\subsection{Function Rings over Reduction Rings}\label{section.right.rr}

The situation becomes more complicated if $\rr$ is not a field.

Let $\rr$ be a non-commutative ring with a reduction relation $\Longrightarrow_B$
associated with  subsets $B \subseteq\rr$ as described in Section \ref{section.reductionrings}.

In a first step let us look at the idea of generalizing the term of special standard
 representations to this setting.
The concept of right reductive standard representations as defined in Definition
 \ref{def.right_reductive} follows the idea of linking standard representations to
 reducibility.
The existence of a right reductive standard representations for a polynomial in terms of 
 a set of polynomials implies that the head monomial of the polynomial is reducible with respect
 to the reduction relation defined in Definition \ref{def.rred} using the same set of polynomials.

One possible generalization in the spirit of these ideas 
 for function rings over reduction rings is as follows:
%
\begin{definition}\label{def.right_reductive_rr}~\\
{\rm
Let $F$ be a set of polynomials in $\f$
 and $g$ a non-zero polynomial in $\ideal{r}{}(F)$.
A representation of the form 
$$g = \sum_{i=1}^n f_i \rmult m_i,
 f_i \in F, m_i \in \monoms(\f), n \in \n$$
 such that
 $\hterm(g) = \hterm(\hterm(f_1) \rmult m_1) =
 \hterm(f_1 \rmult m_1) \geq \hterm(f_1)$ and
 $\hterm(g) \succ \hterm(f_i \rmult m_i)$ 
 for all $2 \leq i \leq n$ is called a
 \betonen{right reductive standard  representation} in terms of $F$.
\dend
}
\end{definition}
%
Notice that it differs from Definition \ref{def.right_reductive} insofar as we
 demand $\hterm(g) \succ \hterm(f_i \rmult m_i)$ 
 for all $2 \leq i \leq n$.
It is directly related to the reduction relation presented in Definition \ref{def.rred}
 generalized to $\f$.
%
\begin{definition}\label{def.rred_rr}~\\
{\rm
Let $f,p$ be two non-zero polynomials in $\f$.
We say $f$ \betonen{right reduces} $p$ \betonen{to} $q$ \betonen{at a monomial}
 $\alpha \skm t$ \betonen{in one step}, denoted by $p \red{}{\myr}{r}{f} q$, if
 there exists $m \in \monoms(\f)$ such that
\begin{enumerate}
\item $t \in \supp(p)$ and $p(t) = \alpha$,
\item $\hterm(\hterm(f)\rmult m) = \hterm(f \rmult m) = t \geq \hterm(f)$,
\item $\hm(f \rmult m) = \alpha \skm t$, and
\item $q = p - f \rmult m$.
\end{enumerate}
We write $p \red{}{\myr}{r}{f}$ if there is a polynomial $q$ as defined
above and $p$ is then called right reducible by $f$. 
%\\
Further, we can define $\red{*}{\myr}{r}{}, \red{+}{\myr}{r}{}$ and
 $\red{n}{\myr}{r}{}$ as usual.
%\\
Right reduction by a set $F \subseteq \f_{\myk}$ is denoted by
 $p \red{}{\myr}{r}{F} q$ and abbreviates $p \red{}{\myr}{r}{f} q$
 for some $f \in F$.
\dend
}
\end{definition}
%
This reduction relation is related to the reduction relation $\R$ on $\rr$ where for
 $\alpha, \beta \in \rr$, $\alpha \R_{\beta}$ if and only if there exists $\gamma \in \rr$
 such that $\alpha = \beta \skm \gamma$.
Item 3 in this definition then can be rephrased as $\alpha \R_{\hc(f \rmult m)}$.

Notice that in contrary to $\f_{\myk}$ now for $g, f \in \f$ and $m \in \monoms(\f)$
 the situation $\hterm(g) = \hterm(f \rmult m) = \hterm(\hterm(f) \rmult m) \geq \hterm(f)$ alone no
 longer implies that $\hm(g)$ is right reducible by $f$.
This is due to the fact that we can no longer modify the coefficients involved in the reduction step
 in the appropriate manner since reduction rings in general will not contain inverse elements.
\begin{lemma}\label{lem.rred.rr}~\\ 
{\sl 
Let $F$ be a finite set of polynomials in $\f \backslash \{ \zero\}$.
\begin{enumerate}
\item For $p,q \in \f$ $p \red{}{\myr}{r}{F} q$ implies $p \succ q$, in particular $\hterm(p)
  \succeq \hterm(q)$.
\item $\red{}{\myr}{r}{F}$ is Noetherian.
\lemend
\end{enumerate}
}
\end{lemma}
\Ba{}
\begin{enumerate}
\item Assuming that the reduction step takes place at a monomial $\alpha \skm t$,
      by Definition \ref{def.rred_rr} we know $\hm(f \rmult m) = \alpha \skm t$ 
      which yields
      $p \succ p  -  f \rmult m $
      since $\hm(f \rmult m) \succ \reductum(f \rmult m)$.
\item This follows from 1. 
\end{enumerate}\renewcommand{\baselinestretch}{1}\small\normalsize
\qed
%
The Translation Lemma no longer holds for this reduction relation.
This is already so for polynomial rings over the integers.
\begin{example}~\\
{\rm
Let $\z[X]$ be the polynomial ring in one indeterminant over $\z$.
Let $p = 2 \skm x$, $q = -3 \skm X$ and $f = 5 \skm X$.
Then $p-q = 5 \skm X \red{}{\myr}{r}{f} 0$ while $p \nred{}{\myr}{r}{f}$
 and $q \nred{}{\myr}{r}{f}$.
\exaend
}
\end{example}
%
The reduction relation $\red{}{\myr}{r}{}$ in polynomial rings over the integers is
 known as Pan's reduction in the literature. 
The generalization of Gr\"obner bases then 
 are weak Gr\"obner bases as by completion one can only achieve
 that all ideal elements reduce to zero.
A proper algebraic characterization of weak right Gr\"obner bases
 related to right reductive standard representations
 and this reduction relation is as follows.
%
\begin{definition}\label{def.gb.rr}~\\
{\rm
A set $F \subseteq \f \backslash \{ \zero\}$ is called a 
 \betonen{weak right Gr\"obner basis} of $\ideal{r}{}(F)$ if
$\hm(\ideal{r}{}(F) \backslash \{ \zero \}) =
 \hm( \{ f \rmult m \mid f \in F, m \in \monoms(\f),
 \hterm(\hterm(f) \rmult m) = \hterm(f \rmult m) \geq \hterm(f) \}
 \backslash \{ \zero \} )$.
\dend
}
\end{definition}
Similar to Lemma \ref{lem.sb=gb} right reductive standard bases and weak right Gr\"obner bases
 coincide.
\begin{lemma}\label{lem.sb=gb_rr}~\\
{\sl
Let $F$ be a subset of $\f \backslash \{ \zero\}$.
Then $F$ is a right reductive standard basis if and only
 if it is a weak right Gr\"obner basis.
\lemend
}
\end{lemma}
\Ba{}~\\
Let us first assume that $F$ is a right reductive standard basis, i.e., every polynomial
 $g$ in $\ideal{r}{}(F)$ has a right reductive standard representation with respect 
 to $F$.
In case $g \neq \zero$ this implies the existence of a polynomial $f \in F$ and a monomial 
 $m \in \monoms(\f)$ such that $\hterm(g) = \hterm(\hterm(f) \rmult m) 
 = \hterm(f \rmult m) \geq \hterm(f)$ and $\hm(g) = \hm(f \rmult m)$.  
Hence $\hm(g) \in \hm( \{  f \rmult m \mid m \in \monoms(\f),  
 f \in F, \hterm(\hterm(f) \rmult m) = \hterm(f \rmult m) \geq \hterm(f) \}\backslash \{ \zero\})$.
As the converse, namely $\hm( \{  f \rmult m \mid m \in \monoms(\f), 
 f \in F, \hterm(\hterm(f) \rmult m) = \hterm(f \rmult m) \geq \hterm(f) \}\backslash \{ \zero\})
 \subseteq \hm(\ideal{r}{}(F) \backslash \{ \zero \})$ trivially holds, 
 $F$ is a weak right Gr\"obner basis.
\\
Now suppose that $F$ is a weak right Gr\"obner basis and again let $g \in \ideal{r}{}(F)$.
We have to show that $g$ has a right reductive standard representation with respect to $F$.
This will be done by induction on $\hterm(g)$.
In case $g = \zero$ the empty sum is our required right reductive standard representation.
Hence let us assume $g \neq \zero$.
Since then $\hm(g) \in \hm(\ideal{r}{}(F)\backslash \{ \zero\})$
 by the definition of weak
 right Gr\"obner bases we know there exists
 a polynomial $f \in F$ and a monomial $m \in \monoms(\f)$ such that
 $\hterm(\hterm(f) \rmult m) = \hterm(f \rmult m) \geq \hterm(f)$ and
 $\hm(g)=\hm(f \rmult m)$.
Let $g_1 = g - f \rmult m$. 
Then $\hterm(g) \succ \hterm(g_1)$ implies the 
 existence of a right reductive standard representation for $g_1$ which can be
 added to the multiple $f \rmult m$ to give
 the desired right reductive standard representation of $g$.
\\ 
\qed 
%
\begin{corollary}~\\
{\sl
Let $F$ a subset of $\f \backslash \{ \zero\}$ be a 
 weak right Gr\"obner basis.
\begin{enumerate}
\item Every $g \in \ideal{r}{}(F)$ has a right reductive standard representation
       in terms of $F$ of the form
        $g = \sum_{i=1}^n f_i \rmult m_i,
         f_i \in F, m_i \in \monoms(\f), n \in \n$
         such that
         $\hterm(g) = \hterm(\hterm(f_1) \rmult m_1) =
         \hterm(f_1 \rmult m_1) \geq \hterm(f_1)$ and
         $\hterm(f_1 \rmult m_1) \succ \hterm(f_2 \rmult m_2) \succ \ldots \succ
          \hterm(f_n \rmult m_n)$.
\item For every $g \in \ideal{r}{}(F)$ we have $g \red{*}{\myr}{r}{F} \zero$.
\end{enumerate}
}
\end{corollary}
\Ba{}~\\
This follows from inspecting the proof of Lemma \ref{lem.sb=gb_rr}.
\\
\qed

Now to find some analogon to s-polynomials in $\f$ we again study what polynomial
 multiples occur when changing arbitrary representations of right ideal elements
 into right reductive standard representations.

Given a generating set $F \subseteq \f$ of a right ideal in $\f$ the key idea 
 in order to characterize weak right Gr\"obner bases is  to distinguish special
 elements of $\ideal{r}{}(F)$ which have representations $\sum_{i=1}^n f_i \rmult h_i$,
 $h_i \in \rr[X]$, $f_i \in F$ such that the head terms $\hterm(f_i \rmult h_i)$
 are all the same within the representation.
Then on one hand the respective coefficients $\hc(f_i \rmult h_i)$ can add up to zero which
means that the sum of the head coefficients is in an appropriate module in $\rr$ --
 m-polynomials
 are related to these situations (see also Definition \ref{def.critical.situations}).
If the result is not zero the sum of the coefficients $\hc(f_i \rmult h_i)$ can be described in terms
 of a (weak) right Gr\"obner basis in $\rr$ -- g-polynomials are related to these situations.
Zero divisors in the reduction ring eliminating the head
 monomial of a polynomial occur as a special instance of m-polynomials
 where $F = \{ f \}$ and $f \skm \alpha$, $\alpha \in \rr$ are considered.

The first problem is related
 to solving linear homogeneous equations in $\rr$ and to the existence of
 possibly finite bases of the respective modules.
In case we want effectiveness, we have to require that these bases are  computable.

The g-polynomials can successfully be treated when possibly 
 finite (weak) right Gr\"obner bases exist for finitely
 generated right ideals in $\rr$.
Here, in case we want effectiveness, we have to require that the (weak) right Gr\"obner bases as well as
  representations for their elements in terms of the generating set are computable.

Using m- and g-polynomials, weak right Gr\"obner bases can be characterized
 similar to the characterizations in terms of syzygies (a direct generalization of 
 the approaches by Kapur and Narendran (1985) respectively M\"oller (1988)).

\begin{definition}\label{def.gpol}~\\
{\rm
Let $P$ be a set of  polynomials in $\f$ and
 $t$ an element in $\myt$
 such that there are  $p_1, \ldots, p_k \in P$ (not necessarily different) and
 $w_1, \ldots, w_k \in \myt$ with
 $\hterm(p_i \rmult w_i) = \hterm(\hterm(p_i) \rmult w_i) = t \geq \hterm(p_i)$, for all $1 \leq i \leq k$.
Further let $\gamma_i = \hc(p_i \rmult w_i)$ for  $1 \leq i \leq k$.
\\
Let $G$ be a (weak)
 right Gr\"obner basis of $\{ \gamma_1, \ldots, \gamma_k \}$ in $\rr$ with respect to $\R$.
Additionally let 
$$\alpha = \gamma_1 \skm \beta_{\alpha,1} + \ldots +  \gamma_k \skm \beta_{\alpha,k}$$
 for $\alpha \in G$, $\beta_{\alpha,j} \in \rr$,  $1 \leq i \leq n$, and $1 \leq j \leq k$.
Then we define the  \betonen{g-polynomials (Gr\"obner polynomials)}
 corresponding to $p_1, \ldots, p_k$ and $t$ by setting
$$ g_{\alpha} = \sum_{j=1}^k  p_j \rmult  w_j \skm  \beta_{\alpha,j}
 \mbox{ for each } 1 \leq i \leq k.$$
Notice that $\hm(g_{\alpha})= \alpha \skm t$.
\\
For the right module $M = \{ (\delta_1, \ldots, \delta_k) \mid  
 \sum_{i=1}^k  \gamma_i \skm \delta_i = 0 \}$, let the set
 $\{A_i \mid i \in I_M \}$ be a basis with
 $A_i = (\alpha_{i,1}, \ldots, \alpha_{i,k})$ for $\alpha_{i,j} \in \rr$
  and $1 \leq j \leq k$.
%\\
Then we define the 
 \betonen{m-polynomials (module polynomials)}
 corresponding to $P$ and $t$ by setting
$$ h_i = \sum_{j=1}^k p_j \rmult w_j \skm  \alpha_{i,j}
 \mbox{ for each } i \in I_M.$$
Notice that $\hterm(h_i) \prec t$ for each $1 \leq i \leq r$.
\dend
}
\end{definition}
%

Notice that while we allow the multiplication of two terms to have influence on
 the coefficients of the result\footnote{Skew-polynomial rings are a classical example,
 see Example \ref{exa.skew}.}, we require that $t \skm \alpha = \alpha \skm t$. 
On the other hand polynomials from $P$ can be involved more than once
 when defining g- and m-polynomials.
This is due to the fact that multiplying one polynomial with {\em different} terms can still
 result in the same term as head term while of course the multiples of
 the reducti are different.
These multiples have to be treated as different.

Given a set of polynomials $F$, the set of corresponding
 g- and m-polynomials contains those which are specified by
 Definition \ref{def.gpol} for each term $t \in \myt$ fulfilling the respective conditions.
For a set consisting of one polynomial the corresponding
m-polynomials reflect the multiplication of the polynomial with
zero-divisors of the head monomial, i.e., by a basis of the annihilator
 of the head monomial.
Notice that given a finite set of polynomials the corresponding sets of
 g- and m-polynomials in general can be infinite.

As in Theorem \ref{theo.rrsb=rgb.k} we can use g- and m-polynomials to characterize special bases
 in function rings.
As before we also have to take into account right multiples of the generating set
 as Example \ref{exa.free.group} does not require a field as coefficient domain.

%
\begin{theorem}\label{theo.rr.cp.i}~\\
{\sl
Let $F$ be a set of polynomials in $\f \backslash \{ \zero \}$.
Then $F$ is a weak right Gr\"obner basis if and only if
\begin{enumerate}
\item for all $f$ in $F$ and for all $m$ in $\monoms(\f)$, 
       $f \rmult m$ has a right reductive standard representation in terms of $F$, and
\item all g- and m-polynomials corresponding to $F$ as specified in
       Definition \ref{def.gpol}
       have right reductive standard representations in terms of $F$.
\end{enumerate}
\theoend
}
\end{theorem}
\Ba{}~\\
In case $F$ is a weak right Gr\"obner basis it is also a right reductive standard basis, and
 since these polynomials are all elements of $\ideal{r}{}(F)$ they must
 have right reductive standard representations.
\\
The converse will be proven by showing that every element in
 $\ideal{r}{}(F)$ has a right reductive standard representation in terms of $F$.
Let $g \in \ideal{r}{}(F)$ have a representation in terms of $F$ of
 the following form:
$g = \sum_{j=1}^m f_{j} \rmult (w_{j} \skm \alpha_j)$  such that
 $f_j \in F$, $w_j \in \myt$ and $\alpha_{j} \in \rr$.
Depending on this  representation of $g$ and the
 well-founded total ordering $\succeq$ on $\myt$ we define
 $t = \max_{\succeq} \{ \hterm(f_{j} \rmult (w_{j} \skm \alpha_j)) \mid 1\leq j \leq m \}$ and
 $K$ as the number of polynomials $f_j \rmult (w_{j} \skm \alpha_j)$ with head term $t$.
We show our claim by induction on $(t,K)$, where
 $(t',K')<(t,K)$ if and only if $t' \prec t$ or $(t'=t$ and
 $K'<K)$.
\\
Since by our first assumption every multiple $f_j \rmult (w_{j} \skm \alpha_j)$ in this
 sum has a right reductive standard representation in terms of $F$,
 we can assume that
 $\hterm(\hterm(f_j) \rmult w_j) = \hterm(f_j \rmult w_j) \geq
 \hterm(f_j)$ holds.
Moreover, without loss of generality we can assume that the polynomial multiples
 with head term $t$ are just $f_1  \rmult w_1, \ldots , f_K \rmult w_K$.
Notice that these assumptions on the representation of $g$ neither change
 $t$ nor $K$.
\\
Obviously, $t \succeq \hterm(g)$ must hold. 
If $K=1$ this gives us $t = \hterm(g)$ and by our assumptions our
 representation is already a right reductive one and we are done.
\\
Hence let us assume $K>1$.
\\
First let $\sum_{j=1}^K \hm(f_j \rmult (w_{j} \skm \alpha_j)) = \zero$.
Then by Definition \ref{def.gpol} there exists a tuple
 $(\alpha_1, \ldots, \alpha_K) \in M$, as 
 $\sum_{j=1}^K \hc(f_j \rmult w_{j}) \skm \alpha_j = 0$.
Hence there are $\delta_1, \ldots, \delta_K \in \rr$ such that
 $\sum_{i=1}^l A_i \skm \delta_i = (\alpha_1, \ldots, \alpha_K)$ for some $l \in \n$,
 $A_i=(\alpha_{i,1}, \ldots, \alpha_{i,K}) \in \{ A_j \mid j \in I_M \}$,
 and $\alpha_j = \sum_{i=1}^l \alpha_{i,j} \skm \delta_i$, $1 \leq j \leq K$.
By our assumption there are module polynomials $h_i = \sum_{j=1}^K f_j \rmult w_j \skm \alpha_{i,j}$,$1 \leq i \leq l$,
 all having right reductive standard representations in terms of $F$.
\\
Then since
\begin{eqnarray}
\sum_{j=1}^K f_j \rmult (w_{j} \skm \alpha_j) & = & \sum_{j=1}^K f_j \rmult w_{j} \skm (\sum_{i=1}^l \alpha_{i,j} \skm \delta_i) \nonumber\\
         &=& \sum_{j=1}^K \sum_{i=1}^l (f_j \rmult w_{j} \skm \alpha_{i,j}) \skm \delta_i \nonumber\\
         &=& \sum_{i=1}^l (\sum_{j=1}^K f_j \rmult w_{j} \skm \alpha_{i,j}) \skm \delta_i \nonumber\\
         &=& \sum_{i=1}^l h_i \skm \delta_i \nonumber
\end{eqnarray}
 we can change the representation of $g$ to
 $\sum_{i=1}^l h_i \skm \delta_i + \sum_{j=K+1}^m f_{j} \rmult (w_{j} \skm \alpha_j)$
 and replace each $h_i$ by its right reductive standard representation in terms of $F$.
Remember that for all $h_i$, $1 \leq i \leq l$ we have $\hterm(h_i) \pred t$.
Hence, for this new representation we now have maximal term smaller than $t$
 and by our induction hypothesis we have a right reductive standard representation
 for $g$ in terms of $F$ and are done.
\\
It remains to study the case where $\sum_{j=1}^K \hm(f_j \rmult (w_j \skm \alpha_j)) \neq 0$.
Then
 we have $\hterm(f_1 \rmult (w_1 \skm \alpha_1) + \dots + f_K \rmult (w_K \skm \alpha_K)) = t = \hterm(g)$,
 $\hc(g) = \hc(f_1 \rmult (w_1 \skm \alpha_1) + \dots + f_K \rmult (w_K \skm \alpha_K)) \in \ideal{r}{}(\{ \hc(f_1 \rmult w_1), \ldots, 
 \hc(f_K \rmult w_K) \})$ and $\hm(f_1 \rmult (w_1 \skm \alpha_1) + \dots + f_K \rmult (w_K \skm \alpha_K)) = \hm(g)$.
Hence $\hc(g) = \alpha \skm \delta$ with $\delta \in \rr$ and $\alpha \in G$, a
 (weak) right Gr\"obner basis of $\ideal{r}{}(\{ \hc(f_1 \rmult w_1), \ldots, 
 \hc(f_K \rmult w_K) \})$
 (compare Definition \ref{def.gpol}).
Let $g_{\alpha}$ be the respective g-polynomial corresponding to $\alpha$.
Then the polynomial $g' = g - g_{\alpha} \skm \delta$ lies in
 $\ideal{r}{}(F)$.
Since the multiple $g_{\alpha} \skm \delta$ has a right reductive standard representation in terms of $F$,
 say $\sum_{j=1}^l f_i \rmult m_j$,
 for the situation $\sum_{j=1}^K f_{j} \rmult (w_{j} \skm \alpha_j)  - g_{\alpha} \skm \delta$
 all multiples involved have head term $t$ and their head monomials add up to $\zero$.
Therefore, this situation again corresponds to an m-polynomial of $F$.
Hence we can apply our results from above and get that the polynomial $g'$ 
 has a smaller representation than $g$,
 especially the maximal term $t'$ is smaller.
Moreover, we
 can assume that $g'$ has a right reductive standard representation in terms
 of $F$, say $g' = \sum_{i=1}^n f_i \rmult \tilde{m}_i$.
Then $g = \sum_{i=1}^n f_i \rmult \tilde{m}_i + g_{\alpha} \skm \delta =
 \sum_{i=1}^n f_i \rmult \tilde{m}_i + \sum_{j=1}^l f_i \rmult m_j \skm \delta$ is a right
 reductive standard representation\footnote{Note that right reductive standard representations are stable
 under multiplication with coefficients which are no zero-divisors of the head coefficient.}
 of $g$ in terms of $F$ and we are done.
\\
\qed
%
Since in general we will have infinitely many g- and m-polynomials related to $F$,
 it is important to look for possible localizations of these situations.
We are looking for concepts similar to those of
 weak saturation and stable localizations in the previous section.
Remember that Lemma \ref{lem.red.reps} is central there.
It describes when the existence of a right reductive standard representation for some
 polynomial implies the existence of a right reductive standard representation
 for a multiple of the polynomial.
Unfortunately we cannot establish an analogon to this lemma for
 right reductive standard representations in $\f$ as defined in Definition \ref{def.right_reductive_rr}.
\begin{example}~\\
{\rm
Let $\f$ be a function ring over the integers with $\myt = \{X_1, \ldots, X_7 \}$
 and multiplication $\rmult : \myt \times \myt \mapsto \f$ defined by the
 following equations:
$X_1 \rmult X_2 = X_4$, $X_4 \rmult X_3 = X_5$, $X_2 \rmult X_3 = X_6 + X_7$,
 $X_1 \rmult X_6 = 3 \skm X_5$, $X_1 \rmult X_7 = -2 \skm X_5$ and else
 $X_i \rmult X_j = \zero$.
Additionally let $X_5 > X_4 > X_1 \succ X_2 \succ X_3 \succ X_6 \succ X_7$.
\\
Then for $p = X_4$, $f = X_1$ and $m = X_3$ we find that
\begin{enumerate}
\item $p$ has a right reductive standard representation in terms of $\{ f \}$, namely $p = f \rmult X_2$.
\item $\hterm( p \rmult m) = \hterm( \hterm(p) \rmult m ) \geq \hterm(p)$ as
      $X_5 = X_4 \rmult X_3 > X_4$ and for all $X_i \pred X_4$ we have $X_i \rmult X_3 \pred X_5$.
\item $p \rmult m = X_5$ has no right reductive standard representation in terms of $\{ f \}$ as only
      $X_1 \rmult X_j \neq \zero$ for $j = \{ 2,6,7 \}$, 
        namely $X_1 \rmult X_2 = X_4$, $X_1 \rmult X_6 = 3 \skm X_5$,
        $X_1 \rmult X_7 = -2 \skm X_5$,  and
      $X_1 \rmult (X_j \skm \alpha) \neq X_5$ for all $j \in \{ 2,6,7 \}$, $\alpha \in \z$.
\end{enumerate}
Notice that these problems are due to the fact that while $(X_1 \rmult X_2) \rmult X_3 =
 X_1 \rmult (X_2 \rmult X_3) = X_5$, $X_1 \rmult (X_2 \rmult X_3) = X_1 \rmult (X_6 + X_7) =
 X_1 \rmult X_6 + X_1 \rmult X_7$ does not give us a right reductive standard representation in terms of $X_1$.
This was the crucial point in the proof of Lemma \ref{lem.red.reps} and it is only fullfilled for the weaker form 
 of right reductive standard representations in $\f_{\myk}$ as defined in Definition \ref{def.right_reductive} where only $\hterm(g) \succeq \hterm(f_i \rmult m_i)$
 is required for $1 \leq i \leq n$.
\exaend
}
\end{example}
As this example shows an analogon to Lemma \ref{lem.red.reps} does not hold in our general case.
Note that the trouble arises from the fact that we allow multiplication of two terms to result
 in a polynomial.
If we restrict ourselves to multiplications where multiples of monomials are again monomials,
 the proof of Lemma \ref{lem.red.reps} carries over and we can look for appropriate localizations.
However, the reduction relation defined in Definition \ref{def.rred_rr} is only one way of
 defining a reduction relation in $\f$ and we stated that the main motivation behind it
 is to link the reduction relation with special standard representations as it is done in the case of $\f_{\myk}$.
The question now arises whether this motivation is as appropriate for $\f$ as it was for $\f_{\myk}$.
In $\f_{\myk}$ any reduction relation based on stable divisibility of terms can be linked
 to right reductive standard representations as defined in Definition \ref{def.right_reductive}
 and hence the approach is very powerful.
It turns out that for different reduction relations in $\f$ based on stable right divisibility
 this is no longer so.
Let us look at another familiar way of generalizing a reduction relation for $\f$ from one 
 defined in the redution ring.
We introduce a (not necessarily Noetherian) partial ordering on $\rr$:
 for $\alpha, \beta \in \rr$, $\alpha >_{\rr} \beta$ if and only if
 there exists a finite set $B \subseteq \rr$ such that $\alpha
 \red{+}{\Longrightarrow}{}{B} \beta$.
This ordering will ensure that reduction in $\f$ is terminating when using a finite set 
 of polynomials.
%
\begin{definition}\label{def.rred.rr}~\\
{\rm
Let $f,p$ be two non-zero polynomials in $\f$.
We say $f$ \betonen{right reduces} $p$ \betonen{to} $q$ \betonen{at a monomial}
 $\alpha \skm t$ \betonen{in one step}, denoted by $p \red{}{\myr}{r}{f} q$, if
 there exists $ m \in \monoms(\f)$ such that
\begin{enumerate}
\item $t \in \supp(p)$ and $p(t) = \alpha$,
\item $\hterm(\hterm(f)\rmult m) = \hterm(f \rmult m) = t \geq \hterm(f)$, 
\item $\alpha \Longrightarrow_{\hc(f \rmult m)} \beta$, with  
      $\alpha = \beta + \hc(f) \skm \gamma$
      for some $\beta, \gamma\in \rr$, and
\item $q = p -  f \rmult m \skm \gamma$.
\end{enumerate}
We write $p \red{}{\myr}{r}{f}$ if there is a polynomial $q$ as defined
above and $p$ is then called right reducible by $f$. 
%\\
Further, we can define $\red{*}{\myr}{r}{}, \red{+}{\myr}{r}{}$ and
 $\red{n}{\myr}{r}{}$ as usual.
%\\
Right reduction by a set $F \subseteq \f \backslash \{ \zero \}$ is denoted by
 $p \red{}{\myr}{r}{F} q$ and abbreviates $p \red{}{\myr}{r}{f} q$
 for some $f \in F$.
\dend
}
\end{definition}
%
Notice that for this reduction relation we can still have $t \in \supp(q)$.
Hence other arguments than used in the proof of Lemma \ref{lem.rred.rr} have
 to be used to show termination.
It turns out that for infinite sets of polynomials $F \subset \f$ the reduction
 relation $\red{}{\myr}{r}{F}$ need not terminate.
\begin{example}\label{exa.reduction.not.terminating}~\\
{\rm
Let $\rr = \q[\{ X_i \mid i \in \n \}]$ with $X_1 \succ X_2 \succ \ldots$
 be the polynomial ring over the rationals with infinitely many indeterminates. 
We associate this ring with the reduction relation based on divisibility of terms.
Let $\f = \rr[Y]$ be our function ring.
Elements of $\f$ are polynomials in $Y^i$, $i \in \n$ with coefficients in $\rr$.
Then for $p = X_1 \skm Y$ and the infinite set $F = \{ f_i = (X_i - X_{i+1}) \skm Y
 \mid i \in \n \}$ we get the infinite reduction sequence
 $p \red{}{\myr}{r}{f_1} X_2 \skm Y \red{}{\myr}{r}{f_2} 
  X_3 \skm Y \red{}{\myr}{r}{f_3} \ldots$
\exaend
}
\end{example}
However, if we restrict ourselves to finite sets of polynomials the reduction
 relation is Noetherian.
%
\begin{lemma}\label{lem.sred.rr}~\\ 
{\sl 
Let $F$ be a finite set of polynomials in $\f \backslash \{ \zero\}$.
\begin{enumerate}
\item For $p,q \in \f$ $p \red{}{\myr}{r}{F} q$ implies $p \succ q$, in particular $\hterm(p)
  \succeq \hterm(q)$.
\item $\red{}{\myr}{r}{F}$ is Noetherian.
\lemend
\end{enumerate}
}
\end{lemma}
\Ba{}
\begin{enumerate}
\item Assuming that the reduction step takes place at a monomial $\alpha \skm t$,
      by Definition \ref{def.rred.rr} we know 
        $\hm(\alpha \skm t -  f \rmult m \skm \gamma) = \beta \skm t$ 
      which yields
      $p \succ p  -  f \rmult m \skm \gamma$
      since $\alpha >_{\rr} \beta$.
\item This follows from 1. and Axiom (A1) as long as only finite sets of polynomials are involved.
\end{enumerate}\renewcommand{\baselinestretch}{1}\small\normalsize
\qed
Now if we try to link the reduction relation in Definition 
 \ref{def.rred.rr} to special standard representations, we find
 that this is no longer as natural as in the cases studied before,
 where for $\f_{\myk}$ we linked the reduction relation from Definition \ref{def.rred}
 to the right reductive standard representations in Definition
 \ref{def.right_reductive} respectively for $\f$ the right reduction relation
 from Definition \ref{def.rred_rr} to right reductive standard 
 representations as defined in Definition \ref{def.right_reductive_rr}.
Hence we claim that for generalizing Gr\"obner bases to $\f$, the rewriting
 approach is more suitable.
\begin{definition}~\\
{\rm
A set $F \subseteq \f \backslash \{ \zero \}$ is called a weak
 right Gr\"obner basis of $\ideal{r}{}(F)$ if for all
 $g \in \ideal{r}{}(F)$ we have $g \red{*}{\myr}{r}{F} \zero$. 
\dend
}
\end{definition}
%
\begin{corollary}~\\
{\sl
Let $F$ be a set of polynomials in $\f_{\myk}$ 
 and $g$ a non-zero polynomial in $\ideal{r}{}(F)$
 such that $g \red{*}{\myr}{r}{F} \zero$.
Then $g$ has a representation of the form 
$$g = \sum_{i=1}^n f_i \rmult m_i,
 f_i \in F, m_i \in \monoms(\f_{\myk}), n \in \n$$
 such that
 $\hterm(g) = \hterm(\hterm(f_1) \rmult m_1) =
 \hterm(f_1 \rmult m_1) \geq \hterm(f_1)$ and
 $\hterm(g) \succeq \hterm(f_i \rmult m_i)$ 
 for all $2 \leq i \leq n$.
}
\end{corollary}
\Ba{}~\\
Since $\hterm(g)$ has to be reduced in the reduction sequence
 $g \red{*}{\myr}{r}{F} \zero$ there exists a polynomial $f \in F$
 and a monomial $m \in \monoms(\f)$
 such that $\hterm(\hterm(f) \rmult m) = \hterm(f \rmult m) = \hterm(g)
 \geq \hterm(f)$ and $\hc(g) \R_{\hc(f \rmult m)} \beta$ with
 $\hc(g) = \beta + \hc(f) \skm \gamma$ for some $\gamma \in \rr$.
Moreover $g - f \rmult m \skm \gamma$ reduces to zero.
Adding up all multiples arising in this sequence together with
 $f \rmult m \skm \gamma$ gives a representation of $g$ of the desired
 form.
\\
\qed 
%
We can characterize weak Gr\"obner bases similar to
 Theorem \ref{theo.rr.cp.i}.
Of course the g-polynomials in Definition \ref{def.gpol} depend on the reduction
 relation $\R$ in $\rr$ which now is  defined according to Definition \ref{def.rred.rr}.
Notice that the characterization will only hold for finite
 sets as the proof needs that the reduction relation is Noetherian.
Additionally we need that the reduction ring fulfills Axiom (A4),
 i.e., for $\alpha, \beta, \gamma, \delta \in \rr$,
             $\alpha \Longrightarrow_{\beta}$ and
             $\beta \Longrightarrow_{\gamma} \delta$ 
             imply $\alpha \Longrightarrow_{\gamma}$ or 
             $\alpha \Longrightarrow_{\delta}$\footnote{Notice that
 (A4) is no basis for localizing test sets, as this would require that
 $\alpha \Longrightarrow_{\beta}$ and $\beta \Longrightarrow_{\gamma} \delta$ 
 imply $\alpha \Longrightarrow_{\gamma}$. Hence even if the reduction relation
 in $\f$ satisfies (A4), this does not substitute Lemma \ref{lem.red.reps}
 or its variants.}.
%
\begin{theorem}\label{theo.gb.reduction}~\\
{\sl
Let $F$ be a finite set of polynomials in $\f \backslash \{ \zero \}$
 where the reduction ring satisfies (A4).
Then $F$ is a weak right Gr\"obner basis if and only if
\begin{enumerate}
\item for all $f$ in $F$ and for all $m$ in $\monoms(\f)$ we have 
       $f \rmult m \red{*}{\myr}{r}{F} \zero$, and
\item all g- and m-polynomials corresponding to $F$ as specified in
       Definition \ref{def.gpol}
       reduce to $\zero$ using $F$.
\end{enumerate}
\theoend
}
\end{theorem}
\Ba{}~\\
In case $F$ is a weak right Gr\"obner basis,
 since these polynomials are all elements of $\ideal{r}{}(F)$ they must
 reduce to zero using $F$.
\\
The converse will be proven by showing that every element in
 $\ideal{r}{}(F)$ is reducible  by $F$.
Then as $g \in \ideal{r}{}(F)$ and $g \red{}{\myr}{r}{F} g'$ implies
 $g' \in \ideal{r}{}(F)$ we have $g\red{*}{\myr}{r}{F} \zero$.
Notice that this only holds in case the reduction relation $\red{}{\myr}{r}{F}$
 is Noetherian.
This follows as by our assumption $F$ is finite (Lemma \ref{lem.sred.rr}).
\\
Let $g \in \ideal{r}{}(F)$ have a representation in terms of $F$ of
 the following form:
$g = \sum_{j=1}^m f_{j} \rmult m_{j}$ such that
 $f_j \in F$ and $m_{j} \in \monoms(\f)$.
Depending on this  representation of $g$ and the
 well-founded total ordering $\succeq$ on $\myt$ we define
 $t = \max_{\succeq} \{ \hterm(f_{j} \rmult m_{j}) \mid 1\leq j \leq m \}$ and
 $K$ as the number of polynomials $f_j \rmult m_j$ with head term $t$.
We show our claim by induction on $(t,K)$, where
 $(t',K')<(t,K)$ if and only if $t' \prec t$ or $(t'=t$ and
 $K'<K)$.
\\
Since by our first assumption every multiple $f_j \rmult m_j$ in this
 sum reduces to zero using $F$ and hence
 has a right reductive standard representation as defined
 in Definition \ref{def.right_reductive}, we can assume that
 $\hterm(\hterm(f_j) \rmult m_j) = \hterm(f_j \rmult m_j) \geq
 \hterm(f_j)$ holds.
Moreover, without loss of generality we can assume that the polynomial multiples
 with head term $t$ are just $f_1  \rmult m_1, \ldots , f_K \rmult m_K$.
Notice that these assumptions neither change $t$ nor $K$ for our representation
 of $g$.
\\
Obviously, $t \succeq \hterm(g)$ must hold. 
If $K=1$ this gives us $t = \hterm(g)$ and even
 $\hm(g) = \hm(f_1 \rmult m_1)$, implying that $g$ is right
 reducible at $\hm(g)$ by $f_1$.
\\
Hence let us assume $K>1$.
\\
First let $\sum_{j=1}^K \hm(f_j \rmult (w_{j} \skm \alpha_j)) = \zero$.
Then by Definition \ref{def.gpol} there is $(\alpha_1, \ldots, \alpha_K) \in M$, as 
 $\sum_{j=1}^K \hc(f_j \rmult w_{j}) \skm \alpha_j = 0$.
Hence there are $\delta_1, \ldots, \delta_K \in \rr$ such that
 $\sum_{i=1}^l A_i \skm \delta_i = (\alpha_1, \ldots, \alpha_K)$ for some $l \in \n$,
 $A_i=(\alpha_{i,1}, \ldots, \alpha_{i,K}) \in \{ A_j \mid j \in I_M \}$,
 and $\alpha_j = \sum_{i=1}^l \alpha_{i,j} \skm \delta_i$, $1 \leq j \leq K$.
By our assumption there are module polynomials $h_i = \sum_{j=1}^K f_j \rmult w_j \skm \alpha_{i,j}$,$1 \leq i \leq l$,
 all having right reductive standard representations in terms of $F$ as defined
 in Definition \ref{def.right_reductive}.
\\
Then since
\begin{eqnarray}
\sum_{j=1}^K f_j \rmult (w_{j} \skm \alpha_j) & = & \sum_{j=1}^K f_j \rmult w_{j} \skm (\sum_{i=1}^l \alpha_{i,j} \skm \delta_i) \nonumber\\
         &=& \sum_{j=1}^K \sum_{i=1}^l (f_j \rmult w_{j} \skm \alpha_{i,j}) \skm \delta_i \nonumber\\
         &=& \sum_{i=1}^l (\sum_{j=1}^K f_j \rmult w_{j} \skm \alpha_{i,j}) \skm \delta_i \nonumber\\
         &=& \sum_{i=1}^l h_i \skm \delta_i \nonumber
\end{eqnarray}
 we can change the representation of $g$ to
 $\sum_{i=1}^l h_i \skm \delta_i + \sum_{j=K+1}^m f_{j} \rmult (w_{j} \skm \alpha_j)$
 and replace each $h_i$ by its right reductive standard representation in terms of $F$.
Remember that for all $h_i$, $1 \leq i \leq l$ we have $\hterm(h_i) \pred t$.
Hence, for this new representation we now have maximal term smaller than $t$
 and by our induction hypothesis $g$ is reducible by $F$ and we are done.
\\
It remains to study the case where $\sum_{j=1}^K \hm(f_j \rmult (w_j \skm \alpha_j)) \neq 0$.
Then
 we have $\hterm(f_1 \rmult (w_1 \skm \alpha_1) + \dots + f_K \rmult (w_K \skm \alpha_K)) = t = \hterm(g)$,
 $\hc(g) = \hc(f_1 \rmult (w_1 \skm \alpha_1) + \dots + f_K \rmult (w_K \skm \alpha_K)) \in \ideal{r}{}(\{ \hc(f_1 \rmult w_1), \ldots, 
 \hc(f_K \rmult w_K) \})$ and 
 even $\hm(f_1 \rmult (w_1 \skm \alpha_1) + \dots + f_K \rmult (w_K \skm \alpha_K)) = \hm(g)$.
Hence $\hc(g)$ is $\R$-reducible  by $\alpha$, $\alpha \in G$, a
 (weak) right Gr\"obner basis of $\ideal{r}{}(\{ \hc(f_1 \rmult w_1), \ldots, 
 \hc(f_K \rmult w_K) \})$ in $\rr$ with respect to the reduction relation $\R$.
Let $g_{\alpha}$ be the respective g-polynomial corresponding to $\alpha$.
Then we know that $g_{\alpha}\red{*}{\myr}{r}{F} \zero$.
Moreover, we know that the head monomial of $g_{\alpha}$ is reducible
 by some polynomial $f \in F$ and we assume
 $\hterm(g_{\alpha}) = \hterm(\hterm(f) \rmult m) = \hterm(f \rmult m) \geq \hterm(f)$
 and $\hc(g_{\alpha}) \R_{\hc(f \rmult m)}$.
Then, as $\hc(g)$ is $\R$-reducible by $\hc(g_{\alpha})$, 
 $\hc(g_{\alpha})$ is $\R$-reducible  and (A4) holds,
 the head monomial of $g$ is also reducible\footnote{Remember that by (A4)
 for $\alpha, \beta_1, \ldots, \beta_k \in \rr$, $\alpha \red{*}{\R}{}{\{
 \beta_1,\ldots, \beta_k \}} 0$ implies $\alpha \R_{\beta_i}$ for some
 $1 \leq i \leq k$.} by some $f´ \in F$ and we are done. 
\\
\qed
%
Of course this theorem is also true for infinite $F$ if we can show
 that for the respective function ring the reduction relation is terminating.

Now the question arises when the critical situations in this characterization
 can be localized to subsets of the respective sets
 as in Theorem \ref{theo.s-pol.2}.
Reviewing the Proof of Theorem \ref{theo.s-pol.2} we find that 
 Lemma \ref{lem.red.reps} is central as it describes when multiples of
 polynomials which have a right reductive standard representation in terms
 of some set $F$ again have such a representation.
As we have seen above, this will not hold for function rings over reduction
 rings in general.
Now one way to introduce localizations would be to restrict the attention
 to those $\f$ satisfying Lemma \ref{lem.red.reps}.
Then appropriate adaptions of Definition \ref{def.weakly.saturated},
 \ref{def.saturator} and \ref{def.s-poly.localization} would allow a localization
 of the critical situations.
However, we have stated that it is not natural to link right reduction as defined
 in Definition \ref{def.rred_rr} to special standard representations.
Hence, to give localizations of Theorem \ref{theo.gb.reduction} another
 property for $\f$ is sufficient:
%
\begin{definition}\label{def.stable.loc}~\\
{\rm
A set $C \subset S \subseteq \f$ is called a \betonen{stable localization} of
 $S$ if for every $g \in S$ there exists $f \in C$
 such that $g \red{}{\myr}{r}{f} \zero$.
% and if $\hm(f)$ is reducible by $F$, so is $\hm(g)$.
\dend
}
\end{definition}
%
In case $\f$ and $\red{}{\myr}{r}{}$ allow such stable localizations,
 we can rephrase Theorem \ref{theo.gb.reduction} as follows:
\begin{theorem}\label{theo.loc}~\\
{\sl
Let $F$ be a finite set of polynomials in $\f \backslash \{ \zero \}$
 where the reduction ring satisfies (A4).
Then $F$ is a weak right Gr\"obner basis if and only if
\begin{enumerate}
\item for all $h$ in a stable localization of 
       $\{ f \rmult m \mid f \in \f, m \in \monoms(\f) \}$ we have 
       $h \red{*}{\myr}{r}{F} \zero$, and
\item for all $h$ in a stable localization of the g- and m-polynomials corresponding to $F$ as specified in
       Definition \ref{def.gpol} we have 
       $h \red{*}{\myr}{r}{F} \zero$.
\end{enumerate}
\theoend
}
\end{theorem} 
%
We have stated that for arbitrary reduction relations in $\f$ it is not natural to
 link them to special standard representations.
Still, when proving Theorem \ref{theo.loc}, we will find that in order to change the
 representation of an arbitrary right ideal element,
 Definition \ref{def.stable.loc} is not enough to ensure reducibility.
However, we can substitute the critical situation using an analogon of
 Lemma \ref{lem.red.reps}, which while not related to reducibility in this case will
 still be sufficient to make the representation smaller.
%
\begin{lemma}\label{lem.red.reps_rr}~\\
{\sl
Let  $F \subseteq \f \backslash \{ \zero\}$ and
 $f$, $p$  non-zero polynomials in $\f$.
If $p \red{}{\myr}{r}{f} \zero$ and $f \red{*}{\myr}{r}{F} \zero$,
 then $p$ has a standard representation of the form
$$p = \sum_{i=1}^n f_i \rmult l_i,
 f_i \in F, l_i \in \monoms(\f), n \in \n$$
 such that
 $\hterm(p) = \hterm(\hterm(f_i) \rmult l_i) =
 \hterm(f_i \rmult l_i) \geq \hterm(f_i)$ for $1 \leq i \leq k$ and
 $\hterm(p) \succeq \hterm(f_i \rmult l_i)$ 
 for all $k+1 \leq i \leq n$ (compare Definition \ref{def.right_reductive}).
\lemend
}
\end{lemma}
\Ba{}~\\
If $p \red{}{\myr}{r}{f} \zero$ then $p = f \rmult m$ with $m \in \monoms(\f)$
 and $\hterm(p) = \hterm(\hterm(f) \rmult m) =
 \hterm(f \rmult m) \geq \hterm(f)$.
Similarly $f \red{*}{\myr}{r}{F} \zero$ implies
 $f = \sum_{i=1}^n f_i \rmult m_i,
 f_i \in F, m_i \in \monoms(\f), n \in \n$
 such that
 $\hterm(f) = \hterm(\hterm(f_1) \rmult m_1) =
 \hterm(f_1 \rmult m_1) \geq \hterm(f_1)$ and
 $\hterm(f) \succeq \hterm(f_i \rmult m_i)$ 
 for all $2 \leq i \leq n$. 
\\
Let us first analyze $f_i \rmult m_i \rmult m$ with $\hterm(f_i \rmult m_i) = \hterm(f)$.
Without loss of generality we can assume $1 \leq i \leq k$.
Since $\hterm(\hterm(f) \rmult m) =
 \hterm(f \rmult m) \geq \hterm(f)$ and for all terms
 $s \neq \hterm(f_i \rmult m_i)$
 in the polynomial $f_i \rmult m_i$
 we have $\hterm(f) \succ s$  we get
 $\hterm(\hterm(f) \rmult m) \succ \hterm(s \rmult m)$ and in particular
 $\hterm(p) =\hterm(f \rmult m) = \hterm(\hterm(f) \rmult m) \succ \hterm(s \rmult m)$.
This now implies
\begin{eqnarray*}
 \hterm(f \rmult m) & = & \hterm(\hterm(f) \rmult m) \\
                    & = & \hterm(\hterm(\hterm(f_i) \rmult m_i) \rmult m), \mbox{ as } 
                          \hterm (f) = \hterm(\hterm(f_i) \rmult m_i) \\
                    & = & \hterm(\hterm(f_i \rmult m_i) \rmult m), \mbox{ as } 
                          \hterm(\hterm(f_i) \rmult m_i) = \hterm(f_i \rmult m_i) \\
                    & = & \hterm(f_i \rmult m_i \rmult m)
\end{eqnarray*}
and
since $\hterm(f \rmult m) \geq \hterm(f) \geq \hterm(f_i)$ we can conclude
 $\hterm(f_i \rmult m_i \rmult m) \geq \hterm(f_i)$.
It remains to show that $(f_i \rmult m_i) \rmult m = f_i \rmult ( m_i \rmult m)$ has
 a standard representation of the desired form in terms of $F$.
\\
In case $m_i \rmult m \in \monoms(\f_{\myk})$ we are done as then $f_i \rmult ( m_i \rmult m)$
 is one.
\\
Hence let us assume
 $m_i \rmult m = \sum_{j=1}^{k_i} \tilde{m}^i_j$, $\tilde{m}^i_j \in \monoms(\f)$.
Then for each $s \in \supp(f_i \rmult m_i) \backslash \{ \hterm(f_i \rmult m_i) \}$
 there exists $t_s \in \supp(f_i)$ such that
 $s \in \supp(t_s \rmult m_i) \backslash \{ \hterm(f_i \rmult m_i) \}$.
Since $\hterm(f) \succ s$ and even $\hterm(f) \succ t_s \rmult m_i$ we find that
 either
 $\hterm(f \rmult m) \succeq \hterm((t_s \rmult m_i) \rmult m) = \hterm(t_s \rmult (m_i \rmult m))$
 in case $\hterm(t_s \rmult m_i) = \hterm(f_1 \rmult m_i)$
 or 
 $\hterm(f \rmult m) \succeq (t_s \rmult m_i) \rmult m = t_s \rmult (m_i \rmult m)$
 as $\f$ is associative.
Hence we can conlude $f_i \rmult \tilde{m}^i_j \predeq \hterm(f \rmult m)$, $1 \leq j \leq k$ and
 for at least one $\tilde{m}^i_j$ we get
 $\hterm(f_i \rmult \tilde{m}^i_j) = \hterm(f_i \rmult m_i \rmult m) \geq \hterm(f_i)$.
\\
It remains to analyze the situation for the function
 $(\sum_{i=k+1}^n f_i \rmult m_i) \rmult m$.
Again we find that for all terms $s$ in the $f_i \rmult m_i$, $k+1 \leq i \leq n$,
 we have $\hterm(f) \succeq s$ and we get 
 $\hterm(f \rmult m) \succeq \hterm(s \rmult m)$.
Hence all polynomial multiples of the $f_i$  in the representation 
 $\sum_{i=k+1}^n \sum_{j=1}^{k_i} f_i \rmult \tilde{m}^i_j$, where
 $m_i \rmult m = \sum_{j=1}^{k_i}\tilde{m}^i_j$, are bounded by 
 $\hterm(f \rmult m)$.
\\
\qed


\Ba{Theorem \ref{theo.loc}}~\\
The proof is basically the same as for Theorem \ref{theo.gb.reduction}.
Due to Lemma \ref{lem.red.reps_rr} we can substitute the multiples $f_j \rmult m_j$
 by appropriate standard representations without changing $(t,K)$.
Hence, we only have to ensure that despite testing less polynomials we are able to
 apply our induction hypothesis.
Taking the notations from the proof of Theorem \ref{theo.gb.reduction},
let us first check the situation for m-polynomials.
\\
Let $\sum_{j=1}^K \hm(f_j \rmult (w_{j} \skm \alpha_j)) = \zero$.
Then by Definition \ref{def.gpol} there exists
 $(\alpha_1, \ldots, \alpha_K) \in M$, as 
 $\sum_{j=1}^K \hc(f_j \rmult w_{j}) \skm \alpha_j = 0$.
Hence there are $\delta_1, \ldots, \delta_K \in \rr$ such that
 $\sum_{i=1}^l A_i \skm \delta_i = (\alpha_1, \ldots, \alpha_K)$ for some $l \in \n$,
 $A_i=(\alpha_{i,1}, \ldots, \alpha_{i,K}) \in \{ A_j \mid j \in I_M \}$,
 and $\alpha_j = \sum_{i=1}^l \alpha_{i,j} \skm \delta_i$, $1 \leq j \leq K$.
There are module polynomials
 $h_i = \sum_{j=1}^K f_j \rmult w_j \skm \alpha_{i,j}$,$1 \leq i \leq l$
 and by our assumption there are polynomials $h_i'$
 in the stable localization such that $h_i \red{}{\myr}{r}{h_i'} \zero$.
Moreover, $h_i' \red{*}{\myr}{r}{F} \zero$.
Then by Lemma \ref{lem.red.reps_rr} the m-polynomials $h_i$ all have standard representations
 bounded by $t$.
Again we get
\begin{eqnarray}
\sum_{j=1}^K f_j \rmult (w_{j} \skm \alpha_j) & = & \sum_{j=1}^K f_j \rmult w_{j} \skm (\sum_{i=1}^l \alpha_{i,j} \skm \delta_i) \nonumber\\
         &=& \sum_{j=1}^K \sum_{i=1}^l (f_j \rmult w_{j} \skm \alpha_{i,j}) \skm \delta_i \nonumber\\
         &=& \sum_{i=1}^l (\sum_{j=1}^K f_j \rmult w_{j} \skm \alpha_{i,j}) \skm \delta_i \nonumber\\
         &=& \sum_{i=1}^l h_i \skm \delta_i \nonumber
\end{eqnarray}
 and we can change the representation of $g$ to
 $\sum_{i=1}^l h_i \skm \delta_i + \sum_{j=K+1}^m f_{j} \rmult (w_{j} \skm \alpha_j)$
 and replace each $h_i$ by the respective special standard representation in terms of $F$.
Remember that for all $h_i$, $1 \leq i \leq l$ we have $\hterm(h_i) \pred t$.
Hence, for this new representation we now have maximal term smaller than $t$
 and by our induction hypothesis $g$ is reducible by $F$ and we are done.
\\
It remains to study the case where $\sum_{j=1}^K \hm(f_j \rmult (w_j \skm \alpha_j)) \neq 0$.
Then
 we have $\hterm(f_1 \rmult (w_1 \skm \alpha_1) + \dots + f_K \rmult (w_K \skm \alpha_K)) = t = \hterm(g)$,
 $\hc(g) = \hc(f_1 \rmult (w_1 \skm \alpha_1) + \dots + f_K \rmult (w_K \skm \alpha_K)) \in \ideal{r}{}(\{ \hc(f_1 \rmult w_1), \ldots, 
 \hc(f_K \rmult w_K) \})$ and 
 even $\hm(f_1 \rmult (w_1 \skm \alpha_1) + \dots + f_K \rmult (w_K \skm \alpha_K)) = \hm(g)$.
Hence $\hc(g)$ is $\R$-reducible  by $\alpha$, $\alpha \in G$, a
 (weak) right Gr\"obner basis of $\ideal{r}{}(\{ \hc(f_1 \rmult w_1), \ldots, 
 \hc(f_K \rmult w_K) \})$ in $\rr$ with respect to the reduction relation $\R$.
Let $g_{\alpha}$ be the respective g-polynomial corresponding to $\alpha$.
Then we know that $g_{\alpha}\red{}{\myr}{r}{g_{\alpha}'} \zero$ for some
 $g_{\alpha}'$ in the stable localization and $g_{\alpha}'\red{*}{\myr}{r}{F} \zero$.
Moreover, we know that the head monomial of $g_{\alpha}'$ is reducible
 by some polynomial $f \in F$ and we assume
 $\hterm(g_{\alpha}) = \hterm(\hterm(f) \rmult m) = \hterm(f \rmult m) \geq \hterm(f)$
 and $\hc(g_{\alpha}) \R_{\hc(f \rmult m)}$.
Then, as $\hc(g)$ is $\R$-reducible by $\hc(g_{\alpha})$, 
 $\hc(g_{\alpha})$ is $\R$-reducible by $\hc(g_{\alpha}')$, 
 $\hc(g_{\alpha}')$ is $\R$-reducible to zero and (A4) holds,
 the head monomial of $g$ is also reducible by some $f´ \in F$ and we are done. 
\\
\qed
Again, if for infinite $F$ we can assure that the reduction relation is Noetherian,
 the proof still holds.