TITLE: Family of Sets Proof for Set Theory
QUESTION [0 upvotes]: Let $A$ be a nonempty set. A ring of subsets of $A$ is a nonempty family $F$ of subsets of $A$ such that for all $S,T, \in F,$
(i) $S \cup T \in F$
(ii) $S \setminus T \in F$ (difference between the sets).
Prove that if $F$ is a ring of subsets of $A$, then for all $S, T \in F, S \cap T \in F.$
My attempt so far:
From the first condition about the union, we have that
$$S \cup T = \{ x: x \in S \; \textbf{or} \; x \in T \} \in F.$$
And from the second condition,
$$S \setminus T  = \{ x: x \in S \; \textbf{and} \; x \not\in T \} \in F.$$
So then we have two cases:
I:
$$\{ x : x \in S \} \in F$$
$$\{ x: x \in S \; \textbf{or} \; x \in T \} \in F.$$
II:
$$\{ x: x \in T \} \in F$$
$$\{ x: x \in S \; \textbf{and} \; x \not\in T \} \in F.$$
What i'm not sure about is how to transform these resulting facts into something about the intersection of arbitrary subsets of $F$.

REPLY [1 votes]: Hint. Show that $S \cap T = S \setminus (S \setminus T)$.


TITLE: Questions about entanglement from a laymen/quantum hobbyist
QUESTION [3 upvotes]: Please note I am not a physicist I just read every article I can on it, I understand a good amount on it though. But no maths. (currently trying to learn the maths)
By what means can we as humans entangle two particles? How does (just examples are fine) it happen in nature? And I would guess technically there could be a particle in me entangled with a particle in another person; maybe my future wife?
I think the answer to the first is... We make them by shooting a photon at a half silvered mirror and somehow it splits and one bounces off the half silver while the other one leaves and we capture the one, and wait for the other and capture it too. Then measure one and measure the other and see they are the same?
EDIT:
One more question. From one of the answers two entangled particles have the same wave function. Does this mean every particle with the same wave function is entangled? Can an electron and photon or any two arbitrary particles or groups of particles(molecules) entangle?

REPLY [1 votes]: The answer is more difficult. This is an introductory answer without mathematics.
Entangled photons is a current research topic, trying to entagle more every time. One "easy" way is to allow to deacy a particle into two other particles, altough you would end with 2 entangled particles rather than photons. With matter antimatter anihilation you could have two entagled photons by the mechanism:
$e^+ + e^- \longrightarrow \gamma + \gamma$
The same mechanism would apply in nature.
You could read also:
More entangled photons
In order to have particles (or photons) entagled, they need to have a common wavefunction, that is usually destroyed with measure, or almost any interaction. From here to your future wife its more complicated.
For more in entaglement read Chris J. Isham's  Lectures on Quantum Mechanics, Mathematical and Structural Foundations.
Keep learning the maths, as it is foundamental in physics, and for it's understanding.


TITLE: Prove a Trigonometric Series
QUESTION [4 upvotes]: Question:
$$\cot^{2}\frac{\pi }{2m+1}+\cot^{2}\frac{2\pi }{2m+1}+\cdots+\cot^{2}\frac{m\pi }{2m+1}=\frac{m(2m-1)}{3}$$
$m$ is a positive integer.
Attempt:
I started by showing that
$$\sin(2m+1)\theta =\binom{2m+1}{1}\cos^{2m}\theta \sin\theta -\binom{2m+1}{3}\cos^{2m-2}\theta \sin^{3}\theta +\cdots+(-1)^{m}\sin^{2m+1}\theta$$
By expanding $\left(\text{cis }\theta\right)^{2m+1}$ and then equating the imaginary part.

REPLY [2 votes]: Divide this equality by $\sin^{2m+1}\theta$ to get
$$
\frac{\sin(2m+1)\theta}{\sin^{2m+1}\theta}=\sum\limits_{k=0}^m (-1)^k{2m+1 \choose 2k+1}\cot^{2k} \theta\tag{1}
$$
Denote
$$
P_m(x)=\sum\limits_{k=0}^m (-1)^k{2m+1 \choose 2k+1} x^m
$$
This is polynomial of degree $m$. From $(1)$ we see that $P_m$ have $m$ roots $x_k=\cot^2\frac{\pi k}{2m+1}$. By Vietas formula their sum equals to
$$
-\frac{{2m+1 \choose 3}}{{2m+1 \choose 1}}=\frac{m(2m-1)}{3}
$$
So we get
$$
\sum\limits_{k=1}^m\cot^2\frac{\pi k}{2m+1}=\frac{(2m-1)m}{3}
$$


TITLE: How exactly Galvanometer scale is calibrated to function as an Ammeter?
QUESTION [4 upvotes]: Today, I read in my textbook that in order to convert a Galvanometer (more precisely a Moving Coil Galvanometer) into an Ammeter we need to attach a small Resistor with it in parallel (of Resistance of the order of 10^-2 ohm or less) for obvious reasons. I understood completely that why we need to do that. So there's no problem with that.
The thing which I'm not getting is how the original scale of the Galvanometer is modified (or as they say "Calibrated" ) in order to use it as a current meter? Kindly explain this with an example.
I'm attaching some screenshots of my book in which this line is highlighted in green

PS: I'm just a high school student so pardon me if you find any mishaps in my description.

REPLY [3 votes]: A galvanometer is an ammeter which is designed to detect and measure small currents.
Like any other ammeter to extend the range of a galvanometer a shunt resistor is connected in parallel with the galvanometer.
Since the potential difference across the galvanometer and the shunt is the same and using the symbols defined in the text
$I_{\rm G}R_{\rm G}  = I_{\rm s}r_{\rm s}$
where $I_{\rm G}$ and $I_{\rm s}$ are the currents through the galvanometer and shunt.
The current $I$ that you wish to measure is the sum of the currents through the galvanometer and the shunt resistor.
$I = I_{\rm G}+I_{\rm s}$
Combining these two equations gives $I = I_{\rm G} \left (1+ \dfrac {R_{\rm G}}{r_{\rm s}} \right)$.
If $R_{\rm G}$ and $r_{\rm s}$ are known and the current through the galvanometer $I_{\rm G}$ is read off its scale the current $I$ can be found.
Another way would be to recalibrate the galvanometer scale so that one could  directly read the current $I$ off the scale.

Another method of calibrating the galvanometer scale would be to pass known currents measured using an already calibrated ammeter through the galvanometer and shunt resistor arrangement.


TITLE: Does the limit of $e^{-1/z^4}$ as $z\to 0$ exist?
QUESTION [3 upvotes]: The question is as follows (note that $z = x + yi$):

Does $\lim_\limits{z \to 0} e^{-1/z^4}$ exist?

I've shown that

$\lim_\limits{x \to 0} e^{-1/x^4} = 0$

is equal to

$\lim_\limits{y \to 0} e^{-1/(iy)^4} = 0$

To prove both parts, but I'm not sure this is how it is done. I've been away from class for surgery, so all I have to work off of is what I've read online and in the textbook.
The problem is, it seems that with problems such as this one:

$\lim_\limits{z \to 2} \frac{z^2 + 3}{iz} = \frac{-7i}{2}$

All you have to do is substitute in 2 directly. I can't seem to figure out how to approach these problems, so hopefully all I have explained makes sense. Thanks for any explanations!

REPLY [6 votes]: Limits in complex plane are not really different from limits in multivariable calculus. (You can find plenty of examples in my  catalog of multivariable limits). The strategy for considering the limit as $z\to a$:

Check if you already know that the function is continuous at $a$ (e.g., it's a rational function with nonzero denominator at $a$). If this is so, then the limit is $f(a)$.
Try some lines of approach. If some of them give different limit values,   the limit does not exist.
If all lines appear to give the same limit value, try some squeeze theorem argument; this step depends very much on what your function is.
If all of the above fails; set the problem aside for a while and try it again later, possibly also considering curves in addition to lines in 2.

In your example $\exp(-1/z^4)$, the horizontal and vertical directions of approach yield $0$. But another line, such as $y=x$, yields $\infty$ because $(x+ix)^4 = (1+i)^4x^4 = -4 x^4$, leading to $\exp(1/(4x^4))$. So the limit at $0$ does not exist.


text:
\begin{document}

\sloppy

\begin{abstract}
Let $p$ be a prime number, $K$ be the henselization of the rational functions over the finite
field $\mathbb{F}_p$ and $R$ be the ring of additive polynomials over K. We show that the field of Laurent series
over $\mathbb{F}_p$ is decidable seen as an R-module. Moreover, we provide a recursively enumerable axiom system
(satisfied by $K$) in the language of $R$-modules together with a unary predicate for the valuation ring,
modulo which every positive primitive formula is equivalent to a universal formula. Consequently the $R$-module
theory of the field of Laurent series is model-complete in this language and admits $K$ as its prime
model.
\end{abstract}
\maketitle
\blfootnote{\textup{2010} \textit{Mathematics Subject Classification}: 11U05 (primary), 12L05, 12J10	03C68 (secondary)}

\section{Introduction}

The decidability and the axiomatization of the field of Laurent
series over the finite field $\mathbb{F}_p$ is a longstanding
problem (cf. for instance \cite{ku-elementary} and \cite{kuhlmanntame}). On the other hand, its characteristic $0$ analog $\mathbb{Q}_p$ is  axiomatized and is decidable by the Ax-Kochen and Ershov Theorem, proved independently by Ax-Kochen in USA and by Ershov in Russia (see \cite{ax-kochen1} and \cite{ersov}) . As a consequence, $\mathbb{Q}_p$ and $\mathbb{F}_p((X))$ have elementary  ultraproducts (over all primes $p$).
The first and maybe the most famous application of Ax-Kochen and Ershov theorem (called also Ax-Kochen theorem after  \cite{ax-kochen1}) is to give a  corrected form of a conjecture by Artin. This corrected form states that
given any integer $d>0$ there is some integer
$m$, such that for any prime $p>m$, every homogeneous polynomial of degree $d$ over $\mathbb{Q}_p$ with $>d^2$ variables has a non-trivial zero in $\mathbb{Q}_p$.
This application uses simply that fact that for every prime $p$ the field $\mathbb{F}_p((X))$  is $C_2$
(i.e. every homogeneous polynomial with strictly more
        variables than the square of its degree has a non-trivial zero)\footnote{The original conjecture of Artin was claiming that $\mathbb{Q}_p$ was $C_2$ for all prime $p$, this is refuted by Ternejian.}.

Despite the (asymptotic -) analogy between $\mathbb{Q}_p$ and $\mathbb{F}_p((X))$, for a fixed prime $p$
the field theory of $\mathbb{F}_p((X))$ remains  unknown. Moreover,  Kuhlmann proved that the naive translation into positive characteristic of the
complete theory of $\mathbb{Q}_p$ is incomplete (see \cite{ku-elementary}, we will give more details on this in the following lines).
Denef and Schoutens showed that the existential theory of $\mathbb{F}_p((X))$, in the language of rings with a constant symbol for $X$, is decidable assuming  resolution of singularities in positive characteristic (see \cite{denefsch}).  Recently Anscombe and Fehm showed unconditionally that the existential theory of $\mathbb{F}_p((X))$, in the language of rings (without the parameter $X$), is decidable \cite{sylvie-fehm}.

After the works of Kuhlmann and van den Dries, incompleteness of this naive theory  can be expressed
using only  properties of   additive polynomials (as we explain in the following lines), that is, in  the language of $S$-modules  where $S$
is the ring of  additive polynomials over $\mathbb{F}_p(X)$.
Rohwer, in his thesis (see \cite{rohwer}), shows that the complete theory of $\mathbb{F}_p((X))$ as an $S$-module is model-complete
in the language of $S$-modules together with a predicate for the valuation ring\footnote{However the meaning of Conjecture A.4. in \cite{rohwer} remains unclear for us.}.
However, he does not provide an axiom system for this theory and does not show a decidability result.

In this article we are considering the same problem (up to passing to the ring of additive polynomials
over the henselization of
$F_p(X))$ and we think that our philosophy is similar to Rohwer’s one.  On the other hand, we have had been rather inspired by  the articles \cite{ddp1}, \cite{vdd-kuhlmann} and \cite{sylvie-fehm}.

Notice that  B\'elair and Point also studied the module theory of
some (valued) fields satisfying  strong divisibility conditions (see
\cite{belair-point10} and \cite{belair-point15}).

\subsection*{Main results}
Let $K$ be the henselization of the field of rational functions over $\mathbb{F}_p$ and  $$\vfi:K\to K,\mesp x\mapsto x^{p^k}$$
    be some (fixed-) power of the Frobenius map. We set $R$ to be the ring of $\vfi$-polynomials, that is, additive polynomials
    whose monomials are of the form $aT^{{p^k}^i}$ ($a \in K$), equipped with the
    composition and the usual addition. Let $M:=\mathbb{F}_p((X))$ be the field of Laurent series over $\mathbb{F}_p$, let $L$ be the language of (right-) $R$-modules
    and finally let $L_{\mathcal{O}}$ be the language $L$ together with the unary predicate $\mathcal{O}$ (for the valuation ring $\mathbb{F}[[X]]$).

    We prove in particular the following results in the present article.

\begin{enumerate}
\item There is a recursively enumerable $L_{\mathcal{O}}$-theory $T_1$ such that $M\models T_1$, and any completion of $T_1$ is model complete. Moreover,  $K$ is the prime model of the complete $L_{\mathcal{O}}$-theory of $M$  (see Theorem \ref{th:pp2universal} and following corollaries).

\item Both the  complete $L$- and $L_{\mathcal{O}}$-theories of $M$ are decidable (see Theorem \ref{thm:finaldecidable}).
\end{enumerate}

\subsection*{Presentation of the problem and our strategy}
Following Kuhlmann (\cite{ku-elementary}), let us first explain why the naive adaptation of the theory
of $p$-adics into positive characteristic is incomplete. We denote this theory by $T_{naive}$, written in
language of rings together with a unary predicate for the valuation ring
and a constant for the uniformizer $X$,

We recall that $(U,v) \models T_{naive}$ if and only if
\begin{enumerate}[-]\label{naivfpt}
\item $(U,v)$ is a  henselian, non trivially valued  field of characteristic $p>0$,
\item the residue field $U/v$ is $\mathbb{F}_p$,
\item the value group $\Gamma$ is a $\mathbb{Z}$-group,
\item $(U,v)$ is defectless,
\item $v(X)=\min \Gamma_{>0}$.
\end{enumerate}

Set $M=\mathbb{F}_p((X))$. We know that:
\begin{equation}\label{as-formula}
M=\bigoplus_{i=1}^{p-1}M^pX^i + \wp(M) + \mathcal{O}_M, \tag{*}
\end{equation}
where $\wp$ is the Artin-Scherier map $x\mapsto x^p -x$  and $\mathcal{O}_M$ is the valuation ring of $M$ (see \cite{ku-elementary}, Lemma 1.2).
But the equality (\ref{as-formula}) fails to hold  for some extension  $N\supset M$, which is a  model of  $T_{naive}$. Hence $T_{naive}$ is incomplete.

Consider the polynomial $F(z_0,\ldots,z_{p-1}):=z_0^p-z_0 +\sum_{i=1}^{p-1} z_i^pX^i$. From the equality above  and using Hensel's Lemma, one can deduce that for any $a$, the set
$\{v(F(x)-a) \tq x \in M\}$ has a maximum in $\mathbb{Z}\cup \{\infty\}$: we say that the
image of $F$ has the {\it optimal approximation property} (see \cite{vdd-kuhlmann}). From this observation Kuhlmann suggests a candidate for a complete axiomatization
of the theory of $M$ in the language of rings,  which is essentially $T_{naive}$ together with sentences
saying that the image of every multi-variable additive polynomial has the optimal approximation property (see \cite{ku-elementary} and \cite{kuhlmanntame}).

Now we present our approach. We work in the language
$L_{\mathcal{O}}$.

Let us illustrate in an example the main ideas of the proof of Theorem \ref{th:pp2universal}, which states that modulo the theory $T_1$
every {\it positive primitive} formula is equivalent to a universal one. This theorem immediately shows that every completion of
$T_1$ is model-complete (see Section \ref{sec:baur-monk} and Corollary \ref{corr:baur-monk}).

Instead of considering the multivariate polynomial $F$ above, we use the equation (\ref{as-formula}) to study the image of the Artin-Scherier map $\wp$ and see the sum
$$C(M):=\sum_{i=1}^{p-1}M^pX^i$$ as {\bf the pseudo-complement}  to the image
$\wp(M)$. Note  that $$\wp(M)\cap C(M) \subseteq \mathcal{O}_M,$$ hence this intersection is {\it small}
with respect to the valuation metric; that is why we call $C$ the pseudo-complement. In addition, using Hensel's lemma we have that
$$\mathcal{O}_M \cap \wp(M)=\mathfrak{m}$$ where
$\mathfrak{m}$ is the maximal ideal $X\mathbb{F}_p[[X]]$.

At this point, we are able to describe the set $\wp(M)$, definable a priori by an existential formula, by the following formula:
\begin{equation}\label{elim-as}
 \phi(x): \forall y  \mesp ((\exists z \mesp \wp(z)=x-y\wedge y\in C + \mathcal{O}) \rightarrow y \in \mathfrak{m}).   \tag{**}
\end{equation}
 It is easy to see that this formula is equivalent to a universal
 formula in the language $L_\mathcal{O}$ (hence also in the language of rings with the parameter $X$) since
 $C$ is existentially definable.

Motivated by the above example, we consider the general case. We introduce the notion of a ball (in this section) as an $L_\mathcal{O}$-formula of the form
$$B(x): x.X^\gamma \in \mathcal{O}$$ for some fixed $\gamma \in \mathbb{Z}$. Let $Q$ be a  finite set of  polynomials (scalars from $R$).  Abusively, we also denote by $Q$ the formula which defines the sum of the images of $q\in Q$.  Our strategy is to assign to $Q$ a positive existential  formula $D$ and balls $B_1,B_2$ in a computable way,  such that in every model $N\models T_1$,
    \begin{enumerate}[(i)]
    \item \label{task1} $D(N)  + Q(N)=N$ and $Q(N)\cap D(N) \subseteq B_2(N)$
    \item \label{task2} $B_1(N) \subseteq  B_2(N)$, such that  $B_1(N)\cap Q(N)$ is definable by a universal formula which depends only on $T_1$.
    \end{enumerate}
 For instance, the above example suggests that for $Q=\{\wp\}$,
 $B_1=\mathcal{O}$, $B_2=\mathcal{O}.X$ and
 $D=C + \mathcal{O}$ suit. Then  prove that, modulo $T_1$, every positive primitive formula is equivalent to a universal one (similar to (\ref{elim-as}) above).

In the general case, for the task (\ref{task2}),  we cannot always ensure that $B_1(N)\cap Q(N)$  is a ball, or more generally quantifier free definable in $L_\mathcal{O}$: to see this it is enough to consider the image of $x\mapsto x^p$. To handle  this fact, we introduce
{\bf $p$-th roots of coordinate functions}, called $\lambda$-functions. For instance, for $x\in \mathbb{F}_p((X))$ we write (using the fact that $1,X,\ldots, X^{p-1}$ is basis of
$\mathbb{F}_p((X))$ over the subfield  consisting of $p$-th powers)
$$ x=x_0^p + x_1^pX+ \dots + x_{p-1}^pX^{p-1}$$
    and define $$\lambda_i(x):=x_i \quad  (i=0,\ldots, p-1).$$

Notice that $\lambda$-functions are both universally and existentially definable in $L$. We prove then $B_1(N)\cap Q(N)$  is positive quantifier-free definable in the language $L$ together with $\lambda$-functions (see Theorem \ref{boule_up}).

This article is organized as follows:
    generalities and definitions about the non-commutative ring $R$, its matrix ring and $R$-modules are given
     Section 2. In the third section we prove an  equivalent version of Hensel's lemma (see Theorem \ref{hensel-onay}) -we do not know if it is really new-
    and use it to axiomatize what we call {\it henselian filtered $R$-modules}. This part  consists of finding a ball (like $\mathfrak{m}$ above) where the trace of the definable set considered can be defined by a positive quantifier-free formula in the language $L$ together with the $\lambda$-functions. In the fourth section, we introduce the notion of a valued module, and study   pseudo-complements in order to satisfy the task (\ref{task1}) above. In the final section we apply the previous results to obtain  the main theorems. Here we use some general model theory of modules (e.g. Baur-Monk elimination) and some recent facts from \cite{koeningsman-sylvie} and \cite{sylvyarno17}

   In the course of this article we introduce several languages and theories.  For the sections 2-4 (i.e. except the last section), it is worth noting that $K$ and $\mathbb{F}_p((X))$, considered as structures with respect to  these languages are  models of all of these theories. When we say that something is {\it computable} we mean that there is a Turing machine which can compute it (see for instance \cite{koenigsmann2014undecidability} for the related definitions about computability).

\subsection*{Acknowledgements}
This is a long-standing work, which began with my Ph.D thesis in early 2007. I would like to thank  my advisor Fran\c{c}oise Point for suggesting me the topic and  for giving the idea to introduce $\lambda$-functions. I thank  Fran\c{c}oise Delon, also my advisor; without her help, this work could not be finished.
I want to thank  Franz-Viktor Kuhlmann for having invited me, as early as  2008 and for sharing his ideas about the topic. Lou van den dries and Matthias Aschenbrenner have kindly discussed with me  some aspects of the present article in last times. Finally, I want to thank  Sylvy Anscombe and Arno Fehm for their encouragement to write this manuscript.

\section{Preliminaries}

 Let $p$ be a prime number and $k$ a positive integer, we set $d:=p^k$. Let $K$ be the henselization of  the field of rational functions over $\mathbb{F}_d$, with respect to the $X$-adic valuation; denoted as $$K=\mathbb{F}_d(X)^{h}.$$ Let
$$\vfi:x\mapsto x^{d} $$ be the $k$-th power of the Frobenius  endomorphism of $K$. Recall that $K$ is a finite extension of $K^{\vfi}$ of dimension $d$. \footnote{Note that the contents of this section can be generalized to any field  and  to any endomorphism
satisfying  similar hypotheses. In particular one can
consider $\mathbb{F}_q(X), \mathbb{F}_q((X)), \ldots etc$. As it is our main interest and for readability reasons,  we preferred to stick to one case.} We fix the notation $v_K$ for the $X$-adic valuation on $K$.

  Set $\alpha(0):= \{1\}$ and let $$\alpha:=\alpha(1)=(\alpha_0,\alpha_1,\ldots, \alpha_{d-1})$$ be  a basis of  the $K^{\vfi}$-vector space of $K$. One can think of the basis $(1,X, \ldots, X^{d-1})$.    It is easy to see that for all $n\geq 1$, $\alpha$  induces canonically
the (ordered -) basis $\alpha(n)$ of the $K^{\varphi^{n}}$-vector space $K$. We identify $d^n$ with the set of functions $\{0,\dots,n-1\} \to \{0,\dots,d-1\}$ ordered lexicographically. We denote by $\alpha_k$ the $k$-th element of the basis $\alpha(n)$.

We define the ring \begin{equation}
R:=K\langle t  \mid ta^\vfi=at \rangle .
\end{equation}
the  (right-) $K$-algebra with the indeterminate $t$, subject to the commutation rule
$ta^\vfi=at$ for all $a \in K$. When
$k=1$, $R$ is isomorphic to the ring of additive polynomials over $K$, equipped with addition and composition. Every non zero  $q\in R$ can be  written as
\begin{equation}\label{R-polynomial}
q=t^na_n + t^{n-1}a_{n-1} + \dots + a_0
\end{equation}
where the $a_i$ are from $K$ and $a_n\neq 0$. Since the image of $q$ in $K[T]$ is
$$Q(T)=a_nT^{d^n}+\dots+a_0T$$ under the aforementioned isomorphism, ``the constant term $a_0$ of $q$" gives rise to the linear term $a_0T$ of $Q(T)$.
\begin{defn}
	An element  $q$ as in (\ref{R-polynomial}) is said to be separable if $a_0\neq 0$.
\end{defn}
\begin{remark}
	Note that $q$ is separable if and only if its image $Q(T) \in K[T]$ is separable.
\end{remark}
The integer $n$ in the expression (\ref{R-polynomial}) is called the degree of $q$ and we set the degree of $0$ as $-1$.
With  the degree function $R$ is right euclidean: for every non zero $r,q \in R$ there exists
$q'$ and a unique $r'$ of degree $<\deg(q)$ such that $r=qq' +r'$. As a consequence,
least common multiple $\lcm(q,r)$ and greatest common divisor $\gcd(q,r)$ are well defined.
In particular $R$ is right Ore. Note that if $\vfi$ is onto then $R$ is also left euclidean and
if it is not (which is the case that we consider here), $R$ is not even left Ore. The reader can see \cite{cohn} or \cite{valmod1} Section 2, for more details.

For our quantifier elimination Theorem \ref{boule_up} we will import some definitions and results from \cite{ddp1} on the ring $R$ and
on $R$-modules. We give the proofs of  some results not to be self-contained but rather, to initiate the reader with the style of the computations that will permit us to refer   easily  to \cite{ddp1}.
\begin{fait}
	The field $K$ is recursively enumerable since $\mathbb{F}_d(X)$ is, and $K$ is formed by adding to $\mathbb{F}_d(X)$ the unique root  of the  each polynomial satifying the hypothesis of the Hensel's Lemma . Consequently the ring $R$ and its matrix rings are recursively enumerable.
\end{fait}

\begin{lemma}\label{lem:decompq}
Let $q\in R$ and $n$ be a positive integer. Let $\alpha$ be a basis of the
$K^\vfi$-vector space $K$. Then
\begin{enumerate}[1.]
\item  \label{qi} $q$ can be uniquely written as:
  \begin{equation}
\sum_{i \in d^n} q_i\alpha_i \quad (q_i \in R),
\end{equation}

\item for all $n>0$ there is an endomorphism $\nqrt{n}{\cdot}:(R,+) \to (R,+)$ such that
    if  $q=\sum_{i \in d^n} q_i\alpha_i$  then
        \begin{equation}
t^nq = \sum_{i \in d^n} \nqrt{n}{q_i}t^n\alpha_i.
    \end{equation}
In addition, if $q$ is separable there exists $i \in d^n$ such that
    $\nqrt{n}{q_i}$ is separable.
\end{enumerate}\medskip
\end{lemma}
\begin{proof}
\item[1.]  Let $0 \leq k\leq s=\deg{q}$. For each monomial $t^ka_k$, by expressing $a_k$ with respect to the basis $\alpha(n)$,  $q$ can be written as
$$t^k(\sum_{i\in d^n} a_{k,i}^{\varphi^n}\alpha_i) .$$ Set
$q_{k,i}=t^ka_{k,i}^{\varphi^n}$ and  $q_i = \sum_{0\leq k \leq s} q_{k,i}.$
So we get
$$q=\sum_{i \in d^n}q_i\alpha_i.$$

\item[2.] Let $a \in K$. Write
$$a=\sum_{i \in d^n} a_i^{\vfi^n}\alpha_i .$$ We define
$$\nqrt{n}{a}:=\sum_{i \in d^n} a_i\alpha_i$$ and extend it to $r \in R$ by applying it to its coefficients. By definition $\nqrt{n}{\cdot}$ preserves the addition.

To see that \begin{equation}\label{nqrt}
t^nq = \sum_{i \in d^n} \nqrt{n}{q_i}t^n\alpha_i
\end{equation} it is enough to show that
$$t^nq_i=\nqrt{n}{q_i}t^n .$$
Note that
 $\nqrt{n}{q_{k,i}}=t^ka_{k,i}$ and hence
 $$\nqrt{n}{q_{k,i}}t^n=t^{n+k}a_{k,i}^{\vfi^{n}}=t^{n}q_{k,i} .$$
Using additivity,
we get $\nqrt{n}{q_i}t^n=t^nq_i$.

Suppose now $q$ is separable. If none of the  $\nqrt{n}{q_i}$ is separable, then  for all  $i$, we have  $\nqrt{n}{q_i}=tq_i'$. But then
$t^nq \in t^{n+1}R$ by the equality (\ref{nqrt}). Hence
$q \in tR$,  which is a contradiction.
\end{proof}

\begin{defn}\label{matriss} An $m\times n$  matrix  $A=(q_{i,j})$ over $R$ is said to be
\begin{enumerate}[1.]
\item  lower triangular if,
    $j>i$ implies $q_{i,j}=0$,
\item   lower triangular diagonally  separable
if it is lower triangular, $n\leq m$ and the  $q_{ii}$ ($i\leq n$) are separable,
\item lower triangular separable  if $A=(A_1,0)$ where $A_1$ is an $m\times k$ lower triangular diagonally separable matrix,
and $0$ is the $m\times l$ null matrix with $k+l=n$.

\end{enumerate}
\end{defn}

\begin{prop}\label{mattri}
For any matrix $A$ there exists an invertible matrix  $P$  with coefficients
in  $\{0,1\}$ and an invertible matrix
$Q$ such that $PAQ$ is lower triangular.
\begin{proof}
See the Proposition $6.1$ in  \cite{ddp1}.
\end{proof}
\end{prop}

\subsection{$R$-modules}
For the rest of the article we will always understand the expression $R$-module as right $R$-module. In an $R$-module $M$, scalar multiplication will be denoted as $x.r$, for $x\in M$ and $r\in R$.
\begin{defn}
Let $M$ be an  $R$-module and $\beta$ a basis of the $K^\vfi$-vector-space $K$.
\begin{enumerate}[1.]
\item $M$ is said to be $t\beta$-decomposable if $x\mapsto x.t$ is injective and we have
\begin{equation}
M=\bigoplus_{i \in d}M.t\beta_i.
\end{equation}
(where $\oplus$ indicates the direct sum as abelian subgroups).
\item $M$ is said to be $t$-decomposable if it is $t\alpha$-decomposable with
$\alpha=(1,X,\dots, X^{d-1})$.  Furthermore, for $i\in d$ we then define
$\lambda_i(x)=x_i$ where
$$x=x_0.t\alpha_0 + \dots + x_i.t\alpha_i + \dots x_{d-1}.t\alpha_{d-1}.$$
\end{enumerate}\medskip
\end{defn}

\begin{remark}\label{lambdascompose}
The direct sum
$$M=\bigoplus_{i \in d}M.t\beta_i$$ induces the direct sum below for every positive $s$
$$M=\bigoplus_{i \in d^s} M.t^s\beta_i.$$
\end{remark}
\begin{nota}
For the rest of the article we set
$\alpha:=\{1,X,\dots,X^{d-1}\}$.
\end{nota}

\begin{remark}\label{extunideflambdas}
The functions $\lambda_i$ are both existentially and universally definable in the language of right $R$-modules:\begin{equation}
y=\lambda_i(x) \iffl \forall x_1, \dots,\forall x_i, \dots, \forall x_d \quad x=\sum_{j=1}^d x_i.t.\alpha_j \longrightarrow x_i=y
\end{equation}
and
\begin{multline}
y=\lambda_i(x) \iffl \exists x_1, \dots, \exists x_{i-1}, \exists x_{i+1}, \dots, \exists x_d \quad \\
x=\sum_{j=1}^{i-1} x_j.t\alpha_j + y.t\alpha_i + \sum_{j=i+1}^d x_j.t\alpha_j.
\end{multline}

For a positive $s$, by Remark \ref{lambdascompose} above, we get canonically the $\lambda$ functions of level $s$,
defined for all $i\in d^s$, in an obvious way.
\end{remark}

{\bf The language $L(\lambda)$.} For the rest of the article we let $L(\lambda)$ be the language of $R$-modules together with  the functions (-symbols) $\lambda_i$.

\begin{defn} We denote by $T_\lambda$ the $L(\lambda)$-theory of $t$-decomposable $R$-modules, that is, the theory of $R$-modules together with the axioms

{\bf $\lambda$-decomposition :}\label{ax:lambdas}
$$\begin{matrix}  \forall x &  x = \sum \lambda_i (x) . t \alpha_i \\
\forall  x \forall (x_i )_{(i\in d)} & (x = \sum_i x_i. t \alpha_i \to  \bigwedge_i x_i = \lambda_i (x)). \end{matrix}$$
\end{defn}

\begin{lemma}\label{Lambdaterms} In any $t$-decomposable $R$-module $M$, any $L(\lambda)$-term can be evaluated on the tuple $(x_i)_i$ from $M$,   as
$$\sum_i \sum_j \lambda_j(x_i).r_{ij}$$ where
$r_{ij} \in R$.
\end{lemma}
\begin{proof}
This is  Corollary 3.3 in \cite{ddp1}.
\end{proof}

\begin{lemma}\label{lem:passagelambdasterms}
 Let  $m > 0$, and $q_ j , q_ j' \in R$ such that  $q_j = t^m .q_j'$ . Then the equation
$\sum y_j.q_j = u$ is equivalent to
$$ \bigwedge_{i \in d^{m}}\sum_j y_i.\nqrt{n}{q'_{j_i}}=\lambda_i(u).$$
 in any $t$-decomposable $R$-module.
\end{lemma}
\begin{proof}
This is  Lemma 3.4 in \cite{ddp1}
\end{proof}

\begin{obs}
Let $q=(q_0, \dots, q_{n-1})$ be a non zero tuple from $R$. We set
$$e:=\min\{k \tq q_i \in t^kR, \mesp \text{for all}\, i\}.$$
Notice that  $e=0$ means that at least one of the $q_i$ is separable.

Suppose $e>0$, one can write
$$q_i=t^eq'_i=\sum_{k \in d^e} \nqrt{e}{q'_i}_{k}t^e\alpha_k .$$
Since at least one of the $q'_i$ is separable, for some $(i,k)$,
the polynomial  $\nqrt{e}{{q'_{i}}_{k}}$ is separable by Lemma \ref{lem:decompq} (2.).
Set $q_{i,k}= \nqrt{e}{q'_i}_k $  and  let $q^e_\lambda$ be the  $n\times d^e$-matrix $(q_{i,k})$ whose $i$-th line consists of
the sequence $(q_{i,k})_{k \in d^e}$.
By  this process we have replaced the tuple $q$ by a matrix which  has at least one separable coefficient. Iterating this process and using  Lemma \ref{lem:passagelambdasterms} above we get the following result.
\end{obs}

\begin{lemma}\label{ddp6.4}
Let  $A$ be a non zero $n\times k$ matrix over $R$. Then,
the system  $y.A = u$ is equivalent to
$$y.PQ = w(u)$$ modulo $T_{\lambda}$,
where $P$ is a permutation matrix (i.e. invertible with coefficient in $\{0,1\}$),  $Q$ is lower triangular separable  and $w$ is a tuple
consisting of $L({\lambda})$-terms.
\end{lemma}
\begin{proof}
This is a reformulation of  Lemma  6.4.4 in \cite{ddp1}.
\end{proof}

\subsection{Baur-Monk Elimination}\label{sec:baur-monk}

The following is a reminder of Theorem A.1.1, Corollary A.1.2 and the discussion which follows in \cite{hodges}, p. 653-656.

Let $\mathcal{L}$ be any language which contains the language $\{+,-,0\}$ of abelian groups.  A {\it positive primitive formula} ($p.p.$) $\phi$ of
$\mathcal{L}$, is the one of the form
$$\exists \bar{y} \mesp \left(\bigwedge_i \psi_i(\bar{x},\bar{y})\right)$$
where the $\psi_i$ are atomic.

A {\it group-like $\mathcal{L}$-structure} $A$, is an $\mathcal{L}$-structure whose base set is a group
with respect to $\{+,-,0\}$. A {\it basic formula} for an $\mathcal{L}$-theory $T$, is a $p.p.$ formula which defines  a subgroup of the corresponding cartesian power of any $\mathcal{L}$-structure $A\models T$.

Note that if $S$ is any ring and $\mathcal{L}$ is the language of $S$-modules then in any $S$-module $N$ any $p.p.$ formula defines a subgroup
of the corresponding cartesian power of $N$, hence any $p.p.$ formula is basic for the theory of $S$-modules.

Let $T$ be an $\mathcal{L}$-theory such that every model of $T$ is group-like,
and every $p.p.$ formula is a basic formula for $T$. An invariant
$\mathcal{L}$-sentence  of $N\models T$ is  an $\mathcal{L}$-sentence
$\Theta$ satisfied by $N$,
such that for some   $p.p.$ formulas $G(x)$ and $H(x)$ of one variable $x$,   for some $m\in \mathbb{N}$,
$$T\models \Theta \Longleftrightarrow  \vert G/G\wedge H \vert =m$$
where the right-hand side of the equivalence is an abbreviation of the formula:
\begin{multline}
\exists  (x_i)_{(i=1..m)}  (\bigwedge_i G(x_i) \wedge \\
\bigwedge_{i\neq j} \neg (G\wedge H)(x_i-x_j)\wedge \forall z \mesp (G(z) \rightarrow \bigvee_i (G\wedge H)(z-x_i))).
\end{multline}
An invariant sentence is an invariant sentence of some $N$.
\begin{theorem}[Baur-Monk]\label{thm:baur-monk}
Every $\mathcal{L}$-formula is equivalent modulo $T$ to a boolean combination of
$p.p.$ formulas and invariant sentences. Hence for all models $N,M \models T$,
$N \equiv M$ if and only if $N$ and $M$ have same invariant sentences.
\end{theorem}

\begin{corr}\label{corr:baur-monk}
A completion of $T$ is model-complete if and only if every $p.p.$ formula is equivalent to a universal formula modulo $T$.
\end{corr}

\section{The tropical action of $R$ on $\mathbb{Z}$ and Filtration}

We recall some elementary facts about henselian valued fields. Let $(F,v)$ be a valued field with value group $\mathfrak{G}$ and valuation ring
$\mathcal{O}$. We set $\Gamma=\mathfrak{G}\cup\{\infty\}$,    extend the usual addition of $\mathfrak{G}$ to $\Gamma$ by letting
$$
\infty+\infty=\infty + a=a+\infty=\infty
$$ for all $a \in \mathfrak{G}$.
 We recall
the {\it tropicalisation} of a one variable polynomial $Q(T)$ over $F$:
Write $$Q(T)=\sum a_iT^i.$$ Then tropicalisation of $Q$ is the map
$$Q_v: \Gamma \to \Gamma$$
$$\gamma \mapsto  \min_i\{i\gamma + v(a_i)\}.$$

A {\it jump value} (or a tropical zero) of $Q$ is some $\gamma \in \Gamma$ such that
$$ \vert \{i \tq  i\gamma + v(a_i)= Q_v(\gamma)\} \vert \geq 2 .$$
We denote by $\jump{Q}$ the set of jump values of $Q$. Note that this set is finite and has at most $n-1$ element if $Q$ is of degree $n$.

\begin{fait}[Newton's Lemma] Let $(F,v)$ be a valued field and $f$ be a polynomial with coefficients from the valuation ring
$\mathcal{O}$. Consider the following property $h(f)$ of $f$,
\begin{multline}
 v(f(0))>2v(f'(0)) \Rightarrow (\exists b \mesp
 f(b)=0 \quad \text{and} \quad v(b)=vf(0)-vf'(0)) \tag{{\it h(f)}}
\end{multline}
Then $(F,v)$ is henselian if and only if $h(f)$ holds for every $f$ over the valuation ring $\mathcal{O}$ of $F$.
\end{fait}
\begin{proof}
See \cite{prestel} Theorem 4.1.3.
\end{proof}

Consider a polynomial $G$,  such that
$G(0)=0$ and $G'(0)\neq 0$. Then $G$ is of the form
$$G(T)=aT + \text{sum of monomials of higher degree}.$$
Consider the set $A_1:=\{ \gamma \tq  G_v(\gamma) <   (G-aT)_v(\gamma) \}$. This is a non-empty final segment of $\Gamma$. Also let
$$A_2:= \{\gamma + v(a) \tq \gamma \in A_1\}.$$
We set
$B_1(G):=v^{-1}(A_1)$ and $B_2(G):=v^{-1}(A_2)$.

Note that $B_i$  $(i=1,2)$ are convex subsets of $F$, that is, inverse images by $v$ of convex subsets of $vF$.

\begin{theorem}\label{hensel-onay} $(F,v)$ is henselian if and only if  each polynomial $G$ with coefficients in the valuation ring  $\mathcal{O}$, such that
$G(0)=0$ and $G'(0)\neq 0$,  induces a bijection $B_1(G) \to B_2(G)$.
\end{theorem}
\begin{proof}
Suppose $(F,v)$ is henselian. By the definitions of $B_1$ and $B_2$ we have
$G(B_1)\subseteq B_2$. We will show the converse inclusion. Let $z \in B_2$ and set $f=G-z$. We have
$vf(0)=v(z)=\gamma + v(a)$ where $\gamma \in A_1$. Hence
$ G_v(\gamma) =\gamma + v(a) <  (G-aT)_v(\gamma)$. Since a monomial of $G-aT$ is at least of degree $2$ and $G$ is over $\mathcal{O}$, we have
$$\gamma + v(a)<2\gamma.$$
It follows that $\gamma > v(a)$ and $v(z)>2v(a)$. By the above fact, there is a root $b$ of
$f$, of valuation $\gamma$. In other words, $G(b)=z$ and hence $G(B_1) \supseteq B_2$.

Now we show that $G \restriction B_1$ is 1-1. Let $z$ and $f$ be as above and $G(x_1)=G(x_2)=z$ with $x_1,x_2 \in B_1$. Then   $x_1,x_2$ have the same  valuation $\gamma$, but it is well-known that $f$ has at most one root of valuation $\gamma$ (or it can be proven by using Taylor expansions that $v(G(x_1-x_2))>G_v(v(x_1-x_2)$ but this is not possible since $v(x_1-x_2)$ is bigger than then all the jump values of $G$).

The converse is similar and easy (and we won't need).
\end{proof}

\begin{corr}
	Let $G$ be any polynomial over $(F,v)$ such that $G(0)=0$ and $G'(0)\neq 0$ then there exist convex subsets $B_1$, $B_2$ such that
$G\restriction B_1:B_1 \to B_2$ is a bijection.
\end{corr}
\begin{proof}
Divide $G$ by the  coefficient which has the minimal valuation among the coefficients of $G$. Let $H$ be the obtained polynomial (which has coefficients over $\mathcal{O}$). Then $B_1:=B_1(H)$ and $B_2:=aB_2(H)$ suits.
\end{proof}

\begin{nota}
	Let $G$ be as above. If the value group of $F$ is discrete of rank 1, then $A_1,A_2$ are closed intervals; in this case, we set $$\h(G):=\min A_1 \quad \text{and} \quad \hens(G):=\min{A_2}.$$
\end{nota}
{\bf The tropical action of $R$.} From now on we set
$$\Gamma:=\mathbb{Z}\cup \{\infty\}$$ and we equip $\Gamma$ with a right action of $R$ using tropicalisations:

Let
$q=\sum_i t^ia_i \in R$.  We define the tropicalisation
of $q$, as the tropicalisation of the $Q(T)=\sum_i a_iT^{d^{i}}$, that  is,  as  the map
\begin{equation}
\cdot q:\Gamma \to \Gamma; \quad \gamma \mapsto \gamma\cdot q= \min_i\{d^i\gamma + v(a_i)\}.
\end{equation}
In particular
$\cdot q$ is strictly increasing  if  $q\neq 0$.
Note that $\gamma \cdot a =  (aT)_v(\gamma) =\gamma + v(a)$ for $a \in K$.

We set  $\jump{q}:=\jump{Q}$, $\h(q):=\h(Q)$ and $\hens(q):=\hens(Q)$. For instance,
if $q=t-1$ then $A_1=A_2:=\{\gamma \in \Gamma \tq \gamma>0\}$. Hence
$\h(q)=\hens(q)=1$.

For our interests, we also introduce the tropicalisations of the $\lambda$-functions. Remark that we have for all $x\in K$,
$x=\sum_{i \in d} \lambda_i(x)^dX^i$,
       $$v_K(x)=\min_i\{v_K(\lambda_i(x)^d)+i\},$$
and the minimum is attained for a unique $i\in d$.
We define
$$ \lambda_i(\gamma):=\begin{cases} \frac{\gamma - i}{d}\quad \text{if} \mesp \gamma \in d\mathbb{Z} + i \\
        0 \quad \text{else}
        \end{cases}$$
and we set
$$\lambda(\gamma):=\sum_i \lambda_i(\gamma).$$

\begin{remark}\label{rem:lambdaregular} For $x\in K$
$$v_K(x)\geq \gamma \Leftrightarrow \quad \text{for all}\mesp i\in d \quad v_K(\lambda_i(x)) \geq \lambda(\gamma).$$
\end{remark}
\begin{proof} Note that $\Leftarrow$ is clear and if $\gamma=v_K(x)$ then the assertion is trivial. Suppose $v_K(x)>\gamma$. Let $j$ be such that
$v_K(\lambda_j(x).t^{d}\alpha_j)=dv_K(\lambda_j(x))+j=v_K(x)$. Let $i$ be such that
$\gamma=d\lambda(\gamma)+i$.  It follows that
$$d(v_K(\lambda_j(x))-\lambda(\gamma))>j-i.$$
Hence $v_K(\lambda_j(x))-\lambda(\gamma)>-1$ and
$v_K(\lambda_j(x))-\lambda(\gamma)\geq 0$.
\end{proof}
\begin{corr}\label{corr:lambas(s)}For all non zero $x\in K$ and $\gamma \in \Gamma$,
$$v_K(x)\geq \gamma \Leftrightarrow \quad \text{for all}\mesp i\in d^s \quad v_K(\lambda_i(x)) \geq \lambda^s(\gamma)$$
where $\lambda^s=\underbrace{\lambda\circ\ldots\circ\lambda}_{s-times}.$
\end{corr}

\begin{remark}Let $\gamma\in \Gamma$ and $r,q \in R$. Then
\begin{enumerate}[1.]
\item $\gamma\cdot rq = (\gamma \cdot r)\cdot q$,
\item $\gamma\cdot (r + q)\geq\min\{\gamma\cdot r, \gamma \cdot q\}$,
\item $\cdot r$ is strictly increasing for all non zero $r$,
\item $\infty \cdot r =  \gamma \cdot 0 = \infty$ for all $r$ and $\gamma$.
\end{enumerate}
\end{remark}
\begin{proof}
This follows by direct computations from the definition. Note that (1) follows more generally from  \cite{gonenc}, Corollary 4.1.14.
\end{proof}

\begin{fait}
	The theory of $\Gamma$ together with the tropical action of $R$ and
	$\lambda$, is decidable since this structure  is definable in the ordered abelian group structure of $\mathbb{Z}$, together with $\infty$ and constants for the elements of $\mathbb{Z}$.
\end{fait}

\begin{defn}
	We call this structure the tropical structure of $\Gamma$.
\end{defn}
\subsection{Quantifier Elimination Near 0}

\subsection*{\bf The languages $L_{\mathcal{O}}$ and $L_{\mathcal{O}}(\lambda)$} The language
$L_{\mathcal{O}}$ is obtained by adding to $L$ a unary predicate $\mathcal{O}$, and the language $L_\mathcal{O}(\lambda)$ is the language $L_\mathcal{O}$ together with the functions (-symbols) $\lambda_i$ added to $L_\mathcal{O}$. We want to study the divisibility conditions for $t$-decomposable $R$-modules which can be seen
as  properties reflecting a kind of henselianity, analog to Theorem \ref{hensel-onay} just above.

\begin{nota}
For the rest of this article, for $\gamma \in \Gamma$,
we write $P_\gamma$
 for the predicate  $\mathcal{O}.X^{\gamma}$, i.e. in any $L_\mathcal{O}$-structure
 $M$, $x\in P_\gamma$ if and only if $x.X^{-\gamma} \in \mathcal{O}$.
\end{nota}
\begin{defn}
An henselian filtered module, is a $t$-decomposable $R$-module which is an $L_\mathcal{O}(\lambda)$-structure satisfying the following axioms:

\item[0. {\bf Balls:}] $P_{\infty}=\{0\}$  and the $P_{\gamma}$ form a chain of subgroups decreasing with $ \gamma$  such that  the  inclusions are proper.
\item[1. {\bf Ultrametric:}]  $\forall x  \forall y  \mesp x \in P_{\gamma}\wedge y \in P_{\delta}  \rightarrow (x.r + y) \in P_{\min\{\gamma\cdot r,\delta\}}$   for all $\gamma,\delta \in \Gamma$ and $r\in R$.
\item[2. {\bf Regularity:}]   $\forall x \mesp  x \in P_{\gamma} \leftrightarrow x.r \in P_{\gamma \cdot r}$, for all $\gamma$ and $r\neq 0$ such that
$\gamma \notin \jump{r}$.
\item [3. $\lambda$-{\bf regularity:}]  $\forall x \, x\in P_{\gamma} \leftrightarrow \bigwedge_i \lambda_i(x) \in P_{\lambda(\gamma)}$, for all $\gamma$.
\item [4. {\bf Henselianity:}] $\forall x  \in P_{\hens(s)}\setminus \{0\} \exists ! \, y \in P_{\h(s)} \mesp y.s=x$ for all separable $s$.

\end{defn}
Note that axiom 3 implies that
$$\forall x \, x\in P_{\gamma} \leftrightarrow \bigwedge_{i\in d^s} \lambda_i(x) \in P_{\lambda^s(\gamma)},$$ a consequence analog to the one expressed in Corollary
\ref{corr:lambas(s)}.
\begin{nota}
	We denote by $T_{hens}$ the theory of henselian filtered modules.

We also isolate some theories of the $R$-modules that are already considered in \cite{ddp1}:
\begin{itemize}
	\item we denote by $T_{free}$ the theory $T_\lambda$ together with the following (scheme of-) axioms:
	\label{ax:separablyclosed} $$\forall x \exists y \mesp x=y.s,$$
	 for all separable $s$,
	\item  we denote by $T^{0}_{free}$, the $L_{\lambda}$-theory of torsion-free non zero models of $T_{free}$.
\end{itemize}
\end{nota}
\begin{remark}\label{P_M} In a henselian filtered module $M$, we denote by
$P_M$ the subgroup defined by  the intersection
$$P_M:=\bigcap_{\gamma\neq \infty} P_\gamma(M).$$
Note that by  ultrametric and regularity axioms, $P_M$ is a
$L(\lambda)$-substructure of $M$ and it is torsion-free as an $R$-module. Moreover, it is straightforward to check that by Hensel's axioms
$$P_M \models T^0_{free}.$$
\end{remark}
\begin{proposition}\label{prop:t0freedecidable}  Given $L$-$p.p.$ formulas	$\mathfrak{a}(x)$ and $\mathfrak{b}(x)$  of one variable $x$,
the quotient $\mathfrak{a}/(\mathfrak{a}\wedge \mathfrak{b})$ is either trivial or infinite in every model of $T^0_{free}$. In particular $T^0_{free}$ is complete and decidable.
	\begin{proof}
This follows from Lemma 6.8 in \cite{ddp1}.
	\end{proof}

\end{proposition}

\begin{defn}
	A ball of $L_\mathcal{O}$, is an atomic formula $W(x_1,\ldots,x_k)$ of the form
	$$W(x_1,\ldots, x_k): \bigwedge_{i=1}^k x_i \in P_{\gamma_i}.$$ We will write it rather as a product
	$$W=P_{\gamma_1}\times \dots \times P_{\gamma_k}$$ of predicates.
	$W$ is said to be proper if none of the ${\gamma_i}$ is equal to $\infty$.
\end{defn}

\begin{remark}
	A  positive primitive  formula $\phi(\bar{x})$ of $L_{\mathcal{O}}$ is equivalent to one in the form
	$$\exists \bar{y} \mesp\bar{x}.B-\bar{y}.A\in \mathcal{O}^k\times\{0\}^n$$ which it self is equivalent to a formula
	$$\exists \bar{y} \mesp\bar{x}.B'-\bar{y}.A'\in W$$
	where $A,B,A',B'$ are matrices over $R$ and $W$ is a ball. Hence any $p.p.$ formula is a basic formula. For our purposes will rather use the latter
	equivalence.
\end{remark}

Note that modulo $T_{hens}$, the set of  definable sets  by an $L_{\mathcal{O}}$-$p.p.$ formula contains the set
of definable sets by an $L$-$p.p.$ formula
since $W$ can be
chosen equal to $P_{\infty}^k$ for some $k$.

In this section, we will prove the following:
\begin{theorem}\label{boule_up}
	Let $\phi(x_1,\dots,x_m)$ be a $p.p.$ formula of
	$L_\mathcal{O}$. Then there is some computable $\delta\neq \infty$ and some positive quantifier free
	$L(\lambda)$-formula $\psi$ which depends only on the theory $T_\lambda$, such that, with
	$V=P_\delta^m$,  we have
	$$T_{hens} \models \phi \wedge V \leftrightarrow \psi \wedge V .$$
	Moreover,
	If $\phi_1$ is another $p.p.$ formula with the same arity, and
	if $\phi$ and $\phi_1$ are equivalent modulo $T^0_{free}$, then for some computable $\gamma$ and with $W:=P_\gamma^m$
	$$ T_{hens} \models \phi\wedge W \leftrightarrow \phi_1 \wedge W .$$
\end{theorem}

\begin{lemma}\label{croissant} Given $\delta\neq \infty$ and an $L(\lambda)$-term $u(\bar{x})$, there exists $\gamma\neq \infty$ such that
\begin{equation}
T_{hens} \models \forall \bar{x} \mesp \bar{x} \in P_\gamma^{\vert \bar{x} \vert} \rightarrow u(\bar{x}) \in P_{\delta}.
\end{equation}
\end{lemma}
\begin{proof} Put $u(\bar{x})$ in the form $\sum_{i,j} \lambda_j(x_i).r_{ij}$ using Lemma \ref{Lambdaterms}.
    By regularity, $\lambda$-regularity and ultrametric axioms, for any $\rho$,
$$\bar{x} \in P_\rho^{|\bar{x}|} \rightarrow u(\bar{x}) \in P_{\min\{\lambda(\rho)\cdot r_{ij}\}}$$ holds. Since all the tropical functions $\lambda$ and $\cdot r_{ij}$ are unbounded and increasing,
    one can choose $\gamma$ such that $\gamma \cdot r_{ij} \geq \delta$ for all $i$.
\end{proof}
\begin{remark}\label{rem:gamma_computable}
	The value $\gamma$  is computable from $\delta$ and $u(\bar{x})$ in the tropical structure of $\Gamma$ .
\end{remark}

As a consequence of the henselianity axioms, we  observe the following.
\begin{lemma}
Given  $\gamma\neq \infty$ and a separable $s$ there exists  $\delta\neq \infty$ such that

$$ T_{hens} \models \forall y \mesp  [(y \in P_\delta) \rightarrow (\exists x  \in P_\gamma \wedge x.s=y)].$$

\end{lemma}
\begin{proof}
If $ \gamma \leq \h(s)$ then set $\delta=\hens(s)$. Otherwise $\gamma \notin \jump{s}$ and by the
regularity axioms $x \in P_{\gamma} $ if and only if $x.s \in P_{\gamma \cdot s}$.
Since $P_\gamma \subseteq P_{\h(s)}$, $P_{\gamma \cdot s} \subseteq  P_{\hens(s)}$.
Hence if $y \in P_{\gamma \cdot s}$ then the unique solution $x$ such that $x.s=y$ lies in
$P_\gamma$.  So $\delta=\gamma \cdot s$ fits for our requirements.
\end{proof}

\begin{remark}\label{rem:delta_computable}
	As above, $\delta$ is computable in the tropical structure $\Gamma$.
\end{remark}

\begin{corr}\label{bouledivisible}
Let $A=(a_{ij})$  be an $m\times k$ lower triangular diagonally separable matrix (in particular  $k \leq m$). Then,
for every proper ball $W$, there exists a proper computable ball $W_1 $ such that
$$T_{hens} \models \forall \bar{x} \mesp (\bar{x}\in W_1 \rightarrow \bar{x}\in  W.A) .$$
\end{corr}
\begin{proof}
Write $W = \prod_{i=1}^{m} P_{\delta_i}$. By Remark \ref{croissant},  for fixed $i$, chose  $\gamma_i  \in \Gamma\setminus\{\infty\}$ such that,
$$\sum_{j \neq i} P_{\gamma_i}.a_{ji} \subset P_{\delta_i}.$$
 Since the $a_{ii}$ are separable, by the above lemma there exist proper balls $U_i$ such that whenever   $z_i \in U_i$,
 there exists $y_i \in P_{\gamma_i}$, such that $z_i = y_i.a_{ii}$; hence
$$z_i - \sum_{j=1}^{n} y_i.a_{ji}= \sum_{j\neq i} y_i.a_{ji} \in P_{\delta_i}$$ for all $1\leq j \leq k$. Take $W_1=
\prod_{i} U_i.$
\end{proof}

{\bf Proof Theorem \ref{boule_up}.}
Let $\bar{y}:=(y_1,\dots,y_k)$, $\bar{x}:=(x_1,\dots,x_m)$,  $\phi(\bar{x})$,
    $W:=\prod P_{\gamma_i}$ and
$$\phi(\bar{x}): \exists \bar{y} \mesp \bar{x}.B -\bar{y}.A \in W.$$

Let $I:=\{i \tq \gamma_i=\infty \}$ and $J:=\{j \tq \gamma_j\neq \infty \}$,
 and $A_I$  be the matrix formed by the columns $C_{i \in I}$ (resp.
     $A_J$ be the matrix formed by the columns $C_{j\in J}$ of $A$). We set
     $u(\bar{x}):=\bar{x}.B$ and  by $u_I(\bar{x})$ (resp.  $u_J(\bar{x})$) we denote the
     tuple formed by $I$-coordinates (resp. $J$-coordinates) of $u(\bar{x})$. We may assume that $\phi$ is of the form
     \begin{equation}\label{firstred}
      \exists \bar{y} \mesp \left( {u_I(\bar{x})} = \bar{y}.A_I \wedge u_J(x) -
\bar{y}.A_J \in W_J \right)
     \end{equation}
where $W_J$ is the obvious projection of $W$ to its non zero coordinates.
By Lemma \ref{ddp6.4}, there exists a lower triangular separable matrix  $\widetilde{A}_I=(S,0)$
 such that the formula
 $$u(\bar{x})=\bar{y}.A_I$$
is equivalent, modulo $T_\lambda$, to
$$
(t_1(u_I(\bar{x})),\dots,t_{n-l}(u_I(\bar{x})))=\bar{y}.PS \wedge (t_{n-l+1}(u_I(\bar{x})), \dots ,t_{n}(u_I(\bar{x})))=0,
$$
 where $P$ is a permutation matrix and the $t_1(u_I(x)),\ldots t_n(u_I(\bar{x}))$ are some $L(\lambda)$-terms.

     By remark \ref{croissant}, chose a proper ball $U'$ such that $U'.A_J \subseteq W_J$
     and by Corollary \ref{bouledivisible} chose a proper ball $V'$ such that $V' \subseteq U'.PS$.

     By remark \ref{croissant} again, we choose $V$ such that for all
     $\bar{a} \in V$,
     $$(t_1(u_I(\bar{a})),\dots,t_{n-l}(u_I(\bar{a}))) \in V'$$ and  $u_J(\bar{a}) \in W_J$.

     Hence:
     \begin{equation} T_{hens} \models \forall \bar{x} \mesp  \left(\phi(\bar{x})\wedge \bar{x}\in V\right) \longleftrightarrow \left((t_{n-l+1}(u_I(\bar{x})), \dots ,t_{n}(u_I(\bar{x})))=0\wedge \bar{x}\in V\right).
     \end{equation}

 Denote by $\psi$ the formula
 $$(t_{n-l+1}(u_I(\bar{x})), \dots ,t_{n}(u_I(\bar{x})))=0.$$

 The first statement is now proved.

 Now if $\phi_1$ is another $p.p.$ formula, modulo $T^0_{free}$, $\phi_1$ is equivalent  to some positive quantifier free formula $\psi_1$. Since $T^{0}_{free}$ is decidable, we have an algorithm which checks if $T^0_{free} \models \psi \leftrightarrow \psi_1$.
 Modifying this algorithm we can remember the finitely many
 non zero $r \in R$, such that algorithm uses the axiom
 $$\forall x \mesp  x\neq 0 \to  x.r\neq 0,$$ and the finitely many separable $s$, such that, the algorithm uses the axiom
 $$\forall x \exists y \mesp y.s=x.$$ Choose $\gamma$ bigger than
 all the $\max \jump{r}$ and $\max\{\h(s),\hens(s)\}$ for all the $r$ and $s$ as above. Set $W:=P_\gamma^m$. Hence for any non zero $x \in P_\gamma$,
 $x.r\neq 0$ and there exists   $y \in P_\gamma$, with $y.s=x$. It follows that the same algorithm computes a proof of
$ W\wedge \psi(\bar{x}) \leftrightarrow W\wedge \psi_1(\bar{x})$ from $T_{hens}$.  In particular,
 we have $\phi_1 \wedge W \leftrightarrow W\wedge \phi(W)$ modulo $T_{hens}$.
 \begin{remark}
 	By passing to an $\omega_1$-saturated model
 	$M$, since $P_M\models T^0_{free} $, it is trivial that if $T^0_{free} \models \phi \leftrightarrow \phi_1$, then for some proper $W$, $$\phi(W) = \phi_1(M).$$ What we show above is that the decidability of $T^0_{free}$ yields the computability of $W$.
 \end{remark}

 \begin{corr}\label{corr:boule_up}
 	Let $\phi$ and $\phi_1$ be $p.p.$ formulas of $L_{\mathcal{O}}$ and $\psi$ and $\psi_1$ are given as in the proof of the above theorem. Then, there is a computable ball $W$ such that
 	$$\vert (\phi\wedge W)/(\phi_1 \wedge W)\vert =1,$$ or
 	for all proper ball $V \subseteq W$,
 	 	$$\vert (\phi\wedge V)/(\phi_1 \wedge V)\vert = \infty.$$
 \end{corr}
 \begin{proof}
 Follows by Proposition \ref{prop:t0freedecidable} and by the above theorem.
 \end{proof}

       \section{pseudo-complements}
       We will introduce the notion of a {\it valued module}  to study henselian filtered modules. These are  $R$-modules  $M$, equipped with a function $v:M\to \Gamma$ inducing the ultrametric topology.  After investigating the elementary properties of valued modules, we will get the consequences that can be expressible in the language $L_{\mathcal{O}}$.

\begin{defn}\label{def:valmod} A valued module  is a $t$-decomposable $R$-module together with a surjective map $v:M \to \Gamma$ such that for all $x,y\in M$,

\begin{enumerate}[1.]
\item  $v(x\pm y)\geq \min\{v(x),v(y)\}$
\item  $v(x)=\infty \leftrightarrow x=0$
\item  $v(x)\notin \jump{r} \to v(x.r)=v(x)\cdot r$,  for all $r\in R$.
\end{enumerate}\medskip
\end{defn}

\begin{remark}\label{rvR} Let $r=t^na + \dots + t^ka_k \in R$ where monomials are written following decreasing degrees,  then
\begin{enumerate}[1.]
\item $v(x.t^ia_i)=v(x)\cdot t^ia_i$, for all $x \in M$,
\item  $v(x.r)=v(x)\cdot t^ka_k <v(x.(r-t^ka_k))$ whenever $v(x)>\max \jump{r}$,
\item  $v(x.r)=v(x)\cdot t^na_n<v(x.(r-t^na_k))$ whenever $v(x)<\min \jump{r}$.
\end{enumerate}
\end{remark}
\begin{proof}
1. Follows from  Definition \ref{def:valmod} (3.) since a monomial has no jump value.

2. By Definition \ref{def:valmod} (3.), if $v(x)>\max \jump{r}$ then $v(x.r)=v(x)\cdot r$.
Let $\gamma > \max \jump{r}$. Then for some $i$, $\gamma\cdot r=\gamma \cdot t^ia_i<\gamma \cdot t^ja_j$ for all $j\neq i$.  In other words the line  $\{(\delta,d^i\delta +v(a_i)\}_\delta$ does not intersect any other line $\{(\delta,d^j\delta +v(a_j)\}_\delta$ in the area $(\max \jump{r}, \infty]\times \Gamma$. This can happen only if $i<j$ for all $j\neq i$. Hence $i=k$.

3. The proof is very similar to (2.)
\end{proof}

 Let $(M,v)$ be a  valued module.
We define the equivalence relation
$\mathbf{RV}$ on $M$ by
$$x \mathbf{RV} y \iffl v(x)=v(y)<v(x-y).$$ We denote the $\mathbf{RV}$-class of an element $x$ by $\rv{x}$ whereas $\rv{A}$ stands for the set $\{\rv{x} \tq x\in A\}$ for $A\subseteq M$.
We also set the notation $P_\gamma$  for the closed ball of radius $\gamma$ centered at $0$.

\begin{defn} For subgroups $A$ and $B$ of $M$, we say that $A$ and $B$
\begin{itemize}
\item   are $m$-immediate (m stands for \emph{mutually}) if $\rv{A\setminus P_\gamma}=\rv{B\setminus P_{\gamma}}$ for some $\gamma \in \Gamma$, and we write
$A \approx B$,
\item   are pseudo-orthogonal if $vA\cap vB \subseteq [\gamma, \infty]$ for some $\gamma \in \Gamma$, and we write $A | | B$.
\end{itemize}\medskip
\end{defn}

\begin{remark}
   $A\cap C \subseteq P_{\gamma}$ for some $\gamma$, whenever $A || C$.
\end{remark}
\begin{remark}\label{rem:modulogamma} If $A\approx B$ and $B\subseteq A$ then for some $\gamma$,
$A+P_\gamma=B+P_\gamma$.
\end{remark}
\begin{proof}
Let $\gamma$ be such that  $\rv{A\setminus P_\gamma}=\rv{B\setminus P_{\gamma}}$.
It is enough to show $\{x\in A \tq v(x)\leq \gamma\} \subseteq B+P_\gamma$.
We proceed by induction w.r.t. the dual order on the initial segment
$(-\infty ,\gamma]$:
Let $a\in A$. If $v(a)=\gamma$ there is nothing to do.
Suppose $v(a)<\gamma$ and for all $a' \in A$ of valuation $>v(a')$ there is some
$b' \in B$ such that $a-b'\in P_\gamma$. Since $A\approx B$, there is $b\in B$ such that
$v(a-b)>v(a)$. Since $b\in B$ and $B\subseteq A$, $a-b \in A$. Now by applying
the induction hypothesis to $a-b$ we have $a-b-b'\in P_\gamma$ for some $b'\in B$. That is, $a \in B+P_\gamma$.
\end{proof}
\begin{defn} A pseudo-complement of a subgroup $A$ is a subgroup $C$ such that,
    for some $\gamma$
    \begin{equation}
    M=A + C + P_\gamma \quad \text{and} \quad A|| C.
    \end{equation}
\end{defn}

\begin{remark}\label{rem:invimage}
	If $A,C$ and $P_\gamma$ are as above and $f:M\to M$ is an additive map such that $f^{-1}(P_\gamma) \subseteq P_\delta$ for some $\delta$, then $f^{-1}(C)$ is a pseudo-complement to $f^{-1}(A)$. In particular, this is the case when
	$f$ is given by a scalar multiplication.
\end{remark}

\begin{remark}\label{rem:+ballpseudocomp}
If $C$ is a pseudo-complement of $A$ then $C$ is a pseudo-complement
	of $A+P_{\gamma}$ for any $\gamma$.
\end{remark}

\begin{remark}\label{rem:samecomplement} It is straightforward to see that if $A$ and $B$ have the same pseudo-complement then $A \approx B$. The following lemma establishes the converse using that the value set is $\mathbb{Z}\cup \{\infty\}$.
\end{remark}
\begin{lemma}\label{changecomp} If $A\approx B$ then $C$ is a pseudo-complement of $A$ if and only if it is a pseudo-complement of
$B$.
\end{lemma}
\begin{proof}
Suppose $C$ is a pseudo-complement of $A$  satisfying
$$   M=A + C + P_\gamma$$
and $A\approx B$. Let $\delta$ be such that  $\rv{A\setminus P_\delta}=\rv{B\setminus P_\delta}$. We may assume that  $\delta \leq \gamma$. We claim that
$$M=B+C+P_\delta .$$

	We proceed by induction on $\{\beta \in \Gamma \tq \beta\leq \delta\}$ as in the proof of Remark \ref{rem:modulogamma}. If $v(z)=\delta$ then trivially $z \in B + C + P_\delta$. Let $z \in M$ and suppose that for all $x$ with $v(x)>v(z)$ there exist
$(b,c,x_\delta) \in B\times C\times P_{\delta} $ such that
$$x=b+c+x_{\delta}.$$
Write $z=a+c+z_\delta$ for some $(a,c,z_\delta) \in A\times C \times P_\delta$ (in fact $z_\delta$ can be chosen in $P_\gamma \subseteq P_\delta$). If $v(a)\geq \delta$ there is nothing to do. If $\delta>v(a)>v(z)$ then by induction hypothesis
$a=b+c'+z'_\delta$ with $(b,c',z_\delta) \in B\times C \times P_\delta$, and hence
$$z=b+c+c'+z_\delta + z'_\delta. $$ The only possibility which remains to be considered is $v(z)=v(a)<\delta$ since $A || C$ implies that
$v(a+c+z_\delta)=\min\{v(a),v(c)\}$ whenever $v(a)<\delta$. Pick  $b\in B$ such that
 $\rv{b}=\rv{a}$. Then, since $v(a-b)>v(a)$, by induction hypothesis the equality
$$a-b=b'+c'+z'_\delta$$ holds for some $ (b,c',z_\delta') \in B\times C \times P_\delta$. Hence $z \in B+C+P_\delta$.

Now to see that $B || C $ we claim that $vB \cap vC \in [\delta, \infty]$. Suppose for a contradiction that
for some $b \in B $ and $c\in C$, $v(b)=v(c)<\delta$. Then we can choose $a \in A$ such that
$v(a)=v(b)$ hence $v(a)=v(c) \in vA \cap vC$. But then $v(a) \in [\gamma, \infty]$.
This is a contradiction since $\gamma \geq \delta$.
\end{proof}

\begin{lemma} Let $A,A',B,B'$ be  such that
$A \approx A'$,  $B\approx B'$ and $A || B$. Then the following holds:
\begin{enumerate}[1.]
\item  $A' || B'$,

\item  every pseudo-complement of $A+B$ is a  pseudo-complement of $A'+B'$.
\end{enumerate}\medskip
\end{lemma}
\begin{proof}
1. Let $\gamma=v(a')=v(b')$ with $a'\in A$ and $b'\in B$. Let $\delta$ be such that $vA\cap vB \subseteq [\delta,\infty]$ and $\gamma_1$, $\gamma_2$ are respectively the values yielding $A\approx A'$ and $B\approx B'$. We claim that
$\gamma\geq \min\{\delta,\gamma_1,\gamma_2\}$. Let $a \in A$ and $b\in B$ be such that $\rv{a}=\rv{a'}$ and
$\rv{b}=\rv{b'}$. Then $\gamma=v(a)=v(b)$. Hence $\gamma$ must be $\geq \delta$.

2. Let $C$ be a pseudo-complement of $A+B$. Let $\delta,\gamma_1,\gamma_2$ be as above. Let
	$z=a+b$ with $a\in A$ and $b \in B$ such that $v(z)<\min\{\delta,\gamma_1,\gamma_2\}$. Then $v(a)\neq v(b)$ and $v(z)=\min\{v(a),v(b)\}$ necessarily. Suppose $v(a)=v(z)$.
Then $\rv{a}=\rv{z}$ hence for some $a'\in A'$, $\rv{z}=\rv{a'}$.  If $v(z)=v(b)$ we can choose  in the same way $b'$ such that $\rv{b'}=\rv{z}$. Hence $A+B \approx A' + B'$.  We apply now Lemma \ref{changecomp}.
\end{proof}

\begin{defn}
	A valuation independent basis $\beta$ of the $K^{\vfi^{n}}$-vector space $K$,  is a basis such that the members of $\beta$ have all different valuations in the finite set $\{0,\ldots,d^{n}-1\}$.
\end{defn}
For example, the basis  $\alpha$ of the $K^{\vfi}$-vector space $K$ is valuation independent.
\begin{remark}
If $\beta$ is a valuation independent basis of the $K^\vfi$-vector space $K$ then for all $s>0$, $\beta(n)$ is a valuation independent basis of $K^{\vfi^n}$-vector space $K$. Moreover,  any valued module is
$t\beta$-decomposable for all valuation independent $\beta$.
\end{remark}

\begin{lemma}
Let $g_1,\dots,g_m$ be all of the same degree $s$  such that the leading coefficients $b_1, \dots, b_m$ have distinct valuations  in
	$\{0,\dots,d^s-1\}$. Then $M.g_i || M.g_j$ whenever $i\neq j$.
\end{lemma}
\begin{proof}
	By \ref{rvR}, we have $v(x.g_i)=v(x)\cdot g_i=v(x)\cdot t^sb_i=p^{d^s}v(x)+v_K(b_i)$ whenever $v(x)<\jump{g_i}$ for $i=1\dots m$. In particular,
if $v(x)<\min\{\jump{g_i},\jump{g_j}\}$ for a $j\neq i$ then $v(x.g_i)\neq v(x.g_j)$.
\end{proof}
\begin{lemma}\label{lem:addtoMta} Let $\beta$ be a valuation independent basis of the $K^{\vfi}$-vector space $K$. Let $q=t^sa+\dots \in R$ be of degree $s$.
Then there is a unique $j=j(q)$ such that $\beta_j \in \beta(s)$ and $M.q \approx M.t^s\beta_j$. As a consequence $C:=\sum_{j' \neq j(q)} M.t^s\beta_{j'}$ is a pseudo-complement for $M.q$.
\end{lemma}
\begin{proof}
Write
$$a=\sum_{j \in d^s} a_j^{\varphi^s}\beta_j. $$

Let $j$ be such that $v_K(a)=v_K(a_j^{\vfi^s}\beta_j)<v_K(a_{j'}^{\vfi^s}\beta_{j'})$ for all $j' \neq j$ (such a $j$ exists since $\beta$ is valuation-independent).
Then $$v(x.t^sa)=v(x)\cdot t^s(a_j^{\vfi^s}\beta_j)<v(x)\cdot t^s(a_{j'}^{\vfi^s}\beta_{j'})$$ for all $x\in M$ and $j'\neq j$.

 Let $\gamma < \min
\jump{q}$. Then by Remark \ref{rvR}, for all $x$ of such that $v(x)\leq \gamma$, we have $\rv{x.q}=\rv{x.t^sa}$, hence
$$\rv{x.q}=\rv{x.t^sa_j^{\vfi^s}\beta_j}=\rv{(x.a_j).t^s\beta_j}.$$ Since $x\mapsto x.a_j$ is a bijection it follows that $M.q \approx M.t^s\beta_j$. The consequence follows by Lemma \ref{changecomp}.
\end{proof}

\begin{lemma}\label{vddku}
Given a non zero matrix $Q$ with coefficient over $R$, say with $k$ rows, there exists a matrix $Q'$ such that
the first column of $Q'$ consists of polynomials which have all the same degree $s$,
	such that the leading coefficients in this column have distinct valuations  in $\{0,\ldots, d^s-1\}$  and
$$M^k.Q =  M^k.Q' \mesp .$$
\end{lemma}
\begin{proof} This proof is essentially a slight generalization  of the proofs of
 Lemma 3 and Lemma 4 in \cite{vdd-kuhlmann}.

 Let $Q=(q_{ij})$ be a matrix over $R$ with $k$-many rows.
	We will proceed by induction on $f=\sum_{\{(i,j)\tq q_{ij}\neq 0\}}(\deg(q_{ij})+1)$.

	Since $Q$ is non zero, $f>0$. Suppose $f=1$. We may  assume
	$q_{11}=c \in K^{\times}$ and all other entries of $Q$ are zero. Then
	$\bar{x}=\bar{y}.Q$ for some $\bar{y}$ if and only if, all the coordinates of $\bar{x}$ except possibly the first one, are $0$. Hence we can take $Q'$ the matrix which has $1$ at the position
	$(1,1)$ and has all other entries equal to zero.

	Now we suppose $f>1$. Let $e:=\max\{\deg{q_{ij}}\}\geq 0$. We may suppose that $q_{11}$ has degree $e$.
Set $$e_{ij}:=\deg{q_{ij}},  \mesp e_i:=e_{i1}$$ and
$$c_{ij}:=\text{the leading coefficient of}\mesp q_{ij},  \mesp c_i=c_{i1}.$$

 \noindent
 {\bf Claim 1}: We may assume that for all $a_1,\ldots a_k \in K$, not all are $0$,
 $\sum_i a_i^{\vfi^{e_i}}c_i \neq 0$.

 Suppose $\sum_i a^{\vfi^{e_i}}c_i = 0$. We may also suppose that $a_1=1$.
 We define for all $j$,
 $$\widetilde{q_{1j}}=\sum_{i=1}^{k} t^{e-e_{i}}a_iq_{ij}.$$

Since $a_1=1$ and $e=e_1$
\begin{equation}\label{eq:matricepassage}
	\widetilde{q_{1j}}=q_{1j} + t^{e-e_2}a_2q_{2j} + \dots + t^{e-e_k}a_kq_{kj}.
\end{equation}
We also define $q'_{ij}$ by the equality
$$q_{ij}=t^{e_i}c_{ij} + q'_{ij}.$$

We claim that $\widetilde{q_{11}}$ has degree  $<e$: We have
\begin{equation}
 \widetilde{q_{11}}=\sum_{i=1}^k t^{e-e_i}t^{e_i}a_i^{\vfi^{e_i}}c_i + t^{e-e_i}a_iq'_{i1} = \sum_{i=1}^k t^ea_i^{\vfi^{e_i}}c_i + t^{e-e_i}a_iq'_{i1}.
\end{equation}
Since each $q'_{i1}$ has degree $<e_i$, each  $t^{e-e_i}a_iq'_{i1}$ has degree $<e$. Hence the coefficient $t^e$ in $\widetilde{q_{11}}$ is
$\sum_i a_i^{\vfi^{e_i}}c_i=0$ and $\deg (\widetilde{q_{11}})<e$.

Let $\tilde{Q}$ be the matrix where we have replaced $q_{1j}$ by  $\widetilde{q_{1j}}$. Now the sum of degrees of the non zero entries of $\tilde{Q}$ is less than the sum of degrees of the non zero entries of $Q$. Hence in order to apply an induction, it is enough to prove that the solvability of the system
$\bar{x}=\bar{y}.Q $ is equivalent to the solvability of $ \bar{x}=\bar{z}.\tilde{Q}$.
But this follows by expressing the equations (\ref{eq:matricepassage}) by the equality
$$\tilde{Q}=PQ$$
where $$P=\begin{pmatrix}
1 & t^{e-e_2}a_2 & t^{e-e_3}a_3 & \cdots & t^{e-e_k}a_k \\
0 & 1            & 0            & \cdots &       0       \\
\vdots & 0 & \ddots \\
0 & \cdots & \cdots & \cdots & 1
\end{pmatrix}					  $$
is clearly invertible in $R$.

{\it Claim 2}: Assume Claim 1. We may assume that the polynomials
$(q_{i1})_i$ of the first  column have same degree and leading coefficients
of the $(q_{i1})_i$ are $K^{\vfi^e}$ linearly independent.

We will show that we can change $Q$ to some $S$, possibly having more rows, such that
the system $\bar{x}=\bar{y}.Q$ is equivalent to $\bar{x}=\bar{z}.S$ with the first column of $S$ has required properties.

Recall that $q_{i1}$ has degree $e_i$ with $e_1=e$. For all  $1\leq i \leq k$,  using the basis $\alpha(e-e_i)$, we write the equality of terms
	$$y_i=\sum_{u \in d^{e-e_i}}\lambda_u(y_i).t^{e-e_i}\alpha_u .$$
Now $$y_i.q_{i1}=y_i.t^{e_i}c_i + y_i.q'_{i1} = \sum_u \lambda_u(y_i).(t^e\alpha_u^{\vfi^{e_i}}c_i + r_{i1}(u))$$
where $r_{i1}(u)= t^{e-e_i}\alpha_uq'_{i1}$ which has degree $<e$. Set $z_i(u)=\lambda_u(y_i)$ and
$$s_{i1}(u):=t^e\alpha_u^{\vfi^{e_i}}c_i + r_{i1}(u).$$
Since the leading coefficient of $s_{i1}(u)$ is $\alpha_u^{\vfi^{e_i}}c_i$, for any $i$ and $u$,
$s_{i1}(u)$ has the degree  $e$.

For $j>1$ and $u\in d^{e-e_i}$, set $s_{ij}(u):=t^{e-e_i}\alpha_u.q_{ij}$.
Note that we keep $e_i=e_{i1}$ but $j$ varies. Let $S$ be the matrix obtained from $Q$, by replacing
$i$-th row by the matrix $(s_{ij}(u))_{u,j}$ where $u$ is the row-index and $j$ is the column index.
Then the system $(x_j=\sum_i y_i.q_{ij})_j$ is equivalent to the system
$(x_j=\sum_i \sum_u \lambda_u(y_i).s_{ij}(u))_j$, which can be written as
$$\bar{x} = \left((\lambda_u(y_1))_u,\ldots,(\lambda_u(y_k))_u \right).S .$$

Now we will show that the leading  coefficients $\alpha_u^{\vfi^{e_i}}c_i$'s are
$K^{\vfi^{e}}$-linearly independent.

Suppose $$\sum_i \sum_u a_{u_i}^{\vfi^e}\alpha_u^{\vfi^{e_i}}c_i= 0$$ for some
tuple $(a_{u_i})_{u,i}$ from $K$. It follows   by Claim 1 that,
$\sum_u (a_{u_i}^{\vfi^{e-e_i}}\alpha_u)=0$ for each $i$. Since the $\alpha_u$ are $K^{\vfi^e}$-linearly independent and $\vfi$ is injective
    $a_{u_i}=0$ for all $i,u$. The Claim 2 is  proved.

We assume now that the first column of $Q$ consists of
polynomials of degree $e$ with leading coefficients being $K^{\vfi^e}$-linearly independent.
By section 3  and by the last paragraph of Lemma 4 of \cite{vdd-kuhlmann}, there exists an invertible matrix over $P$ over $K$
such that $P.Q_1$, where $Q_1$ the first column of $Q$, consists of polynomials with leading coefficients has all different valuations in
$\{0,\ldots,d^{e}-1\}$. Hence considering $P.Q$ finishes the proof.
\end{proof}

For the following lemma, we will use the above result with $Q$ a column matrix and then in the following corollary we will use it in the whole generality.

\begin{lemma}\label{lem:pseudocompbeta}
Let $A \subseteq M$, of the form $A=\sum_i M.q_i$ then for some integer $s$, $A$ has a pseudo-complement of the form
$\oplus_{i\in I} M.t^sb_i$ where $I\subseteq d^s$, and the $b_i$ are valuation independent.
\end{lemma}
\begin{proof}
Use the above lemma to chose $g_j$ such that $\sum_j M.g_j =A$,  all of degree $s$ with leading coefficients $b_j$'s have different valuations in $\{0,\dots,d^s-1\}$. Then
by the lemma \ref{lem:addtoMta}, $M.g_j \approx M.t^sb_j$ for all $j$. Complete the $b_j$ to a valuation independent basis of $K^{\vfi^s}$-vector space $K$. We write the new elements of this basis as  the $b_i$.    Let $C:=\sum_{i} M.t^sb_i$.
Since $C\oplus \sum M.t^sb_j=M$, $C$ is in particular a pseudo-complement for $\sum M.t^jb_j$. Hence by Lemma
\ref{changecomp}, $C$ is a pseudo-complement for $A$.
\end{proof}
\begin{remark}\label{rem:pseudocomplements}
 The pseudo-complement $C$ is $p.p.$ definable  subgroup by an $L$-formula $C(x)$, which does not depend on $M$. In other words, if $\phi(x)$ is the formula
$$\exists y_1, \dots , y_m \mesp x=\sum_i y_i.q_j$$ then in any valued module $(M,v)$,
$C(M)$ is a pseudo-complement to $\phi(M)$.

\end{remark}

\begin{theorem}\label{pseudocomplements}
Let $\phi(\bar{x})$ be a $p.p.$ formula of $L_\mathcal{O}$ of the form
$$\bar{x} - \bar{y}.Q \in W,$$ where $Q$ a matrix with coefficients from $R$ and $W$ a ball. Set $k:=\vert \bar{x} \vert$. Then there exists computable $\gamma$ and an existential $L$-$p.p.$ formula $D(\bar{x})$ such that
$$M^k=\phi(M^k) + D(M^k) + P_\gamma^k$$ $\&$
$$D(M^k)\cap \phi(M^k) \subseteq P_\gamma^k$$
for all valued module $(M,v)$.
\end{theorem}
\begin{proof}

 By Lemma \ref{vddku}
we may assume that the first column $Q_1$ of $Q$ consists of polynomials having same degree $s$ with leading coefficients having distinct valuations in $\{1,\dots, d^s-1\}$. In addition, by Remark \ref{rem:+ballpseudocomp},
we may suppose that $W=\{0\}^k$.

 Take $C$ a pseudo-complement to $M.Q_1$ as in the above lemma.
Let $x=(x_1, \dots, x_k)$ such that $x_1 \in C \cap M.Q_1$. Then $x_1 \in
P_\delta$ for some computable $\delta$. Write $x_1$ also as
$$x_1=y_1.q_{11} + y_2.q_{21} + \dots y_mq_{m1}.$$
Since the leading coefficients of the $q_{1j}$ have different valuations
in  $\{0,\dots, d^s-1\}$,  if
$$v(y_j)\leq \rho:=\min\{\min_{ij} \{\jump{q_{ij}}\}\}-1;$$ we have $v(y_j.q_{1j})\neq v(y_{j'}.q_{1j'})$ for $j\neq j'$. Hence the $y_j$ can not have indefinitely small valuations since $v(x_1)\geq \delta$. Now  for $i>1$, since $x_i=\sum_{j}y_j.q_{ij}$ we have
$v(x_i)\geq \min_j \{\rho\cdot q_{ij}, \delta\}$. Setting $\gamma:=\min_{ij}\{\rho\cdot q_{ij}\}$ and $D:=C\times M^{k-1}$ yields our claim since $D$ and $\gamma$ depends only to $Q$.
\end{proof}

\begin{remark}\label{boundedsolutions}
The above proof shows that whenever a matrix $Q$ with the first column  consists of polynomials whose dominant coefficients are valuation independents and $\gamma \in \Gamma$, as above are given, there is a computable $\delta \in \Gamma$, such that
$$\bar{y}.Q \in P_\gamma^m \Rightarrow \bar{y} \in P_\delta^k  .$$
\end{remark}
\subsection*{\bf  The theory $T_{\Psi}$}

Let $(M,v)$ be a valued module and set
    $$A:=\sum M.q_i$$
  where the $q_i$ are all of degree $s$ such that for a  valuation independent basis $\beta$  of $K^{\vfi}$-vector space $K$, the leading coefficients of the $q_i$ are from the basis $\beta(s)$. Let $\mathbb{A}(\beta)$ be the tree consisting of subgroups $M.t^i\beta_i(n) \, (i,n \in \omega)$, ordered by inclusion. At the level $n$ we have the subgroups
$M.t^i\beta_i(n) \mesp (i \in d^n)$.

\begin{defn}
	We call the pseudo-complement for $A:=\sum_j M.q_j$
	the unique pseudo-complement which can be written as the sum of some elements of level
	$\deg(q_i)$ of the tree $\mathbb{A}(\beta)$.
\end{defn}
This definition only depends on  the decomposition
of the leading coefficients of the $(q_i)_i$ in the basis $\beta(n)$. Hence, given $A$ as above, by Lemmas \ref{vddku} and \ref{lem:pseudocompbeta}, there is a recursive function
\begin{equation}\label{metaf}
\mathfrak{f}: \bigcup_n K^n \to \bigcup_n K^n
\end{equation}
which computes the basis $\beta$ and the pseudo-complement $C$ of  in every valued
$t\beta$-decomposable $R$-module $M$. We  write
$C_{\mathfrak{f}(q_1,\ldots,q_n)}$ for the pseudo-complement computed by $\mathfrak{f}$.

On the other hand, the ball $P_\gamma$ such that $M=A+C + P_\gamma$, can be chosen
by letting $\gamma:=\min\{\min_i\{\jump{q_i}\} \} - 1$. This yields another recursive function
\begin{equation}\label{metaj}
\mathfrak{j}:  \bigcup_n K^n \to \Gamma.
\end{equation}

Hence we can express the statement of Corollary \ref{pseudocomplements} by introducing an axiom $\psi(q_1,\ldots, q_n)$  for any matrix $Q$ which has the first column  $(q_,\ldots,q_n)$, in the language
$L_{\mathcal{O}}$, which says that
 \begin{equation}
 M^k=M^k.Q + C_{\mathfrak{f}(q_1,\ldots,q_n)}\times M^{k-1} + P_{\mathfrak{j}(q_1,\ldots,q_n)}^k \mesp \quad \text{and}
\end{equation}
$$ M^k.Q \cap  C_{\mathfrak{f}(q_1,\ldots,q_n)}\times M^{k-1} \subseteq P_{\mathfrak{j}(q_1,\ldots,q_n.)}^k.$$

 Let $T_\Psi$ be the $L_\mathcal{O}$-theory of  $R$-modules together with the sentences  $\psi(q_1,\ldots, q_n)$. Hence
 $T_\Psi$ is recursively enumerable.

\begin{theorem}\label{approx}
Let $Q$ be a  $m\times k$  matrix over $R$, $W=\prod_{i=1}^k P_{\gamma_i}$  and $\phi$
	be the $L_{\mathcal{O}}$-formula
$$\phi(x_1,\ldots,x_k): \exists y_1\ldots y_m \mesp (x_1,\ldots x_k) - (y_1,\ldots, y_m).Q \in W.$$  Then for some  computable $\gamma$, and some positive primitive  $L_\mathcal{O}$-formula $D(x)$ we have
$$M^k =  \phi(M^k) + D(M^k) + \prod_{i=1}^k P_\gamma   \quad \& \quad D(M) \cap \phi(M^k)\subseteq \prod_{i=1}^k P_\gamma.$$
for all $M\models T_{\Psi}$.
\end{theorem}

\section{Decidability and model completeness of $\mathbb{F}_d((X))$}
We will introduce a new theory $T_1$, containing $T_{\Psi}$, augmented by sentences counting the number of solutions of the $p.p.$ formula
$$\bar{x}-\bar{y}.Q \in B$$ in $B_0$ modulo $B_1$, for proper balls $B_1 \subseteq B_0$.

For  $Q$, an $m\times n$ matrix over $R$, such that its first column consists of
polynomials whose leading coefficients are valuation independent and $B$ a ball, set $$A(\bar{x}):= \exists  \bar{y} \mesp \bar{x}-\bar{y}.Q \in B.$$ Let $\delta$ be the value computed by
Theorem \ref{boule_up}, such that $A\wedge P_\delta^m$ is quantifier-free definable
in language  $L(\lambda)$.

   Let    $B_0:=P_\gamma^m$, where $\gamma$ is given by Theorem \ref{approx}.  $B_1:=P_\delta^m$,  and
   $$k:=\vert (A(K)\cap B_0(K))/A(K)\cap B_1(K)\vert$$ if $\gamma\leq \delta.$ We set the sentence  $\theta(A,\gamma,\delta)$ expressing
    \begin{equation}
     k=  \left| \frac{(A\wedge B_0)(K)}{(A \wedge B_1)(K)}\right| .
    \end{equation}
Let  $\Theta$ the be set of sentences  $\theta(A,\gamma,\delta)$ and we set
 $$T_1:=T'_{hens}\cup T_{\Psi} \cup \Theta$$
 where $T'_{hens}$ is the $L_\mathcal{O}$-theory composed by the axioms of $T_{hens}$, where we have replaced any $L(\lambda)$-term by its equivalent modulo the theory of $R$-modules in language $L$ (recall Remark \ref{extunideflambdas}).

 Note that $T_1$ implies that $P_{\gamma +1}/P_{\gamma}$ has exactly $d$ elements for all $\gamma \neq \infty$.
\begin{remark} $\Theta$ and hence $T_1$ is a recursively enumerable theory. In fact, by Remark \ref{boundedsolutions}, if $\bar{x},\bar{y},\gamma$ are such that
	$\bar{y}.Q=\bar{x}\in P_{\gamma}^n$,  then $\bar{y}$ is in some $P_{\theta}^k$ for a computable $\theta$, hence searching the solutions
	$\bar{y}$, of $\bar{y}.Q=x$ with $\bar{x} \in P_\gamma^n$ can be bounded to searching $y$'s in some ball. Hence
	searching such solutions modulo another ball can be done in some finite $\mathbb{F}_d$-vector space by an algorithm.
\end{remark}
\begin{prop}\label{laurentsatisfyT1}
Let $M:=\mathbb{F}_d((X))$ then as an $L_\mathcal{O}$-structure $M\models \Theta$  and hence $M \models T_1$.
\end{prop}
\begin{proof}
By \cite{kuhlmanntame}, Theorem 5.14, $K$ is existentially closed as a ring  in $M$. In addition, by \cite{sylvyarno17} Corollary 6.18, there exists an  existential ring-formula
without parameters which defines the maximal ideal both in $K$ and in $M$. Since the valuation ring is the complement of the set of inverses of the elements in the maximal ideal, we have a universal ring-formula  which defines uniformly the valuation ring in $K$ and in $M$. Hence the balls
centered at $0$ are definable  universally with the parameter  $X$ both in $K$ and $M$.

Suppose $\vert (A(K)\cap B_0(K))/(A\cap B_1(K))\vert = k$. Consider the sentence $$\sigma: \vert (A\wedge B_0)/(A\wedge B_1)\vert \geq k.$$
Since by theorem \ref{boule_up}$, (A(K)\cap B_1(K))$ is definable both universally
and existentially in $L_{\mathcal{O}}$,  and and the valuation ring is universally definable both in $K$ and $M$,  $\sigma$ is equivalent
to an existential ring-formula with parameters in $K$. Since this quotient is finite, we must have $\vert (A(M)\cap B_0(M))/A(M)\cap B_1(M)\vert = k$.
\end{proof}

To prove Theorem \ref{th:pp2universal}, we will use  a lemma from Rohwer's thesis
(\cite{rohwer}, Lemma 8.2). This lemma is a generalization of the following fact:

\begin{obs} In an abelian group $G$ with existentially definable subgroups $A,B$ such that $A+B=G$, if $A\cap B$ is definable by a universal formula then $A$  is definable by the following universal formula $\psi(x)$:

\begin{equation}
\psi(x): \forall y \mesp (x-y \in A \wedge y \in B) \rightarrow y \in A\wedge B.
\end{equation}
By iterating this observation,  we have the following lemma.
\end{obs}
\begin{lemma}[Rohwer]
	Let $\mathcal{T}$ be a theory expanding  the theory of abelian groups. For each
	$M \models \mathcal{T}$ and for  $A, A_c, A_m, A_s, B_0, B_1$  definable subgroups of $M$ satisfying  the following configuration,
\end{lemma}
\begin{enumerate}[1.]
\item $A + A_c=M$,
\item $A\cap A_c \subseteq B_0$,
\item $A\cap B_1 = A_s\cap B_1$,
\item $A\cap B_0 \subseteq A_m \subseteq A + B_1$.
\end{enumerate}\medskip
where $A,A_c,B_1$ are definable by existential formulas, and $A_m, A_s$
by universal formulas (where all formulas in question do not depend on $M$), $A$ is definable by a universal formula (which does not depend on $M$).
\begin{proof} See \cite{rohwer} Lemma 8.2.
\end{proof}

\begin{theorem}\label{th:pp2universal}
Any $p.p.$ formula of $L_{\mathcal{O}}$ is equivalent modulo $T_1$ to a universal $L_\mathcal{O}$-formula.
\end{theorem}\label{modcomplete}
\begin{proof}
 Let $$A(\bar{x}): \exists \bar{y} \mesp \bar{x}.S -\bar{y}.Q \in W$$ be a $p.p.$ formula. For our purposes, we may assume that
$S$ is the identity matrix since me way replace $A(x)$ by
	$$\forall \bar{z} \mesp (\bar{z}=\bar{x}.S \to \exists \bar{y} \mesp \bar{z}-\bar{y}.Q \in W).$$
	By Lemma \ref{vddku}, we may suppose that the first column of $Q$ consists of polynomials which have coefficients
	in a valuation independent basis.
	By Theorem \ref{boule_up}, there is some proper ball $B_1$ determined by $T_1$ such that the formula $A_s:=A\wedge B_1$ is equivalent
to  a quantifier free $L(\lambda)$-formula. Hence by Remark \ref{extunideflambdas}, $A_s$ is equivalent to a universal
$L$-formula. Now, let  $B_0$ be given by Theorem \ref{approx} such that
$$A \wedge A_c \to B_0$$ where $A_c$ is of the form $A_c=D+P_\gamma^l$ with $D$    as in Theorem \ref{approx}.

\emph{Claim} :  Set $A_m:=(B_0 \wedge  A) + A_s$. Then $A_m$ is equivalent to a  universal
$L_{\mathcal{O}}$-formula modulo $T_1$.

\underline{proof of the claim}:
 Let $M \models T_1$. Let $k$ be the cardinality of $$(B_0(M) \cap A(M))/
 A_s(M)$$ and $y_1, \dots, y_k$ ($y_i$ are tuples of variables)  be  representatives of the classes.
Note that $k$ is determined by a sentence
in $\Theta$, hence depends only on the theory $T_1$.
	Then $$\{y_1, \dots y_k \}  + A_s(M) = A(M)\cap B_0(M)+A_s(M).$$ Moreover,
	for all $z_1,\dots,z_k \in B_0$ satisfying $z_i-z_j \notin A_s(M)$ for $i\neq j$,
	$$\{ z_1, \dots z_k \}  + A_s(M) = A(M)\cap B_0(M) + A_s(M)$$
if and only if,
	$$\{z_1, \dots z_k \}  + A_s(M) \subseteq A(M) \cap B_0(M) + A_s(M).$$ Hence
the formula
\begin{multline}
	x\in B_0 \wedge \forall y_1, \dots, \forall y_k \\ (
  (\bigwedge_{i=1}^k y_i \in B_0 ) \wedge ( \bigwedge_{i\neq j} y_i - y_j \notin A_s) \wedge  \\
	[\exists_{i=1}^{k}z_i (\mesp \bigwedge_{i=1}^k z_i\in A \wedge \bigwedge_{i=1}^k y_i - z_i \in A_s)]) \longrightarrow \bigvee_{i=1}^k y_i -x \in A_s.
\end{multline}
is equivalent to $A_m$, which is equivalent to a universal $L_\mathcal{O}$-formula, thus the claim is proved.

Now $A,A_c,A_m,A_s,B_0,B_1$ are in the configuration of Rohwer's Lemma above.
\end{proof}

\begin{corr}\label{decidable}
	Every completion of $T_{1}$ is model-complete in language $L_{\mathcal{O}}$.  In particular, the complete $L_\mathcal{O}$-theory of $\mathbb{F}_d((X))$ is model-complete.
\end{corr}
\begin{proof}
	It follows by Corollary \ref{corr:baur-monk}.
\end{proof}
Note that
$K$ embeds  (via an $L_{\mathcal{O}}$-embedding) to any model $N\models T_{1}$. In fact, choose any $s \in N\setminus\{0\}$ such that $s.t=s$. Consider
the $L$-embedding $k\mapsto s.k$. We show that it is an $L_{\mathcal{O}}$-embedding:  It is easy to see that the assertion
$$ \mathcal{O} = A \oplus \mathcal{O}.X$$ where  $A$
is the formula $x.t=x$, is a consequence of $T_{1}$ and it is clear that
if $a\in \mathcal{O}_K$ then $s.a \in \mathcal{O}_N$. Suppose now that
$s.a \in \mathcal{O}_N$ for some $a \in K$. Then
$s.a=y  +  m$ where $y.t=y$ and $m\in \mathcal{O}_N.X$. Then
$(s.a).(t-1)=m.(t-1) \in \mathcal{O}_N.X$. Note that
$(s.a).(t-1)=s.ta^{\vfi} -sk=s.(a^{\vfi} -a) \in \mathcal{O}.X$ since $s.t=s$.
This can only happen if $(a^{\vfi} -a) \in \mathcal{O}_K$, only if
$a\in \mathcal{O}_K$.

\begin{corr}\label{K}
The models of $T_1$ in which $K$ is existentially closed as an $L_\mathcal{O}$-structure are elementary equivalent to $K$.
\end{corr}

\begin{proof}

	Let $N\models T_{1}$ in which $K$ is existentially closed.
 Let $A, B$ be $p.p.$ formulas (of the language $L_{\mathcal{O}}$) with one free variable  such that
 $$T_1\models \forall x \mesp A_1(x) \rightarrow A(x).$$

Set $k:=\vert A(K)/A_1(K) \vert$. It is enough to show that
 we have
	$$\vert A(N)/A_1(N) \vert =k.$$

Let $l>k$ and suppose
$$M \models \exists x_1,\ldots \exists x_l \mesp (\bigwedge_i x\in A)\wedge
	(\bigwedge_{i\neq j} x_i-x_j \notin A_1).$$
		Since modulo $T_{1}$ the formula $x \in A_1$ is equivalent to a universal formula, its negation is equivalent to an  existential $L_{\mathcal{O}}$-formula. Since
		$K$ is existentially closed in $N$, there exists at least $l$ element in $A(K)/A_1(K)$. Contradiction.

\end{proof}

In particular, we have:
\begin{corr}
$K$ is the prime model of the complete theory of $\mathbb{F}_d((X))$.
\end{corr}

\begin{proof}
$K$ is existentially closed in $\mathbb{F}_d((X))$ as an $L_{\mathcal{O}}$-structure since it is existentially closed in $\mathbb{F}_d((X))$ as a ring. Hence $$K\equiv \mathbb{F}_d((X)).$$ Since any completion of $T_1$ is model-complete $K$ is an $L_{\mathcal{O}}$ elementary substructure of $\mathbb{F}_d((X))$.
\end{proof}

\subsection*{Decidability of $\mathbb{F}_d((X))$}

The following fact is an easy exercise using Hensel's lemma:
\begin{fait}
	$\mathbb{F}_d[[X]]$ is definable by the $L$-$p.p.$ formula
	$$ \exists y \mesp x.tX=y.(t-1)$$ inside $\mathbb{F}_d((X))$.
\end{fait}

Hence, the decidability of $\mathbb{F}_d((X))$ as an
$L_\mathcal{O}$-structure and as an $L$-structure are equivalent. Recall that the $L$-theory of $K$ (hence of $\mathbb{F}_d((X)))$, are given by the sentences
stating that
$$\vert A(K)/(A(K)\wedge B(K)) \vert =k \quad (k\in \mathbb{N}\cup \{\infty\})$$
where $A, B$ are $L$-$p.p.$ formulas of
with one free variable.

We will show that  a recursively enumerable subset of these sentences forms
a complete axiom system which implies all of them. Hence the $L$-theory of $K$ is decidable.

For $A$ and $B$ as above, we set $D:=A\wedge B$. Let
$A$ be given by $\exists \bar{y} \mesp x.p=\bar{y}\bar{q}$ and
$B$ by $\exists \bar{y} \mesp x.r=\bar{y}\bar{s}$.  We may suppose that
both $p$ and $r$ are unitary. Hence, by Lemma \ref{lem:addtoMta}, for  $k=\deg(p)$ and $s=\deg(r)$  we have
$$K.p \approx K.t^k$$ and $$K.r \approx K.t^s .$$

Now by Remark \ref{rem:samecomplement}, \ref{rem:invimage} and Lemma \ref{changecomp}, $A\approx D$ if and only if  the preimage by $.t^k$ of $C_{\mathfrak{j}(\bar{q})}$(which is a pseudo-complement of $A$) is equal to
the preimage  of $C_{\mathfrak{j}(\bar{s})}$ by $.t^s$ (which is a pseudo-complement of $D$). Hence it is decidable if $A\approx D$.

Notice that  If  $A \not \approx D$ then
$A/D$ is infinite:  Let $a\in A \setminus D$.  Then for  $a' \in A$ with $v(a')<v(a)$  either $a'$ or $a-a'$ is not in $D$.

Now suppose $A\approx D$. It follows by Remark \ref{rem:modulogamma} that
$$ A+P_\alpha=D+P_\alpha $$  for some computable $\alpha$: $\alpha$ can be chosen less than every jump values of all polynomials appearing in the definition of $A$ and $B$.
Hence $\vert A/D \vert = \vert A\cap P_\alpha / D\cap P_\alpha\vert $.

Now by Corollary \ref{corr:boule_up}, there is a computable $\gamma$ such that $A/D$ is finite if and only if
 $$A\cap P_\gamma = D \cap P_\gamma$$

Consider the following algorithm: given $A$ and $B$  setting $D=A\wedge B$, the algorithm check if $A\approx B$, if not it  sets
$\vert A/D \vert =\infty$. Otherwise computes $\alpha $ such that $A+P_\alpha=B+P_\alpha$.  Then it computes the value $\gamma$
 so that it can check whether $A\cap P_{\gamma}/D\cap P_{\gamma}$  is trivial or infinite. If it is infinite it sets
 $\vert A/D \vert = \infty$. If not, by Remark \ref{boundedsolutions} the algorithm can compute the number of the elements of
  \begin{equation}\label{t0}
  \frac{(A\cap P_\alpha)/P_{\gamma}}{(D\cap P_\alpha)/P_{\gamma}}.
  \end{equation}

  Hence the $L$-theory consisting of sentences of type $(\ref{t0})$ can be
  recursively enumerable and implies the $L$-theory of $K$, hence of $\mathbb{F}_d((X))$. We have proved:
  \begin{theorem}\label{thm:finaldecidable}
  	Both the $L$-  and the $L_{\mathcal{O}}$-theories of $\mathbb{F}_d((X))$ are decidable.
  \end{theorem}

  \bibliographystyle{amsalpha}
  \bibliography{onayg}

\end{document}

config: arxiv


TITLE: Minimum length of axes intercept of a line passing through a fixed point
QUESTION [2 upvotes]: I encountered this problem:

A straight line passes through a fixed point $(h,k) \ (h,k>0)$ and intersects the coordinate axes at $P(a,0)$ and $Q(0,b)$. Show that the minimum length of $PQ$ is $(h^{2/3}+k^{2/3})^{3/2}$.

Now, I proceeded like this:
If a line passes through $(h,k)$ then its equation is $$\frac{y-k}{x-h}=m.$$
If it passes through $(a,0)$ and $(0,b)$ then we have $a=h-\frac{k}{m}$ and $b=k+mh$.
Length of $PQ$ is $\sqrt{a^2+b^2}$.
We can put the value of $a$ and $b$ in the expression and then differentiate w.r.t. $m$, to find out the minimum value.
But after equating the first differential to $0$, I found two roots of $m$. As the values of $h$ and $k$ can be positive or negative, I cannot put them in the second differential to find out which one is minimum and which one is maximum.
What am I doing wrong? and, Is there any other, simpler method that can be used?

REPLY [1 votes]: Equation of line $\mathcal{AB}$ is $\displaystyle \frac{x}{a}+\frac{y}{b} =1$
Line passes through fixed point $(h,k)$
Then $\displaystyle \frac{h}{a}+\frac{k}{b} =1$
Use Holder,s Inequality
$$(a^2+b^2)\bigg(\frac{h}{a}+\frac{k}{b}\bigg)\bigg( \frac{h}{a}+\frac{k}{b}\bigg)\geq \bigg(h^{\frac{2}{3}}+k^{\frac{2}{3}}\bigg)^3$$
$$\mathcal{\min{PQ}}=\sqrt{a^2+b^2}\geq \bigg(h^{\frac{2}{3}}+k^{\frac{2}{3}}\bigg)^{\frac{3}{2}}$$


TITLE: Exchange limit and integral sign in front of Heat kernel
QUESTION [2 upvotes]: I have encountered a problem regarding the exchange of limit and integral sign. Currently,I am doing a problem which relates to the heat kernel. Suppose $u(t,x)$ is the solution of heat equation and $u(0,x) = f(x)$ which is bounded and continuous. Therefore, we can express the solution by Heat Kernel:
\begin{equation}
u(t,x) = \frac{1}{(4 \pi t)^{n/2}} \int_{\mathbb{R}^{n}} e^{\frac{-|x-y|}{4t}} f(y) dy
\end{equation}
Then, I need to prove both equalities:
\begin{equation}
\lim_{t \rightarrow \infty} u(t,x) = 0, \quad \quad \lim_{t \rightarrow \infty}(4 \pi t)^{n/2} u(t,x) = \int_{\mathbb{R}^{n}} f(y) dy
\end{equation}
My idea is that by using Lebesgue dominated convergence theorem,  we can switch the order of limit and integral sign and I can prove these two equalities. Suppose I am dealing with second equality, I need to construct a sequence $\{ t_{n} \}$ such that it converges to $t = \infty$. Besides, I need to construct a sequence of function $\{ h_{n} \}$
\begin{equation}
h_{n}  = e^{\frac{-|x-y|}{4t_{n}}} f(y)
\end{equation}
In order to apply Lebesgue dominated convergence theorem, we need to prove $h_{n} \rightarrow h$ pointwisely and $ |h_{n} | \leq g$ where $g$ is a integrable function. My difficulties are that that I do not know how to construct the  a sequence $\{ t_{n} \}$ which
 converges to $t = \infty$ and $h_{n} \rightarrow h$ pointwisely.

REPLY [1 votes]: Note that the second limit implies the first limit. To obtain the second limit, note that
$$(4 \pi t)^{n/2}u(t,x) = \int_{\mathbb{R}^{n}} e^{\frac{-|x-y|}{4t}} f(y) dy.$$
Let $y=x+z.$ Then the integral on the right equals
$$\int_{\mathbb{R}^{n}} e^{\frac{-|z|}{4t}} f(x+z) \,dz.$$
Think of the integrands here as $f_t(z).$ We then have $f_t(z)\to f(x+z)$ pointwise everywhere. We also have  $|f_t(z)|\le |f(x+z)|$ for all $t,z,$ and $f(x+z)\in L^1.$ By the DCT,
$$\int_{\mathbb{R}^{n}} f_t(z)\,dz \to \int_{\mathbb{R}^{n}} f(x+z) \,dz = \int_{\mathbb{R}^{n}} f(y) \,dy.$$


TITLE: Applying the proof of "The subgroup of an infinite cyclic group is cyclic"
QUESTION [0 upvotes]: First of all I want to say question is already asked under the title " A subgroup of cyclic group is cyclic- Understanding the proof" Sorry I dont have the link, but my question is entirely depend on that proof,
My question is-
As there is not mentioned that the cyclic group is finite or infinite , i guess the theorem is for both finite and infinite group.
Now we know that the additive group of integers is infinite cyclic group generated by -1 & +1, thus Z is generated by a=-1 or +1 ( as in the proof generator of group is denoted by a) , then I have subgp of Z as 2Z , right? Then in the proof we have a^n belonging to H for some positive integer n. That means 1^n for some n should belong to 2Z!!!!! But that number comes to be 1 for any n which is not in 2Z!
Please help me where Am I going Wrong?

REPLY [1 votes]: You're confused about notation. For $a \in G$ where $G$ is a group with operation $\star$, we write $a^n$ to mean
$$a^n = \underbrace{a \star \dots \star a}_{n\text{-times}}.$$
But the group $G$ in this situation is $\mathbb{Z}$ with the operation $\star = +$, the usual addition. Instead, then, $a^n$ is $\underbrace{a + \dots + a}_{n\text{-times}} = n \cdot a$ where $\cdot$ is the usual multiplication.
You'll notice that $H = 2\mathbb{Z} = \langle 2 \rangle$ and $2 = 2 \cdot 1$.


TITLE: Directional Derivatives With Respect to Negative Vectors
QUESTION [0 upvotes]: I understand this is probably a silly question but I'm with it struggling nonetheless. Consider the directional derivative of $f(x)=x^2$ at $x=1$ with respect to $u=-1$. I can see that this is equal to $-2$ but I struggle to relate this result with my notion of derivatives as slopes of tangent lines. Could someone explain, in elementary terms, how one should picture this result?

REPLY [1 votes]: If you have the direction $u = -1$, you have to look to the left, usually, you look to the right. So you could just flip your function...
And if you sit at $x = 1$, and go to the left, the function goes downwards, so the slope is negative.


TITLE: Quantum circuit simulation divergence in results
QUESTION [2 upvotes]: I'm learning about quantum computing in order to code a simulator. I tried the following circuit in Quirk

And ran the same circuit using OPENQASM 2.0:

Notice that the input is |11> in both cases, but it seems that the probabilities for 01 and 10 are switched. What I'm missing here? Thank you!

REPLY [3 votes]: Peter Shor nailed the reason. This is because OpenQASM and Quirk have different conventions for how the order of wires maps into the order of bits shown in the kets. The simplest way to check this is to use a circuit that toggles only the top qubit.
Quirk considers the top qubit to be the least significant qubit, and labels kets in big endian order (i.e. the ordering humans use; why 12 represents twelve instead of twenty-one). So the 4-output chance display is showing outputs in the following order: probability of top=bottom=0; probability of top=1 and bottom=0, probability of top=0 and bottom=1; probability of top=bottom=1.
The reason I chose this order when making Quirk is because it makes the displays behave better when resizing. If you resize the chance display from 2-qubits to 3-qubits, and the third qubit is off, the original four outputs stay together in the top half instead of being spread apart. Also it's just the natural thing to do when programming: the wire at index k has weight 1<<k. Quirk's tendency to expand circuits and gates downward pushed it towards a "down means more/bigger" convention in other places as well, e.g. in the arithmetic circuits.


TITLE: Find the parametrization for the curve in which the plane cuts a surface
QUESTION [1 upvotes]: I need to find a parametric equation for the curve for which the plane $z=2$ cuts the surface $z = \sqrt{x^2 + y ^2}$
I'm not sure how to proceed. If the plane cuts the surface at $z =2$ and $z = \sqrt{x^2+y^2}$ then how do I find x, y and z in terms of $t$? Do I need two separate variables? For example...
$$x = s^2 - t^2$$
$$y = s^2 - t^2$$
$$z=2$$
I'm just guessing really, can someone please help me out?

REPLY [1 votes]: You only need one parameter (say $t$), as the intersection of two surfaces is a curve. In this case the projection of the intersection to the $xy$ plane is a circle of radius 2 centered at the origin, so for example: $x=2\cos(t), y=2\sin(t), z=2$ will do.


TITLE: Prove that if $\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=1$ then $\gcd(bc,ac,ab)=1$?
QUESTION [5 upvotes]: So I write 3 Bézout's identities :
$au+bv=1$
$bv+cw=1$
$au+cw=1$
We could have taken different $u,v,w$...
By multiplying I obtain :
$1=ab(v^2ub+u^2va+uvwc)+ac(w^2uc+u^2wa+uvwb)+bc(v^2wb+w^2vc)$
By taking : $v^2ub+u^2va+uvwc=x \in \mathbb{Z}, \ w^2uc+u^2wa+uvwb=y\in\mathbb{Z}, \ v^2wb+w^2vc=z\in\mathbb{Z}$ then $\gcd(bc,ac,ab)=1$.
Now I want to generalize it to a product of $n$ terms without the $i^{th}$ term. Are there any faster methods than developing ?
PS : This statement is a key to start the proof of the CRT !
Thanks in advance !

REPLY [0 votes]: $$au+bv=1 \Rightarrow acu+bcu=c \Rightarrow gcd(ac,bc) |c \Rightarrow gcd(ac,bc, ab) |c$$
Same way $gcd(ac,bc, ab) |b$.
Then $gcd(ac,bc, ab)$ is a common divisor of $b$ and $c$. As $gcd(b,c)=1$ you get
$$gcd(ac,bc, ab) =1$$


TITLE: Solution to a second-order nonlinear ordinary differential equation
QUESTION [1 upvotes]: I am given three functions $z'(t), y'(t), x'(t)$ and I have reduced these functions to the equation:
\begin{equation}
z''=x_{0}\beta z'e^{-\frac{\beta}{\gamma}z} - \gamma z
\end{equation}
where $\beta, \gamma ,x_{0}$ are constants.
Next we introduce a new function:
\begin{equation}
u(t)=e^{-\frac{\beta}{\gamma}z}
\end{equation}
Substitution yields
\begin{equation}
u\frac{d^{2}u}{dt^{2}}- \bigg(\frac{du}{dt}\bigg)^{2}+(\gamma - x_{0}\beta u)u\frac{du}{dt} = 0
\end{equation}
Next a new function is introduced:
\begin{equation}
\phi = \frac{dt}{du}
\end{equation}
Now the equation can be rearranged to:
\begin{equation}
\frac{d\phi}{du}+\frac{1}{u} \phi = (\gamma-x_{0}\beta u )\phi^{2}
\end{equation}
The solution to this equation is:
\begin{equation}
\phi = \frac{1}{u(C_{1}-\gamma \ln u +x_{0}\beta u)}
\end{equation}
So what would be the solution to the original equation in terms of $u$?

REPLY [1 votes]: The last equation od
$$
\frac{dt}{du}=\frac{1}{u(C_{1}-\gamma \ln u +x_{0}\beta\, u)}.
$$
Integrating
$$
t=\int\frac{du}{u(C_{1}-\gamma \ln u +x_{0}\beta\, u)}.
$$
I do not think that there is a closed formula for the integral.


TITLE: Localization at finitely many minimal prime ideals
QUESTION [6 upvotes]: Let $A$ be a commutative ring with finitely many minimal prime ideals $\{p_1,\dots,p_n\}$. Let $A_{p_1,\dots,p_n}$ be the localization of $A$ away from the minimal primes, i.e. $S^{-1}A$ where $S = A-\bigcup_{i=1}^np_i$.
Question 1 (Answered): A paper I am reading claims that $A_{p_1,\dots,p_n} \cong \prod_{i=1}^n A_{p_i}$. I have established by hand that the natural map from left to right is injective, but surjectivity seems harder. What's the best way to see this isomorphism?
Also, the paper uses the notation $\text{Frac}(A)$ for $A_{p_1,\dots,p_n}$ (actually, $\text{Frac}(A)$ is defined to be the ring obtained by inverting in $A$ all non-zero-divisors in $A_\text{red} = A/\sqrt{0}$, but it's easy to see that this is the same thing). This seems odd to me, since, as far as I know, the standard meaning of $\text{Frac}(A)$ is the localization of $A$ at all non-zero-divisors in $A$. Note that if $A$ is not reduced, then there could be zero-divisors which are not in any of the minimal primes, and hence are inverted in $A_{p_1,\dots,p_n}$ (example: $y$ in $\mathbb{Q}[x,y]/(x^2,xy)$).
Question 2 (Still open): Is this definition of $\text{Frac}(A)$ common? Is there a good reason for preferring this definition of $\text{Frac}(A)$ over the other in certain situations (when $A$ is non-reduced)?
One more observation: I've noticed that with this definition of $\text{Frac}(A)$, we have that a map of rings $A\to B$ extends to a map $\text{Frac}(A)\to \text{Frac}(B)$ if and only if the inverse image of every minimal prime ideal in $B$ is a minimal prime ideal of $A$. Geometrically, this says that the map $\text{Spec}(B)\to\text{Spec}(A)$ maps generic points of irreducible components to generic points of irreducible components. Maybe this is relevant...

REPLY [5 votes]: 1) Let $\phi : A \to \prod_{i=1}^n A_{p_i}$ be the natural map (i.e. product of localization maps). Every element of $S$ is sent to a unit under $\phi$, so there is an induced map $\varphi : S^{-1}A \to \prod_{i=1}^n A_{p_i}$. But $\varphi$ is locally an isomorphism (at every maximal ideal $p_i$ of $S^{-1}A$, $\varphi_{p_i} : (S^{-1}A)_{p_i} \cong A_{p_i} \to (\prod_{i=1}^n A_{p_i})_{p_i} \cong A_{p_i}$), so $\varphi$ is globally an isomorphism.
2) For a reduced ring, the set of nonzerodivisors is precisely the complement of the union of the minimal primes (this is easy to see in the Noetherian case, but also holds in general). Thus in this case the two notions of total ring of fractions coincide.


TITLE: Simple Roots of E6 in Coordinates?
QUESTION [5 upvotes]: There are several possibilities how one can write simple roots in terms of coordinates. Firstly, they depend on the numbering of the nodes in the Dynkin diagram. Let's fix the choice for $E_6$ to be

Unfortunately there are still several possibilities. For example, this paper quotes at the second page very different simple roots than Wikipedia, which claims that the simple roots in coordinates are the rows of the following matrix:

Which set is correct? Or are both? My problem is that I get very different result if I use these two different sets. Is there some good resource that lists the different possibilities?

REPLY [1 votes]: It is straightforward to verify $\langle v,v\rangle = 2$ for all rows $v$ and $\langle v,w\rangle=-1$ precisely for rows joined by a line and $\langle v,w\rangle=0$ otherwise.