text:
\section{Introduction}
It is shown in \cite[Theorem 3.1]{CKS} that an infinite field of positive characteristic that is definable
in an NTP$_2$ theory has only finitely many Artin-Schreier extensions. NTP$_2$ is a large class of first order
theories (``without the tree property of the second kind'') defined by S.~Shelah generalizing simple and NIP
theories. Algebraic examples of NTP$_2$ structures are given by ultraproducts of $p$-adic fields and certain
valued difference fields. For the precise definition (which we will not need in this paper) and for further
examples, we refer the reader to \cite{CKS}.

In \cite[Proposition~3.2]{CKS} it is shown that the value group of a valued field of characteristic $p>0$
which has only finitely many Artin-Schreier extensions is $p$-divisible. In this note, assuming throughout that
the valuation is nontrivial, we prove a stronger result,
namely, that such a field is dense in its perfect hull. For a valued field of characteristic $p>0$,
this is equivalent to being a deeply ramified field in the sense of \cite{GR}. The density also
implies that the value group is $p$-divisible and the residue field is perfect. By
Theorem~\ref{charst} in Section~\ref{sectdr},
every deeply ramified field of positive characteristic is a semitame field, which we shall define now.

Take a valued field $(K,v)$ of arbitrary characteristic. Its value group will be denoted by $vK$, and
its residue field by $Kv$. Accordingly, the value of an element $a\in K$ will be denoted by $va$, and its residue
by $av$. We say that $(K,v)$ is \bfind{semitame} if either $\chara Kv=0$, or $\chara Kv=p>0$ and the following two
axioms hold:
\sn
{\bf (DRst)} the value group $vK$ is $p$-divisible,
\sn
{\bf (DRvr)} the homomorphism
\[
\cO_{K^c}/p\cO_{K^c} \ni x\mapsto x^p\in \cO_{K^c}/p\cO_{K^c}
\]
is surjective, where $\cO_{K^c}$ denotes the valuation ring of the completion $K^c$ of $(K,v)$.
Note that this condition implies that the residue field $Kv$ is perfect.

\pars
The following result will be proven in Section~\ref{sectcharppf}:
\begin{theorem}                  \label{MTfinAS}
Take a valued field $(K,v)$ of characteristic $p>0$. If $K$ admits only finitely many distinct Artin-Schreier
extensions, then $(K,v)$ is a semitame field.
\end{theorem}

\begin{corollary}
A nontrivially valued field of positive characteristic that is definable in an NTP$_2$ theory is a semitame field.
\end{corollary}

We will prove Theorem~\ref{MTfinAS} by showing that if $(K,v)$ is not dense in its perfect hull, then it admits
infinitely many distinct Artin-Schreier extensions; see Proposition~\ref{propnotdense}.
However, we will show more than this. We are interested in Galois \bfind{defect extensions} of degree $p$ a prime.
These are immediate Galois extensions $(L|K,v)$ of degree $p$ of valued fields for which $v$ has a unique
extension from $K$ to $L$ (in this case we must have that $p=\chara Kv$). Here, $(L|K,v)$
denotes an extension $L|K$ of fields with $v$ a valuation on $L$ and $K$ endowed with its restriction; the
extension is said to be \bfind{immediate} if the canonical embeddings of $vK$ in $vL$ and of $Kv$ in $Lv$ are
onto. For more details on the defect see Section~\ref{sectfi}, and for further background, see
\cite{BKdrf,Ku1,Ku6}.

In \cite{BKdrf} a classification of Galois defect extensions $\cE=(L|K,v)$ of prime degree $p$ is given as follows.
We show that the set
\[
\Sigma_\sigma\>:=\> \left\{ v\left( \left.\frac{\sigma f-f}{f}\right) \right| \, f\in L^{\times} \right\}
\]
is independent of the choice of a generator $\sigma$ of $\Gal (L|K)$, and we
denote it by $\Sigma_\cE\,$. We say that $\cE$ has \bfind{independent defect} if
\[
\Sigma_{\cE}\>=\> \{\alpha\in vK\mid \alpha >H_\cE\}
\]
for some proper convex subgroup $H_\cE$ of $vK$;  otherwise we say that $\cE$ has \bfind{dependent defect}. If
$vK$ is archimedean ordered, or in other words, $(K,v)$ is of \bfind{rank $1$}, then $H_\cE$ can only be equal
to $\{0\}$.
\pars
In Section~\ref{sectcharppf}, we will prove:
\begin{theorem}                                \label{MTdepAS}
Take a valued field $(K,v)$ of characteristic $p>0$. If it admits an Artin-Schreier extension with dependent
defect, then it admits infinitely many distinct Artin-Schreier extensions with dependent defect.
\end{theorem}

In \cite{Ku6} the classification was originally introduced only for valued fields of positive characteristic.
An Artin-Schreier defect extension $(L|K,v)$ was said to have dependent defect if in a certain way it
depends on immediate purely inseparable extensions of degree $p$ which do not lie in the completion of $(K,v)$.
It was then shown in Section~4.2 of \cite{Ku6} that an Artin-Schreier extension has dependent defect if it is
obtained from such a purely inseparable defect extension by a certain transformation of an inseparable minimal
polynomial which makes it separable. We will describe this transformation in Section~\ref{secttransf} and use it
for the proof of Theorem~\ref{MTdepAS}.

In \cite{BKdrf} we have made the above more precise by showing that {\it every} Artin-Schreier extension with
dependent defect can be obtained by such a transformation. It is also shown that the new definition of the
classification given in \cite{BKdrf} is compatible with the one given in \cite{Ku6}.

\pars
The new definition of the classification became necessary in order to generalize the original definition to the
case of valued fields $(K,v)$ of characteristic $0$ with residue field $Kv$ of positive characteristic $p$
(\bfind{mixed characteristic}), where we cannot rely on nontrivial purely inseparable extensions. We will now
present a partial analogue of Theorem~\ref{MTdepAS} in the mixed characteristic case. To
avoid a number of technical details in the present paper, we will restrict our scope to valued fields $(K,v)$
of rank $1$. We will also assume that $K$ contains a primitive $p$-th root of
unity. We then consider Kummer defect extensions of degree $p$, that is, Galois defect extensions of $(K,v)$
generated by some $\eta\notin K$ such that $\eta^p\in K$. Because such extensions are immediate,
we can show that we can assume $\eta$ to be a \bfind{$1$-unit}, i.e., $v(\eta-1)>0$ (and thus, $v\eta=0$); see
Section~\ref{sectK}.

We need some more preparation. Take an extension $(L|K,v)$ and $a\in L$. We define:
\[
v(a-K)\>:=\>\{v(a-c)\mid c\in K\}\>;
\]
for details on this set, see Section~\ref{sectva-K}.
For $\alpha$ in the divisible hull $\widetilde{vK}$ of the value group $vK$, we will write $v(a-K)<\alpha$
if $v(a-c)<\alpha$ for all $c\in K$, and similarly for ``$\leq$'' in place of ``$<$''. We write
\[
v(a-K)<\!\!|\,\alpha
\]
if $v(a-K)<\alpha$ and $v(a-K)$ is bounded away from $\alpha$ in  $\widetilde{vK}$, i.e.,
there is $\beta\in \widetilde{vK}$ such that $v(a-K)\leq\beta<\alpha$.

A Kummer defect extension of degree $p$ generated by a $1$-unit $\eta$ as above always satisfies
\begin{equation}                             \label{Kde}
v(\eta-K)\><\>\frac{vp}{p-1}\>,
\end{equation}
see Section~\ref{sectK}. In the case of archimedean value groups (where $H_\cE$ cn only be equal to $\{0\}$), the
definition we have given for the extension to have dependent defect is equivalent to the condition that
$v(\eta-K)<\!\!|\, \frac{vp}{p-1}$ (see our discussion at the end of Section~\ref{sectK}).

Note that in \cite{BKdrf} in place of $v(\eta-K)$ a more flexible invariant $\dist(\eta,K)$ is used for our work
with defect extensions. We will give its definition in Section~\ref{sectdist} to enable the reader to relate the results in \cite{BKdrf} and \cite{Ku6} to those in the present paper.

\pars
In view of Theorem~\ref{MTdepAS} it seems reasonable to conjecture that if $(K,v)$ admits a Kummer extension
with dependent defect, then it admits infinitely many distinct Kummer extensions with dependent defect. By
\cite[part 2) of Theorem 1.5]{BKdrf}, this would imply that if $(K,v)$ admits only finitely many Kummer defect
extensions, then it is semitame, in perfect analogy to the equal positive characteristic case. However, so far we
have only been able to prove a weaker result. We will say that a Kummer defect extension as above has
\bfind{super-dependent defect} if
\begin{equation}                                   \label{sd}
v(\eta-K)<\!\!|\, \frac{vp}{p}\>.
\end{equation}
\begin{theorem}                                      \label{MTmixed}
Take a valued field $(K,v)$ of mixed characteristic and rank 1 which contains a primitive $p$-th root of
unity. If it admits a Kummer extension of degree $p$ with super-dependent defect, then it admits infinitely
many distinct Kummer extensions of degree $p$ with super-dependent defect.
\end{theorem}
\n
This leads us to the following
\n
{\bf Open questions:} Does the above theorem also hold with dependent defect in place of super-dependent defect?
What is (possibly) special about a Kummer extension with dependent defect that is not super-dependent?
\sn
We have observed in earlier work already that the threshold $\frac{vp}{p}$ plays a certain role when dealing
with 1-units in mixed characteristic (see
\cite[Corollary 2.11 d)]{Ku8}), but it is not yet sufficiently understood what exactly this role is.

\parm
In the last section of this paper we extend our study of Artin-Schreier extensions of valued fields in order to
fill a gap that occurred in the proof of Lemma 2.31 of \cite{Ku6}, which gives a criterion for such extensions to
have nontrivial defect.

\parm
For general background on valuation theory, we refer the reader to \cite{[En],[EP],[R],[ZS2]}.
\section{Preliminaries}                              \label{sectprel}

\subsection{Deeply ramified and semitame fields}                              \label{sectdr}
For a field $K$ of characteristic $p>0$ we denote by $K^{1/p^\infty}$ the perfect hull
of $K$. Further, we set $K^p=\{a^p\mid a\in K\}$ and $K^{1/p}=\{a^{1/p}\mid a\in K\}$; then $K^p$ is a subfield
of $K$, and $K^{1/p}|K$ is a field extension which is trivial if and only if $K$ is perfect.

\pars
In \cite{BKdrf}, the following is proven:
\begin{theorem}                          \label{charst}
Take a nontrivially valued field $(K,v)$ of characteristic $p>0$. Then the following statements are equivalent:
\n
a) $(K,v)$ is a semitame field,
\n
b) $(K,v)$ is a deeply ramified field,
\n
c) $(K,v)$ satisfies (DRvr),
\n
d) the completion of $(K,v)$ is perfect,
\n
e) $(K,v)$ is dense in its perfect hull, i.e., $K^{1/p^\infty}\subset K^c$,
\n
f) $K^p$ is dense in $(K,v)$.
\end{theorem}
Here, the property ``dense'' refers to the topology induced by the valuation.
\pars
See \cite{BKdrf} for more details and the connection of semitame and deeply ramified fields with the
classification of defect extensions.

\subsection{The set $v(a-K)$}                              \label{sectva-K}
Take a totally ordered set $(T,<)$. For a nonempty subset $S$ of $T$ and an element $t\in T$ we will write $S<t$ if
$s<t$ for every $s\in S$. A set $S\subseteq T$ is called an \bfind{initial segment} of $T$ if for each
$s\in S$ every $t <s$ also lies in $S$. Similarly, $S\subseteq T$ is called
a \bfind{final segment} of $T$ if for each $s\in S$ every $t >s$ also lies in $S$.

If $(T,<)$ is an ordered abelian group, $n\in\N$, $\alpha\in T$, and $S$ is an initial segment of $T$, then
\[
nS+\alpha\>:=\>\{n\beta+\alpha\mid \beta\in S\}
\]
is an initial segment of $nT$.

\pars
Take a valued field extension $(K(a)|K,v)$.  We will now collect various properties of the sets $v(a-K)$ and
$v(a-K)\cap vK$.
\begin{proposition}                          \label{propva-K}
Take a valued field extension $(K(a)|K,v)$.
\sn
1) If $(K(a)|K,v)$ is immediate, then $v(a-K)$ has no largest element.
\sn
2) If $v(a-K)$ has no largest element, then $v(a-K)\subseteq vK$.
\sn
3) The set $v(a-K)\cap vK$ is an initial segment of $vK$.
\sn
4) The set $v(a-K)\setminus (v(a-K)\cap vK)$ has at most one element.
\sn
5) For every $c,d\in K$,
\[
v(da+c-K) \>=\> v(a-K)+vd\>.
\]
\sn
6) If $a,b$ are elements in some valued field extension of $(K,v)$ such that $v(a-b)>v(a-K)$, then
$v(a-c)=v(b-c)$ for all $c\in K$ and $v(a-K)=v(b-K)$.
\end{proposition}
\begin{proof}
1): This follows from \cite[Theorem 1]{[Ka]}.
\sn
2): Take $c\in K$; we wish to show that $v(a-c)\in vK$. By assumption there is $d\in K$ such that $v(a-d)>v(a-c)$.
Hence $v(a-c)=\min\{v(a-c),v(a-d)\}=v(c-d)\in vK$.
\sn
3): Take $\alpha\in v(a-K)\cap vK$ and $\beta\in vK$ such that $\beta<\alpha$. Choose $b,c\in K$ with $vb=\beta$
and $v(a-c)=\alpha$. Then $c-b\in K$ and $vb=\min\{v(a-c),vb\}=v(a-(c-b))\in v(a-K)$.
\sn
4): Assume that $v(a-c),v(a-d)\in v(a-K)\setminus (v(a-K)\cap vK)$. If they were distinct, say $v(a-d)>v(a-c)$,
then as in the proof of 2) we would obtain that $v(a-c)\in vK$, contradiction.
\sn
5): This follows from the equalities $v(a+c-K)=v(a-K)$ and $v(da-K)=v(a-K)+vd$, which are straightforward to prove.
\sn
6): For all $c\in K$, from $v(a-b)>v(a-c)$ we obtain that $v(a-c)=v(b-c)$ as in the proof of 2); hence $v(a-c)\in
v(b-K)$ and $v(b-c)\in v(a-K)$, which shows that $v(a-K)=v(b-K)$.
\end{proof}

We will mostly work with immediate extensions, in which case we have that $v(a-K)\subseteq vK(a)=vK$ for every element $a\notin K$ in the extension.
However, several of our results hold more generally for extensions that are not necessarily immediate. Some of them, such as the following one, remain true when the set $v(a-K)$ is replaced by $v(a-K)\cap vK$.
\begin{lemma}                               \label{distinct}
Take a valued field extension $(K(a)|K,v)$. Take $\alpha\in v(a-K)$ and assume that $d\in K$ is such that
\begin{equation}               \label{alph+vd}
\alpha+vd\> >\> v(a-K)
\end{equation}
(note that $d$ always exists when $v(a-K)$ is bounded by some element from $vK$). Then the sets
\[
v(a-K)-vd^n\>, \;n\in\N\>,
\]
are pairwise distinct.
\end{lemma}
\begin{proof}
Since $\alpha\in v(a-K)$ and $\alpha+vd>v(a-K)$, we know that $vd>0$. Take any natural numbers
$m<n$. Then
\[
\alpha-mvd\> \geq \> \alpha +vd-nvd \> >\> v(a-K)-nvd\>,
\]
which shows that $\alpha-mvd\notin v(a-K)-nvd=v(a-K)-vd^n$. But $\alpha-mvd\in
v(a-K)-mvd=v(a-K)-vd^m$, so the two sets are distinct.
\end{proof}

\subsection{Distances}                              \label{sectdist}
Take again a totally ordered set $(T,<)$.
A pair $(\Lambda^L,\Lambda^R)$  of subsets of $T$ is called a \bfind{cut} in $T$ if $\Lambda^L$ is an initial
segment of $T$ and $\Lambda^R=T\setminus \Lambda^L$; it then follows that $\Lambda^R$ is a final segment of $T$.
To compare cuts in  $(T,<)$ we will use the lower cut sets comparison. That is, for two cuts
$\Lambda_1=(\Lambda_1^L,\Lambda_1^R) ,\, \Lambda_2=(\Lambda_2^L,\Lambda_2^R)$ in $T$ we will write
$\Lambda_1<\Lambda_2$ if $\Lambda^L_1\varsubsetneq\Lambda^L_2$, and $\Lambda_1\leq \Lambda_2$ if
$\Lambda^L_1 \subseteq  \Lambda^L_2$. This defines a linear order on the set of all cuts in $T$.

For a given subset $S$ of $T$ we define $S^+$ to be the cut $(\Lambda^L,\Lambda^R)$ in $T$ such that
$\Lambda^L$ is the least initial segment containing $S$, that is,
\[
S^+\>:=\> (\{t\in T\,|\, \exists\, s\in S:\,t\leq s\}\,,\, \{t\in T\,|\, t>S\} )\>.
\]
Likewise, we denote by $S^-$ the cut $(\Lambda^L,\Lambda^R)$ in $T$ such that $\Lambda^L$ is the largest initial
segment disjoint from $S$, i.e.,
\[
S^-\>:=\> (\{t\in T\,|\, t<S\}\,,\, \{t\in T\,|\, \exists\, s\in S:\,t\geq s\} )\>.
\]
For $s\in T$, we set
\[
s^+\>:=\> \{s\}^+\quad\mbox{ and }\quad s^-\>:=\> \{s\}^-\>.
\]
We note that
\begin{equation}             \label{S+<s-}
S^+\>\leq\>s^- \>\Leftrightarrow\> S<s \;\mbox{ and }\;
S^+\><\>s^- \>\Leftrightarrow\> \exists\, t\in T:\, S\leq t<s\>.
\end{equation}
Indeed, $S^+\leq s^-$ means that the smallest initial segment of $T$ containing $S$ is a subset of the initial
segment $\{t\in T\,|\, t<s\}$, and this is equivalent to $S<s$. Likewise, $S^+< s^-$ means that the smallest
initial segment of $T$ containing $S$ is a proper
subset of $\{t\in T\,|\, t<s\}$, and as both are initial segments, this is equivalent to the existence of some
$t\in T$ such that $S\leq t$.
\pars
We embed $T$ in the linearly ordered set of all cuts in $T$ by identifying each $s\in T$ with $s^-$.

\pars
The set $v(a-K)\cap vK$ is an initial segment of $vK$ and thus the lower cut set of a cut in $vK$. However,
in order to be able to compare $v(a-K)$ with $v(a-L)$ when $L|K$ is algebraic, it is more convenient to work
with the cut $v(a-K)$ induces in the divisible hull $\widetilde{vK}$ of $vK$. Indeed,
$\widetilde{vK}$ is equal to the value group of the algebraic closure $K\ac$ of $K$, and if $L|K$ is algebraic,
then $K\ac=L\ac$ and therefore, $\widetilde{vK}=\widetilde{vL}$. Hence we define:
\[
\dist (a,K)\>:=\>(v(a-K)\cap \widetilde{vK})^+\;\; \textrm{ in the divisible hull $\widetilde{vK}$ of $vK\>$.}
\]
We call this cut the \bfind{distance of $z$ from $K$}. The distance replaces the use of suprema in the case of
non-archimedean value groups, which are not contained in $\R$.

\begin{lemma}                          \label{translate}
Let the situation be as above and assume that $v(a-K)\subseteq \widetilde{vK}$. Take some $\alpha\in
\widetilde{vK}$. Then $\dist (a,K)\leq\alpha^-$ is equivalent to $v(a-K)<\alpha$, and
$\dist (a,K)<\alpha^-$ is equivalent to $v(a-K)<\!\!|\, \alpha$.
\end{lemma}
\begin{proof}
We have:
\begin{eqnarray*}
\dist (a,K)\>\leq\>\alpha^-&\Leftrightarrow& v(a-K)\cap \widetilde{vK}<\alpha \\
&\Leftrightarrow& v(a-K)<\alpha\>,
\end{eqnarray*}
where the first equivalence holds by (\ref{S+<s-}) and the second equivalence holds since $v(a-K)\subseteq \widetilde{vK}$. Similarly,
\begin{eqnarray*}
\dist (a,K)\><\>\alpha^-&\Leftrightarrow& \exists\,\beta\in\widetilde{vK}:\, v(a-K)\cap
\widetilde{vK}\leq\beta<\alpha \\
&\Leftrightarrow& \exists\,\beta\in\widetilde{vK}:\, v(a-K)\leq\beta<\alpha
\>\Leftrightarrow\> v(a-K)<\!\!|\, \alpha\>.
\end{eqnarray*}
\end{proof}

\bn
\subsection{The fundamental equality and the defect}                              \label{sectfi}
Take an extension $(L|K,v)$ of valued fields such that the extension of $v$ from $K$ to $L$ is unique. Assume that
$\chara Kv=p>0$. Then
\begin{equation}                    \label{feuniq}
[L:K]\>=\> p^{\nu }\cdot(vL:vK)[Lv:Kv]\>,
\end{equation}
where by the Lemma of Ostrowski $\nu$ is a nonnegative integer (see \cite[Th\'eor\`eme 2, p.~236]{[R]}
or \cite[Corollary to Theorem 25, p.~78]{[ZS2]}). The factor
$d(L|K,v)=p^{\nu }$ is called the \bfind{defect} of the extension $(L|K,v)$.

\bn
\subsection{Criteria for defect extensions}                              \label{sectde}
\begin{lemma}                         \label{ueGp1}
Take an extension $(K(a)|K,v)$ of valued fields of degree
$p=\mbox{\rm char}(Kv)$ and such that the extension of $v$ from $K$ to $K(a)$ is unique.
If $v(a-K)$ has no maximal element, then $(K(a)|K,v)$ is immediate with defect~$p$.
\end{lemma}
\begin{proof}
By \cite[part (1) of Theorem 2.21]{Ku6}, $(K(a)|K,v)$ is immediate. As the extension of $v$ from $K$ to $K(a)$
is assumed to be unique, it follows that the defect is equal to the degree of the extension.
\end{proof}

\pars

\begin{lemma}	                              	\label{c1}\label{c2}
Assume that $(K(a)|K,v)$ is a normal extension of prime degree $p$.
\sn
1) If $K^h$ is some henselization of $(K,v)$ and $a\notin K^h$, then the extension of $v$ from $K$ to $K(a)$ is
unique.
\sn
2) Assume that $(K,v)$ is of rank 1. If $v(a-K)$ has no maximal element and is bounded in $vK$, then $(K(a)|K,v)$
is immediate and has defect $p$.
\end{lemma}
\begin{proof}
1): Since $K(a)|K$ is normal and of degree $p$, it is linearly disjoint from every other algebraic extension $L$
over $K$ in which it is not contained. This is seen as follows. If $K(a)|K$ is inseparable, then it is purely
inseparable, and so is $L(a)|L$; hence the latter extension can only be of degree $1$ or $p$. If $K(a)|K$ is
separable, then it is Galois with a Galois group of order $p$. Then also $L(a)|L$ is Galois. As restriction embeds
its Galois group in that of $K(a)|K$, again the degree of $L(a)|L$ can only be $1$ or $p$.

Since $a\notin K^h$ by assumption, from what we have just shown we see that $K(a)$ must be linearly disjoint from $K^h$ over $K$, which by
\cite[Lemma~2.1]{BK1} implies our assertion.
\sn
2): Since $(K,v)$ is of rank 1, $K$ lies dense in its henselization $K^h$. By assumption, $v(a-K)$ is bounded in
$vK$, which implies that $a\notin K^h$ and thus by part 1), the extension of $v$ from $K$ to $K(a)$ is unique.
Now our assertion follows from Lemma~\ref{ueGp1}.
\end{proof}

\bn
\subsection{Artin-Schreier extensions of valued fields}                              \label{sectAS}
Throughout this section, we let $K$ be a field of characteristic $p>0$.
Recall that $L|K$ is called an \bfind{Artin-Schreier
extension} if it is generated by a root of a polynomial of the form $X^p-X-b$ with $b\in K$. Note that every
Artin-Schreier extension of $K$ is a Galois extension of degree $p$, and vice versa.
We call $\vartheta$ an \bfind{Artin-Schreier generator} of an Artin--Schreier extension $L|K$ if $L=K(\vartheta)$
with $\vartheta^p-\vartheta\in K$.
\begin{lemma}                               \label{allASg}
Assume that $\vartheta$ is an Artin-Schreier generator of an Artin--Schreier extension $L|K$.
Then $\vartheta'$ is another Artin-Schreier generator
of $L|K$ if and only if $\vartheta' =i\vartheta +c$ for some $i\in\F_p$ and $c\in K$.
\end{lemma}
\begin{proof}
If $\vartheta$ and $\vartheta'$ are roots of the same polynomial $X^p-X-b$,
then $\vartheta-\vartheta'$ is a root of $X^p-X$, whose roots are exactly the elements of
$\F_p\,$. Hence, $\vartheta+i$, $i\in\F_p\,$, are all roots of $X^p-X-b$. Pick a nontrivial $\sigma\in
\Gal L|K$. We then have that $\sigma\vartheta-\vartheta=j$ for some $j\in\F_p^\times$.

If $\vartheta'$ is another Artin-Schreier generator of $L|K$ such that $\sigma\vartheta-\vartheta=j=
\sigma\vartheta'-\vartheta'$, then we have $\sigma(\vartheta-\vartheta')
=\vartheta-\vartheta'$. Since $\sigma$ is a generator of $\Gal L|K\isom
\Z/p\Z$, it follows that $\tau(\vartheta-\vartheta')=\vartheta-
\vartheta'$ for all $\tau\in\Gal L|K$, that is, $\vartheta-\vartheta'\in K$.

If $\vartheta'$ is another Artin-Schreier generator of $L|K$ and
$\sigma\vartheta'-\vartheta'=j'\in\F_p^\times$, then there is some $i\in\F_p^\times$
such that $ij=j'$. Since $\sigma i=i$, we then have that $\sigma i\vartheta-i\vartheta=i(\sigma\vartheta-\vartheta)
=ij=j'$. Then by what we have shown before, $\vartheta'=i\vartheta+c$ for some $c\in K$.

Conversely, if $\vartheta$ is an Artin-Schreier generator of $L|K$ and
if $i\in\F_p^\times$ and $c\in K$, then $(i\vartheta+c)^p-(i
\vartheta+c)=i(\vartheta^p-\vartheta)+c^p-c\in K$. But $i\vartheta+c$
cannot lie in $K$, so $K(i\vartheta+c)=L$ since $[L:K]$ is a prime. This
shows that also $i\vartheta+c$ is an Artin-Schreier generator of $L|K$.
\end{proof}

\begin{corollary}                           \label{ASinv}
Let $(L|K,v)$ be an Artin-Schreier extension of valued fields. Then
$v(\vartheta-K)$ is independent of the choice of the Artin-Schreier generator
$\vartheta$ of the extension, so it is an invariant of the extension.
\end{corollary}
\begin{proof}
Take two Artin-Schreier generators $\vartheta,\vartheta'$ of $L|K$. By Lemma~\ref{allASg} we can write
$\vartheta' =i\vartheta +c$ for some $i\in\F_p^\times$ and $c\in K$. Since $vi=0$ for every valuation, it follows
from assertion 5) of Proposition~\ref{propva-K} that $v(\vartheta'-K)=v(\vartheta-K)$.
\end{proof}

\begin{proposition}                             \label{propve-th}
Assume that $(K(\eta)|K,v)$ is a purely inseparable extension and $(K(\vartheta)|K,v)$ is an
Artin-Schreier extension of valued fields. Then $K(\eta,\vartheta)|K(\vartheta)$ is purely inseparable
and so the extension of $v$ from $K(\vartheta)$ to $K(\eta,\vartheta)$ is unique. If
\begin{equation}                                   \label{ve-th}
v(\eta-\vartheta)\> > \> v(\vartheta-K)\>,
\end{equation}
then the extension of $v$ from $K$ to $K(\vartheta)$ is unique. If in addition, $(K(\eta)|K,v)$ is immediate,
then $(K(\vartheta)|K,v)$ is immediate with defect~$p$.
\end{proposition}
\begin{proof}
Assume that (\ref{ve-th}) holds. Then $\vartheta\notin K^h$ since otherwise it would follow
from Theorem~2 of~\cite{Ku4}
that $K(\eta)|K$ is not purely inseparable, contradicting our assumption. Now our first assertion follows from
part 1) of Lemma~\ref{c1}. By part 6) of Proposition~\ref{propva-K}, (\ref{ve-th}) also implies
that $v(\vartheta-K)=v(\eta-K)$.
\pars
Now assume in addition that $(K(\eta)|K,v)$ is immediate. Then by part 1) of Proposition~\ref{propva-K},
$v(\vartheta-K)=v(\eta-K)$ has no maximal element.  Therefore, our second assertion follows from Lemma~\ref{ueGp1}.
\end{proof}

\subsection{Kummer extensions of prime degree of valued fields}                            \label{sectK}
Take a valued field $(K,v)$ of mixed characteristic, that is, $\chara K=0$ while $\chara Kv=p>0$. We consider
Kummer extensions of degree $p$. Such an extension is generated by an element $\eta$ such that $\eta^p\in K$.
If $(K(\eta)|K,v)$ is immediate, then it can be assumed that $\eta$ and hence also $\eta^p$ is a $1$-unit, i.e.,
$v(\eta-1)>0$. Indeed, since $(K(\eta)|K,v)$ is immediate, we have that $v\eta\in vK(\eta)=vK$,
so there is $c\in K$ such that $vc=-v\eta$. Then $v\eta c=0$, and since $\eta cv\in K(\eta)v=Kv$, there is
$d\in K$ such that $dv=(\eta cv)^{-1}$. Then $v(\eta cd)=0$ and $(\eta cd)v=1$. Hence $\eta cd$ is a $1$-unit.
Furthermore, $K(\eta cd)=K(\eta)$ and $(\eta cd)^p=\eta^pc^pd^p\in K$. Thus we can replace $\eta$ by $\eta cd$
and assume from the start that $\eta$ is a $1$-unit.

\pars
The next proposition follows from \cite[Corollary~3.6 and Proposition 3.7] {BKdrf}:
\begin{proposition}                                       \label{dist_approx}
Take a valued field of mixed characteristic and a Kummer defect extension as detailed above.
Then the distance $\dist(\eta,K)$ does not depend on the choice of the generator $\eta$
of the extension $(K(\eta)|K,v)$ as long as it is a $1$-unit and satisfies $\eta^p\in K$. Moreover,
\begin{equation}                                  \label{disteta1}
0\><\>\dist(\eta,K) \>\leq\> \left(\frac{vp}{p-1}\right)^-\>.
\end{equation}
\end{proposition}
Under the above conditions, the Kummer extension is immediate, hence we have that $v(\eta-K)\subseteq vK$. From the
definition of the distance it then follows that $v(\eta-K)$ is the intersection of the lower cut set of
$\dist(\eta,K)$ with $vK$, hence the distance determines uniquely the set $v(\eta-K)$, showing that this set does
not depend on the choice of the generator $\eta$. Moreover, from Lemma~\ref{translate} we obtain that
inequality (\ref{Kde}) holds.

\parm

In \cite[Proposition 3.7]{BKdrf} we show that a Kummer extension $(K(\eta)|K,v)$, where $\eta$ is a $1$-unit with
$\eta^p\in K$, has dependent defect if and only if
\[
\dist(a,K) \>=\> \frac{vp}{p-1} \,+\, H^-\>,
\]
for some proper convex subgroup $H$ of $\widetilde{vK}$. If $(K,v)$ has rank 1, then $\widetilde{vK}$ is
archimedean ordered and therefore $H=\{0\}$. In this case, the above equation just becomes
\[
\dist(\eta,K)\><\>\left(\frac{vp}{p-1}\right)^-\>.
\]
In view of Lemma~\ref{translate}, this translates to $v(\eta-K)<\!\!|\, \frac{vp}{p-1}$, as mentioned in the
Introduction.

\bn
\subsection{Differences of $p$-th powers in mixed characteristic}                              \label{sectdpp}
\begin{lemma}                           \label{lvep-ap}
Take $\eta$ and $a$ to be two elements of some valued field of characteristic 0 with residue field characteristic $p>0$. If
\begin{equation}                                \label{ve-a}
v(\eta-a)\><\> \frac{vp}{p-1}+v\eta\>,
\end{equation}
then
\begin{equation}                           \label{vep-ap}
v(\eta^p-a^p) \>=\> pv(\eta-a)\>.
\end{equation}
\end{lemma}
\begin{proof}
Let $\zeta$ be a primitive $p$-th root of unity. We have that
\[
\eta^p-a^p \>=\>\prod_{i=1}^p(\zeta^i\eta-a)\>=\>\prod_{i=1}^p(\zeta^i\eta-\eta+\eta-a)\>.
\]
Using the well known fact that
\[
v(\zeta^i-1)\>=\>\frac{vp}{p-1}
\]
(see e.g.\ the proof of Lemma~2.9 of~\cite{Ku8}) and our assumption, we obtain that
\[
v(\zeta^i-1)+v\eta\>=\> \frac{vp}{p-1}+v\eta \> >\> v(\eta-a)\>.
\]
Hence we have that
\[
v(\zeta^i\eta-a)\>=\> v(\zeta^i\eta-\eta+\eta-a)\>=\>\min\{v(\zeta^i-1)+v\eta, v(\eta-a)\}\>=\>v(\eta-a)\>,
\]
which yields equation (\ref{vep-ap}).
\end{proof}

\bn
\section{The case of valued fields of characteristic $p>0$}                              \label{sectcharp}

\subsection{A basic transformation}                              \label{secttransf}
We describe a transformation that was introduced in \cite{Ku6}. Take a valued field $(K,v)$ of characteristic
$p>0$ and a (not necessarily immediate) purely inseparable extension $(K(\eta)|K,v)$ of degree $p$ of valued
fields such that $\eta^p\in K$ and that $v(\eta-K)$ is bounded from above by an element in $vK(\eta)$ (and hence
also by an element in $vK$). We are starting with the minimal polynomial $Y^p-\eta^p$ of $\eta$ over $K$ and turn
it into the separable polynomial
\begin{equation}                         \label{trans1}
Y^p-d^{p-1}Y-\eta^p\>,
\end{equation}
where $d\ne 0$. Setting $Y=dX$ and then dividing the resulting polynomial by $d^p$, we transform this polynomial
into the Artin-Schreier polynomial
\begin{equation}                         \label{trans2}
X^p-X-\frac{\eta^p}{d^p}\>.
\end{equation}
Under the condition that $vd$ is large enough, the following lemma describes the behaviour of the set $v(\eta-K)$
when $\eta$ is replaced by roots of these two polynomials.
\begin{lemma}                               \label{dXeqc}
Take $d\in K$ such that
\begin{equation}              \label{vd}
vd^{p-1}\> >\> pv(\eta-K) -v\eta\>.
\end{equation}
Let $\tilde\vartheta_d$ be a root of the polynomial (\ref{trans1}). Then
\[
\vartheta_d\>:=\> \frac{\tilde\vartheta_d}{d}
\]
is a root of (\ref{trans2}) and
$K(\vartheta_d)|K$ is an Artin-Schreier extension with a unique extension of $v$ from $K$ to
$K(\vartheta_d)$. Furthermore, we have that
\begin{equation}                            \label{dXeqceq1}
v(\tilde\vartheta_d-K)\>=\>v(\eta-K)\>,
\end{equation}
\begin{equation}                            \label{dXeqceq2}
v\left(\frac{\eta}{d}-\vartheta_d\right)\> >\> v(\vartheta_d-K)\>,
\end{equation}
and
\begin{equation}                   \label{vth-K}
v(\vartheta_d-K)\>=\> v(\eta-K)-vd\>.
\end{equation}
\end{lemma}
\begin{proof}
Once we prove that
\begin{equation}                            \label{dXeqceq3}
v(\eta-\tilde\vartheta_d)\> >\> v(\eta-K)\>,
\end{equation}
we obtain equation~(\ref{dXeqceq1}) by part 6) of Proposition~\ref{propva-K}, which in turn implies
equation~(\ref{vth-K}) by part 5) of Proposition~\ref{propva-K}. Further, again using part 5) of
Proposition~\ref{propva-K} again, we obtain
\[
v\left(\frac{\eta}{d}-\vartheta_d\right)\>=\>v(\eta-\tilde\vartheta_d)-vd\> >\> v(\eta-K)-vd\>=\>
v(\tilde\vartheta_d-K)-vd \>=\> v(\vartheta_d-K)\>,
\]
which proves (\ref{dXeqceq2}). Hence we will now prove (\ref{dXeqceq3}). We compute:
\begin{equation}          \label{e-th}
(\eta-\tilde\vartheta_d)^p\>=\> \eta^p-\tilde\vartheta_d^p \>=\> \eta^p - \eta^p - d^{p-1}\tilde\vartheta_d \>=\>
- d^{p-1}\tilde\vartheta_d \>.
\end{equation}

\pars
We set $Y=dX$ to obtain that $\vartheta_d$ is a root of the polynomial $d^pX^p-d^{p-1} dX-\eta^p$ and hence also
of the Artin-Schreier polynomial (\ref{trans2}).

\pars
Since $v\eta\in v(\eta-K)$, from (\ref{vd}) we obtain:
\[
(p-1)vd \>=\> vd^{p-1}\> >\> pv\eta - v\eta = (p-1)v\eta\>,
\]
so that
\[
v\frac{\eta^p}{d^p} \>=\> p(v\eta-vd)\><\>0\>.
\]
Hence we have that $v\vartheta_d<0$ and consequently,
$v\vartheta_d^p=pv\vartheta_d<v\vartheta_d$ and
\[
pv\frac{\eta}{d}\>=\>v\frac{\eta^p}{d^p}\>=\>\min\{v\vartheta_d^p,v\vartheta_d\}\>=\>v\vartheta_d^p
\>=\>pv\vartheta_d\>,
\]
which yields that
\[
v\tilde\vartheta_d\>=\> vd+v\vartheta_d \>=\> v\eta\>.
\]
From this together with (\ref{vd}) and (\ref{e-th}) we obtain:
\[
v(\eta-\tilde\vartheta_d)\>=\> \frac{1}{p}v(d^{p-1}\tilde\vartheta_d)\>=\> \frac{1}{p}(vd^{p-1}+v\eta)
\> >\> v(\eta-K)\>,
\]
as desired.

\pars
It remains to prove that $K(\vartheta_d)|K$ is nontrivial (and hence an Artin-Schreier extension),
and that the extension of $v$ from $K$ to $K(\vartheta_d)$ is unique. Since $v(\eta-K)$ is bounded from above in
$vK$ by assumption, the same holds for $v(\eta-K)-vd$ which by (\ref{vth-K})
is equal to $v(\vartheta_d-K)$. This implies in particular that
$\vartheta_d\notin K$, so the extension $K(\vartheta_d)|K$ is nontrivial. Since $\eta$ is purely
inseparable over $K$ by assumption, the same holds for $\eta/d$. Thus our assertion follows from (\ref{dXeqceq2})
together with Proposition~\ref{propve-th}, where we replace $\eta$ by $\eta/d$ and $\vartheta$ by $\vartheta_d\,$.
\end{proof}

\bn
\subsection{Proof of Theorems~\ref{MTfinAS} and~\ref{MTdepAS}}                  \label{sectcharppf}
Throughout, let $(K,v)$ be a valued field of characteristic $p>0$. First, we prove:
\sn
{\it if $K$ admits only finitely many distinct Artin-Schreier extensions, then $(K,v)$ is a semitame field,}
\sn
which is the assertion of Theorem~\ref{MTfinAS}.
Take a purely inseparable (not necessarily immediate) extension
$K(\eta)|K$ of degree $p$ such that $\eta^p\in K$ and that $v(\eta-K)$ is bounded from above by an element in
$vK$.

\begin{proposition}                    \label{infASdn}
Assume that $\alpha\in v(\eta-K)$ and $d\in K$ are such that $\alpha+vd> v(\eta-K)$ (that is,
(\ref{alph+vd}) holds for $\eta$ in place of $a$) and $vd^{p-1}>pv(\eta-K) -v\eta$ (that is, (\ref{vd}) holds).
Then with the notation from Lemma~\ref{dXeqc}, the Artin-Schreier extensions
\[
K(\vartheta_{d^n})|K\>, \;n\in\N\>,
\]
are pairwise distinct, and the extensions of $v$ from $K$ to $K(\vartheta_{d^n})$ are unique for all $n\in\N$.
\end{proposition}
\begin{proof}
It follows from Lemma~\ref{dXeqc} that for each $n\in\N$, $K(\vartheta_{d^n})|K$ is an Artin-Schreier extension
and the extension of $v$ from $K$ to all $K(\vartheta_{d^n})$ is unique.
From Lemma~\ref{distinct} we infer that the sets $v(\vartheta_{d^n}-K)=v(\eta-K)-vd^n$ are distinct.
Thus by Corollary~\ref{ASinv}, the extensions $K(\vartheta_{d^n})|K$ are distinct.
\end{proof}

\begin{lemma}                        \label{pidegp}
If the perfect hull of $K$ does not lie in the completion
of $(K,v)$, then $(K,v)$ admits a purely inseparable extension $K(\eta)|K$ of degree $p$ such that
$v(\eta-K)$ is bounded from above by an element in $vK$.
\end{lemma}
\begin{proof}
Assume that the perfect hull $K^{1/p^{\infty}}$ does not lie in the completion $K^c$ of $(K,v)$, and take an
element
$\tilde\eta\in K^{1/p^{\infty}} \setminus K^c$. We may assume that $\tilde\eta^p\in K^c$ (otherwise, we replace
$\tilde \eta$ by ${\tilde\eta}^{p^{\nu}}$ for a suitable $\nu\geq 1$). Since $\tilde\eta\notin K^c$, we have that
$v(\tilde\eta-K)$ is bounded from above by some $\alpha\in vK$ and $v(\tilde\eta^p-K^p)= pv(\tilde\eta-K)$ is
bounded from above by $p\alpha$. On the other hand, since $\tilde{\eta}^p\in K^c$, there
is some $b\in K$ such that $v(\tilde\eta^p-b)>p\alpha$. We choose $\eta\in K^{1/p}$ such that $\eta^p=b$.
Then $pv(\tilde\eta-\eta)=v(\tilde\eta^p-\eta^p)=v(\tilde\eta^p-b)>p\alpha\geq pv(\tilde\eta-K)$, which yields that
$v(\tilde\eta-\eta)>v(\tilde\eta-K)$. From this it follows by part 6) of Proposition~\ref{propva-K}
that $v(\eta-K)=v(\tilde\eta-K)$, showing that also $v(\eta-K)$ is bounded from above by $\alpha$.
\end{proof}

\pars
Since $v(\eta-K)$ is assumed to be bounded from above by an element in $vK$, there is some $d\in K$ which satisfies
$vd^{p-1}>pv(\eta-K) -v\eta$.
Then the same is true for every $d'\in K$ with $vd'\geq
vd$. Hence $\alpha\in v(\eta-K)$ and $d\in K$ can be chosen such that $\alpha+vd> v(\eta-K)$.
Thus we can use Lemma~\ref{pidegp} together with Proposition~\ref{infASdn} to obtain:
\begin{proposition}                         \label{propnotdense}
If the perfect hull of $K$ does not lie in the completion
of $(K,v)$, then $K$ admits infinitely many Artin-Schreier extensions.
\end{proposition}
\sn
By the equivalence of assertions a) and e) of Theorem~\ref{charst}, this proposition proves Theorem~\ref{MTfinAS}.

\parm
We will now prove the assertion of Theorem~\ref{MTdepAS}, which states:
\sn
{\it if $(K,v)$ admits an Artin-Schreier extension with dependent
defect, then it admits infinitely many distinct Artin-Schreier extensions with dependent defect.}
\sn
We assume that $(K,v)$ admits an Artin-Schreier extension with dependent defect, which means that it must be
obtained via the transformation described in Section~\ref{secttransf} from a purely inseparable defect extension
$(K(\eta)|K,v)$ of degree $p$. This is only possible if $v(\eta-K)$ is bounded from above by an element in
$vK$. Also, since the extension is of degree $p$ with nontrivial defect, the defect must be $p$ and the extension
must be immediate. Hence by part 1) of Proposition~\ref{propva-K},
$v(\eta-K)$ has no largest element and consequently, the same holds for the sets $v(\eta-K)-vd^n=
v(\vartheta_{d^n}-K)$. Therefore, since the extension of $v$ from $K$ to $K(\vartheta_{d^n})$ is unique by
Proposition~\ref{infASdn}, it follows from Lemma~\ref{ueGp1} that $(K(\vartheta_{d^n})|K,v)$ is immediate with
defect $p$. As this extension is obtained from a purely inseparable defect extension of degree $p$ by the
transformation described in Section~\ref{secttransf}, this defect is dependent by definition. Finally,
Proposition~\ref{infASdn} shows that the extensions $K(\vartheta_{d^n})|K$, $n\in\N$, are distinct. This
completes our proof.

\bn
\section{The case of valued fields of  mixed characteristic}                              \label{sectchar0}
Throughout this section, we assume that $(K,v)$ is a valued field of rank $1$ and of characteristic $0$ with
residue field of characteristic $p>0$. Further, we assume that $K$ contains a primitive $p$-th root of unity,
and that $(K(\eta)|K,v)$ is a Kummer extension, where $\eta$ is a $1$-unit with $\eta^p\in K$.

\subsection{A basic transformation in the mixed characteristic case}            \label{secttransf0}
Given $d\in K$, we transform the minimal polynomial $X^p-\eta^p$ of $\eta$ over $K$ into the polynomial
\begin{equation}                         \label{trans3}
f_{d} (X)\>:=\> X^p+h_d(X)-\eta^p
\end{equation}
with
\[
h_d(X)\>:=\> \sum_{i=1}^{p-1} \binom{p}{i} d^{p-i}X^i\>.
\]

\begin{lemma}                               \label{btrf0}
Assume that $vd<0$ and that
\begin{equation}                             \label{ve-K}
v(\eta-K)\><\>\frac{vp+(p-1)vd}{p}\>.
\end{equation}
Take $\tilde\vartheta_d$ to be a root of the polynomial (\ref{trans3}). Then
\[
v(\eta-K)\>=\>v(\tilde\vartheta_d-K)\>.
\]
\end{lemma}
\begin{proof}
We compute the value of the coefficients of $h_d\,$, using our assumption (\ref{ve-K}):
\begin{equation}
v\left(\binom{p}{i} d^{p-i}\right) \>=\> vp+(p-i)vd\>\geq\> vp+(p-1)vd\> >\>pv(\eta-K)\>.
\end{equation}
Since $0=v(\eta-1)\in v(\eta-K)$, this shows that all coefficients of $h_d$ have positive value, while
$v\eta^p=0$; this forces $\tilde\vartheta_d$ to have value $0$ by the ultrametric triangle law. Consequently,
\begin{equation}                         \label{vtth>}
vh_d(\tilde\vartheta_d) \> >\> pv(\eta-K)\>.
\end{equation}

Suppose that there is $c\in K$ such that $v(\tilde\vartheta_d-\eta)\leq v(\eta-c)$. Combined with our assumption
(\ref{ve-K}), this implies that assumption (\ref{ve-a}) of Lemma~\ref{lvep-ap} holds for $\eta$ and for
$\tilde\vartheta_d$ in place of $a$ since $v\eta=0$. Hence by Lemma~\ref{lvep-ap} we have that
\[
v(\tilde\vartheta_d^p-\eta^p) \>=\> pv(\tilde\vartheta_d-\eta)\>.
\]
Using this together with (\ref{vtth>}) and the fact that
$\tilde\vartheta_d^p=\eta^p+h_d(\tilde\vartheta_d)$ by definition of $\tilde\vartheta_d$, we compute:
\begin{eqnarray*}
v(\tilde\vartheta_d-\eta) &=& \frac 1 p v(\tilde\vartheta_d^p-\eta^p) \>=\> \frac 1 p
v(\eta^p+h_d(\tilde\vartheta_d)-\eta^p)\\
&=& \frac 1 p vh_d(\tilde\vartheta_d)\> >\>v(\eta-K)\>,
\end{eqnarray*}
contradicting our assumption. This shows that $v(\tilde\vartheta_d-\eta)> v(\eta-c)$ for all $c\in K$, whence
$v(\tilde\vartheta_d-\eta)> v(\eta-K)$. Hence by part 6) of Proposition~\ref{propva-K},
$v(\eta-K)=v(\tilde\vartheta_d-K)$.
\end{proof}

\bn
\subsection{Proof of Theorem~\ref{MTmixed}}                              \label{sectchar0pf}
We let $(K,v)$ be a valued field of mixed characteristic and rank $1$ which contains a primitive $p$-th root of
unity. We have to prove the assertion of Theorem~\ref{MTmixed}, which states:
\sn
{\it If $(K,v)$ admits a Kummer extension of degree $p$ with super-dependent defect, then it admits infinitely
many distinct Kummer extensions of degree $p$ with super-dependent defect.}
\sn
We assume that $(K(\eta)|K,v)$ is a Kummer extension of degree $p$ with super-dependent defect, and that
$\eta$ is a $1$-unit with $\eta^p\in K$. Since $0=v\eta\in v(\eta-K)$ and $v(\eta-K)$ does not contain a
maximal element, it must contain positive elements. Since $(K(\eta)|K,v)$ is a super-dependent defect extension,
we know that (\ref{sd}) holds.
Since by assumption, $(K,v)$ is of rank $1$ and $v(\eta-K)$ does not contain a maximal element but is bounded,
$vK$ must be dense in $\widetilde{vK}$. Hence there is some $\td\in K$ such that $v\td<0$ and
\begin{equation}                             \label{ve-Kstrong}
v(\eta-K)\><\>\frac{vp}{p}+2v\td\>.
\end{equation}
Then inequality (\ref{ve-K}) of Lemma~\ref{btrf0} is satisfied, because
\[
\frac{vp}{p}+2v\td \><\>\frac{vp+(p-1)v\td}{p}\>.
\]
We obtain from Lemma~\ref{btrf0} that for a root $\tilde\vartheta_{\td}$ of the
polynomial $f_{\td}(X)$ defined in that lemma,
\[
v(\eta-K)\>=\>v(\tilde\vartheta_{\td}-K)\>.
\]

Now we set $X=\td Y$; dividing the resulting polynomial $f_{\td}(\td Y)$ by $\td^p$, we obtain the polynomial
\[
g_d(Y)\>:=\> Y^p + \sum_{i=2}^{p-1} \binom{p}{i} Y^i-\frac{\eta^p}{\td^p}\>.
\]
We observe that
\[
\vartheta_{\td}\>:=\> \frac{\tilde\vartheta_{\td}}{\td}
\]
is a root of $g_{\td}(Y)$ and that by part 5) of Proposition~\ref{propva-K},
\begin{equation}
v(\vartheta_{\td}-K)\>=\> v(\tilde\vartheta_{\td}-K) -v\td\>.
\end{equation}

Now we set $Y=X-1$, which turns the polynomial $g_{\td}(Y)$ into the polynomial
\[
X^p-\left(\frac{\eta^p}{\td^p} +1\right)
\]
with root
\[
\eta_{\td}\>:=\> \vartheta_{\td} +1\>.
\]
Again by part 5) of Proposition~\ref{propva-K}, we have that
\[
v(\eta_{\td}-K)\>=\>v(\vartheta_{\td}-K)\>=\> v(\tilde\vartheta_{\td}-K) -v\td\>=\> v(\eta-K) -v\td
\>\ne\> v(\eta-K)\>,
\]
where the last inequality holds since $v\td\ne 0$ and $v(\eta-K)$ is a bounded subset of an archimedean ordered abelian group. As $vd<0=v\eta$,
\[
\eta_{\td}^p\>=\> \frac{\eta^p}{\td^p} +1
\]
is a $1$-unit.

\pars
Since $vK$ is dense, there are infinitely many $\td'\in K$ with $v\td'<0$ that satisfy (\ref{ve-Kstrong})
in place of $\td$. With the same argument as above, we see that $v(\eta-K) -v\td\ne v(\eta-K) -v\td'$ if
$v\td\ne v\td'$. In this way we obtain infinitely many Kummer extensions $(K(\eta_{\td})|
K,v)$ with distinct sets $v(\eta_{\td}-K)$, which by Proposition~\ref{dist_approx}
shows that these extensions are pairwise distinct.

\pars
Since $(K(\eta)|K,v)$ is a nontrivial immediate extension, part 1) of Proposition~\ref{propva-K}
shows that the set $v(\eta-K)$ has no maximal element, while being bounded by assumption. The same is consequently
true for the sets $v(\eta-K)-v\td=v(\eta_{\td}-K)$. As the rank of $(K,v)$ is assumed to be $1$, part 2) of
Lemma~\ref{c2} shows that $(K(\eta_{\td})|K,v)$ is immediate and has defect $p$. Finally, this defect is
super-dependent since
\[
v(\eta_{\td}-K)\>=\> v(\eta-K)-v\td \><\>\frac{vp}{p}+2v\td-v\td\>=\>\frac{vp}{p}+v\td\>.
\]

\bn
\section{Filling a gap in \cite{Ku6}}                              \label{sectgap}
In \cite[Lemma 2.31]{Ku6} the following is stated:
\begin{lemma}                               \label{uniqextv}
Assume that $\chara K=p>0$ and $(K(\vartheta)|K,v)$ is an Artin-Schreier extension with Artin-Schreier generator
$\vartheta$. If $\dist (\vartheta,K)\leq 0^-$ and $v(\vartheta-K)$ has no maximal element, then the
extension of $v$ from $K$ to $K(\vartheta)$ is unique and $(K(\vartheta)|K,v)$ is immediate with defect $p$.
\end{lemma}
\n
In the proof it is written: ``In \cite{Ku4} we show that the assumption that $\dist (\vartheta,K)<0$ implies that
the extension of $v$ from $K$ to $K(\vartheta)$ is unique.'' Since $v(\vartheta-K)$ has no maximal element,
Lemma~\ref{ueGp1} shows that this assertion indeed implies that $(K(\vartheta)|K,v)$ is immediate. In addition,
$\dist (\vartheta,K)<\infty$ implies that $\vartheta\notin K$, so the extension is nontrivial and thus has defect
$p$.

However, the assertion was never proven in \cite{Ku4}. In order to complete the proof of Lemma~\ref{uniqextv}, we
will prove it here. As $v(\vartheta-K)$ has no maximal element, we know from part part 2) of
Proposition~\ref{propva-K} that $v(a-K)\subseteq vK\subseteq \widetilde{vK}$. Hence by Lemma~\ref{translate}
the assumption $\dist (\vartheta,K)\leq 0^-$ is equivalent to $v(\vartheta-K)<0$.

\pars
By part 1) of Lemma~\ref{c1}, it suffices to prove that $\vartheta\notin K^h$. Suppose
otherwise. Then by \cite[Theorem~1]{Ku4}, there are $\alpha\in vK$ and a convex subgroup $H\ne\{0\}$ of $vK$
such that the coset $\alpha+H$ is cofinal in $v(\vartheta-K)$. Since $v(\vartheta-K)<0$, we must have that
 $\alpha+H<0$, which yields that $-\alpha>H$ and $-\alpha>0$. As $\alpha\in\alpha+H$, there is some $c\in K$
such that $v(\vartheta-c)\geq\alpha$. Since $\vartheta-c$ is also an Artin-Schreier generator of the extension and
$v(\vartheta-c-K)=v(\vartheta-K)$, we may assume w.l.o.g.\ that $v\vartheta\geq\alpha$.

Let $X^p-X-b$ with $b\in K$ be the minimal polynomial of $\vartheta$ over $K$, and take $d\in K$ with $vd=-\alpha$.
Then by part 5) of Proposition~\ref{propva-K}, $H$ is cofinal in $v(d\vartheta-K)$. Further,
$d\vartheta$ is a root of $d^{-p}X^p-d^{-1}X-b$ and hence also of
\[
X^p-d^{p-1}X-d^pb\>.
\]
Let $\eta$ be the root of $X^p-d^p b$. We wish to show that
\[
v(d\vartheta-K)\>=\>v(\eta-K)\>.
\]
We compute:
\begin{eqnarray*}
pv(d\vartheta-\eta) &=& v((d\vartheta)^p-\eta^p)\>=\>v(d^p\vartheta+d^pb-d^pb)\>=\>vd^p\vartheta\\
&\geq& -p\alpha  +\alpha\>\geq\> -\alpha \> >\> H\>.
\end{eqnarray*}
Since $H$ is a convex subgroup of $vK$, this implies that
$v(d\vartheta-\eta) \>\geq\> -\frac 1 p \alpha \> >\> H\>$, and as $H$ is cofinal in $v(d\vartheta-K)$, we obtain:
\[
v(d\vartheta-\eta) \> >\> v(d\vartheta-K)\>.
\]
By construction, $K(\eta)|K$ is purely inseparable, hence by Theorem~2 of~\cite{Ku4},
$d\vartheta$ cannot lie in the henselization $K^h$. We have thus shown that $\vartheta\notin K^h$, as desired.

redpajama_set_name: RedPajamaArXiv


The following is a guest post by BuzzBuzzHome, the most complete listing of new construction homes. Aside from cataloguing all new construction in Canada and the US, the BuzzHomies maintain a news site, publishing all the real estate news that's fit to print.
Spring is in the air and the real estate market is ready to blossom. Cheesy imagery aside, there is no better season to be a first-time homebuyer than the spring. Not to put down the other three seasons, but spring is known for its abundance of new launches, accompanied by warm weather that gives everyone a much-needed post-winter boost.
We've compiled a list of the five most buzzworthy developments that are already attracting a ton of attention in Spring 2013. Enjoy!
Mid-town Toronto has its fair share of new builds, but 155 Redpath by Freed Developments and CD Capital is turning heads. Located steps from Yonge and Eglinton, the intersection will see a burst of new life. Industry experts are keenly watching the Eglinton corridor now that a Crosstown LRT is in the works. The new transit access will appeal to frequent fliers who'll be able to get out to Pearson International Airport fast. With the Mount Pleasant Road shopping strip and a big revamp coming to the Yonge and Eglinton Centre, 155 Redpath will be right in the centre of the dynamic neighbourhood.
It might be situated in Toronto's downtown core, but this luxury building at 25 Richmond St. E. doesn't sacrifice outdoor space. Designed by architectsAlliance, the entire project emphasizes natural light and outdoor space. The Great Gulf tower features landscaped terraces, big city views and sunlight-friendly suites. And when it comes to getting around the city, the epic, 50-storey skyscraper is about as convenient as they come: Yonge + Rich earned an impressive walk score of 98 and a perfect 100 transit score.
While there are many attractive new home developments in the waterfront community of Southeast False Creek in Vancouver, Pinnacle International's The One is garnering a special amount of attention. Situated at 38 West 1st Avenue between Salt and Ontario streets, The One stands at the heart of this pedestrian, pet and people-friendly community. The multi-building development is just a stone's throw from Vancouver's iconic seawall, meaning residents can walk, jog, bike or rollerblade from their front door to Stanley Park or Kits Beach.
Located at the heart of Surrey's rejuvenated downtown, the micro-suites at Balance will be some of Canada's smallest condos, and perhaps more importantly, some of Metro Vancouver's most affordable homes. The 56-unit, four-storey development will have suites that range in size from 300-to-650 square feet, with space-saving features like foldable wall-beds.
Mark on 10th is a mixed-use high-rise tower currently under construction in Calgary's sought-after Beltline neighbourhood. When it's completed in 2015, the 35-storey tower will be comprised of 270 homes and will boast a massive rooftop amenity space with a gym, yoga and pilates studio, infrared sauna, outdoor hot tub, sunbathing patio, and barbecue and fire pit area. Mark on 10th will also be home to Alberta's first and only large-scale public art installation by renowned artist and author Douglas Coupland.
The image of homebuyers we see most often is of a young couple, maybe expectant, maybe with toddlers in tow, showing off their new bungalow adorned with a white picket fence.


My clan, 1st Tactical Elites, is currently recruiting. Do you want a team working clan? Do you want a active clan? Do you want a clan that will lead you to victory? Well 1st Tactical Elites are for you. We are actively recruiting players like you to join our rosters. 1st Tactical Elites is a fully structured clan that uses the U.S Military ranking system. Now you may say, " This clan is all serious ", but you are wrong. We are all about having fun, and matching and winning at the same time.
1st Tactical Elites is a clan that uses xfire as a primary source of communication located at (www.xfire. com). We also have a fully functional site located at(www. 1stte. ace.st). We deeply encourage EVERYONE to sign up and post on our forums even if you do not want to join our clan. We could always use some new friends around.
You want to join? Come check us! http://1stte.ace.st/ Join Today!


Today was one of those wonderful fall days: warm but not humid, with all the landscape refreshed from yesterday's rain. We savor such weather at this time of year, knowing that the mild temperatures, and the plants that continue to thrive, won't be around much longer.
As I strolled around the garden, relishing the dahlias, Nicotiana mutibilis, and the summer squash that's still producing, I found myself planning for next year. I love how the red and peach dahlias look along the outside of the vegetable garden and I want to plant more of these in Annual Alley next summer.
Sure, plant in the fall. I certainly do, since my plant mania doesn't go dormant after Labor Day. At this time of year, however, I'm also looking forward. It's the perfect season to practice the delicate balance between being totally satisfied with how things are at this moment, while still planning and planting for the future.
This is a complex state of being, filled with practice in staying in the moment, gratitude, and living with paradox. It is as rich and full as my Annual Alley.
Fall is for planting and planning.
Note to self: mark these red and peach dahlias so that I remember which plants to put back along this fence next year. Another note to self: try to keep my mind in the moment as I make those labels and stick them by each plant. Hear the cries of the crows, feel the warmth of the sun and the coolness of the fall breeze. See the brilliant colors of the annuals and the vigorous ingenuity of the weeds. Be in the garden now, while working for the future.


The Domestic Princess: How are we saving for IVF?
How are we saving for IVF?
Since it is now 2011, Christopher and I are creating a new budget and savings plan for IVF. Our goal is to save $1000 a month. We know it's a high goal to meet but think that it is doable if we can start cutting out some of the non-necessities and cutting back on some of the luxuries.
1. Cable was the one blaring right in our face. Currently we are spending $150 a month. We have HBO, Showtime, 1 DVR, cable and internet. I called and cut out HBO and Showtime, bringing our bill to $120. Still outrageous we think and Time Warner was not willing to bring it down any further. We are currently looking at other cable companies to see if we can bring this down further. But this did however save us $30.
2. Insurance (car, home, ring, life): We just switched insurance companies last year and this saved us about $100 a month so I didn't mess with this. I know we are getting a great deal but it's worth a couple calls around to see if you can get a better deal.
1. Eating Out: We are spending about $300 per month on eating out and 'adult beverages'. Ouch, right!?! We have cut our entertainment budget to $50 per week, totaling $200 per month. Saving us $100.
2. Grocery Store: We are currently spending $350 per month on groceries. Usually about $75 at Whole Foods and $100 a Wal-Mart per pay. We need to cut back on the extras we buy when we go to the store. So our plan is to go in with meals planned and nothing else is being bought. Eating organic meats, eggs and milk is still very important to us so we are not going to cut out Whole Foods. But cutting down on what we buy should bring our grocery budget down to $300 per month. Saving us $50.
1.I joined Delve Market Research hoping that I can earn some money in my off time, when it's convenient for me. Since this is not consistent work, I'm not sure I can estimate how much this will bring in.
2. We will be selling items on Ebay. Chris and I like to go to thrift stores and antique stores to find deals then sell them. We are hoping to earn an extra $50 per month.
3. Chris's web design/marketing he does on the side brings in an extra $400 per month. We never include this in our monthly budget because we don't want to ever depend on this money but it's perfect to include in this savings plan.
If we add all of the budget cuts and extra income together, that gives us $680 per month we can save for IVF. So to meet our goal of saving $1000 per month we will need to come up with an extra $320 out of our paycheck. Which I know is doable. I will be doing a monthly post about the money we have saved toward IVF and how we are doing it.
We are doing similar things to help us save for our adoption. 1st, we started couponing. I'm not a super couponer like some people (mostly because we do eat a lot of organic/all natural foods, which don't have coupons), but I do manage to save anywhere from $10-$30 per grocery bill. We are also selling things on eBay, cutting down our cable bill, and trimming our weekly entertainment budget. I think I have us saving around $750 a month!
Katie, isn't it crazy where you find money when it's this important? Good Luck with saving for you adoption. I'll be looking out for your money saving tips too.
What a great budget! I love that you are sharing it with us! I am sure you will save up in no time!
Thank you, I'm hoping that this will help others see they can cut back too. I was amazed at the amount we found we can save.
Isn't it amazing how much extra we can always find in our budgets? I am hoping that you are blessed enough to save even more than you hope so you can get IVF sooner rather than later! We don't have cable at all, just the regular rabbit ears. We get all of the major networks crystal clear thanks to digital tv now, and anything I miss I can just watch on hulu the next day. Just a thought.
Michelle, that does include internet but not phone. I think it's aweful. We need to find something cheaper.
I do love restraunts.com, we need to utilize it more.
Might be the Most excellent topic which I browsed through all month?
Have you ever considered Tutor.com as a way to bring in some extra cash?
I am a new reader to your blog, came across it on the conceive magazine website. Best of luck with this savings, it can be totally stressful but it will be worth it! My husband and I are currently on a somewhat difficult ttc journey and it is refreshing to hear all that you are doing for yourselves and your family!
E Small, first Welcome. I love new readers ;) This is a crazy journey we have been on and you sound like you can totally relate. Thank you for stopping by and I look forward to more comments from you.


People who know me wouldn't believe me when I say that I have an interest in space. I have visited the Kennedy Space Centre in Florida and have a done a lot of reading up on shuttle launches etc. So when I saw that the BBC had a new programme about people going through the astronaut selection programme I really was excited and watched in interest. As I was watching, I thought I wish I could have a go. Then I thought actually I could make my own so here is my no budget version of the BBC's astronaut programme.
Firstly a little bit about me, I am terrible at maths (I had to retake my GCSE!) I do have a degree in Geography (2:1) and I work in Marketing. So in this case I would already not be accepted onto the astronaut programme.
Since I have no budget (and I don't think anyone would let me lose near one) I found an online game! I did quite well to be fair winching all the animals and people.
For this I decided to ask my partner to call out long string numbers that I didn't know beforehand and stand up and down on a step on the stairs.
I did terrible I got the four digit number (I thought I'd better have a test try) but after five I failed miserably. I am not surprised really since numbers are not my strong point.
For this I decided to shut myself in my downstairs cupboard amongst the skis and shoes. I came out at 16 minutes and 55 seconds and was really chuffed with that. I didn't count in my head or anything like that I just thought about the latest book I had just finished and what I was having for tea and things like that. I didn't feel claustrophobic.
I did draw the line at drawing my own blood! So I found this awesome game online where it was like operation, so I decided to play it. I played the heart surgery game and it was quite cute.
This was something I found hard and didn't want to film it. I guess my biggest regret (which is nothing really in the grand scheme of things) was not picking the right A-Levels.
One challenge I did at school! I measured out my garden and it was 10m long. I thought I may have a good chance at this as I am fairly fit. However I was awful, I only managed to get to level… 2.2!
I hope you enjoyed reading my blog on the first episode of the Astronauts: Do you have what it takes programme. I am looking forward to the second one. What did you think of the programme?


text:
RENAME TABLE `t_dba_encode_columns` TO `t_encode_columns`;

ALTER TABLE `t_encode_columns`
DROP  COLUMN `override`;

ALTER TABLE `t_data_schema`
ADD COLUMN  `db_vip` varchar(32) DEFAULT NULL COMMENT 'schema对应数据库虚IP';

ALTER TABLE `t_data_tables`
ADD COLUMN  `fullpull_condition` varchar(512) DEFAULT NULL COMMENT '全量条件';

ALTER TABLE `t_fullpull_history`
ADD COLUMN  `current_shard_offset` bigint(20) DEFAULT NULL;

ALTER TABLE `t_meta_version`
ADD COLUMN  `schema_hash` int(11) DEFAULT NULL COMMENT '该版本数据对应的schema哈希值',
ADD COLUMN  `abandon` TINYINT(4) DEFAULT NULL COMMENT '假删除标记,0或null表示未删除,1表示删除';

ALTER TABLE `t_sink`
ADD  UNIQUE INDEX `idx_sinker_name_unique` (`sink_name`);

ALTER TABLE `t_table_meta`
MODIFY COLUMN  `comments` varchar(1024) DEFAULT NULL COMMENT '列注释';

CREATE TABLE `t_name_alias_mapping` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT COMMENT '主键ID',
  `type` int(2) NOT NULL COMMENT '别名类型1,router拓扑别名;2,增量拓扑别名',
  `name` varchar(64) NOT NULL COMMENT '名称',
  `name_id` int(11) NOT NULL COMMENT '名称对应ID',
  `alias` varchar(64) NOT NULL COMMENT '别名',
  `update_time` timestamp NULL DEFAULT NULL ON UPDATE CURRENT_TIMESTAMP COMMENT '更新时间',
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;

CREATE TABLE `t_sinker_topo_schema` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `sinker_topo_id` int(11) DEFAULT NULL,
  `sinker_name` varchar(64) DEFAULT NULL,
  `ds_id` int(11) DEFAULT NULL,
  `ds_name` varchar(32) DEFAULT NULL,
  `schema_id` int(11) DEFAULT NULL,
  `schema_name` varchar(32) DEFAULT NULL,
  `target_topic` varchar(64) DEFAULT NULL,
  `description` varchar(128) DEFAULT NULL,
  `update_time` timestamp NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  PRIMARY KEY (`id`),
  UNIQUE KEY `idx_schema_id_unique` (`schema_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;

CREATE TABLE `t_sinker_topo` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `sinker_name` varchar(64) DEFAULT NULL,
  `sinker_conf` varchar(1024) DEFAULT NULL,
  `jar_id` int(11) DEFAULT NULL,
  `status` varchar(255) DEFAULT NULL,
  `description` varchar(128) DEFAULT NULL,
  `update_time` timestamp NULL DEFAULT NULL ON UPDATE CURRENT_TIMESTAMP,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;

CREATE TABLE `t_storm_topology` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `topology_name` varchar(128) NOT NULL,
  `ds_id` int(11) NOT NULL,
  `jar_name` varchar(512) DEFAULT NULL,
  `update_time` datetime DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;

CREATE TABLE `t_topology_jar` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `category` varchar(255) DEFAULT NULL,
  `version` varchar(255) DEFAULT NULL,
  `type` varchar(255) DEFAULT NULL,
  `name` varchar(255) DEFAULT NULL,
  `path` varchar(255) DEFAULT NULL,
  `minor_version` varchar(255) DEFAULT NULL,
  `description` varchar(255) DEFAULT NULL,
  `update_time` datetime DEFAULT NULL,
  `create_time` datetime DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;

redpajama_set_name: RedPajamaGithub


text:
Oman ambassador accredited in Turkmenistan
Chairperson of the Turkmen parliament Gulshat Mammadova received credentials from the ambassador of the Sultanate of Oman (with residence in Ankara) Qasim Mohammad Salim Al Salhi, Trend reports with reference to the Turkmen government.
The ambassador stressed his country's interest in enhancing cooperation with Turkmenistan, including in establishing trade, business and cultural contacts, the report said.
A meeting was also held in the Foreign Ministry of Turkmenistan. The sides expressed their commitment to the development of political dialogue and the intensification of economic cooperation between the two countries, the Turkmen Foreign Ministry said.
An agreement to create an international transport and transit corridor between the Islamic Republic of Iran, Qatar, Oman, Turkmenistan and Uzbekistan was signed in Ashgabat in April 2011. This agreement aims to create a reliable link between the countries of Central Asia and the ports of Persian and Oman gulfs.
In accordance with the implementation of Turkmenistan's national program for development of the civil aviation, the country's national air carrier (Turkmenistan Airlines) plans to open air service from Ashgabat to Kuwait, Muscat, Sharm El Sheikh, Hurghada, Cairo.
#TURKMENISTAN
Ashgabat to host int'l telecommunications forum
Turkmenistan, Saudi Arabia exploring joint projects
Ashgabat, Astana preparing meetings at highest level
Mogherini holds negotiations in Turkmenistan
World's first plant to manufacture synthetic fuel from gas opens in Turkmenistan [PHOTO]
New UNICEF representative accredited in Turkmenistan
Turkmenistan, Singapore hold political consultations
Ambassador of Turkmenistan presents credentials to King of Jordan
Kazakhstan, Uzbekistan buy polyethylene in Turkmenistan
Turkmen, Russian presidents mull prospects for trade & economic co-op

redpajama_set_name: RedPajamaCommonCrawl


Curious high school science teacher, Andy Swapp, of Milford, Utah wondered just how well a wind turbine might work in Milford. Given that he has a fresh supply of eager (well, at least some of them) minds, he decided to involve his kids in the experiment. What started as a simple hunch has now caught the eye of Fresh Wind and turned into a 200 MW wind farm for Milford. Andy Swapp and his students, known as the Wind Kids, installed an anemometer from the state of Utah on his property next to his barn simply to take measurements of the wind potential in the area. The initial goal was just to see if there was enough wind to allow Swapp to put up a wind turbine up on his property and power part of his farm. With the Wind Kids to help do the footwork, several wind companies took an interest in the experiment, and more particularly, in the results.
This 200MW installation, worth a quarter of a billion dollars, is also responsible for bringing jobs to Millford. Local business owners also noticed that their sales increased by almost three times thanks to all of the new faces coming to town for the wind farm. Not only did the Wind Kids bring wind to Milford, they also improved the local economy and brought the community together over wind. In addition, several of the students are now interested in continuing careers in renewable energy.


text:
using System;
using NAudio.Dmo;
using System.Diagnostics;

// ReSharper disable once CheckNamespace
namespace NAudio.Wave
{
    /// <summary>
    /// Wave Stream for converting between sample rates
    /// </summary>
    public class ResamplerDmoStream : WaveStream
    {
        private readonly IWaveProvider inputProvider;
        private readonly WaveStream inputStream;
        private readonly WaveFormat outputFormat;
        private DmoOutputDataBuffer outputBuffer;
        private DmoResampler dmoResampler;
        private MediaBuffer inputMediaBuffer;
        private long position;

        /// <summary>
        /// WaveStream to resample using the DMO Resampler
        /// </summary>
        /// <param name="inputProvider">Input Stream</param>
        /// <param name="outputFormat">Desired Output Format</param>
        public ResamplerDmoStream(IWaveProvider inputProvider, WaveFormat outputFormat)
        {
            this.inputProvider = inputProvider;
            inputStream = inputProvider as WaveStream;
            this.outputFormat = outputFormat;
            dmoResampler = new DmoResampler();
            if (!dmoResampler.MediaObject.SupportsInputWaveFormat(0, inputProvider.WaveFormat))
            {
                throw new ArgumentException("Unsupported Input Stream format", nameof(inputProvider));
            }

            dmoResampler.MediaObject.SetInputWaveFormat(0, inputProvider.WaveFormat);
            if (!dmoResampler.MediaObject.SupportsOutputWaveFormat(0, outputFormat))
            {
                throw new ArgumentException("Unsupported Output Stream format", nameof(outputFormat));
            }

            dmoResampler.MediaObject.SetOutputWaveFormat(0, outputFormat);
            if (inputStream != null)
            {
                position = InputToOutputPosition(inputStream.Position);
            }
            inputMediaBuffer = new MediaBuffer(inputProvider.WaveFormat.AverageBytesPerSecond);
            outputBuffer = new DmoOutputDataBuffer(outputFormat.AverageBytesPerSecond);
        }

        /// <summary>
        /// Stream Wave Format
        /// </summary>
        public override WaveFormat WaveFormat => outputFormat;

        private long InputToOutputPosition(long inputPosition)
        {
            double ratio = (double)outputFormat.AverageBytesPerSecond
                / inputProvider.WaveFormat.AverageBytesPerSecond;
            long outputPosition = (long)(inputPosition * ratio);
            if (outputPosition % outputFormat.BlockAlign != 0)
            {
                outputPosition -= outputPosition % outputFormat.BlockAlign;
            }
            return outputPosition;
        }

        private long OutputToInputPosition(long outputPosition)
        {
            double ratio = (double)outputFormat.AverageBytesPerSecond
                / inputProvider.WaveFormat.AverageBytesPerSecond;
            long inputPosition = (long)(outputPosition / ratio);
            if (inputPosition % inputProvider.WaveFormat.BlockAlign != 0)
            {
                inputPosition -= inputPosition % inputProvider.WaveFormat.BlockAlign;
            }
            return inputPosition;
        }

        /// <summary>
        /// Stream length in bytes
        /// </summary>
        public override long Length
        {
            get
            {
                if (inputStream == null)
                {
                    throw new InvalidOperationException("Cannot report length if the input was an IWaveProvider");
                }
                return InputToOutputPosition(inputStream.Length);
            }
        }

        /// <summary>
        /// Stream position in bytes
        /// </summary>
        public override long Position
        {
            get
            {
                return position;
            }
            set
            {
                if (inputStream == null)
                {
                    throw new InvalidOperationException("Cannot set position if the input was not a WaveStream");
                }
                inputStream.Position = OutputToInputPosition(value);
                position = InputToOutputPosition(inputStream.Position);
                dmoResampler.MediaObject.Discontinuity(0);
            }
        }

        /// <summary>
        /// Reads data from input stream
        /// </summary>
        /// <param name="buffer">buffer</param>
        /// <param name="offset">offset into buffer</param>
        /// <param name="count">Bytes required</param>
        /// <returns>Number of bytes read</returns>
        public override int Read(byte[] buffer, int offset, int count)
        {
            int outputBytesProvided = 0;

            while (outputBytesProvided < count)
            {
                if (dmoResampler.MediaObject.IsAcceptingData(0))
                {
                    // 1. Read from the input stream
                    int inputBytesRequired = (int)OutputToInputPosition(count - outputBytesProvided);
                    byte[] inputByteArray = new byte[inputBytesRequired];
                    int inputBytesRead = inputProvider.Read(inputByteArray, 0, inputBytesRequired);
                    if (inputBytesRead == 0)
                    {
                        //Debug.WriteLine("ResamplerDmoStream.Read: No input data available");
                        break;
                    }
                    // 2. copy into our DMO's input buffer
                    inputMediaBuffer.LoadData(inputByteArray, inputBytesRead);

                    // 3. Give the input buffer to the DMO to process
                    dmoResampler.MediaObject.ProcessInput(0, inputMediaBuffer, DmoInputDataBufferFlags.None, 0, 0);

                    outputBuffer.MediaBuffer.SetLength(0);
                    outputBuffer.StatusFlags = DmoOutputDataBufferFlags.None;

                    // 4. Now ask the DMO for some output data
                    dmoResampler.MediaObject.ProcessOutput(DmoProcessOutputFlags.None, 1, new[] { outputBuffer });

                    if (outputBuffer.Length == 0)
                    {
                        Debug.WriteLine("ResamplerDmoStream.Read: No output data available");
                        break;
                    }

                    // 5. Now get the data out of the output buffer
                    outputBuffer.RetrieveData(buffer, offset + outputBytesProvided);
                    outputBytesProvided += outputBuffer.Length;

                    Debug.Assert(!outputBuffer.MoreDataAvailable, "have not implemented more data available yet");
                }
                else
                {
                    Debug.Assert(false, "have not implemented not accepting logic yet");
                }
            }

            position += outputBytesProvided;
            return outputBytesProvided;
        }

        /// <summary>
        /// Dispose
        /// </summary>
        /// <param name="disposing">True if disposing (not from finalizer)</param>
        protected override void Dispose(bool disposing)
        {
            if (inputMediaBuffer != null)
            {
                inputMediaBuffer.Dispose();
                inputMediaBuffer = null;
            }
            outputBuffer.Dispose();
            if (dmoResampler != null)
            {
                //resampler.Dispose(); s
                dmoResampler = null;
            }
            base.Dispose(disposing);
        }
    }
}


Ignite is our Tigerpaw automation package designed to streamline processes, or restrict use of Tigerpaw in ways that don't compliy with company policy or procedure. Purchase of Ignite through our online store includes 1 year of support. For more information see our Ignite product page where the manual is also available for detailed information.
Specify your Cost & Income GL Codes on Service Order Parts Used with granular precision. We found many companies were going to great lengths to apply these GL codes to get the data they needed. Save hours each month or start applying your codes more precisely.
Automatic the EZ Billing of service orders when they move to a closed status. This action can be filtered by SO Type. Less button-clicking makes everyone happier.
Specify an item's optimal on-hand quantity for each inventory location with this module. Then use our Inventory Min-Max Variance report to find out which items need to be orderd, and which items could be redistributed. We like using this report to track inventory on a technician's vehicle: Joe always seems to too many hoard cogs, and we no longer waste travel time driving back to the office to pickup a widget when one should have been on board.
Instantly assign the rep on a service order as it passes through statuses. Was the status changed to "Waiting on Parts"? Assign to the warehouse manager. "Dispatch" goes to the dispatcher for manual technician assignment. Finally, assign to billing personnel by marking the status done.
Put all newly created Tigerpaw accounts on credit hold to establish a line of credit before servicing.


text:
Q: Product Spaces: Tube Lemma The book I am using for my Introduction of Topology course is Principles of Topology by Fred H. Croom.
My professor asked to prove the following Lemma.

The Tube Lemma: Let $K$ be a compact topological space and $Y$ any topological space. Suppose $p\in Y$ and $U$ an open subset of $K\times Y$ containing the entire slice $K\times \{p\}$. Then we can find an open subset $V$ of $Y$ such that $p\in V$ and $K\times V\subset U$.

This is what I understand:

*

*A space X is compact provided that every open cover of X has a finite subcover.

*Given that $K$ is compact, $p\in Y$ and $K\times\{p\}\subset U\subset K\times Y$. I seek to prove $\exists V\subset Y:p\in V, K\times V\subset U$.

Here is my rough attempt at the proof:

Let $p\in Y$ and let $U\subset K\times Y$ be open in the product space such that $K\times \{p\}\subset U$. For every $x\in K$, choose open set $A_x\subset K$ and $B_x\subset Y$ such that $(x,p)\in A_x\times B_x\subset U$. Since $K$ is compact, we can select a finite subcover $A_{x_1},...,A_{x_n}$ such that $K=\bigcup_{i=1}^nA_{x_i}$. Let $V=\bigcap_{i=1}^nB_{x_i}$. Then $V$ is a desired neighborhood of $p$. Suppose that $(x,p)\in K\times V$. Then there is an $j$ such that $x\in A_{x_j}$ as the collection $\{A_{x_i}\}_{i=1}^n$ covers $K$. As $p\in V =\bigcap_{i=1}^nB_{x_i}$, we have $(x,p)\in A_j\times B_j\subset U$. This proves $K\times \{p\}\subset K\times V\subset U$.

Am I on the right track? Did I do anything wrong? Is there anything I need to change regarding the proof? Is there any alternative way to proof the Tube lemma that would be much more efficient or detailed?

I sincerely thank you for taking the time to read this question and my attempt at proving it. I greatly appreciate any assistance you may provide.

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