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\title{Powers of monomial ideals and the Ratliff-Rush operation}

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\author{Oleksandra Gasanova}

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\address{Department of Mathematics, Uppsala University, Box 480, 751 06 Uppsala, Sweden\\} %\addressmark{2}School of Something, University of Something Else, Somewhere}

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\abstract{Powers of (monomial) ideals is a subject that still calls attraction in various ways. In this paper we present a nice presentation of high powers of ideals in a certain class in $\mathbb C[x_1, ..., x_n]$ and $\mathbb C[[x_1, ..., x_n]]$. As an interesting application it leads to an algorithm to compute Ratliff-Rush ideals for that class. The Ratliff-Rush operation itself has several applications, for instance, if $I$ is a regular $\mathfrak m$-primary ideal in a local ring $(R,m)$, then the Ratliff-Rush associated ideal $\tilde I$ is the unique largest ideal containing $I$ with the same Hilbert polynomial as $I$.
}






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\keywords{Ratliff-Rush operation, powers of monomial ideals, polynomial rings}

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\maketitle

\section{Introduction}\label{sec:sec1}

Let $R$ be a commutative Noetherian ring and $I$ a regular ideal in it, that is, an ideal containing a non-zerodivisor. The Ratliff-Rush ideal associated to $I$ is defined as $\tilde I=\cup_{k\ge 0}(I^{k+1}:I^k)$. For simplicity we will call it the Ratliff-Rush operation on $I$, even though it does not preserve inclusion, as shown in \cite{Sw}. In \cite{R-R} it is proved that $\tilde I$ is the unique largest ideal that satisfies $I^l=\tilde I^l$ for all large $l$. An ideal $I$ is called Ratliff-Rush if $I=\tilde I$. Properties of the Ratliff-Rush operation and its interaction with other algebraic operations have been studied by several authors, see \cite{Sw,R-R,He}. In particular, we would like to mention the following two results. If $I$ is an $\mathfrak m$-primary ideal in a local ring $(R,\mathfrak m)$, then $\tilde I$ is the unique largest ideal containing $I$ with the same Hilbert polynomial (the length of ($R/I^l$) for sufficiently large $l$) as $I$. It is also known that the associated graded ring $\oplus_{k\ge 0}I^k/I^{k+1}$ has positive depth if and only if all powers of $I$ are Ratliff-Rush (see \cite{He} for a proof).  Recently there have been discovered connections to Castelnuovo-Mumford regularity (see \cite{Ca}).

In this paper we describe an algorithm for computing the Ratliff-Rush ideal of $\mathfrak m$-primary monomial ideals of a certain class (we will call it a class of good ideals), which is a generalization of algorithms described in \cite{Ve} and \cite{AA}: if we restrict to two variables, the ideals $I_{q_1,0}$ and $I_{0,q_2}$, defined in \cref{sec:sec5}, are exactly $I_T$ and $I_S$, defined in \cite{Ve} and \cite{AA}.

In \cref{sec:sec3} we introduce the notion of a good ideal. The idea is as follows: any $\mathfrak m$-primary monomial ideal has some $x_1^{d_1},\ldots,x_n^{d_n}$ as minimal generators and thus defines a (non-disjoint) covering of $\mathbb N^n$ with rectangular "boxes" of sizes $d_1,\ldots,d_n$.
Then $I$ is called a good ideal if it satisfies the so-called box decomposition principle, namely, if for any positive integer $l$ any minimal generator of $I^l$ belongs to some box $B_{a_1,\ldots,a_n}$ with $a_1+\ldots+a_n=l-1$. We also discuss a necessary and a sufficient condition for being a good ideal. From this point we will work with good ideals, unless stated otherwise.

In \cref{sec:sec4} we associate an ideal to each box in the following way: if $I$ is a good ideal and $B_{a_1,\ldots,a_n}$ is some box, then it contains some of the minimal generators of $I^l$, where $l=a_1+\ldots+a_n+1$. Since they are in $B_{a_1,\ldots,a_n}$, they are divisible by $(x_1^{d_1})^{a_1}\cdots(x_n^{d_n})^{a_n}$. Therefore, we can define 

$$I_{a_1,\ldots,a_n}:=\left\langle{\frac{m}{(x_1^{d_1})^{a_1}\cdots(x_n^{d_n})^{a_n}}\mid m \in B_{a_1,\ldots, a_n} \cap G(I^l) }\right\rangle.$$ 
We will conclude this section by showing that $$I_{a_1,\ldots, a_n}=I^l:\langle(x_1^{d_1})^{a_1}\cdots(x_n^{d_n})^{a_n}\rangle,$$ which immediately implies the following property: if $(a_1,\ldots,a_n)\le(b_1,\ldots,b_n)$, then $I_{a_1,\ldots, a_n}\subseteq I_{b_1,\ldots, b_n}$.
We also study the asymptotic behaviour of $I_{a_1,\ldots,a_n}$. Now that we know that $I_{a_1,\ldots,a_n}$ grows when $(a_1,\ldots,a_n)$ grows, and given that ideals can not grow forever, we are expecting some sort of stabilization in $I_{a_1,\ldots,a_n}$ when $(a_1,\ldots,a_n)$ is large enough. In other words, we are expecting some pattern on $I^l$ for large $l$.

In \cref{sec:sec5} we prove the main theorem of this paper, namely, the following: if $I$ is a good ideal, then 
$\tilde{I}=I_{q_1,0,\ldots,0}\cap I_{0,q_2,\ldots,0}\cap\ldots\cap I_{0,\ldots,0,q_n}$,
where $I_{q_1,0,\ldots,0}$ is the stabilizing ideal of the chain $I_{0,0,\ldots,0}\subseteq I_{1,0,\ldots,0}\subseteq I_{2,0,\ldots,0}\subseteq\ldots$, $I_{0,q_2,\ldots,0}$ is the stabilizing ideal of the chain $I_{0,0,\ldots,0}\subseteq I_{0,1,\ldots,0}\subseteq I_{0,2,\ldots,0}\subseteq\ldots$ and so on. The pattern established in \cref{sec:sec4} plays an important role in the proof of the main theorem.

In \cref{sec:sec6} we show that computation of $I_{0,0,\ldots,q_i,0,\ldots,0}$ is much easier than it seems. In particular, we show that the corresponding chain stabilizes immediately as soon as we have two equal ideals.

\cref{sec:sec7} contains examples and explicit computations of $\tilde{I}$.

%In Section~9 we discuss the following question: are powers of good ideals also good? Unfortunately, the answer is negative in most cases. We also give a sufficient condition for all powers of a good ideal to be Ratliff-Rush and construct a family of examples of such ideals.

%In Section~10 we discuss how to detect whether a given ideal is a good one if it satisfies the necessary condition and does not satisfy the sufficient condition from Section~3.







 
\section{Preliminaries and notation}\label{sec:sec2}

Let $R=\mathbb C[x_1,...,x_n], n\ge2$. We start by listing a few basic properties of monomial ideals in $R$ that will be used later.
\begin{enumerate}
%\item
%For each monomial ideal there is a unique minimal generating set consisting of monomials. For an ideal $I$ we denote $G(I)$ to be its minimal monomial generating set.
%\item
%If $m\in I=\langle m_1,\ldots,m_k\rangle$, where $m$ and all $m_i$ are monomials, then there is some $i$ such that $m_i$ divides $m$.
%\item
%If $I=\langle m_1,\ldots,m_k \rangle$, $J=\langle n_1,\ldots,n_l\rangle$, then $IJ=\langle m_1n_1,\ldots,m_1n_l, \ldots, m_kn_1, \ldots, m_kn_l\rangle$, but this generating set is not minimal in general.
\item
There is a natural bijection between monomials in $\mathbb C[x_1,\ldots,x_n]$ and points in $\mathbb N^n$ via $x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n} \leftrightarrow (\alpha_1,\alpha_2,\ldots, \alpha_n)$. We say that 
$(\beta_1,\beta_2,\ldots, \beta_n) \le (\alpha_1,\alpha_2,\ldots, \alpha_n)$ if $\beta_i\le \alpha_i$ for all $i\in \{1,2,\ldots, n\}$. Clearly, $x_1^{\beta_1}x_2^{\beta_2}\cdots x_n^{\beta_n}$ divides $x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}$ if and only if $(\beta_1,\beta_2,\ldots, \beta_n) \le (\alpha_1,\alpha_2,\ldots, \alpha_n)$. Multiplication of monomials corresponds to addition of points. We will say that a monomial belongs to a subset of $\mathbb N^n$, meaning that the corresponding point belongs to that subset. We will also say that a point belongs to an ideal $I$, meaning that the corresponding monomial belongs to $I$.
\item
$I:(J_1+J_2)=(I:J_1)\cap (I:J_2)$, $(I_1+I_2):\langle m\rangle=I_1:\langle m\rangle+I_2:\langle m\rangle$ and $\langle m_1\rangle:\langle m_2\rangle=\left< \frac{m_1}{\gcd(m_1,m_2)}\right>$.
\end{enumerate}
Let $I$ be an  $\mathfrak m$-primary monomial ideal of $R$, where $\mathfrak m=\langle x_1,x_2,\ldots,x_n\rangle$, that is, for some positive integers $d_1,\ldots,d_n$ we have $\{x_1^{d_1},\ldots,x_n^{d_n}\}\subset G(I)$. Henceforth, by $I$ we always mean an $\mathfrak m$-primary monomial ideal and denote $\mu_i:=x_i^{d_i}$, $1\le i\le n$. In this paper we do not consider any polynomials other than monomials since it will always be sufficient to prove statements for monomials only.
\begin{definition}
Let $I$ be an ideal, let $a_1,\ldots, a_n$ be nonnegative integers and denote
$$B_{a_1,\ldots,a_n}:=([a_1d_1,(a_1+1)d_1]\times\ldots\times[a_nd_n,(a_n+1)d_n])\cap \mathbb N^n.$$
$B_{a_1,\ldots,a_n}$ will be called the {\bf {box}} with coordinates $(a_1,\ldots,a_n)$, associated to $I$. Points of the type $(k_1d_1,\ldots,k_nd_n)$ and the corresponding monomials, where all $k_i$ are nonnegative integers, will be called {\bf {corners}}. 
\end{definition}
Note that all minimal generators of $I$ lie in $B_{0,\ldots,0}$.

\section{Good and bad ideals}\label{sec:sec3}

In this section we will introduce the notion of a good ideal, state a necessary and a sufficient condition for being a good ideal and give some examples.
\begin{definition}
We will say that an ideal $I$ satisfies the {\bf{box decomposition principle}} if the following holds: for every positive integer $l$, every minimal generator of $I^l$ belongs to some box $B_{a_1,\ldots,a_n}$ such that $a_1+\ldots+a_n=l-1.$ Ideals satisfying the box decomposition principle will be called {\bf {good}}, otherwise they will be called {\bf {bad}}.
\end{definition}
\begin{example}
Consider the ideal $I=\langle x^3,y^3,z^3, xyz\rangle$  in $\mathbb C[x,y,z]$. Then $x^2y^2z^2$ is a minimal generator of $I^2$, but it only belongs to $B_{0,0,0}$ and $0+0+0 \not= 1$. Therefore, $I$ is a bad ideal.
\end{example}
\begin{example}
Let $I=\langle x^3,y^3,z^3, x^2y^2z^2\rangle$  in $\mathbb C[x,y,z]$. Then $$G(I^2)=\{x^6,y^6,z^6,x^3y^3,x^3z^3,y^3z^3,x^5y^2z^2, x^2y^5z^2, x^2y^2z^5\}.$$ 
$$G(I^2)\cap B_{1,0,0}=\{x^6,x^3y^3,x^3z^3,x^5y^2z^2\},$$
$$G(I^2)\cap B_{0,1,0}=\{y^6,x^3y^3,y^3z^3,x^2y^5z^2\},$$
$$G(I^2)\cap B_{0,0,1}=\{z^6,x^3z^3,y^3z^3,x^2y^2z^5\}.$$
Note that each minimal generator of $I^2$ belongs to at least one such box. Denote $S_{1,0,0}:=G(I^2)\cap B_{1,0,0}$ and similarly $S_{0,1,0}:=G(I^2)\cap B_{0,1,0}$ and $S_{0,0,1}:=G(I^2)\cap B_{0,0,1}$. We see that $S_{1,0,0}=\mu_1G(I)$, $S_{0,1,0}=\mu_2G(I)$, $S_{0,0,1}=\mu_3G(I)$, that is, $I^2=\langle S_{1,0,0}, S_{0,1,0}, S_{0,0,1}\rangle=\mu_1I+\mu_2I+\mu_3I$. Geometrically it means that $I^2$ is minimally generated by all appropriate shifts of $I$. 
Clearly, the pattern repeats in all powers of $I$:
$$I^l=\sum_{l_1+\ldots+l_n=l-1}\mu_1^{l_1}\ldots \mu_n^{l_n}I,$$
that is, $I$ is a good ideal.
\end{example}
Now we are interested in necessary and sufficient conditions for an ideal to be good.
\begin{theorem}{(A necessary condition)} Let $I$ be an ideal in $\mathbb C[x_1,...,x_n]$. If $I$ is a good ideal, then for any minimal generator $x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}$ of $I$ the following holds:  $$\frac{\alpha_1}{d_1}+\cdots+\frac{\alpha_n}{d_n}\ge1.$$
\end{theorem}
The idea of the proof is the following: assume that there is a minimal generator for which the above condition fails, that is, $m=x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}$ with $\frac{\alpha_1}{d_1}+\cdots+\frac{\alpha_n}{d_n}=1-\epsilon$, $\epsilon>0$. Then it is easy to show that the box decomposition principle fails for any $l>\frac{1}{\epsilon}$.

\begin{theorem}{(A sufficient condition)}
Let $I$ be an ideal in $\mathbb C[x_1,...,x_n]$. Assume that for any minimal generator $x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}$ of $I$ which is not a corner the following holds:
$$\frac{\alpha_1}{d_1}+\cdots+\frac{\alpha_n}{d_n}\ge \frac{n}{2}.$$
Then $I$ is a good ideal.
\end{theorem}
\begin{proof}
Let $m_1=x_1^{\alpha_1}\cdots x_n^{\alpha_n}$, $m_2=x_1^{\beta_1}\cdots x_n^{\beta_n}$ with $\frac{\alpha_1}{d_1}+\cdots+\frac{\alpha_n}{d_n}\ge \frac{n}{2}$ 
and $\frac{\beta_1}{d_1}+\cdots+\frac{\beta_n}{d_n}\ge \frac{n}{2}$. 
It suffices to show that $m_1m_2=\mu_ix_1^{\gamma_1}\cdots x_n^{\gamma_n}$ for some $i$ and with $\frac{\gamma_1}{d_1}+\cdots+\frac{\gamma_n}{d_n}\ge \frac{n}{2}$. Note that $\frac{\alpha_1+\beta_1}{d_1}+\cdots+\frac{\alpha_n+\beta_n}{d_n}\ge n$, thus we must have $\frac{\alpha_i+\beta_i}{d_i}\ge 1$ for some $i$. We can assume $i=1$, then $\frac{\alpha_1+\beta_1-d_1}{d_1}+\cdots+\frac{\alpha_n+\beta_n}{d_n}\ge n-1\ge \frac{n}{2}$. Setting $\gamma_1=\alpha_1+\beta_1-d_1$ and $\gamma_i=\alpha_i+\beta_i$ for $2\le i \le n$ finishes the proof.
\end{proof}
\begin{remark}
For $n=2$ the necessary and sufficient conditions are equivalent.
\end{remark}
\begin{example}(A good ideal that does not satisfy the sufficient condition)

Let $I=\langle \mu_1,\mu_2,\mu_3,m\rangle=\langle x^5,y^5,z^5,xyz^4\rangle\subset \mathbb C[x,y,z]$. The ideal satisfies the necessary condition, but not the sufficient one. For examining $G(I^l)$, we first of all notice that $m^5=x^5y^5z^{20}$ is divisible by $\mu_1\mu_2\mu_3^3\in I^5$, thus $m\not \in G(I^5)$. Therefore, for any $l$, the minimal generators of $I^l$ will be of the form $\mu_1^{k_1}\mu_2^{k_2}\mu_3^{k_3}m^k$, where $k_1+k_2+k_3+k=l$ and $k\le 4$. If $k=0$, the monomial is just a corner and this case is trivial, so let $k\ge 1$. Clearly, such a monomial belongs to a box whose sum of coordinates is $l-1$ if and only if $m^k$ belongs to a box whose sum of coordinates is $k-1$. So the only thing we need to check is whether $m^k$ belongs to a box whose sum of coordinates is $k-1$, $2\le k \le 4$ (this is always true for $k=1$). We see that $m^2=x^2y^2z^8\in B_{0,0,1}$, $m^3=x^3y^3z^{12}\in B_{0,0,2}$, $m^4=x^4y^4z^{16}\in B_{0,0,3}$. Therefore, $I$ is a good ideal.
\end{example}

\begin{example}(A bad ideal that satisfies the necessary condition)

Let $I=\langle x^5,y^5,z^5,x^2y^2z^2\rangle\subset \mathbb C[x,y,z]$. The ideal satisfies the necessary condition, but not the sufficient one. We see that $x^4y^4z^4$ is a minimal generator of $I^2$ and it only belongs to $B_{0,0,0}$. Since $0+0+0 \not= 1$, $I$ is a bad ideal.
\end{example}
We would also like to point out that for any given ideal there exists a way to determine whether it is good or bad, but we do not know of any characterisation.

\section{Ideals inside boxes, their connection to each other and asymptotic behaviour}\label{sec:sec4}

\begin{definition}
Let $I$ be a good ideal and $a_1,\ldots, a_n$ nonnegative integers. We define $$I_{a_1,\ldots,a_n}:=\left\langle{\frac{m}{\mu_1^{a_1}\cdots\mu_n^{a_n}}\mid m \in G(I^l) \cap B_{a_1,\ldots, a_n}}\right\rangle,$$
where $l=a_1+\ldots+a_n+1$. Note that this a minimal generating set of $I_{a_1,\ldots,a_n}$.
\end{definition}
\begin{example}
\label{ex}
Let $I=\langle x^5,y^5,xy^4,x^4y\rangle\subset \mathbb C[x,y]$. $I$ is a good ideal by the sufficient condition. 
The picture below represents powers of $I$ up to $I^4$.

\begin{figure}[htb!]
\centering
\includegraphics
[scale=0.06]{boxes1.eps} 
\end{figure}

Consider the box $B_{1,0}$. Then $$G(I^2)\cap B_{0,1}=\{x^5y^5,x^6y^4,x^8y^2,x^9y,x^{10}\}.$$
%Since they are in $B_{1,0}$, they are divisible by $\mu_1^1\mu_2^0=x^5$. If we divide all these monomials by $\mu_1^1\mu_2^0$, we will get $$\{y^5,xy^4,x^3y^2,x^4y,x^5\}.$$
Therefore, $I_{1,0}=\langle y^5,xy^4,x^3y^2,x^4y,x^5\rangle$. Geometrically, this means viewing monomials in $B_{1,0}$ as if the smallest corner of $B_{1,0}$ was the origin. In this particular example we have $I_{0,0}=I$, $I_{1,0}=\langle y^5,xy^4,x^3y^2,x^4y,x^5\rangle$, $I_{0,1}=\langle y^5,xy^4,x^2y^3,x^4y,x^5\rangle$, $I_{a,b}=\langle y^5,xy^4,x^2y^3,x^3y^2,x^4y,x^5\rangle$ for all other $(a,b)$. 
\end{example}

%This gives rise to a more general definition.
It is easy to show that if $I$ is a good ideal, then any corner $\mu_1^{k_1}\cdots \mu_n^{k_n}$ is a minimal generator of $I^{k_1+\ldots+k_n}$, therefore, $\left\{\mu_j\prod_{i=1}^{n}\mu_i^{a_i}\mid 1\le j\le n\right\}\subseteq I^l\cap B_{a_1,\ldots,a_n}$, where $l=a_1+\ldots+a_n+1$ and therefore $\left\{\mu_1,\ldots,\mu_n\right\}\subseteq G(I_{a_1,\ldots,a_n})$ 
for all $a_1,\ldots,a_n$.



\begin{proposition}
\label{prop1}
Let $I$ be a good ideal and $a_1,\ldots,a_n$ nonnegative integers. Then $$I_{a_1,\ldots, a_n}=I^l:\langle\mu_1^{a_1}\cdots \mu_n^{a_n}\rangle,$$ where $l=a_1+\ldots+a_n+1$.
\end{proposition}
\begin{proof} It is clear from the definition that $I_{a_1,\ldots, a_n}\subseteq I^l:\langle\mu_1^{a_1}\cdots \mu_n^{a_n}\rangle$. For the other inclusion, let $m\in I^l:\langle\mu_1^{a_1}\cdots \mu_n^{a_n}\rangle$. Then $m\mu_1^{a_1}\cdots \mu_n^{a_n}\in I^l$, that is, $m\mu_1^{a_1}\cdots \mu_n^{a_n}$ is a multiple of some $g\in G(I^l)$, say, $m\mu_1^{a_1}\cdots \mu_n^{a_n}=gg_1$. Being a minimal generator of $I^l$, $g$ belongs to some box, say, $B_{b_1,\ldots,b_n}$ with $b_1+\ldots+b_n=l-1=a_1+\ldots+a_n$. If $(a_1,\ldots,a_n)=(b_1,\ldots,b_n)$, then $m$ is a multiple of $\frac{g}{\mu_1^{a_1}\cdots\mu_n^{a_n}}$, which is a generator of $I_{a_1,\ldots, a_n}$ and thus we are done. If $(a_1,\ldots,a_n)\not=(b_1,\ldots,b_n)$, then there is some $a_i<b_i$. Without loss of generality, we assume that $a_1<b_1$. Then the right hand side of $m\mu_1^{a_1}\cdots \mu_n^{a_n}=gg_1$ is divisible by $\mu_1^{b_1}$, thus $m$ is divisible by $\mu_1$, and $\mu_1$ is a minimal generator of $I_{a_1,\ldots, a_n}$ by the discussion before this proposition. Therefore, $m\in I_{a_1,\ldots, a_n}$.
\end{proof}

\begin{corollary}
\label{cor}
Let $I$ be a good ideal and let $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots, b_n$ be nonnegative integers such that $(a_1,\ldots,a_n)\le(b_1,\ldots,b_n)$. Then $I_{a_1,\ldots, a_n}\subseteq I_{b_1,\ldots, b_n}$.
\end{corollary}
Now we know that $I_{a_1,\ldots,a_n}$ grows as $(a_1,\ldots,a_n)$ grows. Since  $I_{a_1,\ldots,a_n}$ can not increase forever, one expects some pattern on high powers of $I$, which is indeed the case. 
%Let us take a closer look at the situation.
\begin{definition}
Let $a_1,\ldots,a_n$ be nonnegative integers. We will use the following notation: $$C_{\underline{a_1},\underline{a_2},\ldots,\underline{a_k},a_{k+1},a_{k+2},\ldots,a_n}:=\{(b_1,\ldots,b_n)\in\mathbb N^n \mid b_1=a_1,\ldots,b_k=a_k,b_{k+1}\ge a_{k+1},\ldots,b_{n}\ge a_{n}\}.$$
We will use a similar notation for any configuration of fixed and non-fixed coordinates. Sets of this type will be called {\bf {cones}}, for any cone the number of non-fixed coordinates will be called its {\bf dimension} and $(a_1,\ldots,a_n)$ will be called its {\bf vertex}. Note that $\mathbb N^n=C_{0,0,\ldots,0}$.
\end{definition}
%\begin{example}
%Let $n=3$ and $a_1=1,a_2=2,a_3=3$. Then $C_{1,\underline{2},3}=\{(b_1,2,b_3)\mid b_1\ge 1,b_3\ge3\}$. 
%\end{example}
\begin{definition}
Let $a_1,\ldots,a_n$ be nonnegative integers. By $A_{a_1,\ldots,a_n}$ we denote the set of all cones that satisfy the following conditions:
\begin{enumerate}


\item 
if $(b_1,\ldots,b_n)$ 
is the vertex of the cone, then $b_i\le a_i$ for all $1\le i\le n$;
\item for all $1\le i\le n$ the following holds: if $b_i=a_i$, then $b_i$ is not underlined and if $b_i<a_i$, then $b_i$ is underlined.
\end{enumerate}
Note that the unique cone of dimension $n$ in $A_{a_1,\ldots,a_n}$ is $C_{a_1,\ldots,a_n}$

\end{definition}

\begin{example}
\label{ex2}
Let $n=2$, $a_1=2$, $a_2=1$. 
%We would like to find all the cones in $A_{2,1}$. For any cone in $A_{2,1}$ the first coordinate of its vertex can only be chosen from the set $\{0,1,2\}$; we underline it if we choose 0 or 1 and do not underline it if we choose 2. Independently, the second coordinate can only be chosen from the set $\{0,1\}$ and we underline it if we choose 0 and do not underline it if we choose 1. Therefore, we will get six cones in total:
Then
$A_{2,1}=\{\textcolor{cyan}{C_{\underline{0},\underline{0}}},\textcolor{green}{C_{\underline{0},1}},\textcolor{othermagenta}{C_{\underline{1},\underline{0}}},\textcolor{blue}{C_{\underline{1},1}},\textcolor{orange}{C_{2,\underline{0}}},\textcolor{red}{C_{2,1}}\}$
\end{example}
\begin{figure}[htb!]
\centering
\includegraphics
[scale=0.1]{boxes2.eps} 
\end{figure}
The picture above represents the six cones from $A_{2,1}$. The boundary lines are only drawn for better visibility. Clearly, the number of boundary lines equals the dimension of the cone.
\begin{lemma}
Let $a_1,\ldots,a_n$ be nonnegative integers. Then cones in $A_{a_1,\ldots,a_n}$ form a disjoint covering of $\mathbb N^n$.
\end{lemma}
%\begin{proof}
%Let $(b_1,\ldots,b_n)$ be a point in $\mathbb N^n$. We will find a unique cone in $A_{a_1,\ldots,a_n}$ that contains this point. First of all, we compare $a_1$ and $b_1$.
%\begin{enumerate}
%\item If $b_1\ge a_1$, then the first coordinate of our future cone containing $(b_1,\ldots,b_n)$ is non-fixed since otherwise $A_{a_1,\ldots,a_n}$ contains $C_{\underline{b_1},\ldots}$, which is a contradiction with $b_1\ge a_1$. Therefore, our first coordinate has to be non-fixed, hence equal to $a_1$.
%\item If $b_1<a_1$, then the first coordinate can only be a fixed one since otherwise it is equal to $a_1$, but $(b_1,\ldots,b_n)$ can not belong to $C_{a_1,\ldots}$ since $b_1<a_1$. Therefore, the first coordinate has to be a fixed one, hence equal to $b_1$.
%\end{enumerate}
%Proceeding in the same way we construct a cone in $A_{a_1,\ldots,a_n}$ that contains $(b_1,\ldots,b_n)$. From the construction it is clear that this cone is unique, which finishes our proof.
%\end{proof}

%We have seen that given $\mathbb N^n=C_{0,0,\ldots,0}$ and a point $(a_1,\ldots,a_n)\in C_{0,0,\ldots,0}$, we can decompose $C_{0,0,\ldots,0}$ into a disjoint union of cones, associated to this point, where the unique cone of dimension $n$ is $C_{a_1,\ldots,a_n}$ and all other cones have strictly lower dimensions. It is not hard to see that we can replace $\mathbb N^n=C_{0,0,\ldots,0}$ with any other cone and replace $(a_1,\ldots,a_n)$ with any point in this cone and have a similar decomposition. First of all, assume that all coordinates of this cone are non-fixed, say, we have $C_{s_1,\ldots,s_n}$ and a point $(s_1+k_1,\ldots,s_n+k_n)\in C_{s_1,\ldots,s_n} $ for some nonnegative integers $k_1,\ldots,k_n$. Clearly, points in $C_{s_1,\ldots,s_n}$ are in bijection with points in $C_{0,0,\ldots,0}$ under the obvious shift. We can find the decomposition of $C_{0,0,\ldots,0}$ with respect to $(k_1,\ldots,k_n)$ as in the proposition above and then shift all the cones in the decomposition by $(s_1,\ldots,s_n)$ to get new cones. This will give us the desired decomposition of $C_{s_1,\ldots,s_n}$. Again, the unique cone of dimension $n$ in this decomposition is $C_{s_1+k_1,\ldots,s_n+k_1}$. Now assume that some coordinates of our cone are fixed, say, we have $C_{s_1,\ldots,s_m,\underline{s_{m+1}}\ldots,\underline{s_n}}$ (without loss of generality, we can assume that fixed coordinates are the last $(n-m)$ coordinates) and $(s_1+k_1,\ldots,s_m+k_m,s_{m+1},\ldots,s_n)\in C_{s_1,\ldots,s_m,\underline{s_{m+1}}\ldots,\underline{s_n}}$. Note that this is an $m$-dimensional cone and points in this cone are in bijection with points in $\mathbb N^m$, in particular, $(s_1+k_1,\ldots,s_m+k_m,s_{m+1},\ldots,s_n)\leftrightarrow(k_1,\ldots,k_m)$. Thus we can find the decomposition of $\mathbb N^m$ with respect to $(k_1,\ldots,k_m)$, then shift all cones by $(s_1\ldots,s_m)$ (this will give us the first $m$ coordinates of each cone) and the last $(n-m)$ coordinates of each cone in this decomposition are $\underline{s_{m+1}},\ldots,\underline{s_n}$. The unique $m$-dimensional cone in this decomposition is $C_{s_1+k_1,\ldots,s_m+k_m,\underline {s_{m+1}},\ldots,\underline{s_n}}$, others have lower dimensions. 
The previous lemma can be restated in a more general context:
\begin{theorem}
\label{th1}
Given any cone $C$ in $\mathbb N^n$ of dimension $k$ and a point $\textbf {a}\in C$, we can decompose $C$ into a disjoint union of finitely many cones, where exactly one cone has dimension $k$ and vertex $\textbf {a}$, and all other cones have strictly lower dimensions.
\end{theorem}

\begin{example}
Let $n=5$ and consider $C_{\underline{5},7,\underline{4},2,\underline{3}}$. Consider $(a_1,\ldots,a_5)=(5,9,4,3,3)\in C_{\underline{5},7,\underline{4},2,\underline{3}}$. The first, the third and the fifth coordinates are fixed once and forever, that is, all cones will have the form $C_{\underline{5},?,\underline{4},?,\underline{3}}$. We are left with the second and the fourth coordinate, that is, $(7,2)$ for the cone and $(9,3)$ for the point. Shifting in the negative direction by $(7,2)$, we will get $(0,0)$ and $(2,1)$ respectively. Thus it is enough to find the decomposition of $\mathbb N^2$ with respect to $(2,1)$, which has been done in \cref{ex2}. We obtained $A_{2,1}=\{C_{\underline{0},\underline{0}},C_{\underline{0},1},C_{\underline{1},\underline{0}},C_{\underline{1},1},C_{2,\underline{0}},C_{2,1}\}$. Shifting in the positive direction by $(7,2)$ gives us $\{C_{\underline{7},\underline{2}},C_{\underline{7},3},C_{\underline{8},\underline{2}},C_{\underline{8},3},C_{9,\underline{2}},C_{9,3}\}$
and inserting back the first, the third and the fifth coordinates gives us
$\{C_{\underline{5},\underline{7},\underline{4},\underline{2},\underline{3}},C_{\underline{5},\underline{7},\underline{4},3,\underline{3}},C_{\underline{5},\underline{8},\underline{4},\underline{2},\underline{3}},C_{\underline{5},\underline{8},\underline{4},3,\underline{3}},C_{\underline{5},9,\underline{4},\underline{2},\underline{3}},C_{\underline{5},9,\underline{4},3,\underline{3}}\}$. Therefore, $C_{\underline{5},7,\underline{4},2,\underline{3}}$ is a disjoint union of these six cones.
\end{example}


Now we will use these results on monomial ideals. Let $I$ be a good ideal. Then for any vector of nonnegative integers $(a_1,\ldots,a_n)$ we have defined a box $B_{a_1,\ldots,a_n}$ and the corresponding ideal $I_{a_1,\ldots,a_n}$. There is a bijection between points in $\mathbb N^n$ and boxes/ideals; recall that if $(a_1,\ldots,a_n)\le(b_1,\ldots,b_n)$, then $I_{a_1,\ldots, a_n}\subseteq I_{b_1,\ldots, b_n}$ by \cref{cor}.
\begin{theorem}
\label{th2}
For any good ideal $I$ there exists a finite coloring of $\mathbb N^n$ such that if $(a_1,\ldots,a_n)$ has the same color as $(b_1,\ldots,b_n)$, then $I_{a_1,\ldots,a_n}=I_{b_1,\ldots,b_n}$ and for each color the set of points with this color forms a cone.
\end{theorem}
\begin{proof}
 We use induction on the highest dimension of uncolored cones. We are starting with an $n$-dimensional cone $\mathbb N^n$. We will show how to obtain finitely many cones of strictly lower dimensions, each of which will then be treated similarly in a recursive way. First of all, note that it is possible to find a point $(a_1,\ldots, a_n)$ such that the following holds: if $(b_1,\ldots,b_n)\ge(a_1,\ldots,a_n)$, then $I_{a_1,\ldots,a_n}=I_{b_1,\ldots,b_n}$. Indeed, if we assume the converse, then for every point of $\mathbb N^n$ there exists a strictly larger point that corresponds to a strictly larger ideal, therefore, we can build an infinite chain of strictly increasing ideals, which is impossible by Noetherianity of the polynomial ring. So existence of such a point $(a_1,\ldots, a_n)$ is justified. Then from the \cref{th1}, $\mathbb N^n$ can be covered with a disjoint union of (finitely many) cones in $A_{a_1,\ldots,a_n}$. The unique $n$-dimensional cone in $A_{a_1,\ldots,a_n}$ is $C_{a_1,\ldots,a_n}$ and, as we have just figured out, we may paint all points in this cone with the same color. Now we are left with a disjoint union of cones of dimension at most $n-1$ which need to be painted and we apply induction on each of them, lowering the maximal dimension by 1 again. Since it is a finite process, in the end we will obtain a finite coloring of $\mathbb N^n$.
\end{proof}

We remark that the coloring described above is not unique since it depends on the choice of $(a_1,\ldots, a_n)$ and its lower dimensional analogues. 

\begin{example}
Let $I$ be the ideal in \cref{ex}. We can choose $(a_1,a_2)=(1,1)$ since $I_{b_1,b_2}=I_{1,1}$ for all $b_1\ge 1$ and $b_2\ge 1$. Then $\mathbb N^2$ is a disjoint union of $\textcolor{red}{C_{1,1}}$, $C_{\underline{0},1}$, $C_{1,\underline{0}}$  and $\textcolor{blue}{C_{\underline{0},\underline{0}}}$. Now consider $C_{\underline{0},1}$. We see that $I_{0,b}=I_{0,2}$ for all $b\ge 2$. Therefore, we consider the decomposition of $C_{\underline{0},1}$ with respect to $(0,2)$: $C_{\underline{0},1}$ is a disjoin union of $\textcolor{green}{C_{\underline{0},2}}$ and $\textcolor{cyan}{C_{\underline{0},\underline{1}}}$. Similarly, $C_{1,\underline{0}}$ is a disjoint union of $\textcolor{orange}{C_{2,\underline{0}}}$ and $\textcolor{othermagenta}{C_{\underline{1},\underline{0}}}$. The left picture below describes the coloring we have just discussed. The picture on the right describes another possible coloring if, for instance, we choose $(a_1,a_2)=(0,2)$.

\begin{figure}[htb!]
\centering
\includegraphics
[scale=0.07]{boxes3.eps} 
\end{figure}
\end{example}

Given a good ideal $I$, any coloring as in \cref{th2} represents a finite disjoint union of cones. Each cone has a vertex. Let $L$ denote the maximum of sums of coordinates of these vertices. This number depends on $I$ and on the coloring we choose, but we will not put any additional indices: as soon as we found some coloring (which exists according to \cref{th2}), we simply work with it henceforth. For example, for both colorings in the picture above we have $L=2$. 
%\begin{remark}
%\label{rem2}
%Note that from the construction in Theorem~\ref{th2} it is clear that $L$ can not be attained at a zero dimensional cone, in other words, for any zero dimensional cone, the sum of coordinates of its vertex is strictly less than $L$.
%\end{remark}
The geometric meaning of this number is the following: starting from $I^{L+1}$, we know exactly how powers of $I$ look like, given that we know the coloring. For instance, for the left coloring in the picture above we know that every power of $I$ starting from $I^3$ consists of a \textcolor{green}{green} box, an \textcolor{orange}{orange} box and several \textcolor{red}{red} boxes and we exactly know where each of them is. This means, there is a pattern on high powers of $I$, and this is a key point for finding the Ratliff-Rush closure of $I$.



\section{The main result}\label{sec:sec5}

Now we are ready to prove our main theorem, but first we need a preliminary lemma.
\begin{lemma}
\label{lem1}
Let $I$ be a good ideal and let $Q$ be any nonnegative integer. Then there exists a number $L(Q)$ such that for any $l\ge L(Q)$ the following holds: for every minimal generator $m$ of $I^l$ there is an $i$ such that $m=m'\mu_i^Q$ and $m'$ is a minimal generator of $I^{l-Q}$.
\end{lemma}
\begin{proof} 
If $Q=0$, the claim is trivial. Let $Q>0$ and let $L$ be the number defined in the end of \cref{sec:sec4}. Take $L(Q)=L+nQ-n+2$ and let $l\ge L(Q)$. Let $m$ be a minimal generator of $I^l$, then it belongs to some box $B_{b_1,\ldots, b_n}$ with $b_1+\ldots+b_n=l-1\ge L+nQ-n+1$. We also know that $(b_1,\ldots, b_n)$ belongs to one of the cones from our coloring; assume that the vertex of this cone is $(a_1,\ldots,a_n)$ (some coordinates are underlined, some are not underlined). Now we want to find a coordinate $b_i$ such that $(b_1,\ldots, b_{i-1}, b_i-Q, b_{i+1},\ldots, b_n)$ belongs to the same cone. Assume that it is not possible. Then it follows that $b_1-Q\le a_1-1, \ldots, b_n-Q\le a_n-1$. These inequalities yield a contradiction $L<b_1+\ldots+b_n-nQ+n\le a_1+\ldots+a_n\le L$, where the last inequality follows from the definition of $L$. So we can find an index $i$ such that $b_i-Q\ge a_i$ (in particular, this implies that $a_i$ is not underlined). Without loss of generality we assume that $i=1$. That means, $(b_1,\ldots, b_n)$ and $(b_1-Q,b_2,\ldots, b_n)$ are both in the same cone. This implies that their colors are equal, which means $I_{b_1,\ldots, b_n}=I_{b_1-Q,b_2,\ldots,b_n}$. In other words, the set of monomials in $B_{b_1,\ldots, b_n}\cap G(I^{l})$ coincides with the set of monomials in $B_{b_1-Q,b_2,\ldots,b_n}\cap G(I^{l-Q})$ up to a shift by $\mu_1^Q$.
Therefore, if $m \in B_{b_1,\ldots, b_n}$ is a minimal generator of $I^l$, then $\frac{m}{\mu_1^Q}\in B_{b_1-Q,b_2,\ldots,b_n}$ is a minimal generator of $I^{l-Q}$, as desired.
\end{proof}

Now let us consider the following line of boxes which is in bijection with nonnegative integer points on the $x$-axis: $B_{0,0,\ldots,0}$, $B_{1,0,\ldots,0}$, $B_{2,0,\ldots,0}$ etc. Let $B_{q_1,0\ldots,0}$ be the stabilizing box of this sequence in a sense that if $t\ge q_1$, then $I_{t,0,\ldots,0}=I_{q_1,0,\ldots,0}$. Similarly, considering lines of boxes going along the other coordinate axes, we will get $q_2,q_3,\ldots, q_n$. Denote $q:=\max\{q_1,\ldots, q_n\}$.
\begin{theorem}
Let $I$ be a good ideal, let $L$ and $q_i$ be as above. Then $\tilde{I}=I_{q_1,0,\ldots,0}\cap I_{0,q_2,\ldots,0}\cap\ldots\cap I_{0,\ldots,0,q_n}$.
\end{theorem}
\begin{proof}
 $\subseteq$ Let $l\ge q$. We will show that $I^{l+1}:I^l\subseteq I_{q_1,0,\ldots,0}\cap I_{0,q_2,\ldots,0}\cap\ldots\cap I_{0,\ldots,0,q_n}$. In fact, we will show that $I^{l+1}:I^l\subseteq I_{q_1,0,\ldots,0}$, other inclusions are analogous. Since $I^{l+1}:I^l\subseteq I^{l+1}:\langle\mu_1^l\rangle$, it is sufficient to show that $I^{l+1}:\langle \mu_1^l\rangle\subseteq I_{q_1,0,\ldots,0}$. By \cref{prop1}, $I^{l+1}:\langle \mu_1^l\rangle=I_{l,0,\ldots,0}$ which equals $I_{q_1,0,\ldots,0}$, given the way $I_{q_1,0,\ldots,0}$ was defined and given that $l\ge q\ge q_1$ . Therefore, everything follows.

 $\supseteq$ Let $m\in I_{q_1,\ldots,0}\cap I_{0,q_2,\ldots,0}\cap\ldots\cap I_{0,\ldots,0,q_n}$, let $l\ge L(q)=L+nq-n+2$ (as in \cref{lem1}). We will show that for every $m_l\in I^l$ we have $mm_l\in I^{l+1}$. It is enough to consider $m_l$ to be minimal generators of $I^l$. First of all, from \cref{lem1} we know that we can factor out some $\mu_i^q$ from $m_l$ and get a minimal generator of $I^{l-q}$, that is, $m_l=\mu_i^qm_{l-q}$ for some index $i$ and $m_{l-q}$ a minimal generator in $I^{l-q}$. Also, since $m$ belongs (in particular) to  $I_{0,\ldots,0,q_i,0\ldots,0}=I^{q_i+1}:\langle\mu_i^{q_i}\rangle$, it means, $m\mu_i^{q_i}\in I^{q_i+1}$. Therefore,
$mm_l=m\mu_i^{q_i}\mu_i^{q-q_i}m_{l-q}\in I^{l+1}$ since $m\mu_i^{q_i}\in I^{q_i+1}$, $\mu_i^{q-q_i}\in I^{q-q_i}$, $m_{l-q}\in I^{l-q}$.
\end{proof} 

 %and since $q_i\le q$, we can multiply further by $\mu_i^{q-q_i}\in I^{q-q_i}$, that is, $m\mu_i^{q_i}\mu_i^{q-q_i}\in I^{q_i+1}I^{q-q_i}$, which is, $m\mu_i^q\in I^{q+1}$.

\vspace{-3ex}
\section{Explicit computation of $I_{0,\ldots,0,q_i,0,\ldots,0}$}\label{sec:sec6}

 We have seen that, given a good ideal $I$, its Ratliff-Rush closure is computed as $\tilde{I}=I_{q_1,0,\ldots,0}\cap I_{0,q_2,\ldots,0}\cap\ldots\cap I_{0,\ldots,0,q_n}$. Therefore, we would like to know more about $I_{0,\ldots,0,q_i,0\ldots,0}$. Let $i=1$, other cases are analogous. So far we only know that $I_{t,0,\ldots,0\ldots,0}=I^{t+1}:\langle\mu_1^t\rangle$. Computation of $I^{t}$ might take much time if $t$ is large enough. In addition, we do not know yet at which moment the line has stabilized. So far the process seems more complicated than it is. We will state a few remarks to make this computation easier.
\begin{remark}
\label{rem1}
 If $I$  is a good ideal, then $I_{t+1,0,\ldots,0}=(I_{t,0\ldots,0}\cdot I):\langle\mu_1\rangle$ for all $t\ge 0$.
 \end{remark}
According to \cref{rem1}, we have

$$I_{t+1,0,\ldots,0}=\left<\frac{fm}{\gcd (fm,\mu_1)} \mid f\in G(I_{t,0,\ldots,0}), m\in G(I)\right>.$$
Let $f\in G(I_{t,0,\ldots,0})$ and $m\in G(I)$. Write $fm=x_1^{\alpha_1}\cdots x_n^{\alpha_n}$. If $\alpha_i\ge d_i$ for any $2\le i \le n$, then $\frac{fm}{\gcd (fm,\mu_1)}$ is a multiple of $\mu_i\in G(I_{t+1,0,\ldots,0})$. Therefore, in the above formula we may force that $deg_{x_i}(fm)<d_i$ for $2\le i\le n$. Moreover, consider $x_1^{\alpha_1}\cdots x_n^{\alpha_n}\mu_1^t=f\mu_1^tm\in I^{t+2}$ since $f\mu_1^t\in I^{t+1}$ according to \cref{prop1}. Since $\alpha_i<d_i$ for all $2\le i\le n$, we must have $\alpha_1\ge d_1$, otherwise $f\mu_1^tm$ would belong to a box with the sum of coordinates at most $t$. Thus $\gcd(fm,\mu_1)=\mu_1$. Therefore, we conclude that 
$$I_{t+1,0,\ldots,0}=\left<\frac{fm}{\mu_1} \mid f\in G(I_{t,0,\ldots,0}), m\in G(I), deg_{x_i}(fm)<d_i \text{ for } 2\le i\le n\right>.$$
 
 \begin{remark}
If $I_{t,0,\ldots,0}=I_{t+1,0,\ldots,0}$, then the line has stabilized, that is, $I_{k,0,\ldots,0}=I_{t,0,\ldots,0}$ for all $k\ge t$. This is a direct corollary of \cref{rem1}.
 \end{remark}
 \begin{remark}
Assume that we have computed $I_{t,0,\ldots,0}$ and $I_{t+1,0,\ldots,0}$ and let $E_t:=G(I_{t,0,\ldots,0})$ and $F_{t+1}:=\{\text{minimal generators of } I_{t+1,\ldots,0} \text{ which are not in } I_{t,\ldots,0}\}$. Clearly, $E_{t+1}$ is the reduced union of $E_t$ and $F_{t+1}$.  From \cref{rem1} we remember that
$I_{t+2,0,\ldots,0}=(I_{t+1,0,\ldots,0}\cdot I):\langle\mu_1\rangle=(\langle E_t \cup F_{t+1}\rangle\cdot I):\langle\mu_1\rangle=((I_{t,\ldots,0}+\langle F_{t+1}\rangle)\cdot I):\langle\mu_1\rangle=(I_{t,\ldots,0}\cdot I+\langle F_{t+1}\rangle\cdot I):\langle\mu_1\rangle=(I_{t,\ldots,0}\cdot I):\langle\mu_1\rangle+(\langle F_{t+1}\rangle\cdot I):\langle\mu_1\rangle=I_{t+1,0,\ldots,0}+(\langle F_{t+1}\rangle\cdot I):\langle\mu_1\rangle$. Therefore, we conclude that minimal generators of $I_{t+2,0,\ldots,0}$ which are not in $I_{t+1,0,\ldots,0}$  (our future $F_{t+2}$) could only be among $(\langle F_{t+1}\rangle\cdot I):\langle\mu_1\rangle$, that is, only new monomials from the previous iteration can give rise to new monomials in the next iteration. Therefore, in order to compute $F_{t+2}$ we need to compute $(\langle F_{t+1}\rangle\cdot I):\langle\mu_1\rangle$, reduce this set and throw away monomials that are already in $E_{t+1}$.  We can start with $E_{-1}=\emptyset$, $F_0=G(I)$.
 \end{remark}

\begin{remark}
We can exclude all $\mu_i$ from all sets and it will not affect the algorithm. In other words, we can replace $G(I)$ by $P(I):=G(I)\backslash\{\mu_1,\ldots,\mu_n\}$ everywhere and then include the corners in the very end of the algorithm. We have already discussed why we can exclude $\mu_i$ for $2\le i\le n$ (we want $deg_{x_i}(fm)<d_i \text{ for } 2\le i\le n$). We can also exclude $\mu_1$ since multiplying and then dividing by $\mu_1$ does not give us any new monomials. 


\end{remark}
 
\vspace{-3ex}
\section{Examples}\label{sec:sec7}

 \begin{example}
 Let $I=\langle \mu_1,\mu_2,\mu_3,m_1,m_2,m_3\rangle=\langle x^{29}, y^{29}, z^{29}, x^{28}y^8z^8, x^8y^{28}z^8, x^8y^8z^{28}\rangle\subset\mathbb C[x,y,z]$. Since $I$ satisfies the sufficient condition, it is a good ideal. Computations in Singular show that $$I^2:I=I+\langle x^{27}y^{27}z^{27}\rangle,$$
 $$I^3:I^2=I^4:I^3=I+\langle x^{26}y^{27}z^{27}, x^{27}y^{26}z^{27}, x^{27}y^{27}z^{26}\rangle,$$
 $$I^5:I^4=I^6:I^5=\cdots=I^{10}:I^9=I+\langle x^{26}y^{26}z^{26}\rangle.$$
 It is natural to conjecture that $\tilde{I}=I+\langle x^{26}y^{26}z^{26}\rangle.$
 Now let us see what we get if we apply the algorithm above. We start with $E_{-1}=\emptyset$, $F_0=P(I)=\{m_1,m_2,m_3\}$. Then we obtain $E_0$ by reducing $E_{-1}\cup F_0$, that is, $E_0=P(I)$. In order to compute $F_1$, we take all products of $F_0=P(I)$ with $P(I)$, keeping in mind that $y-$ and $z-$ coordinates of each product need to be less than 29, and divide each such product by $\mu_1$. The only such monomial is $\frac{m_1^2}{\mu_1}=\frac{x^{56}y^{16}z^{16}}{x^{29}}=x^{27}y^{16}z^{16}$. This monomial is not in $\langle E_0\rangle$, therefore, we add it to our set $F_1$ (and this set is already reduced). Thus $E_{0}=P(I)$, $F_1=\{x^{27}y^{16}z^{16}\}$. Now $E_1=E_0\cup F_1=P(I)\cup \{x^{27}y^{16}z^{16}\}$ (this union is already reduced), and in order to compute $F_2$ we need to multiply $x^{27}y^{16}z^{16}$ with monomials from $P(I)$ (keeping in mind the condition on $y-$ and $z-$ coordinates) and divide the products by $\mu_1$. The only possible monomial is $\frac{x^{27}y^{16}z^{16}\cdot m_1}{\mu_1}=x^{26}y^{24}z^{24}$. This monomial is not in $\langle E_1\rangle$, therefore, $F_2=\{x^{26}y^{24}z^{24}\}$. $E_2=E_1\cup F_2=P(I)\cup \{x^{27}y^{16}z^{16},x^{26}y^{24}z^{24}\}$ (this set is already reduced) and if we try to compute $F_3$, we see that we can not get any new monomials. Therefore, $F_3=\emptyset$ and the stabilizing point is $I_{2,0,0}=\langle E_2\cup\{\mu_1,\ldots,\mu_n\}\rangle=I+\langle x^{27}y^{16}z^{16},x^{26}y^{24}z^{24}\rangle$.
By symmetry, $I_{0,2,0}=I+\langle x^{16}y^{27}z^{16}, x^{24}y^{26}z^{24}\rangle$ and $I_{0,0,2}=I+\langle x^{16}y^{16}z^{27}, x^{24}y^{24}z^{26}\rangle$. According to the theorem, $\tilde{I}=I_{2,0,0}\cap I_{0,2,0}\cap I_{0,0,2}=I+\langle x^{26}y^{26}z^{26}\rangle$, just as expected.
 \end{example}
 %\begin{example}
 %Let $I=\langle x^{53},y^{56},z^{59},w^{61},x^{50}y^{18}z^{20}w^{25},x^{15}y^{54}z^{22}w^{24},x^{18}y^{20}z^{56}w^{22}, x^{16}y^{19}z^{23}w^{60}\rangle\subset\mathbb C[x,y,z,w]$. Since $I$ satisfies the sufficient condition, it is a good ideal.
 %Then we will get
 %$$I_{1,0,0,0}=I+x^{47}y^{36}z^{40}w^{50}=I_{2,0,0,0}=I_{3,0,0,0}=I_{4,0,0,0}=\cdots$$
 %$$I_{0,1,0,0}=I+x^{30}y^{52}z^{44}w^{48}=I_{0,2,0,0}=I_{0,3,0,0}=I_{0,4,0,0}=\cdots$$
 %$$I_{0,0,1,0}=I+x^{36}y^{40}z^{53}w^{44}=I_{0,0,2,0}=I_{0,0,3,0}=I_{0,0,4,0}=\cdots$$
 %$$I_{0,0,0,1}=I+x^{32}y^{38}z^{46}w^{59}=I_{0,0,0,2}=I_{0,0,0,3}=I_{0,0,0,4}=\cdots$$
%Then $\tilde{I}=I_{1,0,0,0}\cap I_{0,1,0,0}\cap I_{0,0,1,0}\cap I_{0,0,0,1}=I+\langle x^{47}y^{52}z^{53}w^{59}\rangle$ which coincides with computations in Singular.
 %\end{example}
 %\begin{example}
 %Let $I=\langle x^{10}, x^9y,xy^9,y^{10}\rangle\subset\mathbb C[x,y]$. Since $I$ satisfies the sufficient condition, it is a good ideal. It is obvious that both $I_{q_1,0}$ and $I_{0,q_2}$ as well as their intersection are generated by all possible monomials of degree 10. Computations in Singular show that $I^7:I^6\subsetneq I^8:I^7=\tilde I$, that is, in order to see some stabilization happening, one needs to compute successive quotients up to $I^8:I^7$. In general, if we consider $I=\langle x^d, xy^{d-1}, x^{d-1}y, y^d\rangle$, then $\tilde I$ is generated by all possible monomials of degree $d$ and the reduction number of $I$ grows as $d$ grows. This example shows that if one wants to compute quotients explicitly with the help of computer algebra, it can take much time, while computing $\tilde I$ with the help of the formula is much faster.
 %\end{example}
 \begin{example}
 Let $I=\langle x^{41},y^{41},z^{41},x^{40}y^5z^5,x^5y^{40}z^5,x^5y^5z^{40}\rangle\subset\mathbb C[x,y,z]$. It can be shown that $I$ is a good ideal. All the new monomials can only be obtained from powers of non-corners:
 $I_{1,0,0}=I+x^{39}y^{10}z^{10},I_{2,0,0}=I_{1,0,0}+x^{38}y^{15}z^{15}, \ldots,I_{6,0,0}=I_{5,0,0}+x^{34}y^{35}z^{35}$.
 Here the line stabilizes. We similarly get $I_{0,6,0}$ and $I_{0,0,6}$. Intersecting them we will get $$\tilde I=I_{6,0,0}\cap I_{0,6,0}\cap I_{0,0,6}=I+\langle x^{34}y^{35}z^{35},x^{35}y^{34}z^{35},x^{35}y^{35}z^{34}\rangle.$$ 
Computing successive quotients via computer algebra gives is the following:
$I^2:I^1$ has 7 minimal generators, that is, $|G(I^2:I^1)|=7$; $|G(I^3:I^2)|=9$; $|G(I^4:I^3)|=12$; $|G(I^5:I^4)|=16$; $|G(I^6:I^5)|=21$; $|G(I^7:I^6)|=27$; $|G(I^8:I^7)|=31$; $|G(I^9:I^8)|=33$; $|G(I^{10}:I^9)|=33$; $|G(I^{11}:I^{10})|=31$; $|G(I^{12}:I^{11})|=24$; $|G(I^{13}:I^{12})|=18$; $|G(I^{14}:I^{13})|=13$; $|G(I^{15}:I^{14})|=9$ and it finally coincides with the ideal obtained above. It takes much time to perform these computations using computer algebra, whereas the computation of $I_{6,0,0}$, $I_{0,6,0}$ and $I_{0,0,6}$ and their intersection is much easier. 
 \end{example}

 



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%\lipsum[17]

%Here is a citation~\cite{greenwade93} with URL. Here is another
%citation~\cite{MR4}, which is the earliest-numbered Math Review with a
%functioning DOI: MR0000004. Finally, a reference to
%equation~\eqref{eqn:eq1}. And a reference to Figure~\ref{fig:plot}.

%\acknowledgements{\lipsum[20]}

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