]]^{\prec 1}$ of $C[[\fM]]$
onto the multiplicative subgroup $1+C[[\fM]]^{\prec 1}$ of $C[[\fM]]^\times$, with inverse
\[1+\varepsilon\ \mapsto\ \log (1+\varepsilon)\ =\ \sum_{j=1}^\infty (-1)^{j-1}\varepsilon^j/j\ :\ 1+C[[\fM]]^{\prec 1}\to C[[\fM]]^{\prec 1}.\]

\subsection*{Hahn fields} We now consider a coefficient field
$\smallk$ instead of just a coefficient ring $C$ as before.
 For
nonzero $f\in \smallk[[\fM]]$ we have $f=a\fd(f)(1-\varepsilon)$ with $a\in \smallk^\times$ and $\varepsilon \prec 1$, so $f$ has a multiplicative inverse in $\smallk[[\fM]]$, namely $a^{-1}\fd(f)^{-1}\sum_n \varepsilon^n$. In particular,
$\smallk[[\fM]]$ is again a field with $\smallk$ as a subfield and
$\fM$ as a subgroup of its multiplicative group
$\smallk[[\fM]]^\times$.

Suppose $\smallk$ is an ordered field. Then we construe
$\smallk[[\fM]]$ as an ordered field extension of $\smallk$ by requiring that for nonzero $f\in \smallk[[\fM]]$ with $\fm=\fd(f)$ we have $f>0 \Leftrightarrow f_{\fm}>0$;
in this role as ordered field we call $\smallk[[\fM]]$ an \textbf{ordered Hahn field} (over $\smallk$).

\medskip\noindent
Often we prefer
additive notation. To explain this, let
$\Gamma$ be an additively written ordered abelian group (with zero element $0_\Gamma$ if we need to indicate the dependence on $\Gamma$); then $\le$ rather than $\preceq$
denotes the ordering of $\Gamma$.  When $\Gamma$ is clear from the context, we let $\alpha, \beta, \gamma$ range over $\Gamma$.
We now have the monomial group
\[\fM\ =\ t^\Gamma\]
where $t$ is just a symbol, and
$\gamma\mapsto t^\gamma: \Gamma\to t^\Gamma=\fM$ is an {\bf order-reversing\/} group
isomorphism of $\Gamma$ onto $\fM$. With this $\fM$ we denote $\smallk[[\fM]]$
also by $\smallk((t^\Gamma))$, and write
$f=\sum_{\fm} f_{\fm}\fm\in \smallk[[\fM]]$ as
$f=\sum_\gamma f_{\gamma}t^\gamma$, with $f_\gamma:= f_{\fm}$ for $\fm=t^\gamma$.
In this situation we prefer to take $\supp f$ as a subset of $\Gamma$
rather than of $\fM=t^\Gamma$: $\supp f =\{\gamma:\ f_{\gamma}\ne 0\}$, and the
well-based requirement turns into the requirement that
$\supp f$ is a well-ordered subset of $\Gamma$. The order type of the support of $f$ is by definition the unique ordinal isomorphic to $\supp(f)$ with respect to the ordering induced by $\Gamma$, and will be denoted by $o(f)$. Of course, all this is only
a matter of notation, and we shall freely apply results for
Hahn fields $\smallk((t^\Gamma))$ to Hahn fields $\smallk[[\fM]]$, since we
can pretend that any monomial group $\fM$ is of the form
$t^\Gamma$ for a $\Gamma$ as above.  Thus for $f$ in a Hahn field $\smallk[[\fM]]$, the ordinal $o(f)$ is isomorphic to $\supp(f)\subseteq \fM$ with respect
to the ordering induced by the {\em reverse\/} ordering on $\fM$.

\section{Valued Fields}

 \subsection*{Valuation rings and valued fields}
 A \textbf{valuation ring} is an integral domain in which the set of ideals is totally ordered by inclusion. Here are a few characterizations of valuation rings:

\begin{lemma} Let $\bigO$ be an integral domain. The following are equivalent:
\begin{enumerate}
\item[\rm{(i)}] The ring $\bigO$ is a valuation ring.
\item[\rm{(ii)}] For every element $x\neq 0$ in the field of fractions $K$ of $\bigO$, either $x$ or $x\inv$ is in $\bigO$.
\item[\rm{(iii)}] For all $a,b \in \bigO$, either $b\in a\bigO$ or $a\in b\bigO$.
\end{enumerate}
\end{lemma}

\medskip\noindent The set of non-units of a  valuation ring is an ideal. In fact, it is the unique maximal ideal of the valuation ring. Thus any valuation ring $\bigO$ is a local ring.

\medskip\noindent A valued field is a field $K$ equipped with a valuation ring $\bigO$ such that $K$ is the field of fractions of $\bigO$.

The following are examples of valued fields:
\begin{itemize}
\item Any field $K$ with $K$ itself as the valuation ring. In this case we say that the valuation of $K$ is trivial.
\item The field of Laurent series in $x\inv$ over $\mathbb{C}$ with valuation ring
\[\bigO\ :=\ \left\{f=\sum_{n\in \mathbb{N}}f_n x^{-n}\right\}\]
\item In fact, any Hahn field $\smallk((t^\Gamma))$ equipped with
\[\bigO\ :=\ \{f\in \smallk((t^\Gamma)) :\ \supp(f) \subseteq \Gamma^{\geq}\}\].
\item Any Hausdorff field $K$ (a field of germs at $+\infty$ of continuous real functions) together with
\[\bigO\ :=\ \left\{f\in K:\ f \text{ is eventually bounded}\right\}\]

\item The field $\Q$, equipped with the valuation ring
\[\bigO\ :=\ \left\{\frac{a}{b}:\ a,b\in \Z, \text{ and } b\notin p\Z\right\}\]
where $p$ is a fixed prime number.
\end{itemize}

\subsection*{Residue fields} Given a local ring $R$ with maximal ideal $\smallo$ we define the \textbf{residue field of} $R$ to be the quotient $\smallk = \bigO/\smallo$. We call the quotient map $\bigO \rightarrow \smallk$ the residue map. For a valued field $K$, we define the \textbf{residue field of} $K$ to be the residue field of its valuation ring. The residue fields corresponding to the valued fields mentioned above are naturally isomorphic to $K$, $\C$, $\smallk$, $\R$, and $\Z/p\Z$ respectively.

\medskip\noindent It is sometimes useful to distinguish valued fields by the characteristic of both itself and its residue field. We say that $K$ is of \textbf{equicharacteristic zero}, \textbf{equicharacteristic} $p$, or of \textbf{mixed characteristic} if $(\cha(K),\cha(\smallk))$ is $(0,0)$, $(p,p)$, or $(0,p)$, respectively, where $\smallk$ is the residue field of $K$ and $p$ is a prime number.   In this dissertation we are mainly concerned with the equicharacteristic zero case, especially when there is also a derivation in play.

\medskip\noindent We say that the residue field $\smallk$ of $K$ lifts to $K$ if there is a field embedding $\iota:\smallk\rightarrow K$. All the previously mentioned residue fields lift to their respective valued fields except for $\Z/p\Z$ to $\mathbb{Q}$. In fact there is no hope of lifting the residue field in the mixed characteristic case.

\subsection*{Dominance relations}
A \textbf{dominance relation} on a field $K$ is a binary relation $\preceq$ on $K$ such that for all $f,g,h\in K$:
\begin{itemize}
	\item[(DR1)] $1\not\preceq 0$,
    \item[(DR2)] $f\preceq f$,
    \item[(DR3)] $f\preceq g$,  $g\preceq h$ $\Rightarrow$ $f\preceq h$,
    \item[(DR4)] $f\preceq g$ or $g\preceq f$,
    \item[(DR5)] $f\preceq g$ $\Rightarrow $ $hf\preceq hg$,
    \item[(DR6)] $f\preceq h$, $g\preceq h$ $\Rightarrow$ $f+g\preceq h$.
\end{itemize}

\medskip\noindent
Let $\preceq$ be a dominance relation on the field $K$.  We define some further (asymptotic) binary  relations
$\succeq, \prec, \succ, \asymp, \sim$ on $K$ in terms of $\preceq$ as follows:
\begin{align*} f \succeq g &:\iff g\preceq f; \qquad f \prec g :\iff f\preceq g \text{ and }g\not\preceq f; \qquad f\succ g :\iff g\prec f;\\
 f\asymp g &:\iff f\preceq g \text { and }g\preceq f; \qquad f\sim g :\iff f-g\prec f.
 \end{align*}
We also say that $f$ \textbf{dominates} $g$ if $f\succeq g$, that $f$ \textbf{strictly dominates} $g$ if $f\succ g$, and that $f$ is \textbf{asymptotic} to $g$ if $f\asymp g$. The relation $\asymp$ is an equivalence relation on $K$; reflexivity and symmetry are straightforward from the definition and transitivity follows from DR3 and symmetry.
Alternatively we note that by DR2 and DR3 we have that $\preceq$ is a preorder on $K$. If $f\sim g$, then $f,g\ne 0$. It is easy to check that
$\sim$ is an equivalence relation on $K^\times$.

\subsection*{Valuations}
A \textbf{valuation} on a field $K$ is a map $v:K^\times \rightarrow \Gamma$ onto an ordered abelian group $\Gamma$ such that for all $x,y\in K^\times$,
\begin{itemize}
	\item[(VF1)] $v(x+y)\geq \min\{v(x),v(y)\}$ for $x+y\neq 0$, and
    \item[(VF2)] $v(xy) = v(x) + v(y)$.
\end{itemize}

\noindent
As for valued abelian groups it follows that if $v(x)< v(y)$, then $v(x+y) = v(x)$. We extend $+$ and $<$ to a binary operation $+$ on $\Gamma_\infty= \Gamma \cup \{\infty\}$ and a (total) ordering $<$ on $\Gamma_\infty$ so that for all $\gamma\in \Gamma$
\begin{itemize}
	\item $\infty > \gamma$ ,
    \item $\infty + \gamma = \gamma + \infty =  \infty$, and
    \item $\infty + \infty = \infty$,
\end{itemize}
an extend $v$ to $v:K\rightarrow \Gamma_\infty$ by $v(0)=\infty$ all of $K$ by setting $v(0)= \infty$ so that $v$ is a valuation on the additive group of $K$ as defined on Section \ref{VAG}.

\begin{proposition}\label{propeqval} The following are equivalent for a subring $\bigO$ of a field $K$:
\begin{enumerate}
	\item[\rm{(i)}] $K$ equipped with $\bigO$ is a valued field;
    \item[\rm{(ii)}] the equivalence $f\preceq g :\iff f\in \bigO g$ defines a dominance relation $\preceq$ on $K$;
    \item[\rm{(iii)}] there is a valuation $v:K\rightarrow \Gamma_{\infty}$ on $K$ such that $ \bigO=\{f\in K:\ v(f)\ge 0\}$.
\end{enumerate}
\end{proposition}
\begin{proof}
Assume (i) and define the binary relation $\preceq$ by:  $f\preceq g: \iff f\in \bigO g$. Then DR1-DR3, DR5, DR6 follow from $\bigO$ being a ring with $1\ne 0$, and DR4 holds since for $f\neq 0\neq g$, either $f/g \in \bigO$ or $g/f\in \bigO$, so $\preceq$ is a dominance relation.
Assume (ii), and  let $v:K^\times \rightarrow \Gamma:=K^\times\hspace{-1 mm}/\hspace{-1 mm}\asymp$ be the map sending $f\in K^\times$ to its equivalence class
$f_{\asymp}$, and consider the quotient group $\Gamma$ of the multiplicative group $K^\times$ as an additive group. (This is just a notational convention.)
Equip $\Gamma$ with the reverse order: $f_{\asymp}\leq g_{\asymp}$ iff $g\preceq f$. It is easy to check  then that $v$ is a valuation on the field $K$, with $\bigO=\{f\in K:\ v(f)\ge 0\}$.  Assuming $v$ is as in (iii) we have $v(f)\ge 0$ or $v(f^{-1})\ge 0$ for all $f\in K^\times$, so $\bigO$ is a valuation ring of $K$.
\end{proof}

\noindent
Let $K$ be a valued field.  By Proposition~\ref{propeqval}
and its proof we have a valuation $v: K^\times\to \Gamma$ such that $\bigO=\{f\in K:\ v(f)\ge 0\}$, and such that if $v^*: K^\times\to \Gamma^*$ is also a valuation with
$\bigO=\{f\in K:\ v^*(f)\ge 0\}$, then $v^*=\phi\circ v$ for a unique ordered group isomorphism $\phi: \Gamma\to \Gamma^*$. We equip $K$ with
such a valuation $v$; by the fact just mentioned, all ways of doing this are equivalent. The dominance relation $\preceq$
on $K$ defined in (ii) of Proposition~\ref{propeqval} is then given in terms of $v$ by: $f\preceq g \iff v(f)\ge v(g)$; likewise: $f\prec g \iff v(f)> v(g)$, $f\asymp g \iff v(f)=v(g) $, and $f\sim g \iff  v(f-g)>v(f)$.
Given $S\subseteq K$ and $f\in K$ we set $S^{\preceq f}:=\{g\in K:\ g\preceq f\}$, and similarly with $S^{ \prec f}$, and the like.

We make $K$ into a {\em topological field\/} by giving it the {\bf valuation topology}: a neighborhood basis of $0\in K$ is given by the sets $\{f\in K: v(f)>\gamma\}$ with $\gamma\in \Gamma$.

\medskip\noindent We always consider the Hahn field $\smallk((t^\Gamma))$ to be equipped with the  valuation
\[v\ :=\ v_\Gamma:\smallk((t^\Gamma))^\times\rightarrow \Gamma, \qquad v(f)\ :=\ \min \supp f.\] The dominance relation $\preceq$ associated to this valuation
then agrees with the relation $\preceq$ that we defined earlier on this Hahn field in terms of  leading monomials. Likewise for $\prec, \succ$, and $\asymp$.

A Hahn field $\smallk[[\fM]]$ is accordingly considered equipped with a valuation $v: \smallk[[\fM]]^\times \to \Gamma_{\fM}$
where the restriction of $v$ to the ordered multiplicative group $\fM\subseteq \smallk[[\fM]]^\times$ is an order-reversing group isomorphism onto the ordered (additive) group $\Gamma_{\fM}$.

A \textbf{monomial group of} a valued field $K$ is a multiplicative subgroup $\fM$ of $K^\times$ that is bijectively mapped onto the value group
$\Gamma$ by $v$. Thus a Hahn field $\smallk[[\fM]]$ (with the usual valuation) has $\fM$ as a monomial group.

\begin{example} The same valued field can have many monomial groups: For example,
$K = \R[[x^\Z]]$ has the set $x^\Z$ as a monomial group, but  has also for every $r\in \R^\times$ the set $\{r^kx^k:\ k\in \Z\}$ as a monomial group.
\end{example}

\subsection*{Valued field extensions} Let $K$ and $L$ be valued fields. We say that $L$ is \textbf{valued field extension of} $K$  (or $K$ is a \textbf{valued subfield of} $L$) if $L$ is a field extension such that $\bigO_L\cap K = \bigO$. (When two valued fields $K$ and $L$ are in play, it is understood that $\bigO$ is the valuation ring of $K$ and $\bigO_L$ is the valuation ring of $L$.)
Let $L$ be a valued field extension of $K$. Then we have a natural field embedding $\smallk=\bigO/\smallo\to \smallk_L=\bigO_L/\smallo_L$ of the residue fields, and we identify
$\smallk$ with a subfield of $\smallk_L$ via this embedding. Likewise, we have a natural ordered group embedding $\Gamma \to \Gamma_L$ of the value groups, and we identify
$\Gamma$ with an ordered subgroup of $\Gamma_L$ via this embedding. This extension is said to be \textbf{immediate} if $\smallk=\smallk_L$ and $\Gamma=\Gamma_L$.
A valued field $K$ is \textbf{maximal} if it has no proper immediate valued field extension. A valued field is \textbf{algebraically maximal} if it has no proper immediate algebraic valued field extension. Hahn fields $\smallk[[\fM]$ are maximal, and every valued field has a maximal immediate extension.

\subsection*{Henselian fields}
A valued field $K$ is said to be \textbf{henselian} if for every polynomial $P\in \bigO[X]$ and every $a\in \bigO$ with $P(a)\prec 1$ and $(dP/dX) (a) \asymp 1$ there exists $b\in \bigO$ such that $b-a\prec 1$ and $P(b) = 0$. Hahn fields are henselian. Here are some useful characterizations of henselianity:

\begin{lemma}Let $K$ be a valued field. The following are equivalent:
\begin{enumerate}
	\item[\rm{(i)}] $K$ is henselian.
    \item[\rm{(ii)}] For every $P(X) = 1 + X + c_2X^2 +\cdots + c_nX^n \in K[X]$ with $c_2,\ldots, c_n \prec 1$, there is $x\in \bigO$ such that $P(x) = 0$.
    \item[\rm{(iii)}] For every $Q(Y) = Y^n + Y^{n-1} + c_2Y^{n-2} + \cdots + c_n\in K[Y]$ with $c_2, \ldots, c_n\prec 1$, there is $y\in \bigO$ such that $Q(y) = 0$.
    \item[\rm{(iv)}] For every $P(X) \in \bigO[X]$ and $a\in \bigO$ with $P(a) \prec (dP/dX)(a)^2$, there is $b\in \bigO$ such that $a-b\prec (dP/dX)(a)$ and $P(b)=0$.
\end{enumerate}
\end{lemma}

\begin{lemma} If a valued field is algebraically maximal, then it is henselian.
The converse holds in the equicharacteristic zero case.
\end{lemma}

\noindent
A \textbf{henselization} of a valued field $K$ is a valued field extension of $K$ that embeds uniquely over $K$ into every henselian valued field extension of $K$.

\begin{lemma}
Every valued field $K$ has a henselization. Such a henselization is unique up to unique isomorphism over $K$, and is an immediate
algebraic valued field extension of $K$.
\end{lemma}

\noindent
Let $E$ be a subfield of a Hahn field $\smallk[[\fM]]$. We consider $E$ as a valued subfield of $\smallk[[\fM]]$. Since
$\smallk[[\fM]]$ is henselian, there is a unique embedding
$E^h\to \smallk[[\fM]]$ over $E$, and the image of this embedding is called the \textbf{henselization of $E$ inside $\smallk[[\fM]]$}.

\section{Differential Rings}
\noindent
Let $R$ be a commutative ring.  A \textbf{derivation} on
$R$ is an additive map  $\der: R \to R$
such that for all $a,b\in R$ we have  $\der(ab) = \der(a)b + a\der(b)$ (Leibniz rule).

A \textbf{differential ring} is a ring together with a derivation on it.

Let $R$ be a differential ring. Unless we specify otherwise, we let
$\der$ be the derivation of $R$, and for $a\in R$ we let $a'$ denote
$\der(a)$. The
\textbf{ ring of constants of $R$} is the subring
\[C_R\ :=\ \{a\in R:\ a'=0\}\]
of $R$. A \textbf{differential ring extension of} $R$ is a differential ring $R^*$ that has $R$ as a subring and whose derivation extends the derivation on $R$; in that case we also call $R$ a \textbf{differential subring of} $R^*$.
Let $R^*$ be a differential ring extension of $R$.  If $A$ is a subset of $R^*$ and $a'\in R[A]$ for all $a\in A$, then the subring $R[A]$ of $R^*$ is closed under the derivation of $R^*$, and is accordingly viewed as a differential subring of $R^*$. Given $a\in R^*$, we let $R\{a\}:= R[a,a',a'',\ldots]$ denote the smallest differential
subring of $R^*$ that contains $R$ and $a$, and we call $R\{a\}$  the \textbf{differential ring generated by} $a$ \textbf{over} $R$.

\medskip\noindent
The differential ring $R\{Y\}$ of differential
polynomials in the indeterminate $Y$ over $R$ is defined as follows: as a ring, this is just the
polynomial ring
\[R[Y, Y', Y'',\dots]\ =\ R[Y^{(n)}:n=0,1,2,\dots]\]
in the distinct
indeterminates $Y^{(n)}$ over $R$, and it is made into a differential ring extension of $R$ by requiring that $(Y^{(n)})'= Y^{(n+1)}$ for all $n$.  If $R$ is a domain, then $R\{Y\}$ is also a domain.

A \textbf{differential field} is a differential ring whose underlying ring is a field of characteristic zero; then the ring of constants is also a field.
Given a differential field denoted by $K$ we generally denote its field of constants by $C$, while if a differential field is denoted by another letter, say $L$, we let $C_L$ be its field of constants. If $R$ is a differential ring that is also an integral domain
of characteristic zero, then its ring of fractions is made into a differential field by setting
\[\der\left(\frac{a}{b}\right)\ =\ \frac{\der(a)b - a\der(b)}{b^2}\ \text{ for } a,b\in R,\ b\ne 0.  \]
Let $K$ be a differential field. Then for $f\in K^\times$ we let
 $f^\dagger$ denote the logarithmic derivative $\der(f)/f$, so
$(fg)^\dagger=f^\dagger + g^\dagger$ for $f,g\in K^\times$.
For a subset $S$ of $K^\times$ we
set $S^\dagger:=\{f^\dagger:\ f\in S\}$. Also $K^\dagger:=(K^\times)^\dagger$, an additive subgroup of $K$.
A \textbf{differential field extension of $K$} is a differential ring extension $L$ of $K$ that is also a field.
Let $L$ be a differential field extension of $K$ and $a\in L$.
We let $K\<a\>$ be the field of fractions of $K\{a\}$ inside $L$ and we call it the \textbf{differential field generated by $a$ over $K$}. We say that $a$ is \textbf{differentially-algebraic over $K$}, abbreviated $\d$-algebraic over $K$, if $P(a) = 0$ for some nonzero differential polynomial $P(Y)$ over $K$. If $a$ is not $\d$-algebraic over $K$ we say that $a$ is \textbf{differentially transcendental}, or $\d$-transcendental, over $K$. The field extension $L$ is \textbf{$\d$-algebraic over $K$} if every element in $L$ is $\d$-algebraic over $K$. Let $E\subseteq L$. We let $K\<E\>$ be the differential field generated by $E$ over $K$, that is, the smallest differential subfield of $L$ containing $K$ and $E$. We say that $E$ is \textbf{$\d$-algebraically-independent over $K$} if  $a$ is $\d$-transcendental over $K\<E\setminus\{a\}\>$
for all $a\in E$. The subset $E$ of $L$ is a \textbf{$\d$-transcendence basis of $L$ over $K$} if $E$ is $\d$-algebraically independent over $K$ and $L$ is $\d$-algebraic over $K\<E\>$. All $\d$-transcendence bases of $L$ over $K$ have the same cardinality, and we call this cardinality the \textbf{differential-transcendence degree}, or $\d$-transcendence degree, of $L$ over $K$.

\section{Valued Differential Fields}

\noindent
A \textbf{valued differential field} is a differential field $K$ equipped with a valuation ring $\bigO \supseteq \Q$ of $K$. In this section we fix a valued differential field $K$; as usual we denote its valuation ring by $\bigO$ and its residue field $\bigO/\smallo$ by $\smallk$. The next result is part of \cite[Lemma 4.4.2]{ADH}.

\begin{lemma} Assume that $\der(\smallo) \subseteq \smallo$. Then $\der(\bigO)\subseteq \bigO$.
\end{lemma}
\begin{proof}
Let $x\in \bigO$ and suppose that $x'\notin \bigO$. Then $1/x'\in \smallo$, so $x/x'\in \smallo$, $(1/x')'\in \smallo$, hence $(x/x')'\in \smallo$ . But $(x/x')' = x(1/x')' + 1\in 1+\smallo$, a contradiction.
\end{proof}
\noindent
If $\der(\smallo)\subseteq \smallo$, we equip $\smallk$ with the derivation  $a+\smallo \mapsto \der(a) + \smallo:\bigO/\smallo\rightarrow \bigO/\smallo$. The condition
 $\der(\smallo) \subseteq \smallo$ means that derivatives of ``small'' elements (elements of $\smallo$) are small, so we say that $\der$ is {\bf small} (or that $K$ has \textbf{small derivation}) if this condition is satisfied.
If $K$ has small derivation, then we refer to $\smallk$ with the induced derivation as the \textbf{differential residue field} of $K$.

\medskip\noindent
\subsection*{Continuous derivations} We now equip $K$ with the valuation topology. Even though it is not part of the definition of ``valued differential field'',  the
only case of interest to us is when the derivation is continuous. If the derivation on $K$ is continuous, then for every $P\in K\{Y\}$ the function $y\mapsto P(y): K\rightarrow K$ is continuous.

If $\der$ is small, then $\der$ is continuous. In fact, continuous derivations are just  ``small derivations in disguise'' as shown by the next  result  (\cite[Lemma 4.4.7]{ADH}):

\begin{lemma}
The following conditions on $K$ are equivalent:
\begin{enumerate}
\item[\rm{(i)}] $\der:K\rightarrow K$ is continuous;
\item[\rm{(ii)}] for some $a\in K^\times$ we have $\der\smallo \subseteq a\smallo$;
\item[\rm{(iii)}] for some $a\in K^\times$ the derivation $\derdelta:=  a^{-1}\der$ is small.
\end{enumerate}
\end{lemma}

\subsection*{Asymptotic fields and differential-valued fields} The material from this subsection can be found in more detail in \cite[Section 9.1]{ADH}. Let $K$ be a valued differential field. We say that $K$ is \textbf{asymptotic} if for all $f,g\in K^\times$ with $f,g\prec 1$,
\[f\prec g\  \Longleftrightarrow\  f'\prec g'.\]
The derivation of an asymptotic field is continuous (with respect to the valuation topology). We say that $K$ is a \textbf{differential-valued field}, or just \textbf{$\d$-valued}, if $K$ is asymptotic and $\bigO = C +\smallo$. If $K$ is $\d$-valued, then its constant field is naturally isomorphic to its residue field via the residue map. For the next lemma, see for example \cite[Proposition 9.1.3]{ADH}.

\begin{lemma}
The following are equivalent:
\begin{enumerate}
\item[\rm{(i)}] $K$ is $\d$-valued;
\item[\rm{(ii)}] $\bigO = C +\smallo$ and for all $f,g\in K^\times$, $f,g\prec 1\Rightarrow f^\dagger\succ g'$.
\end{enumerate}
\end{lemma}

\subsection*{H-fields}  For more details about $H$-fields, see \cite[Chapter 10]{ADH}. An $H$-field is by definition an ordered differential field $K$ such that:
\begin{enumerate}
	\item[\rm{(i)}] for all $f\in K$, if $f>C$, then $f'>0$;
    \item[\rm{(ii)}] $\bigO=C+\smallo$, where $\bigO = \{g\in K\ :\ |g|\leq c\text{ for some } c\in C\}$ and $\smallo$ is the maximal ideal of the convex subring $\bigO$ of $K$.
\end{enumerate}

\noindent
An $H$-field $K$ is considered also as a valued differential field by taking the above $\bigO$ as the valuation ring. This makes $H$-fields into differential-valued fields.
If $K$ is an $H$-field with $C\ne K$, then its valuation topology coincides with its order topology.
Hardy fields containing $\R$ as a subfield are $H$-fields.
The $H$ in ``$H$-field'' is in honor of the pioneers Borel, Hahn, Hardy, and Hausdorff, three of which share the initial $H$.

An $H$-field $K$ is said to be \textbf{Liouville closed} if it is a real closed $H$-field and for all $a\in K$ there exist $y,z\in K^\times$ such that $y'=a$ and $z^\dagger = a$.
An $H$-field $K$ is Liouville closed if and only if $K$ is real closed and every equation  $y'+ay=b$ with $a,b\in K$ has a nonzero solution in $K$.

config: arxiv


TITLE: Constructing the asymptotes of a hyperbola by compass and straightedge.
QUESTION [2 upvotes]: Is it possible to construct the asymptotes of a hyperbola from the points on the hyperbola by compass and straightedge alone? And if so, how to construct them?
I have no idea how to approach the first question. It seems it should be possible as it is similar to constructing a tangent to an ellipse, but I haven't been able to adjust such a construction to work for the asymptotes of a hyperbola. An illustration or a reference would be welcome.

REPLY [2 votes]: See in https://fr.answers.yahoo.com/question/index?qid=20130810183100AArTMlg Warning, not the first solution but the second one, by Pope, which constructs the hyperbola's centre first, which looks compulsory. For fully understanding it, you need to have seen a minimum of properties of what is called a diameter in a conic curve, whose definition I recall:
When a line is moved paralelly to itsef, i.e., keeping the same direction (D), the midpoints of its points of intersection with the conic section belong to a same line (D') ; if the conic section possesses a centre (this is the case here, or for an ellipse, but not for a parabola) the centre belongs to this diameter. The line with direction (D) passing through the centre is called the conjugate diameter of (D').


TITLE: Cokernel in abelian category is epic?
QUESTION [0 upvotes]: Prove that cokernel in abelian Category is epic.
$\mathbf{My\  attempt}$
Let $A,B \in \mathcal{C}$ and $f:A \rightarrow $ be a morphism, and and $g:B \rightarrow C$ such that $gf=0$ for some $B \in \mathcal{C}.$ Let $h_1,h_2: C \rightarrow D$ such that $h_1 g=h_2 g$. We know that $h_1gf=h_1(0)=h_2gf=h_2(0),$ but this is not enough to show $g_1=g_2$. I could not go forward from this, Thanks for any help.

REPLY [2 votes]: The definition of the cokernel of $f:A \to B$ is the following : a morphism $g : B \to C$ such that for any $g^\prime : B \to D$ with $g^\prime \circ  f=0$, there exists a $\textbf{unique}$ $\phi : C \to D$ such that $g^\prime= \phi \circ g$.
So for any $h:C \to D$, if $h \circ g=0$ then letting $g^\prime=h \circ g: B \to D$, we have  $g ^\prime \circ f=0$. Since $g$ is the cokernel of $f$, there is a unique $\phi : C \to D$ such that $\phi \circ g=g^\prime$. But $\phi=0$ and $\phi=h$ are possible solutions, so they are equal, that is $h=0$. We show that way that $g$ is an epimorphism.


text:
\section{Editing Hardness Results}
\label{section:negative_results}

\ifdefined\isBDeEhard
    \subsection{Degeneracy: $o(\log (n/\Tr) )$-Inapproximability of Vertex and Edge Editing}
    \label{section:negative_bounded_degeneracy_editing}
    \input{./negative_results/bounded_degeneracy_editing.tex}
\fi

\ifdefined\iswcolhard
    \subsection{Weak Coloring Numbers: $o(\Tt)$-Inapproximability of Editing}
    \label{section:negative_weak_coloring}
    \input{./negative_results/weak_coloring.tex}
\fi

\ifdefined\istwhard
    \subsection{Treewidth and Clique Number: $o(\log n)$-Inapproximability of Vertex Editing}
    \label{section:negative_treewidth_and_clique}
    \input{./negative_results/treewidth_and_clique.tex}
\fi

\ifdefined\isBDVhard
    \subsection{Bounded Degree: $(\ln \Td - C\cdot\ln \ln \Td)$-Inapproximability of Vertex Editing}
    \label{section:negative_bounded_degree}
    \input{./negative_results/bounded_degree.tex}
\fi

\ifdefined\isstarforesthard
    \subsection{Treedepth 2: $(2-\epsilon)$-Inapproximability of Vertex Editing }
    \label{section:negative_starforest_v}
    \input{./negative_results/starforest_vertex.tex}

    \subsection{Treedepth 2: APX-hardness of Edge Editing}
    \label{section:negative_starforest_e}
    \input{./negative_results/starforest_edge.tex}

\fi


TITLE: About the maximal condition on subgroups and normality
QUESTION [1 upvotes]: Let $G$ be a group which satisfies the maximal condition on subgroups. Now, let $H$ be a subgroup containing a non trivial conjugate (i.e. there exists an element $x\in G,\,\notin H $ such that $H^x\leq H$). Is it true that, in this case, $H$ is normal in $G$? How can we prove that? Is solvability of any help?
Note
A group $G$ is said to satisfy the maximal condition on subgroups iff it is impossibile to find an infinite strictly ascendant chain of subgroups.

REPLY [4 votes]: Observe that any finite nilpotent group $\;G\;$ both fulfills the maximal condition for subgroups and also for any proper subgroup $\;H\;$ we have that $\;H\lneqq N_G(H)\;$ which means there exists $\;x\in G\setminus H\;\;s.t.\;\;H^x\le H\;$ ...and this doesn't make the subgroup $\;H\;$ normal.


TITLE: Integrate the radial term of Laplace's eqn in cylindrical coords
QUESTION [0 upvotes]: I'm reading a paper in plasma physics and I can't understand how they go from this:
$$
\frac{\delta}{\delta r}(r \frac{\delta \phi}{\delta r})=0
$$
to
$$
\phi = A+B\ln (r)
$$
where $A$ and $B$ are constants.
The way I was looking at it, you integrate both sides, the right side stays $0$, and then you divide by $r$, the right side stays $0$. And so you get that the derivative of $\phi$ is $0$, meaning it is a constant. I know I'm missing something here.

REPLY [0 votes]: Consider the equation
\begin{align}
\frac{d}{dr} \left( r \ \frac{d f}{dr} \right) = 0.
\end{align}
Integrating both sides leads to
\begin{align}
\frac{d f}{dr} = \frac{a}{r},
\end{align}
where $a$ is a constant. Now integration again leads to
\begin{align}
f(r) = a \ln(r) + b
\end{align}
where $b$ is also a constant. If the condition is that $f << \infty$ as $r \rightarrow \infty$ then it is required that $a = 0$.


TITLE: Prove that $A^{(\omega)}\nleq_T A^{(n)}$
QUESTION [2 upvotes]: I am trying to solve Exercise 7.1.24 (i) of Computability Theory by Rebecca Weber. $A^{(n)}$ denotes the $n$-th Turing jump and $A^{(\omega)}=\{\langle x,n\rangle: x\in A^{(n)}\}$ the $\omega$-jump.
I am able to prove that $A^{(n)}\leq_T A^{(\omega)}$. Since $x\in A^{(n)}\Leftrightarrow \langle x,n\rangle\in A^{(\omega)}$ we get that $\chi_{A^{(n)}}(x)=\chi_{A^{(\omega)}}(x,n)$, which is computable. Hence, $A^{(n)}$ is $A^{(\omega)}$-computable, i.e. $A^{(n)}\leq_T A^{(\omega)}$.
I now want to show that $A^{(\omega)}\nleq_T A^{(n)}$. My idea was to use the Jumping Theorem which says that $A^{(n+1)}\nleq_T A^{(n)}$ but I couldn't find a way to do so.

REPLY [2 votes]: This can also be done from first principles, but the easiest way is to use what you've already shown. Since you've shown that $A^{(n)}\leq_TA^{(\omega)}$ for arbitrary $n$, you also know that $A^{(n+1)}\leq_TA^{(\omega)}$. Now, if it were the case that $A^{(\omega)}\leq_TA^{(n)}$, then by transitivity of $\leq_T$ you would have $A^{(n+1)}\leq_TA^{(n)}$, which you already know to be false.