\section{Introduction}
This paper is motivated by wanting to better understand concepts, conjectures and methods introduced in
\cite{BH1}, \cite{GHVT}, \cite{G}, \cite{DFMS} and \cite{DD}
involving various approaches to the containment problem.
It makes particular use of the groundbreaking methods of
\cite{DFMS} and \cite{DD}.

Let $ \ensuremath{\mathbb{K}}$ be a field and let $N\geq1$. Then $R= \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ denotes the polynomial ring $ \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]= \ensuremath{\mathbb{K}}[x_0,\ldots,x_N]$.
Now let $(0)\neq I\subsetneq  \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ be a homogeneous ideal;
thus $I=\oplus_{t\geq0} I_t$, where $I_t$ is the $ \ensuremath{\mathbb{K}}$ vector space span of all homogeneous polynomials of degree $t$ in $I$.
The symbolic power $I^{(m)}$ is defined as
$$I^{(m)}=R\cap (\cap_{P\in{\rm Ass}(R/I)}I^mR_P)$$
where the intersections take place in $ \ensuremath{\mathbb{K}}( \ensuremath{\mathbb{P}}^N)$.

While $I^r\subseteq I^{(m)}$ holds if and only if $m\leq r$, the containment problem of
determining for which $m$ and $r$
the containment $I^{(m)}\subseteq I^r$ holds is much more subtle. 
If $h_I$ is the minimum of $N$ and the bigheight of $I$ (i.e., the maximum of the heights of associated 
primes of $I$), it is known that
\begin{equation}\label{ELSHH}
I^{(rh_I)}\subseteq I^r
\end{equation}
\cite{ELS, HoHu}, so given $r$, the issue is for which $m$ with
$m<rh_I$ do we have $I^{(m)}\subseteq I^r$.
The {\it resurgence} $\rho(I)$, introduced in \cite{BH1},
gives some notion of how small the ratio $m/r$ can be and still be sure to have
$I^{(m)}\subseteq I^r$; specifically,
$$\rho(I)=\sup\Big\{\frac{m}{r} : I^{(m)}\not\subseteq I^r\Big\}.$$

A case of particular interest is that of ideals of {\em fat points}.
Given distinct points $p_1,\ldots,p_s\in \ensuremath{\mathbb{P}}^N$ and nonnegative integers $m_i$
(not all 0), let $Z=m_1p_1+\cdots+m_sp_s$ 
denote the scheme (called a fat point scheme) defined by the ideal
$$I(Z)=\cap_{i=1}^s(I(p_i)^{m_i})\subseteq  \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N],$$
where $I(p_i)$ is the ideal generated by all homogeneous polynomials
vanishing at $p_i$. Note that $I(Z)$ is always nontrivial (i.e., not $(0)$ nor $(1)$).
Symbolic powers of $I(Z)$ take the form
$I(Z)^{(m)}=I(mZ)=\cap_{i=1}^s(I(p_i)^{mm_i})$.
We say $Z$ is {\it reduced} if $m_i$ is either 0 or 1 for each $i$
(i.e., if $I(Z)$ is a radical ideal).

Subsequent to \cite{BH1}, two asymptotic notions of the resurgence were introduced by \cite{GHVT}.
The first is 
$$\rho'(I)=\limsup_t\rho(I,t)=\lim_{t\to\infty}\rho(I,t),$$
where $\rho(I,t)=\sup\Big\{\frac{m}{r} : I^{(m)}\not\subseteq I^r, m\geq t, r\geq t\Big\}$.
The second is 
$$\widehat{\rho}(I)=\sup\Big\{\frac{m}{r} : I^{(mt)}\not\subseteq I^{rt}, t\gg0\Big\}.$$
A useful new perspective on $\widehat{\rho}(I)$ is given by \cite[Corollary 4.14]{DFMS}, which shows that
$$\widehat{\rho}(I)=\sup\Big\{\frac{m}{r} : I^{(m)}\not\subseteq \overline{I^{r}}\Big\},$$
where $\overline{I^{r}}$ is the integral closure of ${I^{r}}$ (defined below).
Since we always have $I^r\subseteq \overline{I^r}$, this new perspective makes it
clear that we always have $\widehat{\rho}(I)\leq\rho(I)$, and that we have $\widehat{\rho}(I)=\rho(I)$
if $I^r= \overline{I^r}$ for all $r\geq1$.

Our major results are to show that $\widehat{\rho}(I) = \rho'(I)$ (Theorem \ref{MT1}),
that $\widehat{\rho}(I)=h_I$ if and only if $\rho(I)=h_I$ (Theorem \ref{MT2}), and that
$\widehat{\rho}(I(Z))=1$ if and only if $\rho(I(Z))=1$ when $Z\subset \ensuremath{\mathbb{P}}^N$ is a reduced scheme of points (Theorem \ref{MT3})
and for every fat point subscheme $Z\subset \ensuremath{\mathbb{P}}^2$ (Corollary \ref{CorIntRho1}).
We also introduce a new version of the resurgence, $\rho_{int}$, based on integral closure,
and relate it to the original resurgence. We then discuss
the relevance of our results to a conjecture of Grifo.
Finally we provide some examples and raise some questions,
and include a discussion of the computability of the resurgence.

\subsection{Background}\label{bkgrnd}
Let $I\subseteq  \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ be a nontrivial homogeneous ideal. 
Given the comments and definitions above we have (see \cite{GHVT}) 
\begin{equation}\label{ELSHHbounds}
1\leq\widehat{\rho}(I)\leq \rho'(I)\leq \rho(I)\leq h_I.
\end{equation}
By \cite{BH1} and \cite{GHVT} we also have
\begin{equation}\label{BHbounds}
\frac{\alpha(I)}{\widehat{\alpha}(I)}\leq\widehat{\rho}(I)\leq \rho'(I)\leq \rho(I),
\end{equation}
where $\alpha(I)$ is the least degree of a nonzero element of $I$ and
$\widehat{\alpha}(I)$ is the {\it Waldschmidt constant}, defined as
$$\widehat{\alpha}(I)=\lim_{m\to\infty}\frac{\alpha(I^{(m)})}{m}.$$
For a nontrivial fat point subscheme $Z\subseteq \ensuremath{\mathbb{P}}^N$, by \cite{BH1} we have in addition
\begin{equation}\label{BHbounds2}
\rho(I(Z))\leq \frac{{\rm reg}(I(Z))}{\widehat{\alpha}(I(Z))},
\end{equation}
where ${\rm reg}(I(Z))$ is the Castelnuovo-Mumford regularity of $I(Z)$.

A version of resurgence can be defined with integral closure replacing symbolic powers.
We pause to briefly discuss the concept of integral closure.
Given an ideal $I\subseteq R= \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$, we recall (see \cite{HuSw}) that the {\it integral closure} $\overline{I}$ of $I$
consists of all elements $c\in R$ satisfying for some $n\geq1$ a polynomial equation
$$c^n+a_1c^{n-1}+\cdots+a_n=0$$
where $a_j\in I^j$. We say $I$ is {\it integrally closed} if $I=\overline{I}$. 
We note that $\overline{I}$ is monomial (resp. homogeneous) if $I$ is \cite[Proposition 1.4.2]{HuSw} 
(resp. \cite[Corollary 5.2.3]{HuSw}).
If $I^r$ is integrally closed for all $r\geq1$, we say $I$ is {\it normal}.

For example, the ideal $I(p_i)$ of a point $p_i\in \ensuremath{\mathbb{P}}^N$ is normal. Likewise,
$M=(x_0,\ldots,x_N)$ is normal. This is because $M$ is a monomial prime ideal
and $I(p_i)$ is also, up to choice of coordinates, but monomial primes are normal.
(Apply the usual criterion for integral closure for monomial ideals, that 
the integral closure of a monomial ideal $I$ is the monomial ideal
associated to the convex hull of the Newton polygon of $I$ \cite{HuSw}.)
As further examples of integrally closed ideals, 
we note $I(Z)$ is integrally closed for all $Z$, as is $M^t\cap I(Z)$
for every $t$; this is because intersections of integrally closed ideals are integrally closed.
Now assume $\alpha(I(Z))={\rm reg}(I(Z))$; then
$(I(Z)^r)_t=(I(rZ))_t$ for $t\geq \alpha(I(Z)^r)$ (apply \cite[Lemma 2.3.3(c)]{BH1}
using the fact that $\alpha(I(Z)^r)=r\alpha(I(Z))$).
Thus we have $I(Z)^r=M^{r\alpha(I)}\cap I(rZ)$ and hence
$I(Z)$ is normal, if $\alpha(I(Z))={\rm reg}(I(Z))$. Thus, when $\alpha(I(Z))={\rm reg}(I(Z))$,
we have $\widehat{\rho}(I(Z)) =\rho(I(Z))$ by the normality, but in fact
\eqref{BHbounds} and \eqref{BHbounds2} give us more, namely 
$$\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}=\widehat{\rho}(I(Z)) =\rho(I(Z)).$$

Now we define the integral closure resurgence.
Given any nontrivial homogeneous ideal $I\subset \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ but
replacing symbolic power by integral closure in the definition of resurgence
gives us the {\it integral closure resurgence},
$$\rho_{int}(I)=\sup\Big\{\frac{m}{r} : \overline{I^m}\not\subseteq I^r\Big\}.$$
If symbolic powers of $I$ are integrally closed
(as is the case when $I$ is the ideal of a fat point subscheme), we have
$\rho_{int}(I)\leq\rho(I)$, and if moreover 
$I^{(m)}=\overline{I^m}$ for all $m\geq1$, it follows that $\rho_{int}(I)=\rho(I)$
(see Theorem \ref{BSkoda}).
A lower bound such as $\rho_{int}(I)\leq\rho(I)$ is of interest since for any $c>1$
(as the proof of Theorem \ref{BSkoda} shows) it is in principle a finite
calculation (although not necessarily an easy one)
to verify whether or not $\rho_{int}(I)\geq c$, and if so
to compute $\rho_{int}(I)$ exactly.
In a further analogy of $\rho_{int}$ with $\rho$, by Lemma \ref{rhoIntLem}
we have $1\leq\rho_{int}(I)\leq N$, and, if $I=\overline{I}$, we have
$\overline{I^m}\subseteq I^r$ for all $m\geq Nr$.

\section{Main results}
We begin with the result that started this paper.

\begin{proposition}
Let $Z$ be a fat point subscheme of $ \ensuremath{\mathbb{P}}^N$ with an integer $c$ such that $I(cZ)^t=I(ctZ)$ for all $t\geq 1$.
Then $\widehat{\rho}(I(Z)) =\rho'(I(Z))$.
\end{proposition}

\begin{proof}
Let $b$ be a rational such that $\widehat{\rho}(I(Z))<c/b$. Pick any integer $d>0$ such that $db$ is an integer.
Since $\widehat{\rho}(I(Z))<c/b=cd/(db)$, we have by definition of $\widehat{\rho}(I(Z))$ that $I(cdtZ)\subseteq I(Z)^{dbt}$ for $t\gg0$,
Note that $I(cdZ)^t=I(cZ)^{dt}=I(cdtZ)$ for all $t\geq1$. 
Hence by \cite[Theorem 1.2(3)]{GHVT} we have
$\rho'(I(Z))\leq cdt/(dbt)=c/b$. Since this holds for all $b$ with $c/b>\widehat{\rho}(I(Z))$, we have
$\widehat{\rho}(I(Z))\geq\rho'(I(Z))$. But by \cite[Theorem 1.2(1)]{GHVT} we have
$\widehat{\rho}(I(Z))\leq\rho'(I(Z))$, hence $\widehat{\rho}(I(Z))=\rho'(I(Z))$.
\end{proof}

A much stronger result can be proved based on an argument in \cite{DD}.

\begin{theorem}\label{MT1}
Let $I$ be a nontrivial homogeneous ideal of $ \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$.
Then $\widehat{\rho}(I) =\rho'(I)$.
\end{theorem}

\begin{proof}
If $\widehat{\rho}(I) =\rho(I)$, then \eqref{ELSHHbounds} gives $\widehat{\rho}(I) =\rho'(I)$,
so assume $\widehat{\rho}(I) < \rho(I)$.
Thus there is an $\epsilon>0$
such that $\widehat{\rho}(I) + \epsilon < \rho(I)$.
As in the proof of \cite[Proposition 2.6]{DD}
(which in turn is a consequence of \cite[Lemma 4.12]{DFMS}),
we have 
$$\frac{s}{r+N} < \widehat{\rho}(I)$$
whenever $I^{(s)}\not\subseteq I^r$.

This means there are only finitely many $s$ and $r$ for which 
$\widehat{\rho}(I)+\epsilon\leq s/r$ holds but
$I^{(s)}\subseteq I^r$ fails.
(This is because $\widehat{\rho}(I) + \epsilon \leq s/r$ and $\frac{s}{r+N} < \widehat{\rho}(I)$
implies $r(\widehat{\rho}(I)+\epsilon)\leq s< \widehat{\rho}(I)(r+N)$ and $r < N\widehat{\rho}(I)/\epsilon$.)

Hence for all $s$ and $r$ sufficiently
large which have $I^{(s)}$ not contained in $I^r$, we will have 
$s/r < \widehat{\rho}(I)+\epsilon$ and hence 
$\rho'(I) \leq \widehat{\rho}(I)+\epsilon$. This is true for every $\epsilon > 0$, 
and so we get $\rho'(I) \leq \widehat{\rho}(I)$.
Since we already have $\rho'(I) \geq \widehat{\rho}(I)$, we conclude 
that $\rho'(I) = \widehat{\rho}(I)$.
\end{proof}

An alternate statement of the next result is that $\widehat{\rho}(I)=h_I$ if and only if $\rho(I)=h_I$.
We have learned that Theorem \ref{MT2} was also obtained independently by DiPasquale and Drabkin, 
but not included in \cite{DD}.

\begin{theorem}\label{MT2}
Let $I$ be a nontrivial homogeneous ideal of $ \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$.
Then $\widehat{\rho}(I)<h_I$ if and only if $\rho(I)<h_I$.
\end{theorem}

\begin{proof}
If $\rho(I) < h_I$, then we have $\widehat{\rho}(I) \leq \rho(I) < h_I$ by \eqref{ELSHHbounds}.
So assume $\widehat{\rho}(I) < h_I$. If $\widehat{\rho}(I)=\rho(I)$, then $\rho(I) < h_I$.
If $\widehat{\rho}(I) < \rho(I)$, then by \cite[Proposition 2.6]{DD}, $\rho(I)$ is the maximum
of finitely many ratios $s/r$ with $I^{(s)}\not\subseteq I^r$.
Thus $\rho(I) = h_I$ would imply that $s/r = h_I$, which contradicts
the result of \cite{ELS, HoHu} that $I^{(m)}\subseteq I^r$
whenever $m\geq rh_I$.
\end{proof}

The other extreme is also of interest. Let $Z\subset \ensuremath{\mathbb{P}}^N$ be a fat point subscheme. 
The next result, Theorem \ref{MT3}, shows that the question of when $\rho(I(Z))=1$
is related to two concepts: to analytic spreads (see \cite{HuSw}) and to
symbolic defects (see \cite{GGSV}). 

We can define the {\it analytic spread} $\ell(I)$ of a homogeneous ideal $I\subseteq R= \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ as being the minimum
number of elements of $I$ such that, after localizing at the irrelevant ideal, the ideal $J$ they generate
has $\overline{J}=\overline{I}$ \cite[Corollary 1.2.5, Proposition 8.3.7]{HuSw}. 
The analytic spread $\ell(I)$ of an ideal $I\subseteq R$ is at least the height of $I$, since
$J$ and $I$ have the same minimal primes, and the minimal number of generators of an ideal 
is at least the height of its minimal prime of minimal height.
By \cite[Proposition 5.1.6]{HuSw}, it is also at most the dimension of $R$ (which is $N+1$).
For $I(Z)\subset  \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ we thus have $N\leq\ell(I(Z))\leq N+1$.

We say an ideal $B\subseteq \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ is a {\it complete intersection}
if $B$ is generated by a regular sequence; equivalently, $B$ is a complete intersection
if all associated primes of $B$ have height $c$, where $c$ is the minimal number of
generators of $B$. If for a fat point subscheme $Z\subset \ensuremath{\mathbb{P}}^N$ its ideal $I(Z)$ is a 
complete intersection (i.e., $I(Z)$ has $N$ generators), 
then $\ell(I(Z))=N$, but $\ell(I(Z))=N$ can occur even when $I(Z)$ is not a complete intersection
(for example, $\ell(I(mZ))=N$ when $I(Z)$ is a complete intersection but $m>1$).

The {\it symbolic defect} ${\rm sdefect}(I(Z),m)$ is the minimum number of generators of the module
$I(mZ)/I(Z)^m$ \cite{GGSV}. Thus ${\rm sdefect}(I(Z),m)=0$ if and only if $I(mZ)=I(Z)^m$.
If $I(Z)$ is a complete intersection, then ${\rm sdefect}(I(Z),m)=0$ for all $m\geq1$
since we have $I(mZ)=I(Z)^m$ (indeed, if $I$ is generated by a regular sequence, then
$I^r=I^{(r)}$ for all $r\geq 0$ by \cite[Lemma 5, Appendix 6]{ZS}). 
When $N=2$ and $Z$ is reduced (i.e., $I(Z)$ is radical), \cite[Theorem 2.6]{GGSV}
gives a converse:
if ${\rm sdefect}(I(Z),m)=0$ for all $m\geq1$, then $I(Z)$ is a complete intersection
(see also \cite[Remark 2.5]{CFGLMNSSV} and \cite[Theorem 2.8]{HU}). 

Our next result gives a number of equivalent conditions for ${\rm sdefect}(I(Z),m)=0$ for all $m\geq1$ for any 
reduced fat point subscheme $Z\subset  \ensuremath{\mathbb{P}}^N$, thereby extending from $N=2$ to all $N$ 
the result that $I(mZ)=I(Z)^m$ for all $m\geq1$ if and only if $I(Z)$ is a complete intersection.
We thank Seceleanu and Huneke for the implications (d) $\Rightarrow$ (e), and, for $Z$ reduced, (e) $\Rightarrow$ (a).
We do not know if (e) $\Rightarrow$ (a) holds for nonreduced fat point schemes $Z$ (i.e., when $I(Z)$ is not radical),
but see Corollaries \ref{CorIntRho1} and \ref{CorIntRho2}, and also Examples \ref{Ex1} and \ref{Ex1b} and Question \ref{Q4b}.

\begin{theorem}\label{MT3}
Let $Z$ be a nontrivial fat point subscheme of $ \ensuremath{\mathbb{P}}^N$. 
Then each of the following criteria implies the next.
\begin{enumerate}
\item[(a)] $I(Z)^m=I(mZ)$ (i.e., ${\rm sdefect}(I(Z),m)=0$) for all $m\geq1$.
\item[(b)] $\rho(I(Z))=1$.
\item[(c)] $\widehat{\rho}(I(Z))=1$.
\vskip.3\baselineskip
\item[(d)] $\overline{I(Z)^m}=I(mZ)$ for all $m\geq1$.
\item[(e)] The analytic spread of $I(Z)$ is $N$.
\end{enumerate}
Moreover, if $Z$ is reduced or $N=1$, then (e) implies (a).
In fact, if $Z$ is reduced or $N=1$, then each of the conditions (a)-(e) 
is equivalent to $I(Z)$ being a complete intersection.
\end{theorem}

\begin{proof}
That (a) implies (b) is clear (since $I(Z)^m\subseteq I(Z)^r$ if and only if $m\geq r$), 
and (b) implies (c) since $1\leq\widehat{\rho}(I(Z))\leq\rho(I(Z))$.
Next (c) is equivalent to (d) by \cite[Corollary 4.16]{DFMS}.
Now we show (d) implies (e). By
\cite[Proposition 5.4.7]{HuSw}, we have that (d) implies $\ell(I(Z))\neq N+1$, hence
$\ell(I(Z)) = N$. 

If $N=1$, then $I(Z)$ is principal, hence (a) always holds and $I(Z)$ is a complete intersection.
Finally assume that $Z$ is reduced. Thus the primary components of $I(Z)$
are ideals of points of multiplicity 1, hence are complete intersections
(i.e., $I(Z)$ is locally a complete intersection). By \cite{CN}, $\ell(I(Z)) = N$
implies that the localization $I(Z)_M$ of $I(Z)$ at the irrelevant ideal $M\subset R= \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ 
is a complete intersection, and hence $(I(Z)^m)_M=(I(Z)_M)^m$ is saturated for all $m\geq1$, so 
$I(Z)^m$ itself is saturated so (a) holds.
Moreover, since (e) implies that $I(Z)_M$ is a complete intersection when $Z$ is reduced, 
the number of generators of
$I(Z)_M$ is $N=\dim I(Z)_M/I(Z)_MM_M=\dim I(Z)/I(Z)M$, hence $I(Z)$ 
also has $N$ generators, so is itself is a complete intersection and hence (a) holds.
\end{proof}

We can now give a characterization of those $Z$ for which $I(mZ)=I(Z)^m$ for all $m\geq1$.
This characterization is not so interesting in itself, but it does raise the question
of whether the normality hypothesis can be dropped; see Question \ref{Q4b}.
Although $I(Z)$ is not always normal (see Example \ref{Ex2}),
we do not know any examples with $\widehat{\rho}(I(Z))=1$ where $I(Z)$ is not normal. 

\begin{corollary}\label{MT3Cor}
Let $Z\subset \ensuremath{\mathbb{P}}^N$ be a fat point subscheme. 
Then $I(mZ)=I(Z)^m$ for all $m\geq1$ if and only if $\widehat{\rho}(I(Z))=1$ and $I(Z)$ is normal.
\end{corollary}

\begin{proof}
Having $I(mZ)=I(Z)^m$ for all $m\geq1$ implies $\widehat{\rho}(I(Z))=1$ by Theorem \ref{MT3},
and it implies $I(Z)$ is normal since $I(Z)^m\subseteq \overline{I(Z)^m}\subseteq I(mZ)$. Conversely,
$\widehat{\rho}(I(Z))=1$ implies $I(mZ)=\overline{I(Z)^m}$ for all $m\geq1$ 
by Theorem \ref{MT3} and normality implies $\overline{I(Z)^m}=I(Z)^m$ for all $m\geq1$,
hence $I(mZ)=I(Z)^m$ holds for all $m\geq1$.
\end{proof}

We now consider $\rho_{int}(I)$. Note for a nontrivial homogeneous ideal $I\subset \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$
that an ideal independent version of \eqref{ELSHH}
can be stated as $I^{(Nr)}\subseteq I^r$, and the corresponding bounds \eqref{ELSHHbounds} 
on $\rho(I)$ are $1\leq \rho(I)\leq N$. In analogy with this, we have the following lemma.
(See Example \ref{Ex4} for an example showing that the assumption $I=\overline{I}$ is needed both
for the second part of Lemma \ref{rhoIntLem} and for Theorem \ref{BSkoda}(b).)

\begin{lemma}\label{rhoIntLem}
Let $I\subset \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ be a nontrivial homogeneous ideal.
Then $1\leq \rho_{int}(I)\leq N$. Moreover,
if $N>1$ and $t>1$, or if $I=\overline{I}$  and $t\geq1$, then $\overline{I^{Nt}}\subseteq I^t$.
\end{lemma}

\begin{proof}
Since $I^t\subseteq \overline{I^{t}}$ but $I^t\not\subseteq I^{t+1}$,
we see that $\overline{I^t}\not\subseteq I^{t+1}$, so $1\leq\rho_{int}(I)$.

The Brian{\c c}on-Skoda Theorem \cite{HuSw} asserts
$\overline{I^{t+N}}\subseteq I^t$ for each $t\geq1$.
For $t,N\geq2$ we have $tN\geq t+N$, hence
$I^{Nt}\subseteq I^{t+N}$ and thus
$\overline{I^{Nt}}\subseteq\overline{I^{t+N}}\subseteq I^t$.
This also tells us that if $\overline{I^m}\not\subseteq I^t$, we must have either
$m<Nt$ (and so $m/t<N$) or $t$ or $N$ must be equal to 1.
By Brian{\c c}on-Skoda, $\overline{I^m}\not\subseteq I^t$ implies $m<t+N$, so if $t=1$, then $m/t = m\leq N$,
while if $N=1$, then we have $m<t+1$, so $m/t\leq t/t=1=N$.
Thus in all cases we have $m/t\leq N$, hence $\rho_{int}(I)\leq N$.

We already saw that $\overline{I^{Nt}}\subseteq I^t$ if $N,t>1$. Assume $N>1$ but $t=1$.
Then $I^N\subseteq I$ so if $I=\overline{I}$ we have $\overline{I^{N}}\subseteq I$.
Finally assume $N=1$ and $I\subset  \ensuremath{\mathbb{K}}[x_0,x_1]$ is a nontrivial homogeneous ideal.
Thus $I=(G_1,\ldots,G_s)$ for some nonzero homogeneous generators $G_i$.
Let $F$ be the greatest common divisor of the $G_i$, and for each $i$ let $H_iF=G_i$.
Then $I=(F)Q$ where $Q=(H_1,\ldots,H_s)$. If $\deg H_i=0$ for some $i$, we have $Q=(1)$ and so $I=(F)$.
If $\deg H_i>0$ for all $i$, then $Q$ is primary for $(x_0,x_1)$.

If $I=(F)$, then $I$ is normal, so we have $\overline{I^t}=I^t$ for all $t\geq1$. 
So say $I=(F)Q$ where $Q$ is primary for $(x_0,x_1)$, hence $I^t=(F^t)Q^t$ for each $t\geq1$.

Note that $I^t=\overline{I^t}$ if and only if $Q^t=\overline{Q^t}$.
(Here's why. Assume $I^t=\overline{I^t}$.
By \cite[Remark 1.3.2(2)]{HuSw}, we have $\overline{I^t:J}\subseteq \overline{I^t}:\overline{J}$
for any ideal $J$. Take $J=(F^r)$. Then $Q^t\subseteq \overline{Q^t}=
\overline{I^t:J}\subseteq \overline{I^t}:\overline{J}=I^t:(F^r)=Q^t$.
Alternatively, say $x\in \overline{Q^t}$. 
Then $x^n+a_1x^{n-1}+\cdots+a_n=0$ for some $n$ and some $a_i\in Q^{ti}$.
Multiplying by $F^{tn}$ gives $(F^tx)^n+F^ta_1(F^tx)^{n-1}+\cdots+(F^t)^na_n=0$ so
$F^tx\in \overline{(F^t)Q^t}=\overline{I^t}=I^t=(F^t)Q^t$, so $x\in Q^t$.
Now assume $Q^t = \overline{Q^t}$. Say $x\in \overline{I^t}$.
Then $x^n+F^ta_1x^{n-1}+\cdots+(F^t)^na_n=0$ for some $n$ with $a_i\in Q^{ti}$.
Say $F^t=AP^m$ where $P$ is an irreducible factor of $F$ and $P$ does not divide $A$. Then $P$ divides $x$.
Let $yP=x$. Dividing out gives
$y^n+AP^{m-1}a_1y^{n-1}+\cdots+A^nP^{n(m-1)}a_n=0$.
We can keep dividing out until we have
$z^n+Aa_1z^{n-1}+\cdots+A^na_n=0$, where $zP^m=x$ and $z\in\overline{(A)Q^t}$.
Continuing in this way, dividing out irreducible factors of $F^t$,
we eventually see that we get an element $w$ with $wF^t=x$,
where $w\in\overline{Q^t}=Q^t$, hence $x\in (F^t)Q^t=I^t$.) 

Now we claim that $Q$ is normal if and only if $Q=\overline{Q}$.
Given this, if $I=\overline{I}$, then $Q=\overline{Q}$, hence $Q^t=\overline{Q^t}$
so $I^t=\overline{I^t}$ (i.e., $I$ is normal), which is what we needed to show.
But note that $Q$ normal implies by definition that $Q=\overline{Q}$.
Conversely, assume $Q=\overline{Q}$. It suffices to show
$Q^t=\overline{Q^t}$ for each $t$. 
By \cite[Proposition 4.8]{AM} and \cite[Proposition 1.1.4]{HuSw}, 
it is enough to check this after localizing at $(x_0,x_1)$.
But by results of Zariski, $Q=\overline{Q}$ implies 
$Q^t=\overline{Q^t}$ for each $t$ in the local case
\cite[Theorem 14.4.4]{HuSw}.
\end{proof}

\begin{theorem}\label{BSkoda}
Let $I\subset  \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ be a nontrivial homogeneous ideal, $N\geq1$.
\begin{enumerate}
\item[(a)] We have 
$$1\leq\rho_{int}(I)=\max\Big\{\big\{\frac{m}{r}: \overline{I^m}\not\subseteq I^r\big\}\cup\{1\}\Big\}.$$
\item[(b)] If $I=\overline{I}$ and $N>1$, then $\rho_{int}(I)<N$.
\item[(c)] If $I^{(m)}=\overline{I^{(m)}}$ for all $m\geq1$ (as for example is the case for 
$I=I(Z)$ for a fat point subscheme  $Z\subset \ensuremath{\mathbb{P}}^N$), then
$\rho_{int}(I)\leq\rho(I)$. 
\item[(d)] If $I^{(m)}=\overline{I^m}$ for all $m\geq1$, then
$\rho_{int}(I)=\rho(I)$. 
\end{enumerate}
\end{theorem}

\begin{proof}
(a) By Lemma \ref{rhoIntLem} we have $1\leq \rho_{int}(I)$.
In order for $\rho_{int}(I) > 1$ there must be a pair of positive integers $(r_0,m_0)$
with both $m_0/r_0 > 1$ and $\overline{I^{m_0}}\not\subseteq I^{r_0}$ (and hence
$m_0<r_0+N$ by Brian{\c c}on-Skoda). Set $c=m_0/r_0$.
If $\rho_{int}(I)>c$, then as before we have $(r,m)$
with both $m/r > c$ and $\overline{I^m}\not\subseteq I^r$ (and hence $m<r+N$).
But there are only finitely many pairs $(r,m)$ with $cr<m$ and $m<r+N$
(in particular, we have $r< N/(c-1)$ and $cr<m<r+N$).
Thus either $\rho_{int}(I)=1$ or $\rho_{int}(I)=\max\{\frac{m}{r}: \overline{I^m}\not\subseteq I^r\}$,
hence $\rho_{int}(I)=\max\Big\{\big\{\frac{m}{r}: \overline{I^m}\not\subseteq I^r\big\}\cup\{1\}\Big\}.$

(b) Now assume $I=\overline{I}$ and $N>1$. By Lemma \ref{rhoIntLem} we have $\overline{I^{Nr}}\subseteq I^r$ for $r\geq1$,
so $\overline{I^m}\not\subseteq I^r$ implies $m/r<N$, hence $\rho_{int}(I)$, being either 1 or
a maximum of values $m/r$ less than $N$, is less than $N$.

(c) Here we assume $I^{(m)}=\overline{I^{(m)}}$ for all $m\geq1$. 
Then since $I^m\subseteq I^{(m)}$ we have $\overline{I^m}\subseteq I^{(m)}$,
so $\overline{I^m}\not\subseteq I^r$ implies $I^{(m)}\not\subseteq I^r$,
and hence $\rho_{int}(I)\leq\rho(I)$.

(d) Assuming $I^{m}=\overline{I^{(m)}}$ for all $m\geq1$, we have
$$\rho_{int}(I) =\sup\Big\{\frac{m}{r} : \overline{I^m}\not\subseteq I^r\Big\}=
\sup\Big\{\frac{m}{r} : I^{(m)}\not\subseteq I^r\Big\}=\rho(I).$$
\end{proof}

We do not know any examples with $\rho_{int}(I(Z))>1$. For $Z\subset \ensuremath{\mathbb{P}}^2$, there are none, by the next result,
which is an immediate consequence of \cite[Theorem 3.3]{AH1}.
We thank Huneke for alerting us to this result.

\begin{corollary}\label{CorIntRho3}
Let $Z\subset \ensuremath{\mathbb{P}}^N$ be a nontrivial fat point subscheme and let $I=I(Z)$.
Then we have the following:
\begin{enumerate}
\item[(a)] $\overline{I^{N+m-1}}\subseteq I^m$ for $m\geq1$;
\item[(b)] $\rho_{int}(I(Z))\leq N/2$ for $N\geq2$; and
\item[(c)] $\rho_{int}(I(Z))=1$ for $N=1,2$.
\end{enumerate}
\end{corollary}

\begin{proof}
(a) Let $M=(x_0,\ldots,x_N)\subset  \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$.
Using $I$ as a reduction for $I$ and $\ell$ for the analytic spread of $I$,
\cite[Theorem 3.3]{AH1} states (after localizing at the irrelevant ideal $M$)
that $\overline{I^{\ell+m}}\subseteq I^{\ell-N+m+1}$ for $m\geq0$.
But $N\leq \ell\leq N+1$, so for $\ell=N$ we have
$\overline{I^{N+m}}\subseteq I^{m+1}$, while for $\ell=N+1$ we have
$\overline{I^{N+m+1}}\subseteq I^{m+2}$, both for $m\geq0$. Either way, we have
$\overline{I^{N+m-1}}\subseteq I^m$ for $m\geq1$, and hence 
$\overline{I^{N+m-1}}\subseteq I^m$ holds without localizing, since all of the ideals
are homogeneous.

(b) Assume $N\geq 2$. For $r=1$ we have $\overline{I^m}\not\subseteq I^r$ for all $m\geq 1$,
so consider $r>1$. Then we have $\overline{I^m}\subseteq I^r$ for all $m\geq N+r-1$,
so the fractions $m/r$ for which we have $\overline{I^m}\not\subseteq I^r$ 
are contained in the set $\{m/r : 1\leq m\leq N+r-2, r\geq2\}$. The supremum occurs for
$m=N+r-2$ and $r=2$, hence the supremum is $N/2$, so $\rho_{int}(I(Z))\leq N/2$.

(c) When $N=1$ we have $\rho_{int}(I(Z))=1$ from Lemma \ref{rhoIntLem}, and
when $N=2$ we have $\rho_{int}(I(Z))=1$ from (b).
\end{proof}

If $\rho_{int}(I(Z))$ is always 1, then the next result would imply that 
$\widehat{\rho}(I(Z))=1$ if and only if $\rho(I(Z))=1$.
In particular, it shows that $\widehat{\rho}(I(Z))=1$ if and only if $\rho(I(Z))=1$ for every fat point subscheme $Z\subset \ensuremath{\mathbb{P}}^2$.

\begin{corollary}\label{CorIntRho1}
Let $Z\subset \ensuremath{\mathbb{P}}^N$ be a nontrivial fat point subscheme.
If $\widehat{\rho}(I(Z))=1$, then $\rho_{int}(I(Z))=\rho(I(Z))$,
hence $\widehat{\rho}(I(Z))=1$ if and only if $\rho(I(Z))=1$ when $N=2$.
\end{corollary}

\begin{proof}
By Theorem \ref{MT3}, $\widehat{\rho}(I(Z))=1$ implies $I(mZ) = \overline{I(Z)^m}$,
so $\rho_{int}(I(Z))=\rho(I(Z))$. Since $\rho_{int}(I(Z))=1$ when $N=2$ by Corollary \ref{CorIntRho3},
and since $\rho(I(Z))=1$ implies $\widehat{\rho}(I(Z))=1$, we have
$\widehat{\rho}(I(Z))=1$ if and only if $\rho(I(Z))=1$ when $N=2$.
\end{proof}

We now recover a version of \cite[Corollary 4.17]{DFMS}.

\begin{corollary}\label{CorIntRho2}
Let $Z\subset \ensuremath{\mathbb{P}}^N$ be a nontrivial fat point subscheme.
Then $\rho(I(Z))=1$ if and only if $\widehat{\rho}(I(Z))=\rho_{int}(I(Z))=1$.
\end{corollary}

\begin{proof}
The proof is immediate from 
$$1\leq\rho_{int}(I(Z))\leq\rho(I(Z)),$$
$$1\leq\widehat{\rho}(I(Z))\leq\rho(I(Z))$$
and Corollary \ref{CorIntRho1}.
\end{proof}

\begin{remark}\label{rhointRem}
For a nontrivial homogeneous ideal $I\subset \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$, it is also of interest to define
$$\rho^{int}(I)=\sup\Big\{\frac{m+1}{r} : \overline{I^m}\not\subseteq I^r\Big\}.$$
This is exactly what \cite{DFMS} denotes as $K(I)$. 
Clearly we have $\rho_{int}(I)\leq \rho^{int}(I)$, and by applying Theorem \ref{BSkoda}
we see we have equality if and only if $\rho^{int}(I)=1$, in which case $I$ is normal. By
\cite[Proposition 4.19]{DFMS} we have $\rho(I)\leq \widehat{\rho}(I)\rho^{int}(I)$.
Thus we have 
$$\rho_{int}(I(Z))\leq \rho(I(Z))\leq \widehat{\rho}(I(Z))\rho^{int}(I(Z))$$
for every fat point subscheme $Z\subset \ensuremath{\mathbb{P}}^N$.
\end{remark}

\section{Grifo's Conjecture}
We now discuss Grifo's containment conjecture \cite[Conjecture 2.1]{G}. 
In our context it says the following
(we note it is true and easy to prove for $N=1$).

\begin{conjecture}\label{GrifoConj}
Let $I\subseteq \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ be a radical homogeneous ideal.
Then $I^{(h_Ir-h_I+1)}\subseteq I^r$ for all $r\gg0$.
\end{conjecture}

Remark 2.7 of \cite{G} shows the conjecture holds for $I$ if $\rho(I)<h_I$
(whether $I$ is radical or not), or if $\rho'(I)<h_I$,
and raises the question of whether it holds when $\widehat{\rho}(I)<h_I$.
This was answered affirmatively by \cite[Proposition 2.11]{GHM} with a direct proof
(we thank Grifo and Huneke for bringing this result to our attention).
Our results also answer this question affirmatively, in two ways.
By Theorem \ref{MT1}, $\widehat{\rho}(I)<h_I$ implies $\rho(I)'<h_I$,
hence Conjecture \ref{GrifoConj} holds for $I$ by the results of \cite{G}.
And by Theorem \ref{MT2}, $\widehat{\rho}(I)<h_I$ implies $\rho(I)<h_I$,
so again Conjecture \ref{GrifoConj} holds for $I$ by the results of \cite{G}.

\begin{remark}\label{GrifoCorRem}
When $I=I(Z)$ for a fat point scheme $Z\subset \ensuremath{\mathbb{P}}^N$ we have $h_I=N$.
No examples of a fat point scheme $Z\subset \ensuremath{\mathbb{P}}^N$ (radical or not) are known
for which it is not true that $I(Z)^{(Nr-N+1)}\subseteq I(Z)^r$ for all $r\gg0$.
By Remark 2.7 of \cite{G}, one approach to proving that $I(Z)^{(Nr-N+1)}\subseteq I(Z)^r$ for all $r\gg0$
holds for all $Z$ is to show $\rho(I(Z))<N$ whenever $N>1$. 
This raises the question of: for which $Z$ is it known that $\rho(I(Z))<N$?

Let $Z=m_1p_1+\cdots+m_sp_s$. If ${\rm gcd}(m_1,\ldots,m_s)>1$,
then $\rho(I(Z))<N$ by \cite[Proposition 2.1(2)]{TX}.
Thus it is the cases with ${\rm gcd}(m_1,\ldots,m_s)=1$ that remain of interest.

Another approach is to apply our Theorem \ref{MT2}: $\rho(I(Z))<N$ holds
if $\widehat{\rho}(I(Z))<N$.
Aiming to show $\widehat{\rho}(I(Z))<N$ has the advantage that the results of \cite{DFMS,DD} suggest
that $\widehat{\rho}(I(Z))$ is more accessible computationally than is $\rho(I(Z))$.
Another advantage is that in most cases where $\widehat{\rho}(I(Z))$ is known we have
$\widehat{\rho}(I(Z))=\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}$
(but see Example \ref{Ex3})
and in all known cases we have $\widehat{\alpha}(I(Z))\geq \frac{\alpha(I(mZ))+N-1}{m+N-1}$ for all $m\geq 1$.
Assuming both we have
$$\widehat{\rho}(I(Z))=\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}\leq \frac{\alpha(I(Z))}{\frac{\alpha(I(Z))+N-1}{N}}
= N\frac{\alpha(I(Z))}{\alpha(I(Z))+N-1}<N$$
and thus we would have $\rho(I(Z))<N$.
\end{remark}

The previous paragraph merits further discussion.
First we recall \cite[Conjecture 2.1]{HaHu}, which if true would refine the containment
$I(rNZ)\subseteq I(Z)^r$ of \cite{ELS, HoHu}:

\begin{conjecture}\label{HaHu}
Let $Z\subset \ensuremath{\mathbb{P}}^N$ be a fat point scheme. Let $M=(x_0,\ldots,x_N)$. Then
$$I(rNZ)\subseteq M^{r(N-1)}I(Z)^r$$ 
holds for all $r>0$.
\end{conjecture}

A further refinement of $I(rNZ)\subseteq I(Z)^r$ is that
$I(r(m+N-1)Z)\subseteq I(mZ)^r$ \cite{ELS,HoHu}. This suggests
a refinement of Conjecture \ref{HaHu} (cf. \cite[Question 4.2.3]{HaHu}), namely
\begin{equation}\label{HaHuRefinement}
I(r(m+N-1)Z)\subseteq M^{r(N-1)}I(mZ)^r.
\end{equation}

If \eqref{HaHuRefinement} were true, then the following conjecture would also be true:

\begin{conjecture}\label{Chud}
Let $Z\subset \ensuremath{\mathbb{P}}^N$ be a fat point scheme. Then
$$\widehat{\alpha}(I(Z))\geq \frac{\alpha(I(mZ))+N-1}{m+N-1}$$ 
for all $m\geq 1$.
\end{conjecture}

The proof that Conjecture \ref{HaHu} implies Conjecture \ref{Chud} when $m=1$ is 
given in \cite{HaHu}. The same argument shows that \eqref{HaHuRefinement}
implies Conjecture \ref{Chud}.

The first version of Conjecture \ref{Chud} was posed by Chudnovsky \cite{Ch} 
over the complex numbers for the case that $m=1$ and $Z$ is reduced.
He also sketched a proof of his conjecture for $N=2$ which works over any algebraically closed field
(see \cite{HaHu} for a proof).
Conjecture \ref{Chud} assuming $Z$ reduced was posed by Demailly \cite{Dem}.

The best result currently known is by Esnault and Viehweg \cite{EV}. It is over the complex numbers
and says that if $Z$ is reduced with $N>1$, then
$$\widehat{\alpha}(I(Z))\geq \frac{\alpha(I(mZ))+1}{m+N-1}$$
holds for all $m\geq 1$.
Thus for reduced $Z$ over the complex numbers, taking $m=1$, we have
$$\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}\leq \frac{\alpha(I(Z))}{\frac{\alpha(I(Z))+1}{N}}
= N\frac{\alpha(I(Z))}{\alpha(I(Z))+1}<N.$$
Hence $\rho(I(Z))<N$ holds over the complex numbers 
whenever $Z$ is reduced and $\widehat{\rho}(I(Z)) = \frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}$.

\section{Examples and questions}\label{ExSect}
For this section assume $Z$ is a fat point subscheme of $ \ensuremath{\mathbb{P}}^N$ 
(so $I(Z)$ is a nontrivial ideal of $ \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$).

\begin{example}\label{Ex1}
If $I(Z)$ is a complete intersection or a power thereof, then 
$I(mZ)=I(Z)^m$ for all $m\geq1$, hence $\rho(I(Z))=1$ by Theorem \ref{MT3}.
Again by Theorem \ref{MT3}, if $Z$ is reduced, then 
$\rho(I(Z))=1$ if and only if $I(Z)$ is a complete intersection.
However, when $Z$ is not reduced, $\rho(I(Z))=1$ does not imply
$I(Z)$ is a complete intersection, or even a power of a complete intersection
ideal (homogeneous or not),
as is shown by Example \ref{verticesEx} and Remark \ref{FatPtsAndCI1}.
(Additionally, let $p_1,p_2,p_3\in \ensuremath{\mathbb{P}}^N$ be noncollinear points and let
$Z=m_1p_1+m_2p_2+m_3p_3$ with $1\leq m_1\leq m_2\leq m_3$.
If either $m_1+m_2\leq m_3$ or if $m_1+m_2+m_3$ is even,
then by \cite[Theorem 2]{BZ} we have
$I(mZ)=I(Z)^m$ for all $m\geq1$ and hence $\rho(I(Z))=1$.
See \cite[Example 5.1]{BH2} for additional examples in $ \ensuremath{\mathbb{P}}^2$ of $Z$
for which all powers of $I(Z)$ are symbolic.
By Remark \ref{FatPtsAndCI2}, in none of these cases
is there a homogeneous complete intersection ideal $J$ 
such that $I(Z)=J^r$ for some $r\geq1$.
See \cite[Proposition 3.5]{HaHu} for a criterion for $Z\subset \ensuremath{\mathbb{P}}^2$ 
such that $I(mZ)=I(Z)^m$ for all $m\geq1$; this gives further examples
for which $I(Z)$ is not a power of a homogeneous complete intersection.)
\end{example}

It is an interesting problem to clarify which fat point subschemes $Z\subset \ensuremath{\mathbb{P}}^N$ 
have $I(mZ)=I(Z)^m$ for all $m\geq1$. 
We do not know any $Z$ for which the analytic spread $\ell(I(Z))=N$ but for which
$I(mZ)=I(Z)^m$ fails for some $m\geq1$.
Nor do we know any $Z$ for which either $\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$ but for which
$I(mZ)=I(Z)^m$ fails for some $m\geq1$.
Thus we have the following question.

\begin{question}\label{Q-NonreducedZCI}
Is it true that all powers of $I(Z)$ 
are symbolic (i.e., $I(mZ)=I(Z)^m$ for all $m\geq1$)
if the analytic spread of $I(Z)$ is $N$, or if
$\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$? 
\end{question}

\begin{example}\label{Ex1b}
It is worth noting here that there is a monomial ideal $I$ with $\rho(I)=\widehat{\rho}(I)=1$ 
where $I^m\subsetneq I^{(m)}$ for every $m>1$ 
(\cite[Remark 3.5]{DD}; we thank DiPasquale and Grifo for bringing this to our attention).
Since the ideal $I$ given in \cite[Remark 3.5]{DD} is a radical monomial ideal and powers of monomial primes are primary,
we have $I^{(m)}=\cap_{P\in{\rm Ass(I)}}P^m$. But monomial primes are normal,
so $I^{(m)}$ is integrally closed. Thus $\overline{I^m}\subseteq I^{(m)}$, and since
$\widehat{\rho}(I)=1$ we have by \cite[Corollary 4.16]{DFMS} that 
$I^{(m)}\subseteq \overline{I^m}$, so $I^{(m)}=\overline{I^m}$ for all $m\geq 1$.
Thus $I^m$ is integrally closed only for $m=1$, but nonetheless $\rho_{int}(I)=\rho(I)=1$.
\end{example}

Answering Question \ref{Q-NonreducedZCI} is closely related to whether
having $\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$ gives a 
complete solution to the containment problem for $I(Z)$.
If $\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$,
then by Theorem \ref{MT3} when $I(Z)$ is radical, we do have a complete solution to the containment 
problem for $I(Z)$: $I(mZ)\not\subseteq I(Z)^r$ for $m<r$ and otherwise we have 
$I(mZ)\subseteq I(Z)^r$.
When $I(Z)$ is not radical, we do not know if having either $\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$
solves the containment problem for $I(Z)$. For example, having $\rho(I(Z))=1$ means
$\widehat{\rho}(I(Z))=1$ and it means $I(mZ)\not\subseteq I(Z)^r$ for $m<r$ and 
$I(mZ)\subseteq I(Z)^r$ for $m>r$, but we do not know for which $m\geq1$ that we have 
$I(mZ)\subseteq I(Z)^m$.

We also do not know any examples with $\widehat{\rho}(I(Z))=1$ but $\rho(I(Z))>1$.
This raises the following question.

\begin{question}\label{Q4b}
Does $\widehat{\rho}(I(Z))=1$ always imply $\rho(I(Z))=1$?
\end{question}

\begin{example}\label{Ex4}
The assumption $I=\overline{I}$ is needed in both Lemma \ref{rhoIntLem} and Theorem \ref{BSkoda}(b).
For example, take $t=1$ and any $N\geq1$.
Let $I=(x_0^{N+1},x_1^{N+1},\ldots,x_N^{N+1})\subset \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$.
Then $x_0^N\cdots x_N^N\in\overline{I^N}=(x_0,\ldots,x_N)^{N(N+1)}$ but 
$x_0^N\cdots x_N^N\not\in I$, so $\overline{I^{tN}}\not\subseteq I^t$ and $\rho_{int}(I)=N$.
\end{example}

Although $I(Z)$ is not always normal
(see Example \ref{Ex2}) and hence $I(Z)^m=\overline{I(Z)^m}$ can fail, 
we do not have an example with $\overline{I(Z)^m}\not\subseteq I(Z)^r$
when $m>r$, so we do not know of any $Z$ for which $\rho_{int}(I(Z))\neq1$. 
If $\rho_{int}(I(Z))=1$ were always true, then 
by Corollary \ref{CorIntRho2} we would have that $\rho(I(Z))=1$
if and only if $\widehat{\rho}(I(Z))=1$, thus answering Question \ref{Q4b}.
This raises the following question. 

\begin{question}\label{FinalQuest2}
Is it ever true that
$\rho_{int}(I(Z)) >1$?
\end{question}

The next example shows that $\rho_{int}(I(Z))=1$ does not force $\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$,
even if $Z$ is reduced. In contrast, 
we know that (b) (and hence (a)) of Theorem \ref{MT3} implies
$\rho_{int}(I(Z))=1$, but we do not know if any of the other criteria of 
Theorem \ref{MT3} imply $\rho_{int}(I(Z))=1$, unless $Z$ is reduced or $N=2$.

\begin{example}\label{Ex5}
Examples occur with $\rho_{int}(I(Z))=1$ but with $\widehat{\rho}(I(Z))>1$.
Let $Z\subset \ensuremath{\mathbb{P}}^N$ be a {\it star configuration}, meaning we have 
$s>N$ general hyperplanes, and $Z$ is the reduced scheme
consisting of $\binom{s}{N}$ points, where each point is the intersection of $N$ of 
the $s$ hyperplanes; see \cite[Definition 3.8]{HaHu}.
Then the ideal $I(Z)$ has $\alpha(I(Z))={\rm reg}(I(Z))=s+N-1$ \cite[Lemma 2.4.2]{BH1}
and $\widehat{\alpha}(I(Z))=s/N$ \cite[Lemma 2.4.1]{BH1}.
As noted in \S\ref{bkgrnd}, $\alpha(I(Z))={\rm reg}(I(Z))$ 
implies that $I(Z)$ is normal and hence $\rho_{int}(I(Z))=1$,
but $\alpha(I(Z))={\rm reg}(I(Z))$ also implies that 
$\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}=\widehat{\rho}(I(Z))=\rho(I(Z))$,
and in the case of a star configuration we have 
$\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}=N(s-N+1)/s>1$ when $N>1$.
However we do not know of any $Z$ with
$\widehat{\rho}(I(Z))<\rho_{int}(I(Z))$.
\end{example}

\begin{question}\label{FinalQuest}
Is it ever true that
$\rho_{int}(I(Z)) > \widehat{\rho}(I(Z))$?
Do any of (c), (d) or (e) of Theorem \ref{MT3} imply $\rho_{int}(I(Z))=1$, when $Z$ is not reduced and $N>2$?
\end{question}

\begin{example}\label{Ex2}
When $\widehat{\rho}(I(Z)) >1$, it is known that $\widehat{\rho}(I(Z)) < \rho(I(Z))$ can occur.
For example, let $I = (x(y^n-z^n),y(z^n-x^n),z(x^n-y^n))\subset  \ensuremath{\mathbb{C}}[x,y,z]= \ensuremath{\mathbb{C}}[ \ensuremath{\mathbb{P}}^2]$,
so $I=I(Z)$ where $Z$ is a certain set of $n^2+3$ points.
Then by \cite[Theorem 2.1]{DHNSST}, we have $\widehat{\rho}(I(Z)) =
\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))} =(n+1)/n < 3/2=\rho(I(Z))$ for $n\geq3$.
Moreover, since $\widehat{\rho}(I(Z)) < \rho(I(Z))$, it follows that
$I(Z)$ cannot be normal. 
\end{example}

\begin{example}\label{Ex3}
Moreover, examples of $Z$ occur with $\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))} < \widehat{\rho}(I(Z))$. 
The results of \cite{DFMS} suggest that this should occur, but up to now
no explicit examples have been given.
For one such explicit example (indeed, the first we are aware of), let 
$Z$ consist of 8 points in the plane, where 3 of the points (say $p_1,p_2,p_3$)
are general (we may as well assume they are the coordinate vertices)
and the other 5 (say $p_4,\ldots,p_8$) are
on a general line $L$ (defined by a linear form $F$) and are general on that line. 
Then one can show that $\alpha(I(Z))=3$, $\widehat{\alpha}(I(Z))=5/2$
and $I(25sZ)\subsetneq I(Z)^{19s+1}$ for all $s\geq1$, and hence that 
$\alpha(I(Z))/\widehat{\alpha}(I(Z)) =6/5 < 25/19\leq \widehat{\rho}(I(Z))$.
(We now sketch the justification of these claims. 
The key is that $(I(25sZ))_{65s}$ vanishes on $L$ with order $15s$,
but $(I(Z)^{19s+1})_{65s}$ vanishes on $L$ with order $15s+3$, hence 
$(I(25sZ))_{65s}\not\subseteq (I(Z)^{19s+1})_{65s}$ and so $I(25sZ)\not\subseteq I(Z)^{19s+1}$.
It is easy to check that $\alpha(I(Z))=3$.
Using Bezout's Theorem, one can show that 
$(I(2tZ))_{5t}=(xyzF^2)^t \ensuremath{\mathbb{K}}\subset \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^2]= \ensuremath{\mathbb{K}}[x,y,z]$
and hence (since $\dim ((I(mZ))_i)>0$ implies $\dim ((I(mZ))_j)>1$ for all $j>i$)
that $\alpha(I(2tZ))=5t$.
(In more detail, note that $(xyzF^2)^t\in (I(2tZ))_{5t}$.
If $(I(2tZ))_{5t}$ contained anything more than scalar multiples of
$(xyzF^2)^t$, then $(I(8tZ))_{20t}$ would contain more than scalar multiples of
$(xyzF^2)^{4t}$, so it's enough to consider $(I(8tZ))_{20t}$.
But by Bezout, we have 
\begin{align*}
(I(8tZ))_{20t}&=F^{5t}((I(8t(p_1+p_2+p_3)+3t(p_4+\cdots+p_8)))_{15t})\\
&=(xyz)^tF^{5t}((I(3t(2(p_1+p_2+p_3)+(p_4+\cdots+p_8))))_{12t})\\
&=(xyz)^tF^{5t+1}((I((3t(2(p_1+p_2+p_3))+(3t-1)(p_4+\cdots+p_8))))_{12t-1})\\
&=(xyz)^{t+1}F^{5t+1}((I((3t-1)(2(p_1+p_2+p_3)+(p_4+\cdots+p_8))))_{12t-4})\\
&=\cdots=(xyz)^{4t}F^{8t} \ensuremath{\mathbb{K}}.)\\
\end{align*}
\vskip-\baselineskip
\noindent It follows that $\widehat{\alpha}(I(Z))=5/2$.
We also have by Bezout that $(I(25sZ))_{65s}=F^{15s}(I(5sZ'))_{50s}$, where
$Z'$ is the fat point subscheme obtained by taking the three coordinate vertices
with multiplicity 5 and the five points on $L$ with multiplicity 2. Moreover, one can
check that $(I(Z'))_{10}$ has greatest common divisor 1 (i.e., it defines a linear system
which is fixed component free); to see this, write elements of $(I(Z'))_{10}$ in two different ways
using conics and lines, where the two ways have no factors in common.
Thus $(I(5sZ'))_{50s}$ also defines a linear system that is fixed component free,
so $(I(25sZ))_{65s}=F^{15s}(I(5sZ'))_{50s}$ has $15sL$ as the divisorial
part of its base locus. But $(I(Z))_t$ has $L$ in its base locus for $t=3,4$
and is fixed component free for $t\geq5$. Now $(I(Z))^{19s+1}$ is spanned by 
products of $a$ homogeneous elements of $I(Z)$ of degree 5 or more and $b$ elements
of degree 4 or less with $a+b=19s+1$. Such an element vanishes on $L$ to order at least $b$.
To minimize $b$ we want $a$ as large as possible, and hence we want
to take all $a$ elements to have degree 5 and as many of the $b$ elements as possible
to have degree 3, and thus we have the inequality $5a+3b\leq 65s$. Solving
$5a+3b\leq 65s$ given $a+b=19s+1$ gives $2a\leq 8s-3$ and hence,
taking $a$ as large as possible we have $a=4s-2$, so $b=15s+3$.
I.e., every element of $((I(Z))^{19s+1})_{65s}$ vanishes on 
$L$ to order at least $15s+3$.) We do not know if
$\widehat{\rho}(I(Z))=25/19$, nor do we know the value of $\rho(I(Z))$.
\end{example}

\begin{question}\label{rhohatValue}
For the example $Z$ in the previous paragraph, what are the exact values of 
$\widehat{\rho}(I(Z))$, $\rho(I(Z))$ and $\rho_{int}(I(Z))$?
\end{question}

As Remark \ref{GrifoCorRem} explains, if the next question
has a negative answer, then Conjecture \ref{GrifoConj} holds
for $I=I(Z)$ whenever $\widehat{\rho}(I(Z))=\alpha(I(Z))/\widehat{\alpha}(I(Z))$.

\begin{question} 
Is it ever false that
$$\widehat{\alpha}(I(Z))\geq \frac{\alpha(I(Z))+1}{N}?$$
\end{question}

By \cite{DFMS}, $\widehat{\rho}(I(Z))$ is equal to the maximum
of $v(I(Z))/\widehat{v}(I(Z))$ for valuations $v$ supported on $I(Z)$, and this maximum always
occurs for what is known as a Rees valuation. Thus we have
$$\frac{v(I(Z))}{\widehat{v}(I(Z))}\leq \widehat{\rho}(I(Z))\leq N,$$ 
hence
$$\frac{v(I(Z))}{N}\leq \widehat{v}(I(Z)).$$ 
This raises the question of whether Chudnovsky-like bounds occur for valuations:

\begin{question}\label{FinalQuest3}
Is it always true for $N>1$ that
$$\frac{v(I(Z))+1}{N}\leq \widehat{v}(I(Z))?$$ 
\end{question}

If the answer is affirmative, then for some $v$ we would have 
$$\widehat{\rho}(I(Z))=\frac{v(I(Z))}{\widehat{v}(I(Z))} \leq \frac{v(I(Z))}{\frac{v(I(Z))+1}{N}}<N$$
which would confirm Grifo's Conjecture for $I(Z)$.

\begin{example}\label{verticesEx}
Consider $Z=p_1+\cdots+p_N+2p_{N+1}\subset \ensuremath{\mathbb{P}}^N$ where the points $p_i$ are 
the coordinate vertices. Here we show that $I(Z)^m=I(mZ)$ for all $m\geq1$.
For $N=1$, $I(Z)^m=I(mZ)$ holds for every $Z$ (since fat point ideals are principal), so assume $N>1$.
(We note that the case $N=2$ is covered by the results of  \cite{BZ}.)
Since $I(Z)$ is a monomial ideal, we merely have to check for each monomial 
$f=x_0^{a_0}\cdots x_N^{a_N}\in I(mZ)$ that $f\in I(Z)^m$; i.e., that $f$ can 
is divisible by a product of $m$ monomials, each in $I(Z)$.
But if we choose coordinates $x_i$ such that $x_j\neq0$ at $p_j$ for each $j$,
then $f\in I(mZ)$ is equivalent to $a_1+\cdots+a_N\geq 2m$ and $a_0+a_1+\cdots+a_N-a_i\geq m$
for each $0<i\leq N$. Without loss of generality we may assume
$a_1\leq a_2\leq \cdots\leq a_N$, so the inequalities $a_0+a_1+\cdots+a_N-a_i\geq m$ reduce to the single
inequality $a_0+a_1+\cdots+a_{N-1}\geq m$. Let us set $b=a_1+\cdots+a_{N-1}$.
Then $f\in I(mZ)$ if and only if $a_0+b\geq m$ and $b+a_N\geq 2m$,
so we want to show $a_0+b\geq m$ and $b+a_N\geq 2m$ implies $f\in I(Z)^m$.

First suppose $b\geq m$. We have two cases: $b+a_N=2m$ and $b+a_N>2m$.
First assume $b+a_N=2m$, so $0<b/(N-1)\leq a_{N-1}\leq a_N\leq b$. Let 
$e_0=(e_{01},\ldots,e_{0N})=(a_1,\cdots,a_N)$ so $e_0$ is the exponent vector of 
$g_0=x_1^{e_{01}}\cdots x_N^{e_{0N}}=x_1^{a_1}\cdots x_N^{a_N}$.
Let $b_0=b=e_{01}+\cdots+e_{0,N-1}$.
Note that $g_0\in I(mZ)$ and $g_0$ divides $f$, so it's enough to show $g_0\in I(Z)^m$. 

Starting with $e_0$ and $b_0$, we will recursively define a sequence $e_i$ of exponent vectors 
$e_i=(e_{i1},\ldots,e_{iN})$ and nonnegative integers $b_i$ for $0\leq i\leq \omega$ with 
$b_i>0$ for $0\leq i<\omega$ and $b_\omega=0$,
satisfying the following conditions:
\begin{enumerate}
\item[(i)] the entries of $e_i$ are nonnegative and nondecreasing;
\item[(ii)] $b_i+e_{i,N}$ is even, where $b_i=e_{i1}+\cdots+e_{i,N-1}$;
\item[(iii)] $b_i\geq e_{iN}$.
\end{enumerate}

For $e_0$ we have $b_0=b\geq m>0$.
Moreover, conditions (i)-(iii) hold for $e_0$: 
(i) holds by assumption; 
(ii) holds since $b_0+e_{0,N}=b+a_N=2m$ by assumption;
(iii) holds since we noted above that $a_N\leq b$, but $e_{0N}=a_N$ and $b_0=b$.

Given that conditions (i)-(iii) hold for some $e_i$ with $b_i>0$, we now
define $e_{i+1}=(e_{i+1,1},\ldots,e_{i+1,N})$ with $b_i>b_{i+1}$.
In brief, $j_i$ and $k_i$ are chosen to be as large as possible such that $j_i<k_i$
and so that the entries of $e_{i+1}=(e_{i+1,1},\ldots,e_{i+1,N})$ are nondecreasing,
where $e_{i+1,l}=e_{il}$ for all $l$ except that $e_{i+1,j_i}=e_{ij_i}-1$ and $e_{i+1,k_i}=e_{ik_i}-1$.
More precisely, $j_i$ is the least index $t$ such that $e_{it}=e_{i,N-1}$.
Then $k_i=N$ if $e_{iN}>e_{i,N-1}$ and $k_i=j_i+1$ if $e_{iN}=e_{i,N-1}$.
Since $b_{i+1}$ is either $b_i-1$ or $b_i-2$ we have $b_i>b_{i+1}$. 

It is easy to check that the construction ensures that the
entries of $e_{i+1}$ are nonnegative and nondecreasing, so (i) holds for $e_{i+1}$.
Since $b_i+e_{i,N}$ is even and $b_{i+1}+e_{i+1,N}=b_i+e_{i,N}-2$, (ii) holds.
For (iii), if $e_{i+1,N}=e_{iN}$, then $k_i<N$ so $0<e_{i,N-2} = e_{i,N-1}=e_{i,N}$,
and we cannot have $e_{i,N-2}=1$ with $e_{i,N-3}=0$ since then the sum of the entries of
$e_i$ would be odd. Thus either $e_{i,N-2}>1$ or $e_{i,N-3}>0$; either way
$b_i\geq 2+e_{i,N-1}=2+e_{iN}$, so $b_{i+1}=b_i-2\geq e_{iN}=e_{i+1,N}$.
If $e_{i+1,N}=e_{iN}-1$, then $b_{i+1}=b_i-1$, so $b_i\geq e_{iN}$ implies
$b_{i+1}\geq e_{i+1,N}$ (and hence $b_{i+1}\geq0$).

Let $g_i$ be the monomial whose
exponent vector is $e_i$, so $g_i=x_1^{e_{i1}}\cdots x_N^{e_{iN}}$.
We have $g_\omega=1$, since $0=b_\omega\geq e_{\omega N}$,
so $e_\omega=(0,\ldots,0)$.
If $i<\omega$, then $g_i=g_{i+1}x_{j_i}x_{k_i}$.
Thus $g_0=(x_{j_0}x_{k_0})\cdots (x_{j_{\omega-1}}x_{k_{\omega-1}})$,
but each factor $x_{j_i}x_{k_i}$ is in $I(Z)$, and since
$\deg g_0=b_0+e_{0N}=2m$, we see that there are $m$ factors, so
$g_0\in I(Z)^m$, as we wanted to show.

We now consider case 2, $b+a_N>2m$ (still under the assumption that $b\geq m$). 
Starting with $(a_1,\ldots,a_{N-1},a_N)$, 
replace $a_N$ by the smallest integer $a$ satisfying $a\geq a_{N-1}$ and 
$b+a\geq 2m$. We then have that $g=x_0^{a_0}\cdots x_{N-1}^{a_{N-1}}x_N^a$
divides $f$ and is still in $I(mZ)$. Thus it is enough to show that $g\in I(Z)^m$.
If $b+a=2m$, we are done by case 1, so assume $b+a>2m$. 
Then by construction we have $a=a_{N-1}$. Since $b\geq m$,
we must have $a\leq m$. It is now not hard to find new exponents
$0\leq a_1'\leq a_2'\leq \cdots\leq a_{N-1}'$ such that 
$a_i'\leq a_i$ and $b'+a=2m$, where $b'=a_1'+\cdots+a_{N-1}'$.
Let $g_0=x_1^{a_1'}\cdots x_{N-1}^{a_{N-1}'}x_N^a$; then $g$ divides $f$
and by case 1 we have $g\in I(Z)^m$.

We are left with considering the case that $b<m$. We have
$a_0+b\geq m$ and $b+a_N\geq 2m$. Clearly we can reduce
$a_0$ and $a_N$ so that $a_0+b = m$ and $b+a_N = 2m$
(since the associated monomial $g$ divides $f$ and still is in $I(mZ)$).
So we may assume $a_0+b = m$ and $b+a_N = 2m$, in which case
$f=(x_1^{a_1}\cdots x_{N-1}^{a_{N-1}}x_N^b)x_0^{a_0}x_N^{a_N-b}
=(x_1^{a_1}\cdots x_{N-1}^{a_{N-1}}x_N^b)x_0^{m-b}x_N^{2(m-b)}
=(x_1^{a_1}\cdots x_{N-1}^{a_{N-1}}x_N^b)(x_0x_N^2)^{m-b}$.
But $x_0X_N^2\in I(Z)$ and $x_1^{a_1}\cdots x_{N-1}^{a_{N-1}}x_N^b$
is a product of $b$ factors of the form $x_ix_N$ for $1\leq i\leq N-1$,
and each of these is in $I(Z)$. Thus $f$ is a product of $m=b+(m-b)$
elements of $I(Z)$, hence $f\in I(Z)^m$.
\end{example}

\begin{remark}\label{FatPtsAndCI1}
Again consider $Z=p_1+\cdots+p_N+2p_{N+1}\subset \ensuremath{\mathbb{P}}^N$ where the points $p_i$ are 
the coordinate vertices and $N>1$. Then there is no $N$-generated ideal 
$J$, homogeneous or not, such that $I(Z)=J^s$ for some $s\geq1$. 
Suppose there were; say $J=(F_1,\dots,F_N)$. 
The ideal $I(Z)$ defines the $N+1$ coordinate lines
in affine $N+1$ space, all taken with multiplicity 1 
except one taken with multiplicity 2.
Since each $F_i$ vanishes on each coordinate line, none 
of the $F_i$ can have any terms which are a power of
a single variable (i.e., every term involves a product of two or more variables).
Therefore, none of the $F_i$ have terms of degree less than 2. 
Since $(F_1,\dots,F_N)^s=J^s=I(Z)$ and since $I(Z)$ has elements with terms of degree 2,
we see that $s=1$ (otherwise the least degree of a term of an element of $J^s$
would be at least $2s\geq4$). But $s=1$ implies that $J=I(Z)$ is homogeneous,
and an easy argument shows that the least number of homogeneous generators
of a homogeneous ideal is the least number of generators possible,
which in this case is $\binom{N}{2}+N>N$ (since, for a particular ordering of the points $p_i$
we have $I(Z)=(x_ix_j:i>0, j>0)+(x_0x_i^2:i>0)$).
\end{remark}

\begin{remark}\label{FatPtsAndCI2}
Let $Z=m_1p_1+\cdots+m_rp_r\subset \ensuremath{\mathbb{P}}^N$ with $N>1$, $m_i>0$ for all $i$ and
$ \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]= \ensuremath{\mathbb{K}}[x_0,\ldots,x_N]$.
Here we show that there is an ideal $J$ generated by $N$ forms such that
$I(Z)=J^m$ for some $m\geq1$
if and only if $m_1=\cdots=m_r=m$ where $I(p_1+\cdots+p_r)$
is generated by $N$ forms. The reverse implication is known \cite[Lemma 5, Appendix 6]{ZS},
so assume $I(Z)=J^m$ for some $m\geq1$ with forms $F_i$ such that
$J=(F_1,\ldots,F_N)$.
Choose coordinates $x_0,\ldots,x_N$ such that none of the points lies on $x_0=0$
and $p_1=(1,0,\ldots,0)$.
Let $U_0$ be the affine neighborhood defined by $x_0\neq0$,
and let $f_i(x_1,\ldots,x_N)=F_i(1,x_1,\ldots,x_N)$ be the polynomial obtained by
setting $x_0$ to 1 in $F_i$. Then on $U_0$, $Z$ is defined by
the ideal $I'(Z)$ obtained from $I(Z)$ by setting $x_0=1$ in each
element of $I(Z)$, so $I'(Z)=(J')^m$ where $J'=(f_1,\ldots,f_N)$. 
Localizing at $P_1=I(p_1)=(x_1,\ldots,x_N)$ gives
$(P_1)_{P_1}^{m_1}=(I'(Z))_{P_1}=(J')^m_{P_1}$
and modding out by $P_1^{m_1+1}$ gives graded isomorphisms
$$\frac{P_1^{m_1}}{P_1^{m_1+1}}\cong
\frac{(P_1)_{P_1}^{m_1}}{(P_1)_{P_1}^{m_1+1}}=
\frac{(J')^m_{P_1}}{(P_1)_{P_1}^{m_1+1}}\cong
\frac{(J')^m+P_1^{m_1+1}}{(P_1)^{m_1+1}}.$$
Thus each $f_i$ can have no terms of degree less than
$d=m_1/m$ and there must be terms of degree exactly $d$, so $d$ is an integer,
hence $m_1=dm$ and $m\leq m_1$.
But the vector space dimension of $\frac{P_1^{m_1}}{P_1^{m_1+1}}$
is $\binom{m_1+N-1}{N-1}$, and the vector space dimension of
$\frac{(J')^m+P_1^{m_1+1}}{(P_1)^{m_1+1}}$ is at most
$\binom{m+N-1}{N-1}$, so we have $\binom{m_1+N-1}{N-1}\leq \binom{m+N-1}{N-1}$,
which implies $m_1\leq m$, hence $m=m_1$ and $d=1$.
With a change of coordinates, the 
same argument works for each point $p_i$, so $m_i=m$ for all $i$,
and, at each point $p_i$, the linear terms of the $f_j$ must
be linearly independent (otherwise we would have
$\dim_ \ensuremath{\mathbb{K}}\frac{(J')^m+P_1^{m_1+1}}{(P_1)^{m_1+1}}<\binom{m+N-1}{N-1}$).
Thus $J'=I(p_1+\cdots+p_r)$ on $U_0$, hence 
$J=I(p_1+\cdots+p_r)\subseteq \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$.
\end{remark}

\begin{question}
In Remark \ref{FatPtsAndCI2}, do we need to assume a priori that $J$ is homogeneous?
\end{question}

\section{Computational estimates of resurgences}

It might be possible to address some of the foregoing questions computationally.
More generally, it is of interest to consider to what extent quantities like resurgences can be computed.
 
\subsection{Denkert's thesis}
In an unpublished part of her thesis \cite{Den}, Denkert 
gives an algorithm for computing $\rho(I(Z))$ arbitrarily accurately
when the symbolic Rees algebra 
$$R_s(I(Z))=\oplus_t I(tZ)x^t\subseteq  \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N][x]$$
is Noetherian (equivalently, when for some $a\geq 1$, all
powers of $I(aZ)$ are symbolic \cite[Theorem 2.1]{HHT}).

Let $I=I(Z)\subseteq  \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ be nontrivial and assume $I^{(at)} = (I^{(a)})^t$
for all $t > 0$. For each $s\geq1$, let $b_s$ be the largest $b$ such that
$I^{(asN)}\subseteq I^{b}$ and let $\epsilon>0$. Since $b_s\geq as$, 
we have $\frac{asN}{b_s}-\frac{asN}{b_s+1}=\frac{asN}{b_s(b_s+1)}\leq\frac{N}{as+1}$.
So, by picking $s\gg0$ and $\epsilon$ small, we can make
$\frac{N}{as+1}+\epsilon$ arbitrarily small. Denkert's algorithm either computes
$\rho(I)$ exactly or gives an estimate which is accurate to less than 
$\frac{N}{as+1}+\epsilon$.

Assume we have picked $s$ and $\epsilon$. Let $A=asN$ and let $B=b_s$. 
Thus we have 
$$I^{(At)}=(I^{(A)})^t\subseteq (I^{B})^t=I^{Bt}$$ 
for all $t\geq1$, and
for $m\geq At$ and $r\leq Bt$ we have 
$$I^{(m)}\subseteq I^{(At)}\subseteq I^{Bt}\subseteq I^r.$$

If $r\geq B\lceil\frac{A}{B\epsilon}\rceil$ and $\frac{m}{r}\geq\frac{A}{B}+\epsilon$, we now show for
the least $t$ such that $r<Bt$ that $m\geq At$, and hence $I^{(m)}\subseteq I^r$.
Indeed, we have $B(t-1)\leq r<Bt$, so $t-1=\lfloor\frac{r}{B}\rfloor\geq\lceil\frac{A}{B\epsilon}\rceil$.
Thus $(t-1)B\epsilon\geq A$, which is equivalent to
$(\frac{A}{B}+\epsilon)(t-1)B\geq At$, hence we have
$m\geq r(\frac{A}{B}+\epsilon)\geq B(t-1)(\frac{A}{B}+\epsilon)\geq At$.

In summary, we have $I^{(m)}\subseteq I^r$ for all 
$r\geq B\lceil\frac{A}{B\epsilon}\rceil$ and $\frac{m}{r}\geq\frac{A}{B}+\epsilon$,
and we have $I^{(A)}\not\subseteq I^{B+1}$.
Thus $\frac{A}{B+1}\leq \rho(I)$ and either $\rho(I)< \frac{A}{B}+\epsilon$
or $\rho(I)=m/r$ for some $r< B\lceil\frac{A}{B\epsilon}\rceil$
with $N>m/r\geq\frac{A}{B}+\epsilon$ (and there are only finitely many
such $r$ and $m$).

Moreover, since we have $I^{(m)}\subseteq I^r$ for all but finitely many $m$ and $r$
with $m/r\geq\frac{A}{B}+\epsilon$, we have $\widehat{\rho}(I) \leq \frac{A}{B}+\epsilon$
for each $\epsilon>0$, and hence we have $\widehat{\rho}(I) \leq \frac{A}{B}$.
This also follows by Theorem 1.2(3) of \cite{GHVT}.

\subsection{DiPasquale-Drabkin method}
An alternate approach is based on \cite{DD}.
Again assume the symbolic Rees algebra of a homogeneous ideal $I\subset \ensuremath{\mathbb{K}}[ \ensuremath{\mathbb{P}}^N]$ is finitely 
generated. Then \cite{DD} shows that we can compute $\widehat{\rho}(I)$ exactly
(assuming we know the Rees valuations). Let $\epsilon>0$. Then, as in the proof of Theorem \ref{MT1},
whenever $I^{(m)}\not\subseteq I^r$, we have $\frac{s}{r+N} < \widehat{\rho}(I)$.
But there are only finitely many $r$ for which there is an $s$ such that both
$\frac{s}{r+N} < \widehat{\rho}(I)$ and
$\widehat{\rho}(I)+\epsilon \leq s/r$ hold.
For each of these $s$ and $r$ we check if $I^{(s)}$ is not contained in $I^r$.
For all such $r$ and $s$ which occur (if any), then $\rho(I)$ is the maximum of their ratios $s/r$. If none occur, then 
$\widehat{\rho}(I) \leq \rho(I) < \widehat{\rho}(I)+ \epsilon$. 

The estimates on $\rho(I)$ given by using \cite{DD} versus those of \cite{Den} are somewhat
similar, but each starts with different input data. Also, \cite{DD} gives 
us $\widehat{\rho}(I)$ exactly (if we know the Rees valuations), whereas 
\cite{Den} just gives us an upper bound on $\widehat{\rho}(I)$.


