text:
\section{Introduction}
Let  $R={\Bbbk}[x_1,\dots, x_n]/I$ be a standard graded algebra over a field ${\Bbbk}$. Suppose $R$ is a domain. How many independent quadrics can $I$ contain? Such a question has long been studied in algebraic geometry, and some beautiful answers are known. For example, if ${\Bbbk}$ is algebraically closed, and $I$ is prime and contains no linear forms, then the number of quadratic minimal generators of $I$ is at most  $\binom{n-d+1}{2}$ where $d=\dim R$. Furthermore, equality happens if and only if $R$ is a variety of minimal degree (see Theorem \ref{h2} for a generalization of this result to non-reduced schemes).

In another vein, a striking result by Bertram-Ein-Lazarsfeld (\cite[Corollary 2]{BEL}) states that if ${\mathrm{Proj\; }} R $ is smooth and the sum of $n-d$ highest degrees among the minimal generators of $I$ is less than $n$, then $R$ is Cohen-Macaulay. We will also extend this result to mild singularities.

The common theme in the aforementioned statements is how the singularities of $R$ affect its Hilbert function. This article grew out of an attempt to better understand this phenomenon. It turns out that such numerical information is most conveniently expressed via the $h$-vector. Recall that we can write the Hilbert series of $R$
 \[H_R(t)= \sum_{i\in\N} (\dim_{{\Bbbk}}R_i)t^i = \frac{h_0+h_1t+ \cdots + h_st^s}{(1-t)^d}. \]
We will call the vector of coefficients $(h_0,h_1,\ldots ,h_s)$ the $h$-vector of $R$. The bound on the number of quadrics mentioned above can be easily seen to be equivalent to the statement that $h_2\geq 0$. Furthermore, it is well-known and easy to prove that if $R$ is Cohen-Macaulay then $h_i\geq 0$ for all $i$.  A very concrete motivation for our work comes from the following question:

\begin{question}\label{mainques}
Suppose $R$ satisfies Serre's condition $(S_r)$. When is it true that $h_i \geq 0$ for $i=0,\ldots , r$?
\end{question}

An answer for this question when $I$ is a square-free monomial ideal is contained in a beautiful result of  Murai-Terai  (the $h$-vector of a simplicial complex $\Delta$ equals the $h$-vector of its Stanley-Reisner ring ${\Bbbk}[\Delta]$).

\begin{theorem*}[\cite{MT}]
Let $\Delta$ be a simplicial complex with $h$-vector $(h_0,h_1,\ldots ,h_s)$. If ${\Bbbk}[\Delta]$ satisfies $(S_r)$ for some field ${\Bbbk}$, then $h_i\geq 0$ for $i=0,\ldots , r$.
\end{theorem*}

In general, some additional assumptions are needed as shown in Remark \ref{rem:negh2} where, for any $n\geq 3$, it is constructed a standard graded ${\Bbbk}$-algebra $R$ satisfying $(S_{n-1})$ and having $h$-vector $(1,n,-1)$. It even turns out that such an $R$ has  Castelnuovo-Mumford regularity 1 and is Buchsbaum. Contrary to this, we can surprisingly show the following, which is our main result:

\begin{theorem}\label{main0}
Let $R$ be a standard graded algebra over a field and $(h_0,\ldots ,h_s)$ be the  $h$-vector of $R$. Assume $R$ satisfies $(S_r)$ and either:
\begin{enumerate}

\item $\chara {\Bbbk}=0$ and $X={\mathrm{Proj\; }} R$ is Du Bois.
\item $\chara {\Bbbk}=p>0$ and $R$ is $F$-pure.
\end{enumerate}
Then $h_i\geq 0$ for $i=0,\ldots , r$. Also, $h_r+ h_{r+1}+\dots+h_s \geq 0$, or equivalently $R$ has multiplicity at least $h_0+h_1+\dots +h_{r-1}$.

If furthermore $R$ has Castelnuovo-Mumford regularity less than $r$ or $h_i=0$ for some $i\leq r$,  then $R$ is Cohen-Macaulay.
\end{theorem}

Note that Theorem \ref{main0} generalizes the work of Murai-Terai, since a Stanley-Reisner ring is Du Bois in characteristic $0$ and $F$-pure in characteristic $p$. The key point is to establish good bounds on certain $\operatorname{Ext}$ modules, just like what Murai and Terai did, but for ideals that are not necessarily square-free.

One can deduce other surprising consequences from our results. We discuss one of them now.  There is a well-known but mysterious theme in algebraic geometry that ``nice singularities of small codimension" should be Cohen-Macaulay. Perhaps the most famous example is Hartshorne's conjecture that a smooth subvariety in $\mathbb P_{{\Bbbk}}^n$ of codimension $e$ has to be a complete intersection provided $e< n/3$. Although this conjecture is wide-open, numerous partial results has been obtained, and many of them assert that such variety has to be projectively Cohen-Macaulay.  One or our corollaries can be viewed as another confirmation of this viewpoint (extending \cite[Corollary 1.3]{dFE}).

\begin{corollary}[Corollary \ref{cor2}]
Let $R={\Bbbk}[x_1,\dots, x_n]/I$ be a standard graded algebra over a field ${\Bbbk}$ of characteristic $0$ with $e=\operatorname{ht} I$ and $d=\dim R$. Let $d_1 \geq d_2 \geq \dots $ be the degree sequence of a minimal set of generators for $I$. Assume that $R$ is unmixed, equidimensional, Cohen-Macaulay in codimension $l$ and $X={\mathrm{Proj\; }} R$ has only MJ-log canonical singularities. If $e+l \geq d_1+\dots +d_e$, then $R$ is Cohen-Macaulay.
\end{corollary}

For instance, the Corollary asserts that if $I$ is generated by quadrics, then $R$ is Cohen-Macaulay provided that it is Cohen-Macaulay in codimension $e$. Under stronger assumptions, it follows that $R$ is Gorenstein or even complete intersection, see Corollary \ref{cor3}.

We now briefly describe the content and organization of the paper. Section \ref{prelim} establishes all the important preparatory results. It contains a careful analysis of a property which we call $(MT_r)$. This is a condition on regularity of certain $\operatorname{Ext}$ modules studied by Murai-Terai (these modules are sometimes called deficiency modules in the literature, see \cite{Sc}). It turns out that this regularity condition is precisely what we need to prove all the statements of the main theorem. We give a self-contained and independent proof that $(MT_r)$ implies non-negativity of $h_i$ for $i\leq r$, when equalities occur,  and also that $(MT_r)$ is preserved under generic hyperplane section. Our proofs work also for modules.

What remains, which is our main technical contribution, is to establish the condition $(MT_r)$ for algebras with nice singularities. We achieve this goal in Sections \ref{main} (characteristic $0$) and \ref{mainp} (positive characteristic). Our results there give strong bounds on regularity of the Matlis dual of the local cohomology modules of such algebras, and are perhaps of independent interests. As mentioned above, they can be viewed as Kodaira-vanishing type statements.

In Section \ref{apps} we put everything together to prove Theorem \ref{main0} and give several applications. Our bounds imply stringent conditions on Hilbert functions of algebras satisfying assumptions of \ref{main0}, see Corollary \ref{cor1}. One can also show that certain algebras with nice singularities and small codimension must be Cohen-Macaulay, see Corollary \ref{cor2}. We also write down perhaps the strongest possible statement for $h_2(R)$, which can be viewed as generalization of the classical bound $e(R) \geq 1+ e$ where $e$ is the embedding codimension, and the equality case which forces $R$ to be rings of minimal multiplicity.

In the last section, we propose an open question which can be viewed as a very strong generalization of our result and prove a first step toward such problem, which generalized one of the main results of Huneke and Smith in \cite{HuSm}.

\subsection*{Acknowledgement} The first author is partially supported by Simons Foundation Collaboration Grant 527316. The second author is partially supported by an NSF Grant \#1836867/1600198. The second author would like to thank Karl Schwede for many discussions on Du Bois singularities. The first and third authors would like to thank Kangjin Han for useful discussions on $h$-vectors and singularities, which started this project. We also thank Tommaso de Fernex and
S\'andor  Kov\'acs for some helpful email exchanges.

\section{Preliminaries}\label{prelim}

\begin{setup}\label{setup}
We will use the following notations and conventions throughout the paper:

\begin{enumerate}[(i)]
\item $n$ is a positive integer;
\item  ${\Bbbk}$ is a field;
\item $S$ is the standard graded polynomial ring in $n$ variables over ${\Bbbk}$;
\item $\mm$ is the irrelevant ideal of $S$;
\item $I\subset S$ is a homogeneous ideal of height $e$;
\item  $R=S/I$ is the quotient of dimension $d=n-e$;
\item  $X={\mathrm{Proj\; }} R$ is the projective scheme associated to $R$ ($\dim X=d-1$).
\item For a ${\Bbbk}$-scheme $Y$  we say that $Y$ has  $P$ singularities if and only if $Y \otimes_{{\Bbbk}} \bar{{\Bbbk}}$ has $P$ singularities, where $\bar {\Bbbk}$ is the algebraic closure of ${\Bbbk}$. Here $P$ can be non-singular, Du Bois, $MJ$-log canonical, $F$-pure, etc.
\item We say that $R$ has  $P$ singularities if and only if ${\mathrm{Spec\; }} R$ has  $P$ singularities.
\item For a nonzero finitely generated graded $S$-module $M$ of dimension $d$, let $H_M(t)$ denote the Hilbert series of $M$, and $(h_a(M),h_{a+1}(M),\ldots ,h_s(M))$ denote the $h$-vector of $M$, given by:
\[H_M(t)= \sum_{i\in\Z} (\dim_{{\Bbbk}}M_i)t^i = \frac{h_at^a+h_{a+1}t^{a+1}+ \cdots + h_st^s}{(1-t)^d},  \]
We also write $p_M(t)$ for the numerator and $c_r(M)$ for the quantity $h_r(M) +h_{r+1}(M)+\dots + h_s(M)$. Note that  $c_r(M)= e(M)-\sum_{i<r} h_i(M)$,where $e(M)$ denotes the Hilbert-Samuel multiplicity.
\end{enumerate}

\end{setup}

Let us recall the following result of Murai and Terai (\cite[Theorem 1.4]{MT}):
\begin{theorem}\label{t:MT}
If $\reg(\operatorname{Ext}_S^{n-i}(R,\omega_S))\leq i-r$ $ \forall \ i=0,\ldots ,d-1$, then $h_i(R)\geq 0$ for $i\leq r$.
\end{theorem}

Motivated by this result, we make the following definition.
\begin{definition}\label{defMT}
Consider a graded module $M$ over $S$ of dimension $d$. We say that $M$  satisfies condition $(MT_r)$ if  $\reg(\operatorname{Ext}_S^{n-i}(M,\omega_S))\leq i-r$ $ \forall \ i=0,\ldots ,d-1$.
\end{definition}

The analysis of this notion led us to a more direct proof of Theorem \ref{t:MT}. We first note a few preparatory results.
The following facts are trivial consequences of graded duality:
\begin{proposition}\label{easy}
\begin{enumerate}
\item  The condition $(MT_r)$ depends only on $R$ and does not depend on the presentation $R=S/I$ (we won't need this).
\item If $M$ is Cohen-Macaulay, then it is $(MT_r)$ for any $r$.
\item If $N = H^i_{\m}(M)$ has finite length, then $\reg(\operatorname{Ext}_S^{n-i}(M,\omega_S))\leq i-r$ if and only if $N_{<r-i} =0$.
 \end{enumerate}

\end{proposition}

\begin{proposition}\label{linear}
Suppose ${\Bbbk}$ is infinite. Let $M$ be a graded $S$-module. For a generic linear form $l$:
$$ \reg(M) \geq \max\{\reg{0:_Ml}, \reg M/lM\}$$
\end{proposition}
\begin{proof}
This is well-known and easy, but we give a proof for convenience. Let $N=H_\m^0(M)$. Then for generic $l$, $0:_Ml \subseteq N$, so it's regularity is at most $\reg(N)\leq \reg(M)$.
Consider the short exact sequence $0\to N\to M\to M'\to 0$. We can assume $l$ is regular on $M'$, which leads to $0\to N/lN\to M/lM\to M'/lM'\to 0$. Since $\reg(N/lN)\leq \reg(N)$ and $\reg(M'/lM') = \reg(M') \leq \reg (M)$, the result follows.
\end{proof}

\begin{proposition}\label{mtr1}
Suppose ${\Bbbk}$ is infinite. Let $M$ be a graded $S$-module. Assume that  $M$ is $(MT_r)$, then so are the following modules:
$M'= M/H^0_{\m}(M), M'/lM', M/lM$,  for a generic linear form $l$.
\end{proposition}

\begin{proof}
Let $K_i(M)$ denote the module $\operatorname{Ext}_S^{n-i}(M,\omega_S)$. The short exact sequence $0 \to H_{\m}^0(M) \to M \to M' \to 0$ tells us that $K_0(M')=0$ and $K_i(M') = K_i(M)$ for $i\geq 1$. So $M'$ is $(MT_r)$.
For the rest, we assume $d=\dim M\geq 1$. As $l$ is generic, it is regular on $M'$, so from the short exact sequence $0 \to M'(-1) \to M' \to M'/lM' \to 0$ we obtained for each $i\leq d-1=\dim M'/lM'$:
$$0 \to K_{i+1}(M'(-1))/l K_{i+1}(M'(-1)) \to K_i(M'/lM') \to 0:_{K_i(M')}l \to 0    $$
The bound on regularity of $K_i(M'/lM')$ follows from the property $(MT_r)$ for $M'$ and Proposition \ref{linear}.
For the module $M/lM$, note that we have the short exact sequence $0\to N/lN\to M/lM\to M'/lM'\to 0$ where $N=H_\m^0(M)$. Another application of the long exact sequence of $\operatorname{Ext}$ and the fact that $M'/lM'$ is $(MT_r)$ proves what we need.
\end{proof}

The next proposition tracks the behavior of $h$-vector  modulo a generic linear form for a module satisfying $(MT_r)$. Let $c_r(M)$ be the quantity: $h_r(M)+ h_{r+1}(M)+... = e(M)-\sum_{i=0}^{r-1}h_i(M)$.
\begin{proposition}\label{mtr2}
Suppose ${\Bbbk}$ is infinite. Let $M$ be a graded $S$-module of dimension $d\geq 1$. Let $l$ be a generic linear form. Let $N=H_\m^0(M)$ and $M'=M/N$.  Assume that $N_{<r}=0$ (for example, if $M$ is $(MT_r)$). Then:
\begin{enumerate}
\item $h_i(M) =h_i(M/lM)$ for $i<r$.
\item $h_r(M) \geq h_r(M/lM)$.
\item $c_r(M) =c_r(M'/lM')$.
\item $c_r(M) =   c_r(M/lM)$ if $d>1$.
\end{enumerate}
\end{proposition}

\begin{proof}
First of all $M$ is $(MT_r)$ implies $N_{<r}=0$, see Proposition \ref{easy}. Since $p_M(t) = p_{N}(t)(1-t)^d + p_{M'}(t)$, we have $h_i(M) = h_i(M')$ for $i<r$ and $h_r(M) = h_r(M')+h_r(N)$, as well as $c_r(M)=c_r(M')$. As $l$ is generic, it is regular on $M'$ and so $h_i(M'/lM') = h_i(M')$ for all $i$. We also have the short exact sequence $0\to N/lN\to M/lM\to M'/lM'\to 0$, which gives:
\begin{itemize}
\item $h_i(M/lM) = h_i(M'/lM') = h_i(M') = h_i(M)$ for $i<r$.
\item $h_r(M/lM) =h_r(M'/lM') +h_r(N/lN) \leq h_r(M)$.
\item $c_r(M)=c_r(M') = c_r(M'/lM')$.
\item $c_r(M/lM) = c_r(M'/lM') = c_r(M)$ if $d>1$. \qedhere
\end{itemize}
\end{proof}

We can now give a direct and simple proof of Theorem \ref{t:MT}, extended to the module case.

\begin{theorem}\label{mtr3}
Let $M$ be a module satisfying $(MT_r)$. Then $h_i(M)\geq 0$ for $i\leq r$ and also $c_r(M)\geq 0$.
\end{theorem}

\begin{proof}
We can extend ${\Bbbk}$ if necessary and hence assume it is infinite. Since $M$ is $(MT_r)$, we have $N_{<r}=0$ where $N=H_\m^0(M)$, see Proposition \ref{easy}. By Propositions \ref{mtr1} and \ref{mtr2}, we can cut down by a generic linear system of parameters of length $d$ without increasing any of the relevant invariants (for the statement about $c_r(M)$, at the step $d=1$ we need to use $M'/lM'$ and not $M/lM$). But for Artinian modules all inequalities are trivial.\end{proof}

\begin{proposition}\label{mtr4}
If $M$ is $(MT_r)$ and either:
\begin{enumerate}
\item $\reg(M)<r$, or
\item $h_i(M)=0$ for some $i\leq r$ and $M$ is generated in degree $0$.
\end{enumerate}
Then $M$ is Cohen-Macaulay.
\end{proposition}

\begin{proof}
We can extend ${\Bbbk}$ if necessary and hence assume it is infinite. We use induction on $d=\dim M$ and therefore assume $d\geq 1$.
Assume first that $\reg(M)<r$.  Let $N =H_\m^0(M)$, then $N_{<r}=0$. But $\reg(N)\leq \reg(M) <r$, so $N=0$. Thus $\depth M\geq 1$.
By Propositions \ref{linear} and \ref{mtr1} as well as induction hypotheses, $M/lM$ is Cohen-Macaulay for a generic $l$, which implies that $M$ is Cohen-Macaulay.

Now assume that $h_i(M)=0$ for some $i\leq r$. Again, we shall show that $N= H_\m^0(M)=0$. Let  $M'=M/N$. Then as in the proof of Proposition \ref{mtr1} and \ref{mtr2}, $M'$ is still $(MT_r)$ and $h_i(M') = h_i(M)$. Passing to $M'/lM'$ preserves everything, so by induction $M'/lM'$ is Cohen-Macaulay, and so is $M'$. It follows that $h_j(M')=0$ for $j\geq i$ and $\reg M'\leq i-1$.

Write $M= F/P$ where $P$ is the first syzygy of $M$. Then $N=Q/P$ for $Q\subseteq F$, and $M'=F/Q$. Since $\reg(M')\leq i-1$, $Q$ is generated in degree at most $i$. But $N_{\leq i}=0$ (we already know that $N_{<r}=0$, if $i=r$ then the equality $h_r(M)= h_r(N)+h_r(M')$ forces $h_r(N)=N_r=0$). Thus $Q= Q_{\leq i} \subseteq P$, and $N=0$. The rest follows as in the first part.
\end{proof}

\begin{eg}
\begin{enumerate}
  \item Let $R={\Bbbk}[x,y]/(x^3,x^2y)$. Then $R$ is $(MT_2)$ and $c_2(R)=0$ but $R$ is not Cohen-Macaulay. The $h$-vector is $(1,1,1,-1)$.
  \item Let $R={\Bbbk}[x,y,u,v]/(xuv,yu)$. Then the $h$-vector of $R$ is $(1,1,0,-1)$. in other words, $h_2(R)=0$ but $R$ is not Cohen-Macaulay. Of course, $R$ fails $(MT_2)$.
  \item Let $R= {\Bbbk}[a^4,a^3b,ab^3,b^4]$. Then $R$ is $(MT_2)$ but not $(MT_3)$. The $h$-vector of $R$ is $(1,2,2,-1)$.
  \item Let $S={\Bbbk}[x]$, $M=S\oplus {\Bbbk}(-3)$. Then $M$ is $(MT_2)$, $h_2(M)=0$, but $M$ is not Cohen-Macaulay.
\end{enumerate}
\end{eg}

The following connects Serre's condition $(S_i)$ to dimension of $\operatorname{Ext}$, it appears without proof in \cite[Proposition 3.51]{Vas}:

\begin{proposition}\label{algSl}
If $R$ is equidimensional, it satisfies $(S_r)$ if and only if
\[\dim \operatorname{Ext}_S^{n-i}(R,S) \leq i-r \ \ \ \forall \ i<\dim R\]
(the Krull dimension of the zero-module is $-\infty$).
\end{proposition}
\begin{proof}
Recall that $\dim R=d$ and $\operatorname{ht} I=e$. Assume first that $R$ satisfies $(S_r)$. If $i<r\leq \depth R$, there is nothing to prove because $\operatorname{Ext}_S^{n-i}(R,S)=0$. Otherwise, let $\mathfrak{p}$ be a prime ideal of $S$ containing $I$ of height $h$. Notice that, because $R$ is equidimensional, $\dim R_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}}=h-e$. If $h<n-i+r$, then
\[\operatorname{Ext}_S^{n-i}(R,S)_{\mathfrak{p}}\cong \operatorname{Ext}_{S_{\mathfrak{p}}}^{n-i}(R_{\mathfrak{p}},S_{\mathfrak{p}})=0,\]
because $n-i=h-(h-n+i)$ and $\depth R_{\mathfrak{p}}>h-n+i$. To see it, note that $h-n+i<n-i+r-n+i=r$ and $\dim R_{\mathfrak{p}}>h-n+i$, so because $R$ satisfies $(S_r)$
\[\depth R_{\mathfrak{p}}=\min\{r,\dim R_{\mathfrak{p}}\}>h-n+i.\]
So $\mathfrak{p}\notin \mathrm{Supp} \operatorname{Ext}_S^{n-i}(R,S)$ whenever $\operatorname{ht} \mathfrak{p}<n-(i-r)$, i. e. $\dim \operatorname{Ext}_S^{n-i}(R,S)\leq i-r$.

For the other direction, for all $h=e,\ldots ,n$ let us denote by $V_h$ the set of prime ideals of $S$ of height $h$ containing $I$. We have the following chain of equivalences:
\begin{eqnarray*}
\mbox{$R$ satisfies $(S_r)$ } & \iff \\
\depth R_{\mathfrak{p}}=\min\{r,\dim R_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}}\}=:b \ \ \ \forall \ h=e,\ldots ,n, \ \forall \ {\mathfrak{p}}}\def\mm{{\mathfrak{m}} \in V_h & \iff \\
\depth R_{\mathfrak{p}}=\min\{r,h-e\}=:b \ \ \ \forall \ h=e,\ldots ,n, \ \forall \ {\mathfrak{p}}}\def\mm{{\mathfrak{m}} \in V_h & \iff \\
H_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}} S_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}}}^i(R_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}})=0 \ \ \ \forall \ h=e,\ldots ,n, \ \forall \ {\mathfrak{p}}}\def\mm{{\mathfrak{m}} \in V_h, \ \forall \ i<b & \iff \\
\operatorname{Ext}_{S_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}}}^{h-i}(R_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}},S_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}})=0 \ \ \ \forall \ h=e,\ldots ,n, \ \forall \ {\mathfrak{p}}}\def\mm{{\mathfrak{m}} \in V_h, \ \forall \ i<b & \iff \\
\dim \operatorname{Ext}_{S}^{h-i}(R,S)<n-h \ \ \ \forall \ h=e,\ldots ,n, \ \forall \ i<b & \iff \\
\end{eqnarray*}
(where in the equivalence between the second and third line we used the equidimensionality of $R$). Since $i<b\leq h-e$, we have $n-h+i<n-e=d$. If $\dim \operatorname{Ext}_S^{n-i}(R,S) \leq i-r \ \ \ \forall \ i=0,\ldots ,d-1$, then
\[\dim \operatorname{Ext}_{S}^{h-i}(R,S)=\dim \operatorname{Ext}_{S}^{n-(n-h+i)}(R,S)\leq n-h+i-\ell< n-h \]
for all $h=e,\ldots ,n$ (because $i<b\leq r$).
\end{proof}

\begin{remark}
In Proposition \ref{algSl}, the equidimensionality of $R$ is needed only in the case `` $r=1$". Indeed, if $R$ satisfies $S_r$ with $r\geq 1$ then it has no embedded prime, and if $R$ satisfies $S_r$ with $r\geq 2$ then it is equidimensional by \cite[Remark 2.4.1]{Ha}.

On the other hand, an argument similar to  the one used to prove Proposition \ref{algSl} shows that  $\dim R/{\mathfrak{p}}}\def\mm{{\mathfrak{m}}=\dim R$ for all associated prime ideals if and only if $\dim \operatorname{Ext}_S^{n-i}(R,S) <i \ \ \ \forall \ i<\dim R$.
\end{remark}

We conclude this section with the following remark that shows that we cannot hope to answering ``always!'' to Questions \ref{mainques}.

\begin{remark}\label{rem:negh2}
Let $S={\Bbbk}[x_i,y_i: i=1,\ldots ,n]$ and $I\subseteq S$ the following ideal:
\[I=(x_1,\ldots ,x_n)^2+(x_1y_1+\ldots +x_ny_n).\]
Setting $R=S/I$, we claim the following properties:
\begin{itemize}
\item[(i)] $R$ is an $n$-dimensional ring satisfying $(S_{n-1})$;
\item[(ii)] $\reg R=1$;
\item[(iii)] $R$ is not Cohen-Macaulay;
\item[(iv)] $H_R(t)= \displaystyle \frac{1+nt-t^2}{(1-t)^n}$;
\item[(v)] $H_{\mathfrak{m}}^{n-1}(R)=H_{\mathfrak{m}}^{n-1}(R)_{2-n}={\Bbbk}$, thus $R$ is Buchsbaum.
\end{itemize}
For it, let us notice that, for any term order on $S$, the set $\{x_ix_j:1\leq i \leq j\leq n\}\cup \{x_1y_1+\ldots +x_ny_n\}$ is a Gr\"obner basis by Buchberger's criterion. So we can choose a term order such that the initial ideal of $I$ is
\[J=(x_1,\ldots ,x_n)^2+(x_1y_1).\]
So $\depth R\geq \depth S/J=n-1$ and $H_R(t)=H_{S/J}(t)=H_{P/J\cap P}(t)/(1-t)^{n-1}$, where $P={\Bbbk}[x_1,\ldots ,x_n,y_1]$. Note that $\dim_{{\Bbbk}} (P/J\cap P)_0=1$, $\dim_{{\Bbbk}} (P/J\cap P)_1=n+1$ and $\dim_k (P/J\cap P)_d=n$ for any $d>1$. So $H_{P/J\cap P}(t)=(1+nt-t^2)/(1-t)$, yielding
\[H_R(t)= \frac{1+nt-t^2}{(1-t)^n}.\]
In particular, $R$ is not Cohen-Macaulay, therefore $\depth R=n-1$. We have $1\leq \reg R\leq \reg S/J$. Since it is a monomial ideal, it is simple to check that $J$ has linear quotients, so $\reg S/J=1$.
Finally, take a prime ideal ${\mathfrak{p}}}\def\mm{{\mathfrak{m}}\in{\mathrm{Proj\; }} S$ containing $I$ and consider the ring $S_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}}/IS_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}}$. Notice that there exists $i$ such that $y_i\notin {\mathfrak{p}}}\def\mm{{\mathfrak{m}}$, so
\[IS_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}}=(x_1,\ldots ,x_n)^2+(x_i+1/y_i\sum_{j\neq i}x_jy_j). \ \]
Therefore, by denoting $S'={\Bbbk}[x_j,y_k:j\neq i]$ and ${\mathfrak{p}}}\def\mm{{\mathfrak{m}}'={\mathfrak{p}}}\def\mm{{\mathfrak{m}}\cap S'$, we get
\[\frac{S_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}}}{IS_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}}}\cong \frac{S'_{{\mathfrak{p}}}\def\mm{{\mathfrak{m}}'}}{(x_j, \ :j\neq i)^2},\] that is certainly Cohen-Macaulay.

For the last point, notice that we have a short exact sequence:
$$
0\to \frac{S}{(x_1,\dots,x_n)}(-2) \xrightarrow{\cdot (x_1y_1+\ldots +x_ny_n)} \frac{S}{(x_1,\ldots ,x_n)^2} \to \frac{S}{I}\to 0
$$
which induces
$$
0\to H_{\mathfrak{m}}^{n-1}(S/I)\to H_{\mathfrak{m}}^{n}\left(\frac{S}{(x_1,\dots,x_n)}(-2)\right) \xrightarrow{\cdot (x_1y_1+\ldots +x_ny_n)} H_{\mathfrak{m}}^{n}\left(\frac{S}{(x_1,\ldots ,x_n)^2}\right)
$$
So $H_{\mathfrak{m}}^{n-1}(S/I)$ is the kernel, but it is easy to compute that, up to scalar, the only element in the kernel is the element $\frac{1}{y_1\cdots y_n}$, which has degree $2-n$ in $H_{\mathfrak{m}}^{n}(\frac{S}{(x_1,\dots,x_n)}(-2))$. Thus we are done.
\end{remark}

\section{Regularity bounds in characteristic 0}\label{main}

In this section we establish the key vanishing result in the characteristic $0$ that is needed in Theorem \ref{main0} stated in the introduction. As explained in Section \ref{prelim}, we need to establish condition $(MT_r)$ (Definition \ref{defMT}) for nice singularities satisfying Serre's condition $(S_r)$. The following gives a bound on regularity of $\operatorname{Ext}_S^{i}(R, \omega_S)$ which is crucial. We begin by recalling the definition of Du Bois singularities.

Suppose that $X$ is a reduced scheme essentially of finite type over a field of characteristic $0$.  Associated to $X$ is an object $\underline{\Omega}_X^0 \in D^b(X)$ with a map $O_X \to \underline{\Omega}_X^0$. Following \cite[2.1]{Sch}, if $X\subseteq Y$ is an embedding with $Y$ smooth (which is always the case if $X$ is affine or projective), then $\underline{\Omega}_X^0 \cong R\pi_*O_E$ where $E$ is the reduced pre-image of $X$ in a strong log resolution of the pair $(Y, X)$.\footnote{This means $\pi$ is a log resolution of $(Y, X)$ that is an isomorphism outside of $X$. We note that this is not the original definition of $\underline{\Omega}_X^0$ but it is equivalent by the main result of \cite{Sch}.} $X$ is Du Bois if $O_X\to \underline{\Omega}_X^0$ is an isomorphism.

\begin{proposition}\label{p:koda}
If $X = {\mathrm{Proj\; }} R$ is Du Bois, then  $$H_\m^j(\operatorname{Ext}_S^{n-i}(R, \omega_S))_{>0}=0$$ for all $i, j$. In particular,
\[\reg \operatorname{Ext}_S^{i}(R, \omega_S)\leq \dim \operatorname{Ext}_S^{i}(R, \omega_S) \ \ \ \forall \ i.\]
\end{proposition}

To this purpose, we will use $\omega_R^\bullet$ to denote the normalized dualizing complex of $R$, thus $h^{-i}(\omega_R^\bullet)\cong \operatorname{Ext}_S^{n-i}(R, \omega_S)$. We use $\underline{\omega}_R^\bullet$ to denote $R{\mathrm{Hom}}} \def\Ker{{\mathrm{Ker\;}}} \def\hight{{\mathrm{ht\; }}_R(\underline{\Omega}_R^0, \omega_R^\bullet)$. Similarly we have $\omega_X^\bullet$, $\underline{\omega}_X^\bullet$ for $X$. Note that when $R$ (resp. $X$) is Du Bois, we have $\underline{\omega}_R^\bullet={\omega}_R^\bullet$ (resp. $\underline{\omega}_X^\bullet=\omega_X^\bullet$). We begin by recalling some vanishing/injectivity result that we will need:

\begin{theorem}[Lemma 3.3 in \cite{KS}]
\label{theorem--key injectivity}
$h^{-i}(\underline{\omega}_R^\bullet)\hookrightarrow h^{-i}({\omega}_R^\bullet)$ is injective for all $i$.
\end{theorem}

\begin{theorem}[Lemma 3.3 in \cite{KSS} and Theorem 3.2 in \cite{Am}]
\label{theorem--Ambro's vanishing}
$h^{-i}(\underline{\omega}_X^\bullet)$ satisfies the Kodaira vanishing: $H^j(X, h^{-i}(\underline{\omega}_X^\bullet)\otimes L^m)=0$ for all $i$, $j\geq 1$, $m\geq 1$ and $L$ ample line bundle on $X$.
\end{theorem}

\begin{theorem}[Proposition 4.4 and Theorem 4.5 in \cite{MSS}]
\label{theorem--MSS injectivity}
Suppose $X$ is Du Bois, then:
\begin{enumerate}
\item $h^j(\underline{\Omega}_R^0)\cong H_\m^{j+1}(R)_{>0}$ for all $j\geq 1$, and $h^0(\underline{\Omega}_R^0)/R\cong H_\m^1(R)_{>0}$.
\item We have a degree-preserving injection $\operatorname{Ext}_S^{n-i}(R, \omega_S)_{\geq 0} \hookrightarrow H_I^{n-i}(\omega_S)_{\geq 0}$ for all $i$.
\end{enumerate}
\end{theorem}

We are now ready to prove Proposition \ref{p:koda}:

\begin{proof}
First we note that $h^{-i}(\omega_R^\bullet)\cong \operatorname{Ext}_S^{n-i}(R, \omega_S)$. When $j\geq 2$, we have
$$H_\m^j(h^{-i}({\omega}_R^\bullet))_{m}=H^{j-1}(X, \widetilde{h^{-i}({\omega}_R^\bullet)}\otimes L^m)=H^{j-1}(X, h^{-i+1}(\omega_X^\bullet)\otimes L^m)=0$$
for all $m>0$ by Theorem \ref{theorem--Ambro's vanishing} (here $L=O_X(1)$ is very ample, and since $X$ has Du Bois singularities $\omega_X^\bullet=\underline{\omega}_X^\bullet$). Thus $H_\m^j(h^{-i}({\omega}_R^\bullet))_{>0}=0$ for all $i$ and $j\geq 2$.

We next consider the case $j=0$. We note that by Theorem \ref{theorem--MSS injectivity} we have $$H_\m^0(\operatorname{Ext}_S^{n-i}(R, \omega_S))_{>0}\hookrightarrow\operatorname{Ext}_S^{n-i}(R, \omega_S)_{>0}\hookrightarrow H_I^{n-i}(\omega_S)_{> 0}\hookrightarrow H_I^{n-i}(\omega_S).$$
Because $H_\m^0(\operatorname{Ext}_S^{n-i}(R, \omega_S))_{>0}$ is clearly $\m$-torsion, the above injection factors through $H_\m^0H_I^{n-i}(\omega_S)$. Thus we have an induced injection $$H_\m^0(\operatorname{Ext}_S^{n-i}(R, \omega_S))_{>0}\hookrightarrow H_\m^0H_I^{n-i}(\omega_S)_{>0}.$$
Now we note that $H_\m^0H_I^{n-i}(\omega_S)_{>0}\cong H_\m^0H_I^{n-i}(S)_{>-n}$. But by the main result of \cite{MZ}, we have $H_\m^0H_I^{n-i}(S)_{>-n}=0$ and thus the injection above implies $$H_\m^0(h^{-i}({\omega}_R^\bullet))_{>0}=H_\m^0(\operatorname{Ext}_S^{n-i}(R, \omega_S))_{>0}=0$$ for all $i$.

It remains to prove the case $j=1$. By Theorem \ref{theorem--key injectivity} we have
\begin{equation}
0\to h^{-i}(\underline{\omega}_R^\bullet)\to h^{-i}({\omega}_R^\bullet) \to C\to 0
\end{equation}
such that $\Supp C=\{\m\}$ because $X$ is Du Bois. It follows from this sequence that $H_\m^0(h^{-i}(\underline{\omega}_R^\bullet))_{>0}=0$ for all $i$ since it injects into $H_\m^0(h^{-i}({\omega}_R^\bullet))_{>0}=0$.

At this point, we note that the long exact sequence of local cohomology of (1) implies that $H_\m^j(h^{-i}(\underline{\omega}_R^\bullet))\cong H_\m^j(h^{-i}({\omega}_R^\bullet))=0$ for all $j\geq 2$ and that $H_\m^1(h^{-i}(\underline{\omega}_R^\bullet))\twoheadrightarrow H_\m^1(h^{-i}({\omega}_R^\bullet))$. Thus it suffices to show $H_\m^1(h^{-i}(\underline{\omega}_R^\bullet))_{>0}=0$. Since $\omega_R^\bullet$ is the normalized dualizing complex, by the definition of $\underline{\omega}_R^\bullet$, we have
$$\underline{\Omega}_R^0\cong R{\mathrm{Hom}}} \def\Ker{{\mathrm{Ker\;}}} \def\hight{{\mathrm{ht\; }}_R(\underline{\omega}_R^\bullet, \omega_R^\bullet).$$ Thus we have a spectral sequence $$E_2^{ij}=h^{-j}(R{\mathrm{Hom}}} \def\Ker{{\mathrm{Ker\;}}} \def\hight{{\mathrm{ht\; }}_R(h^{-i}(\underline{\omega}_R^\bullet), \omega_R^\bullet))\Rightarrow h^{i-j}(\underline{\Omega}_R^0).$$
By graded local duality, $(E_2^{ij})^\vee\cong H_\m^j(h^{-i}(\underline{\omega}_R^\bullet))$, and we already know the latter one vanishes in degree $>0$ for all $i$ and $j\neq 1$, thus $(E_2^{ij})_{<0}=0$ for all $i$ and $j\neq 1$. Therefore the spectral sequence in degree $<0$ degenerates at $E_2$-page. It follows that $$h^{-1}(R{\mathrm{Hom}}} \def\Ker{{\mathrm{Ker\;}}} \def\hight{{\mathrm{ht\; }}_R(h^{-i}(\underline{\omega}_R^\bullet), \omega_R^\bullet))_{<0}=(E_2^{i1})_{<0}\cong h^{i-1}(\underline{\Omega}_R^0)_{<0} .$$ By Theorem \ref{theorem--MSS injectivity}, $h^{i-1}(\underline{\Omega}_R^0)_{<0}=0$ for all $i$. This implies $$h^{-1}(R{\mathrm{Hom}}} \def\Ker{{\mathrm{Ker\;}}} \def\hight{{\mathrm{ht\; }}_R(h^{-i}(\underline{\omega}_R^\bullet), \omega_R^\bullet))_{<0}=0$$ and thus by graded local duality $H_\m^1(h^{-i}(\underline{\omega}_R^\bullet))_{>0}=0$.
\end{proof}

\section{Regularity bounds in positive characteristic}\label{mainp}

The purpose of this section is to prove  the analog of Proposition \ref{p:koda} in characteristic $p>0$. The natural analogous assumption would be to assume $X={\mathrm{Proj\; }} R$ is $F$-injective. Albeit we do not know any counterexample to the thesis of Theorem \ref{main0} under the assumption that $X$ is $F$-injective, we know that Proposition \ref{p:koda} is not true in positive characteristic (as Kodaira vanishing fails even if $X$ is smooth). As usual for this business, we can settle the problem  assuming that $R$ itself is $F$-pure.

\begin{proposition}\label{sp}
If $\chara({\Bbbk})=p>0$ and $R$ is $F$-pure, then
\[H_{\mm}^j(\operatorname{Ext}_S^i(R,\omega_S))_{>0}=0 \ \ \ \forall \ i,j\]
In particular, $\reg(\operatorname{Ext}_S^{i}(R,\omega_S))\leq \dim \operatorname{Ext}_S^{i}(R,\omega_S)$ for all $i$.
\end{proposition}
\begin{proof}
Let $q=p^t$ for some $t\in\N$. Given an $R$-module $M$, we denote by $^q \!M$ the $R$-module that is $M$ as an abelian group and has $R$-action given by $r\cdot m=r^{q}m$. The statement allows to enlarge the field, thus we can assume that ${\Bbbk}$ is perfect.
Under the assumptions, by \cite[Corollary 5.3]{HR} the $t$th iterated Frobenius power $F^t:R \ \rightarrow \ ^q \!R$ \ splits as a map of $R$-modules, so for any $j=0,\ldots ,n$ the induced map on local cohomology
\[H_{\mm}^j(R)\xrightarrow{H_{\mm}^j(F^t)} H_{\mm}^j(^q \! R)\]
is an injective splitting of $R$-modules as well. Notice that
a degree $s$ element in $H_{\mm}^j(R)$ is mapped by $H_{\mm}^j(F^t)$ to a degree $sq$ element of $H_{\mm}^j(^q \! R)$.
By abuse of notation, from now on we will write $F^t$ for $H_{\mm}^j(F^t)$. Let us apply to such map the functor $-^{\vee}={\mathrm{Hom}}} \def\Ker{{\mathrm{Ker\;}}} \def\hight{{\mathrm{ht\; }}_{{\Bbbk}}(-,{\Bbbk})$, obtaining the surjective splitting of $R$-modules:
\[H_{\mm}^j(R)^{\vee}\xleftarrow{(F^t)^{\vee}} H_{\mm}^j(^q \!R)^{\vee}.\]
Furthermore, notice that if $\eta^*$ is a degree $u$ element in $H_{\mm}^j(^q \!R)^{\vee}$, then $(F^t)^{\vee}(\eta^*)=\eta^*\circ F^t$ is 0 if $q$ does not divide $u$, while it has degree $s$ if $u=qs$. By graded Grothendieck's duality, we therefore have a surjective splitting of $R$-modules
\[\operatorname{Ext}_S^{n-j}(R,\omega_S)\xleftarrow{(F^t)^{\vee}}\operatorname{Ext}_S^{n-j}(^q \!R,\omega_S)\cong \ ^q \!\operatorname{Ext}_S^{n-j}(R,\omega_S)\]
which ``divides" the degrees by $q$. By applying the local cohomology functor $H_{\mm}^k(-)$ to the above splitting, we get a surjective map of graded $R$-modules
\[H_{\mm}^k(\operatorname{Ext}_S^{n-j}(R,\omega_S))\xleftarrow{H_{\mm}^k((F^t)^{\vee})}H_{\mm}^k(^q\operatorname{Ext}_S^{n-j}(R,\omega_S))\cong \ ^q \!H_{\mm}^k(\operatorname{Ext}_S^{n-j}(R,\omega_S))\]
dividing the degrees by $q$.
Hence, because $q$ can be chosen arbitrarily high, the graded surjection of $R$-modules $H_{\mm}^k((F^t)^{\vee})$ yields that $H_{\mm}^k((\operatorname{Ext}_S^{n-j}(R,\omega_S)))_s$ is actually 0 whenever $s>0$.
\end{proof}

\section{Applications and examples}\label{apps}
In this section, we first put together our technical results to prove main Theorem \ref{main0}.
\begin{proof}[Proof of Theorem \ref{main0}]
Assume $R$ satisfies $(S_r)$ and one of the assumptions: $X={\mathrm{Proj\; }} R$ is Du Bois (characteristic $0$) or $R$ is $F$-pure (characteristic $p$). We can extend ${\Bbbk}$ if necessary and hence assume it is algebraically closed. By Theorem \ref{mtr3} and Proposition \ref{mtr4}, all the conclusions  follow if we can establish that $R$ is $(MT_r)$. But  Propositions \ref{algSl} tells us that $\dim  \operatorname{Ext}_S^{n-i}(R,\omega_S) \leq i-r $, and Propositions \ref{p:koda} and \ref{sp} assert that  $\reg  \operatorname{Ext}_S^{n-i}(R,\omega_S) \leq \dim \operatorname{Ext}_S^{n-i}(R, \omega_S)$, which gives precisely what we want.
\end{proof}

The $h$-vector inequalities can be translated to information about the Hilbert functions. Here is the precise statement. Recall that $R=S/I$, $e= \operatorname{ht} I, d=\dim R$. Let $s_i =  \dim_k I_i$.
\begin{proposition}
For each $l\geq 0$, $$h_l = {e+l-1 \choose l} - \sum_{j=0}^l (-1)^js_{l-j}{d \choose j}$$
\end{proposition}

\begin{proof}
The Hilbert series of $R$ can be written as:
$$H_R(t)= \sum r_it^i = \frac{h_0+h_1t+ \cdots + h_lt^l}{(1-t)^d}  \cdots (*) $$
where $r_i=\dim_{{\Bbbk}}R_i$. By comparing coefficients in $(*)$ we always have:
$$h_l =  \sum_{j=0}^l (-1)^jr_{l-j}{d \choose j} $$
Since $s_i= {n+i-1 \choose i} - r_i$, what we need to prove amount to:
$$\sum_{j=0}^l (-1)^j{n+l-j-1 \choose l-j}{d \choose j} =   {e+l-1 \choose l}$$
But we have $$\sum_{i\geq 0} {n+i-1 \choose i}t^i = \frac{1}{(1-t)^n}$$
Since $n=d+e$, we have
$$\left(\sum_{i\geq 0} {n+i-1 \choose i}t^i\right)(1-t)^d = \frac{1}{(1-t)^e}$$
The $l$-th coefficient on both sides give exactly the equality we seek.
\end{proof}

From the equality above, the following is an immediate consequence of our main Theorem \ref{main0}.

\begin{corollary}\label{cor1}
Let $R={\Bbbk}[x_1,\dots, x_n]/I$ be a standard graded algebra over a field ${\Bbbk}$ with $d=\dim R$ and $e=\operatorname{ht} I$. Let $s_i= \dim_k I_i$. Assume $R$ satisfies $(S_r)$ and either:
\begin{enumerate}

\item $\chara({\Bbbk})=0$ and $X={\mathrm{Proj\; }} R$ is Du Bois.
\item $\chara({\Bbbk})=p>0$ and $R$ is $F$-pure.
\end{enumerate}
Then for each $l\leq r$, we have:
$$\sum_{j=0}^l (-1)^js_{l-j}{d \choose j}\leq {e+l-1 \choose l}$$
In particular, if $I$ contains no elements of degree less than $l$, then $s_l \leq {e+l-1 \choose l}$. Moreover, if equality happens for any such $l$, then $R$ is Cohen-Macaulay.
\end{corollary}

\begin{remark}
In \cite{KM}, it is established that for a monomial ideal $I$ (not necessarily square-free), the deficiency modules of $R=S/I$ satisfies the same regularity bounds as in Propositions \ref{p:koda} and \ref{sp}, so our main results above applied for this situation as well.
\end{remark}

The next corollary fits the well-known but mysterious theme that ``nice singularities of small codimension" should be Cohen-Macaulay.

\begin{corollary}\label{cor2}
Let $R={\Bbbk}[x_1,\dots, x_n]/I$ be a standard graded algebra over a field ${\Bbbk}$ of characteristic $0$ with $e=\operatorname{ht} I$ and $d=\dim R$. Let $d_1 \geq d_2 \geq \dots $ be the degree sequence of a minimal set of generators for $I$. Assume that $R$ is unmixed, equidimensional, Cohen-Macaulay in codimension $l$ and $X={\mathrm{Proj\; }} R$ has only MJ-log canonical singularities.\footnote{We refer to \cite{EI} for detailed definition and properties of MJ-log canonical singularities.} If $e+l \geq d_1+\dots +d_e$, then $R$ is Cohen-Macaulay.
\end{corollary}
\begin{proof}
By \cite[Proposition 2.7]{Niu} (see also \cite{Ish}), $X$ is MJ-log canonical is equivalent to saying that the pair $(\mathbb{P}^{n-1}, \widetilde{I}^e)$ is log canonical. Let $r= d_1+\dots + d_e$. Now applying \cite[Corollary 5.1]{dFE}, we have
$$H_\m^i(I)_j=0 \text{ for all $i>1$ and $j\geq r-n+1$. }$$
Since we have $0\to I\to S\to R\to 0$, the long exact sequence of local cohomology tells us that $H_\m^i(R)_j=0$ for $0<i<d$ and $j\geq r-n+1$. This is  equivalent to saying that  for each such $i$, ${[K_i]}_{\leq n-r-1}=0$ with $K_i = \operatorname{Ext}_S^{n-i}(R,\omega_S)$. However, since MJ-log canonical singularities are Du Bois by \cite[Theorem 7.7]{dFD}, Proposition \ref{p:koda} implies that $\reg K_i\leq \dim K_i \leq d-l-1$. Our assumption says that $d-l-1\leq  n-r-1$, thus $K_i=0$ for all $i<d$ and hence $R$ is Cohen-Macaulay.
\end{proof}

\begin{corollary}\label{cor3}
Let $R={\Bbbk}[x_1,\dots, x_n]/I$ be a standard graded algebra over a field ${\Bbbk}$ of characteristic $0$. Let $e=\operatorname{ht} I$ and $d_1 \geq d_2 \geq \dots $ be the degree sequence of a minimal set of generators for $I$. Assume that ${\mathrm{Proj\; }} R$ is a  smooth variety, $n\geq 2e+3$ and $n> d_1+\dots +d_e$. Then $R$ is Gorenstein. In particular, if $e=2$ then $R$ is a complete intersection.
\end{corollary}

\begin{proof}
By Corollary \ref{cor2}, $R$ is Cohen-Macaulay. Let $X={\mathrm{Proj\; }} R$. The assumption of smoothness and $n\geq 2e+3$ implies that $\operatorname{Pic}(X) = \Z$, generated by $\mathcal O_X(1)$, see \cite[Theorem 11.4]{Lyu}.  It follows that the class group of $R$ is trivial, and therefore the canonical module is free.
\end{proof}

Our methods also give streamlined proof and sometimes strengthen known results. Here is a statement on $h_2$ that can be seen as a modest extension of \cite[Theorem 4.2]{EG}.

\begin{theorem}\label{h2}
Let $R={\Bbbk}[x_1,\dots, x_n]/I$ be a standard graded algebra over an algebraically closed field ${\Bbbk}$ with $e=\operatorname{ht} I$. Suppose that ${\mathrm{Proj\; }} R$ is connected in codimension one and the radical of $I$ contains no linear forms. Then $h_2(R)\geq 0$ and $e(R)\geq 1+ e$. If equality happens in either case, then $R_{\text{red}}$ is Cohen-Macaulay of minimal multiplicity.
\end{theorem}

\begin{proof}
Let $J= \sqrt{I}$ and $T=R_{\text{red}}=S/J$. By assumption, $J, I$ contains no linear forms, so $h_2(R) ={e+1 \choose 2}-\dim_{{\Bbbk}}I_2$ and $h_2(T) ={e+1 \choose 2}-\dim_{{\Bbbk}}J_2$. It follows that $h_2(R)\geq h_2(T)$. One also easily see  that $e(R) \geq e(T)$. So we may replace $R$ by $T$ and assume $R$ is reduced. We may assume now that $d =\dim R\geq 2$.

Let $l$ be a generic linear form, $\bar R= R/lR$ and $R'={\bar R}/N$ with $N=H^0_{\m}(\bar R)$.
Bertini Theorem tells us that one ${\mathrm{Proj\; }} R'$ is still connected in codimension one and reduced. Since $R$ is connected and reduced, $H^1_{\m}(R)_{\leq 0}=0$, so $H_{\m}^0(N)_{\leq 1}=0$. It follows as in the proof of Proposition \ref{mtr2} that $h_i(R')=h_i(R)$ for $i<2$, $h_2(R')\leq h_2(R)$ and $c_2(R)=c_2(R')$. Thus by induction on dimension, we can assume $\dim R'=2$. But here $H_{\mm}^0(\bar R)_{\leq 1}= H_{\mm}^1(\bar R)_{\leq 0} =0$ say precisely that $\bar R$ is $(MT_2)$ (see Proposition \ref{easy}). Our Theorem \ref{mtr3} now applies to show the needed inequalities (note that $e+1= h_0(R) + h_1(R)$).  If any equality happens, then Theorem \ref{mtr4} asserts that $\bar R$ is Cohen-Macaulay (necessarily of regularity $2$). To finish we now claim that if $\dim R\geq 2$ and $R/lR$ is Cohen-Macaulay for a generic $l$, then so is $R$. Let $N =H_\m^0(R)$ and $R' = R/N$. We obtain an exact sequence $0\to N/lN \to R/lR \to R'/lR' \to 0$. But since $R/lR$ is Cohen-Macaulay of dimension at least $1$, $N/lN=0$, which forces $N=0$, thus $l$ is regular on $R$ and we are done.
\end{proof}

\begin{eg} It is important that ${\Bbbk}$ is algebraically closed in the previous theorem. Take $R = \R[s,t,is,it]$ which is a domain. Then $R \cong S/I$ with $S =\R[a,b,c,d]/I$ and $I= (a^2+c^2, b^2+d^2, ad-bc, ab+cd)$ and $h_2(R)=-1$. Note that $R\otimes_{\R} \C  \cong \C[x,y,u,v]/(xu,xv,yu,yv)$.
\end{eg}
Next, we give a formula relating the $h$-vectors of the $R$ and the (deficiency) $\operatorname{Ext}$ modules. Recall the notations at the beginning of section \ref{prelim}.

\begin{proposition}\label{abs}
Let $R=S/I$ be a standard graded algebra of dimension $d$. For $i=0,\ldots ,d$ set $K_i=\operatorname{Ext}_S^{n-i}(R,\omega_S)$. Then we have the following relations between the numerators of Hilbert series:
$$p_R(1/t)t^d = \sum_{i=0}^d(-1)^{d-i}p_{K_i}(t)(1-t)^{d-\dim K_i} $$

\end{proposition}

\begin{proof}
Applying the main Theorem of \cite{ABS} with $M=R, N=\omega_S = S(-n)$, we obtain an equality of rational functions:
$$\sum_i (-1)^{n-i} H_{K_i}(t) = \frac{H_R(1/t)H_{\omega_S}(t)}{H_S(1/t)} $$
Since $H_{K_i}(t)= \frac{p_{K_i}(t)}{(1-t)^{\dim K_i}}, H_{\omega_S}(t) = \frac{t^n}{(1-t)^n}, H_S(1/t) = \frac{t^n}{(t-1)^n}$, the assertion follows.
\end{proof}

\begin{corollary}
\label{cor4}
Retain the notations of Proposition \ref{abs}. Suppose that for $i<d$, we have $H_{m}^i(R)_{>0} =0$ (for example, this holds if $R$ is $F$-pure or if $R$ is Du Bois \cite[Theorem 4.4]{Ma}). Then  $$h_d(R) = \sum_{0\leq i \leq d}(-1)^{d-i}\dim_{{\Bbbk}} H^i_{\m}(R)_0.$$
\end{corollary}

\begin{proof}
By (graded) local duality, our assumption implies that for each $i<d$, $[K_i]_{<0}=0$, so the polynomial $p_{K_i}(t)$ contains no negative powers of $t$. What we need to prove follows immediately from Proposition \ref{abs}.
\end{proof}

Next we point out that in certain situations, the negativity of  $h_d(R)$ is guaranteed if $R$ is not Cohen-Macaulay. This class of examples shows that one can not hope to improve of our main theorem too  much.

\begin{corollary}\label{corh_d}
Let $R={\Bbbk}[x_1,\dots, x_n]/I$ be a standard graded algebra with $\dim R=d$. Suppose that $\depth R=d-1 = \reg (R)=d-1$ and $H_{\m}^d(R)_{-1}=0$. Then $h_d(R)<0$.
\end{corollary}
\begin{proof}
Since $\depth R=d-1$, $\reg R=d-1$ is equivalent to the condition $H_{\m}^{d-1}(R)_{>0}=H_{\m}^d(R)_{\geq 0} =0$. Since $H_\m^d(R)_{-1}=0$ and $\reg R=d-1$, we have $H^{d-1}_{\m}(R)_0\neq 0$. Now by Corollary \ref{cor4}, $h_d(R) = - \dim_{{\Bbbk}} H^{d-1}_{\m}(R)_0<0$.
\end{proof}

\begin{eg}
Let $A$ be a standard graded Cohen-Macaulay  ${\Bbbk}$-algebra of dimension at least $2$ with zero $a$-invariant and $B={\Bbbk}[x,y]$. Then the Segre product $R= A\sharp B$ satisfies the assumptions of Corollary \ref{corh_d}. It is even $(S_{d-1})$.
\end{eg}
\section{Some further results and open questions}\label{speculate}

Our results and examples suggest the following natural question, which can be viewed as a vast generalization of everything discussed so far:

\begin{question}\label{openques}
Assume that $\chara({\Bbbk})=0$, $R$ satisfies $(S_r)$ and is Du Bois in codimension $r-2$. If $(h_0,\ldots ,h_s)$ is the  $h$-vector of $R$, is it true that $h_i\geq 0$ whenever $i=0,\ldots ,r$? If furthermore $R$ has Castelnuovo-Mumford regularity less than $r$, is it true that $R$ is Cohen-Macaulay?
\end{question}

\begin{remark}
Question \ref{openques} is true if $r=2$: in fact we can assume ${\Bbbk}=\bar {\Bbbk}$ and that $I$ contains no linear forms. The condition that $R$ is Du Bois in codimension 0, together with $(S_1)$, says that $I=\sqrt{I}$. Further, the condition that $R$ satisfies $(S_2)$ yields that ${\mathrm{Proj\; }} R$ is connected in codimension 1 by \cite[Proposition 2.1, Theorem 2.2]{Ha}. So we can apply Theorem \ref{h2}, in which we proved that $R/lR$ is ($MT_2$) for a generic linear form $l$. We can now conclude like in that proof.
\end{remark}

The first step to answer \ref{openques} would be to relax the condition $X ={\mathrm{Proj\; }} R$ is Du Bois  (i.e., $R$ is Du Bois in codimension $\leq d-1$) in key Proposition \ref{p:koda} to $R$ being Du Bois in lower codimension. This amounts to strong Kodaira-vanishing type statements.   At this point we can prove the following, which is a generalization of \cite[Theorem 3.15]{HuSm}.

\begin{proposition}
Let $0<k\leq d-1$. Suppose $X$ is $S_{d-k}$ and is Du Bois in codimension $d-k-1$. Then we have $H_\m^{j}(R)_{<0}=0$ for all $j\leq d-k$, or equivalently, $\operatorname{Ext}_S^{n-j}(R,\omega_S)_{>0}=0$ for all $j\leq d-k$.
\end{proposition}
\begin{proof}
If $k=d-1$, we know $X$ is $S_1$ and generically reduced and thus $X$ is reduced. Then $H_\m^1(R)_{<0}$ vanishes. For the case $k<d-1$, notice that it is enough to show $H_\m^{d-k}(R)_{<0}=0$ (because the assumption remains valid for any $k'\geq k$). Hence it suffices to prove $H^{d-k-1}(X, L^{-m})=0$ for all $m\geq 1$ (here we used $k<d-1$). Since $X$ is $S_{d-k}$, the possible non-vanishing cohomology of $\omega_X^\bullet$ are:
$$h^{-(d-1)}(\omega_X^\bullet), h^{-(d-1)+1}(\omega_X^\bullet),\dots, h^{-(d-1)+(k-1)}(\omega_X^\bullet),$$ and we have $$\dim h^{-(d-1)+1}(\omega_X^\bullet)\leq k-2, \dim h^{-(d-1)+2}(\omega_X^\bullet)\leq k-3, \dots, \dim h^{-(d-1)+(k-1)}(\omega_X^\bullet)\leq 0.$$
By Serre duality, $H^{d-k-1}(X, L^{-m})$ is dual to $\mathbb{H}^{-d+k+1}(X, \omega_X^\bullet\otimes L^m)$. Using spectral sequence, the possible $E_2$-page contributions to $\mathbb{H}^{-d+k+1}(X, \omega_X^\bullet\otimes L^m)$ are: $$H^k(X, h^{-(d-1)}(\omega_X^\bullet)\otimes L^m), H^{k-1}(X, h^{-(d-1)+1}(\omega_X^\bullet)\otimes L^m),\dots, H^1(X, h^{-(d-1)+(k-1)}(\omega_X^\bullet)\otimes L^m).$$ The latter ones $$H^{k-1}(X, h^{-(d-1)+1}(\omega_X^\bullet)\otimes L^m),\dots, H^1(X, h^{-(d-1)+(k-1)}(\omega_X^\bullet)\otimes L^m)$$ all vanish because dimension reasons (for example, $H^{k-1}(X, h^{-(d-1)+1}(\omega_X^\bullet)\otimes L^m)=0$ since $\dim h^{-(d-1)+1}(\omega_X^\bullet)\leq k-2$).

For $H^k(X, h^{-(d-1)}(\omega_X^\bullet)\otimes L^m)$, note that by Theorem \ref{theorem--key injectivity} we have a short exact sequence $$0\to h^{-(d-1)}(\underline{\omega}_X^\bullet)\to h^{-(d-1)}(\omega_X^\bullet)\to C\to 0$$ such that $\dim \Supp C\leq k-1$ since $X$ is Du Bois in codimension $(d-1)-k$. The induced long exact sequence of sheaf cohomology: $$0=H^k(X, h^{-(d-1)}(\underline{\omega}_X^\bullet)\otimes L^m)\to H^k(X, h^{-(d-1)}(\omega_X^\bullet)\otimes L^m)\to H^k(X, C\otimes L^m)=0,$$ where the first $0$ is by Theorem \ref{theorem--Ambro's vanishing}, shows that $H^k(X, h^{-(d-1)}(\omega_X^\bullet)\otimes L^m)=0$ for all $m\geq 1$. Hence all $E_2$-page contributions to $\mathbb{H}^{-d+k+1}(X, \omega_X^\bullet\otimes L^m)$ are $0$, it follows that $\mathbb{H}^{-d+k+1}(X, \omega_X^\bullet\otimes L^m)=0$ for all $m\geq 1$ and thus $H^{d-k-1}(X, L^{-m})=0$ for all $m\geq 1$.
\end{proof}

\begin{comment}

If $k$ is a field of characteristic $0$, we say that $R$ of \emph{$F$-pure type} if there exists a Noetherian domain $A$ and a standard graded $A$-algebra $R_A$ such that:
\begin{itemize}
\item $R_A\otimes_{A}k \cong R$.
\item For any $a\in A$, there is a field $\kappa$ of positive characteristic and a ring homomorphism $A\xrightarrow{\phi} \kappa$ such that $\phi(a)\neq 0$ and $R_A\otimes_A\kappa$ is $F$-pure.
\end{itemize}

\begin{lemma}
Let $M$ be a finitely generated graded $S_A$-modules, where $A$ is a Noetherian domain of characteristic 0 and $S_A=A[x_1,\ldots ,x_n]$. Then, for any $m\in{\mathbb{Z}}$, there exists $0\neq a\in A$ such that $\dim_{\kappa}\operatorname{Ext}_{S\otimes_A\kappa}^i(M\otimes_A\kappa,S_A\otimes_A\kappa)_m$ is constant for all $i\in\mathbb{N}$, whenever $\kappa=A/\mm$ for a maximal ideal $\mm$ not containing $a$.
\end{lemma}

\begin{proof}
We will write $N_{\kappa}$ for $N\otimes_A\kappa$ whenever $N$ is an $S_A$-module. Let $Q$ be the field of fractions of $A$, and let
\[{\bf F}_{\bullet ,Q}\rightarrow M_Q\rightarrow 0\]
be a free resolution of $M_Q$ as an $S_Q$-module. By multiplying for some nonzero element $a_1\in A$, we can assume that the entries of the matrices in ${\bf F}_{\bullet ,Q}$ are in $S_A$.
To compute $\operatorname{Ext}_{S_Q}^i(M_Q,S_Q)$ we simply dualize the exact sequence $F_{i+1,Q}\xrightarrow{\sigma}F_{i,Q}\xrightarrow{\tau}F_{i-1,Q}$, getting the complex
\[F_{i+1,Q}^*\xleftarrow{\sigma^*}F_{i,Q}^*\xleftarrow{\tau^*}F_{i-1,Q}^*,\]
where $\sigma^*$ and $\tau^*$ are the transposes of $\sigma$ and $\tau$ (so, still entries in $S_A$) and compute the homology. For a fixed $m\in{\mathbb{Z}}$, the dimension of the $Q$-vector space $\operatorname{Ext}_{S_Q}^i(M_Q,S_Q)_m$ is, thus, just the rank of $\alpha$ minus the rank of $\beta$, where
\[(F_{i+1,Q}^*)_{m}\xleftarrow{\alpha}(F_{i,Q}^*)_{m}\xleftarrow{\beta}(F_{i-1,Q}^*)_{m}\]
is the piece of degree $m$ of the above complex and $\alpha$ and $\beta$ are the matrices between finite $Q$-vector spaces induced by $\sigma^*$ and $\tau^*$. In particular, they have entries in $A$, and there exist $a_2\in A$ such that their ranks stay the same in any field $\kappa$ such that $a_2$ is nonzero in $\kappa$. We conclude by letting $a=a_1a_2$.
\end{proof}

\begin{corollary}\label{sptype}
If $k$ has characteristic 0 and $R$ is of $F$-split type, then \[H_{\mm}^j(\operatorname{Ext}_S^i(R,\omega_S))_s=0 \ \ \ \forall \ s>0, \ i,j=0,\ldots ,n.\]
In particular, $\reg(\operatorname{Ext}_S^{i}(R,\omega_S))\leq \dim \operatorname{Ext}_S^{i}(R,\omega_S)$ for all $i=0,\ldots ,n$.
\end{corollary}

We are now ready to give the proof of \ref{main}.

\begin{proof} (of Theorem \ref{main}). Part $(1)$:
We will use the result of Murai and Terai, namely that it is enough to show that
\[\reg(\operatorname{Ext}_S^{n-i}(R,\omega_S))\leq i-\ell \ \ \ \forall \ i=0,\ldots ,d-1,\]
where $S=k[x_1,\ldots ,x_n]$. Because $X={\mathrm{Proj\; }} R$ is Cohen-Macaulay, the local cohomology modules $H_{\mathfrak{m}}^i(R)$ have finite length for all $i=0,\ldots ,d-1$. Being Matlis' dual of $H_{\mathfrak{m}}^i(R)$, so $\operatorname{Ext}_S^{n-i}(R,\omega_S)$ has finite length for any $i=0,\ldots ,d-1$. So
\[\reg(\operatorname{Ext}_S^{n-i}(R,\omega_S))=\max\{m\in\mathbb{Z}:\operatorname{Ext}_S^{n-i}(R,\omega_S)_m\neq 0\}.\]
Because $\depth R\geq \ell$,  $\operatorname{Ext}_S^{n-i}(R,\omega_S)=0$ for each $i<\ell$. For $i=\ell, \ldots ,d-1$, by Kodaira vanishing (which holds as long as $X$ has Du Bois sungularities \cite{KSS})
\[\operatorname{Ext}_S^{n-i}(R,\omega_S)_m=H_{\mathfrak{m}}^i(R)_{-m}\neq 0 \implies m\leq 0\leq i-\ell.\]
Part $(2)$: it follows from \ref{sp} and \ref{sptype}.

\end{proof}

The above proof has an unexpected consequence concerning an interesting relation between singularities and the shape of the generic initial ideal:

\begin{corollary}\label{cor1}
If $X={\mathrm{Proj\; }} R$ is smooth, $k$ has characteristic 0 and $\depth R=\ell$, then the minimal generators of degrees $\leq \ell$ of the generic initial ideal $\mathrm{gin}(I)$ with respect to the deg-revlex term order are monomials in $k[x_1,\ldots ,x_{n-d}]$.
\end{corollary}

This has a two further nice consequences:

\begin{corollary}\label{cor2}
If $X={\mathrm{Proj\; }} R$ is smooth, $k$ has characteristic 0 and $\depth R=\ell$, then $\beta_{i,i+j}(R)=0$ for all $i>n-d$ and $j<\ell$. In particular, if $\mathrm{reg}(R)<\ell$, then $R$ is Cohen-Macaulay.
\end{corollary}
\begin{proof}
This follows immediately by Corollary \ref{cor1} and the Elihaou-Kervaire formula.
\end{proof}

\begin{corollary}\label{cor3}
Assume that ${\mathrm{Proj\; }} R$ is smooth of codimension $e$, $I$ is generated in degrees $\leq r$ and $k$ has characteristic 0. If $\depth R\geq e(r-1)$, then $R$ is Cohen-Macaulay.
\end{corollary}

\begin{proof}
We can certainly assume that $I$ is not a complete intersection. By the main result in \cite{CU}, then, $\mathrm{reg}(R)<e(r-1)$, so the conclusion follows by Corollary \ref{cor2}.
\end{proof}

\section{Segre product of CM rings}
In this section we check Question \ref{mainques} above for Segre products. They are natural source for rings which are $S_l$ but not CM.  From the K\"unneth formula for local cohomology it is easy to see that if $A,B$ are CM graded ring of dimension $l\leq l'$ respectively, then $A\sharp B$ is $S_l$ but not $S_{l+1}$. However in this situation the $h$-coefficients are still non-negative due to  Theorem 2.4 of \cite{MN}  (\tv{CHECK!}).

\bigskip

MATTEO: I don't know if it is possible to deduce it by \cite{MN}. However we can prove more in another way: let us say that a standard graded $d$-dimensional algebra $R$ satisfies {\it week Kodaira vanishing} if
\[H_{R_+}^i(R)_m=0 \ \ \ \forall \ m<0, i<d\]
(for example if $R$ is Cohen-Macaulay then it obviously satisfies week Kodaira vanishing).

\begin{proposition}
Let $R_1$ and $R_2$ be standard graded algebras of depth at least 2 satisfying weak Kodaira vanishing. If $R=R_1\sharp R_2$ satisfies $(S_{\ell})$, then
\[h_0,\ldots ,h_{\ell}\geq 0.\]
\end{proposition}
\begin{proof}
The point is that $R$ satisfies the weak Kodaira vanishing by the K\"unneth formula for local cohomology.
\end{proof}

\section{Things to do}

\begin{enumerate}

\item Section 2: the best conjecture perhaps should look some thing like: if $R$ is $S_l$ and is $G_{l'}$, where $G_{l'}$ means good singularity in codimension $l'$ (for example, Du Bois in codimension $l-2$), then the $h_0,...,h_l$ are non-negative. Perhaps one should explore this. Even when we assume $X$ is Du Bois, as in Remark \ref{2.2}.  Perhaps it is best to check the following first: if $R$ is regular in codimension $d-2$ (so $X$ has isolated singularities), do our inequality hold?

\item The first case to consider is when $R$ has dimension $4$, $S_3$, and is normal but does not have isolated singularity.

\item Here is another thought. Start with a smooth projective $X$. Some  positive enough line bundle $L$ gives you a normal embedding. If $H^1(X)=0$, then high enough line bundle embedding gives you $S_3$, etc. So our result should says something about the hilbert function for such line bundle. Would be interesting to write down exactly what it says.

\item Add more details to Corollaries \ref{cor1}, \ref{cor2},\ref{cor3}. Compare them with the recent preprint \url{http://front.math.ucdavis.edu/1507.05459}, especially Section 7.

\item There is another new preprint about secant varieties are Du Bois, \url{http://arxiv.org/abs/1503.01099}. It would be very interesting to see what our results imply about the numerical date of secant varieties. For example, I think the main result say that if you start with a projective smooth $Z$ such that $H^i=0$ for $i<\dim Z$, then the secant variety is Cohen-Macaulay and Du Bois.

\item Section 4: how does the Segre products fit in all of these? Can we use them to give examples?

\end{enumerate}

\end{comment}

redpajama_set_name: RedPajamaArXiv


text:
The Colt Python .357 Magnum Revolver
By Chuck Hawks
Illustration courtesy of Colt's Mfg. Co., Inc.
The Colt Python was officially introduced in 1955 as Colt's top-of the-line revolver. It was Originally intended to be a large frame, double action (DA), .38 Special target revolver, but at the last minute the chambering was changed to .357 Magnum. This fortuitous decision resulted in what is arguably the best all-around handgun in the world. Almost immediately the Python gained a reputation as the premium American revolver.
It is built on a .41 caliber size frame for extra durability. The action is fitted and hand-honed in the Colt Custom Shop to insure superior fit, smoothness, and a good trigger pull.
The Python has many special features. Like all Colt revolvers, the cylinder rotates into the frame for an extremely tight lock-up. The Python barrel has a ventilated rib on top and a full length underlug. Inside, Python barrels are bored with a very slight, full length, taper toward the muzzle for superior accuracy.
All Pythons come with excellent adjustable sights. The front sight is pinned to the full-length ramp atop the ventilated rib so that it can be changed (at the factory or by a gunsmith) if desired. The rear sight is an Accro target sight. These sights are durable and very accurate.
The full-length lug beneath the barrel was originally hollowed-out. Colt soon began leaving the underlug solid.
Even the finish on the Python is superior. Colt's Royal Blue is the ultimate polished blue finish for steel guns. No other blued production revolver can compare to the beauty of Colt's Royal Blue Python.
For many years the apex in beauty and protection was the bright nickel finish. This was a very smooth, very durable, nickel plating. The nickel finish was eventually retired after Colt perfected the mirror polished Ultimate stainless steel Python.
Colt offers two finishes for stainless steel Pythons, a conventional "brushed" finish, or the mirror-like Ultimate finish. The Ultimate is a stainless steel Python polished to degree similar to a Royal Blue gun.
Pythons have been produced with four standard barrel lengths, 2.5 inches, 4 inches, 6 inches, and 8 inches. At the 1998 SHOT Show I was told by one of the Colt factory reps that all Python revolvers would henceforth be produced entirely in the Custom Shop. In 2005 the Python was discontinued altogether.
The 2.5-inch barrel is the easiest to conceal. the 4-inch barrel is probably the optimum length for uniformed personnel to carry on duty, and also makes a very fine civilian self-defense weapon. The 6 inch barrel lets hot .357 loads achieve higher velocity, and is the most popular barrel length. A 6-inch Python is the ultimate all-around handgun, useful for target shooting, plinking, hunting, and self-defense.
The hunter's special is the 8-inch barrel. Scope mounts are available for mounting optical sights on all Pythons.
The accuracy of the Python revolver is legendary. I have owned Pythons with 4 inch, 6 inch, and 8 inch barrels, and all have been, literally, more accurate than I can shoot. Custom Shop Pythons come with targets fired on the Colt range. The target that came with my 4" Ultimate Python was a 3/4-inch, 6 shot group fired by hand at 25 yards, from a rest, with factory loads.
Subjectively, this finest of all DA revolvers is a "soft" shooter. By which I mean that most shooters feel that the Python kicks less than other DA magnum revolvers of comparable weight. I use the genuine Colt/Pachmayr target style rubber grips on all of my Pythons, which also helps to minimize the effect of recoil.
The style and features of the Python have been extensively copied by other gun makers, particularly the frame size and the full-length barrel underlug. For example, an "L" frame S&W will fit perfectly in a holster designed for a Python. However, no one has succeeded in copying all of the features of the Python.
Note: A full length review of the Python revolver can be found on the Product Reviews page.
Back to Collector's Corner
Back to Handgun Information
Copyright 2003, 2016 by Chuck Hawks. All rights reserved.

redpajama_set_name: RedPajamaCommonCrawl


text:
/**
 * @venus-library jasmine
 * @venus-code ../src/ReflectionClass.js
 * @venus-include ../src/ReflectionMethod.js
 * @venus-include ../src/ReflectionConstant.js
 * @venus-include ../src/ReflectionProperty.js
 */

describe('ReflectionClass', function () {

    var Mock = function () {
            this.property = null;
        };

    Mock.prototype = Object.create(Object.prototype, {
        CONSTANT: {
            value: 42
        },

        method: {
            value: function (arg) {}
        }
    });

    it('should throw if not a function passed to the constructor', function () {
        expect(function () {
            new ReflectionClass({});
        }).toThrow('You must pass a function as argument');
    });

    it('should return the list of properties', function () {
        var refl = new ReflectionClass(Mock),
            props = refl.getProperties();

        expect(props instanceof Array).toBe(true);
        expect(props.length).toEqual(1);
        expect(props[0] instanceof ReflectionProperty).toBe(true);
    });

    it('should return a ReflectionProperty object', function () {
        var refl = new ReflectionClass(Mock),
            prop = refl.getProperty('property');

        expect(prop instanceof ReflectionProperty).toBe(true);
    });

    it('should throw if trying to get unknown property', function () {
        var refl = new ReflectionClass(Mock);

        expect(function () {
            refl.getProperty('unknown');
        }).toThrow('Unknown property');
    });

    it('should say that the class has the property', function () {
        var refl = new ReflectionClass(Mock);

        expect(refl.hasProperty('property')).toBe(true);
    });

    it('should say that the class does not have the property', function () {
        var refl = new ReflectionClass(Mock);

        expect(refl.hasProperty('unknown')).toBe(false);
    });

    it('should throw if trying to get unknown method', function () {
        var refl = new ReflectionClass(Mock);

        expect(function () {
            refl.getMethod('unknown');
        }).toThrow('Unknown method');
    });

    it('should return a ReflectionMethod object', function () {
        var refl = new ReflectionClass(Mock),
            method = refl.getMethod('method');

        expect(method instanceof ReflectionMethod).toBe(true);
    });

    it('should say that the method exist', function () {
        var refl = new ReflectionClass(Mock);

        expect(refl.hasMethod('method')).toBe(true);
    });

    it('should say that the method does not exist', function () {
        var refl = new ReflectionClass(Mock);

        expect(refl.hasMethod('unknown')).toBe(false);
    });

    it('should throw if trying to get unknown constant', function () {
        var refl = new ReflectionClass(Mock);

        expect(function () {
            refl.getConstant('unknown');
        }).toThrow('Unknown constant');
    });

    it('should return a ReflectionConstant object', function () {
        var refl = new ReflectionClass(Mock),
            constant = refl.getConstant('CONSTANT');

        expect(constant instanceof ReflectionConstant).toBe(true);
    });

    it('should say that the constant exist', function () {
        var refl = new ReflectionClass(Mock);

        expect(refl.hasConstant('CONSTANT')).toBe(true);
    });

    it('should say that the constant does not exist', function () {
        var refl = new ReflectionClass(Mock);

        expect(refl.hasConstant('unknown')).toBe(false);
    });

    it('should create a new instance of the class', function () {
        var refl = new ReflectionClass(Mock);

        expect(refl.create() instanceof Mock).toBe(true);
    });

    it('should return the prototype', function () {
        var refl = new ReflectionClass(Mock);

        expect(refl.getPrototype()).toEqual(Mock.prototype);
    });

});

redpajama_set_name: RedPajamaGithub


Canon MAXIFY MB5150 Driver Download – Hi there, fellow Canon MAXIFY MB5150 user! As soon as all the printer driver files are extracted, you still need to install the printer driver. comply with these Technodrivers procedures, as you may employ the Add Printer Wizard to aid you do this.
On this page, you must absolutely choose the port that the printer will use to communicate with the laptop computer. Many laptops work with LTP slots. If your printer is attached to a USB port, choose the USB port option in the following port list, and click Next.
In the Open box, type Control colorprinters], and press OK. The "Printers and Faxes" window sets up.
On the following page, click on Local Printer of Canon MAXIFY MB5150 connected to this computer or laptop. Essential note: Make clear that the Automatically discover and install my Plug and Play printer check pack is truly not chosen.
Next off, instruct the wizard where to discover the printer driver files that you extracted:.
In the Install From Disk dialog box, press Browse, and situate the directory to which you extracted the printer driver files.
When you have actually located the exact directory, click Open.
The path from your directory must appear in the copy producer's files from box in the Install From Disk dialog box.Select OK.
The color printers that you might install needs to be actually shown in the Install Printer Software dialog box. Select the Canon MAXIFY MB5150 printer to link, and press Next.
If you are prompted to make the most of an existing driver, select Replace existing driver, and press Next. When you pick to upgrade the existing driver, Windows will try to replace the files on your system using the fresh ones that you downloaded and install.
On the following page, press Next if you need to accept the suggested printer name and also employ the Canon MAXIFY MB5150 driver as your default printer. If you don't, change these setups to what exactly you want, and click Next.
If you find the Printer Sharing page next in the wizard, you may share your printer to ensure that various other personal computers on your home network could use it to print. Select Share name, and define a name for thecommon printer.
Else, click on "Do not share this printer". After you have already made your choice, click on Next. You may possibly be inquired to enter optional location info for the printer. Type the location of the printer such as a Home Office, and press Next.
On the Print Test Page page from the wizard, press Yes to print a test page. Subsequently press Next, and press Finish.
If you may print a test page properly, you are simply done. At the same time, if the test page was actually not printed properly, you still get some troubleshooting to perform that is generally beyond the span of this write-up.
To begin with, click on Troubleshooting in the Add Printer Wizard to discover whether it might offer you valuable information. If that does not deal with your Canon MAXIFY MB5150 printer situation, head to Next Steps for some subsequent moves that you may intend to utilize.


Here's an easy fun idea for making a simple wooden serving tray decorative, yet suitable for any holiday or special occasion. For this project we are going to use an aniline wood dye powder rather than paint or stains. One fun thing about wood dye powders is that you can play around with colors, mix various shades to get the perfect hue. Today Mohawk Finishing Products is the leading manufacturer and distributor of professional wood touch up, repair and finishing products. PENOFIN WOOD FINISHES MEET AND EXCEED THE NORTHEAST AIR QUALITY MANAGEMENT DISTRICT'S CLEAN AIR REGULATIONS.
For hundreds of years, enhancing the beauty of wood has been essential to the woodworker's craft. If you're using the immersion method, you'll need the same amount of dye with 2 ''quarts'' of water (depending on the size of your product). It's not a bad idea to dip in one side and set the wood down to dry on a side that isn't dyed yet. If you like the current shade, consider sealing it with polyurethane spray or a wood varnish. These water-base stains are formulated with a proprietary blend of pigments and dyes to provide rich colors.
Dyeing wood can be done in a number of ways, often with materials that you may have lying around the house. If you haven't yet purchased wood and are looking to do so from a home improvement store, ask if they'll sand it for you. Mix the dye according to the package instructions – you'll probably need ? cup liquid dye or 1 box of powder dye with 2 cups of very hot water. Dip a piece of scrap wood (or use an area of the wood you're using that can't be seen) into the bowl of dye.
It'll also show you how the dye spreads and how you have to apply the dye to get the look you want. Dip a foam or bristle brush or old cloth in the dye and spread it evenly over your wood surface.
Before doing anything with the wood, you should have a suitable place to work in, where you can make a mess and it doesn't matter. With rubber gloves on to prevent staining your hands and fingers, add a package of drink powder to water to create a dye. Using a foam brush, spread the paint everywhere on the wood, wherever you need the dye to be applied. Be sure to coat the entirety of the wood before you move onto coating it a second time to keep the color even.
The more dye that you add, the more saturated the color will be (and the less water you use, too). A resealable plastic bag is ideal for immersing the wood, depending on how large the piece of wood is. For the record, this isn't a very strong dye and is only suitable for light colored wood, such as pine. These will be used as part of the dye, making it richer and deeper – and that translates to fewer coats for you to spread.
While the coffee is still warm (not hot), use a paintbrush or rag dipped into the coffee and apply it to the wood. Use a glass or ceramic bowl to avoid reactions with the dye colors in the microwave and stir well. If dye drops splash onto your wood, you can get rid of them by sanding the wood immediately.
Use a table or other surface that is comfortable to work on and that you're okay getting a splash of dye on.
Based, clear polyurethane finish for interior just position the board, swivel most frustrating.
Doing so requires some knowledge shutterstock the world's largest.
The ability to use two different gTS1031 may have.


text:
Integration & Physician Issues
A Seat at the Table: How Physician Advisory Councils Contribute to Hospital-Physician Alignment
Staff - Tuesday, May 20th, 2014
In the Becker's Hospital Review 5th Annual Meeting in Chicago, Jerome Levine, MD, senior physician executive, consultant, Coors Healthcare Solutions and Phillip Wright, CEO of Santa Rosa Medical Center in Milton, Fla., discussed the importance of creating physician leadership and how physician advisory councils can go a long way in creating hospital-physician alignment that lasts.
Physician recruitment and retention are extremely important aspects of a successful healthcare organization, according to Dr. Levine. Physician expectations regarding their jobs has changed and leadership development is an important driver of physician alignment with a hospital
"Physicians are a hospital's partner and have to be involved in every aspect of the organization," said Dr. Levine. "How can an organization be successful if the physicians are not sitting at all the decision-making tables? Organizations have to help develop and create physician leadership."
A tried and tested way to create effective physician leadership at a hospital is the formation of a physician advisory council. A PAC is designed to give physicians a forum for transparent communication. It allows them to give CEOs direct input and feedback and gives them a seat at the table. "Early discussion of initiatives and ideas with physicians through the PAC also creates alignment with an organization's strategic goals," said Dr. Levine.
According to Mr. Wright, whose hospital has a PAC, it is a key aspect of hospital-physician partnership. Here are four steps for creating an effective PAC:
1. Selecting 12 to 15 physicians from different specialties — an equal mix of employed and independent physicians.
2. Organizing monthly dinner meetings with the PAC and hospital administration.
3. Appointing a physician leader to chair the council and run the meetings.
4. Setting meeting for the CEO and PAC chair 24 hours before the meeting.
It is important that the hospital administration follow up on the ideas and actions discussed at meetings with the PAC. "You will get physician participation, attention and commitment if they see action when they give you input," said Mr. Wright.
After the creation of the PAC at Santa Rosa Medical Center, there was a 10-point improvement in physician satisfaction. There have also been changes in equipment purchase guidance and the design of the emergency room as a result of PAC meetings
"It just goes to show, if you build it, they will come," said Dr. Levine.
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Burnished silvertone 2 1/4" heart pendant accented with designer-look detail and sparkling clear crystals on a unique 18"L adjustable necklace formed of silvertone links and natural brown cotton cords. Matching pierced wire earrings. I LOVE this set!


I am the third generation to run the family travel business originally started by my grandfather in the 1950's. I've been skiing throughout Europe and North America since the age of four, culminating in a winter tour of duty in the French Alps in the late eighties. Before joining the company in 2002, I worked in Edinburgh for the property company Rettie & Co.


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Business | Sports
TinCaps Game Notes: July 8 vs. South Bend (Game 82)
Fort Wayne TinCaps (8-5, 40-41) vs. South Bend Cubs (7-6, 36-44)
RHP Ernesto Montas (7-5, 3.33) vs. RHP Erick Leal (6-5, 5.27)
Wednesday, July 8, 2015 — Parkview Field — First Pitch 7:05 p.m. — Game 82
ABOUT LAST GAME: Fort Wayne was foiled at South Bend, 2-1. The TinCaps took a 1-0 lead in the second on an Edwin Moreno RBI single and stayed in front until the seventh. Starter Chris Huffman allowed only two baserunners through the first six innings, but gave up a leadoff single to Jason Vosler and then a two-run homer to Yasiel Balaguert. Zach Hedges threw eight innings to get the win.
HOME-FIELD ADVANTAGE: The TinCaps are 6-0-1 at Parkview Field over their last seven series here. Fort Wayne's record is 15-5 during the run that goes back to May 26.
GETT-ING HOT: Center fielder Michael Gettys is on a season-best seven-game hitting streak. He extended it Tuesday with two singles. During the stretch, Gettys is 12-for-30 (.400) with five doubles. His 22 doubles on the year are tied for third most in the MWL. (Gettys' teammate Nick Torres is tied with Lansing's Ryan McBroom for the league lead with 27.)
PITCHING WINS: Well, most of the time, if not on Tuesday night. The TinCaps are 12-5 over their last 17 games. In that span, they've allowed more than three runs in a game only three times. In 157 2/3 innings of work since June 18, Fort Wayne has yielded just 35 earned runs. That equates to a 2.00 ERA.
AGE IS BUT A NUMBER: The three youngest players in the Midwest League are at Parkview Field tonight. South Bend's Gleyber Torres, an 18-year-old from Venezuela(born Dec. 13, 1996) began the year as the league's youngest until Ruddy Giron (Jan. 4, 1997) joined the TinCaps in May. Luis Urias (June 3, 1997) now has the distinction since his arrival in Fort Wayne on Monday.
THE OTHER GUYS: South Bend's roster features the following Chicago Cubs prospects, per MLB.com: No. 4 SS Gleyber Torres, No. 10 RHP Jake Stinnett, and No. 26 RHP Trevor Clifton.
TONIGHT'S STARTING LINEUP:
1. #1 Michael Gettys CF — Had a 12-game hitting streak last July 21-August 6.
2. #9 Luis Urias 2B — Making his Parkview Field debut after being called up Mon.
3. #12 Ruddy Giron SS — Says he's not a power hitter, just looks to hit line drives.
4. #10 Nick Torres RF — 13th game this year cleaning up. (Has hit 3rd 56 times.)
5. #16 Trae Santos 1B* — Has walked in 12.7% of PA (9th highest in MWL).
6. #34 Franmil Reyes DH — Turned 20 yesterday.
7. #15 Edwin Moreno LF* — Drove in Fort Wayne's lone run on Tuesday at SB.
8. #21 Jose Ruiz — Featured in article in Wednesday's The Journal Gazette.
9. #3 Mitch Morales 3B* — Parents (Clare & Alex) here to see him as pro for 1st time.
* Left-Handed Batter
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Back in my graduate student days, I was a managing editor for an online academic journal. One of my duties was to maintain the website. So I was a "webmaster," even though I had absolutely no mastery of the web.
Nevertheless, I fumbled along and learned as I went. Then, one day, I received an inquiry from Some Professor at Very Important University asking me a few technical questions. I had no clue. So I forwarded it along to a friend, saying something along the lines of, "Hey, I left my Klingon-to-English dictionary at home. Please translate."
You know... that whole Computer Guys and Star Trek thing. Get it? I mean, it's funny at different levels.
Anyway, a few minutes later the aforementioned professor's name once again appeared in my inbox. "What is this Klingon-to-English dictionary of which you speak?" he asked.
With that, a surge of white, hot, and holy terror went through my body. Indeed, I had hit "reply" instead of "forward."
I'm a pretty neurotic guy to begin with, but as a graduate student this particular character trait was amplified to intolerable levels. Every little bump in the road from typos to rejected manuscripts had me convinced that I had no future in academia.
So this whole thing threw me deep into the pit of obsessive fretting. I imagined the insulted professor composing an email to some secret "All Professors Throughout The Universe" listserv. He would declare me unworthy of employment in this or any profession until the end of time. No public shaming. Just a short and miserable life in a ditch in southwest Alabama (it's always southwest Alabama).
I immediately sent an apology. After five minutes, I sent another. I'm pretty sure that I sent a third. I never heard back. And I don't blame him.
Needless to say, life went on. But a memory of this episode remains instructive. For starters, it reminds me to be careful with my digital communications. But still, when I'm rushed, I can be careless. And I have been careless.
I suppose, then, that this is ultimately about the value of humility. I got busted being a jerk. That made me feel really, really bad. But what if I had hit forward? Would I have still felt guilty and ashamed?
The fact is that here and pretty much always, I am only aware of my sins when they are shoved in my face. Otherwise, I'm pretty good at making excuses and shrugging off my shortcomings. "Oh, I'm just kidding around." But if the wrong (right?) person hears the joke, it stops being funny.
Since Klingongate, I have been on the receiving end of a few missent emails. I usually get a frantic apology shortly after. In response, I always tell my own story. I then admit that I still make poor decisions with my words and, therefore, have no room to judge anyone.
I don't like saying that I am "forgiving" anyone when I have done this. To me, that almost implies that I have some moral advantage in the situation. I don't. Instead, what I intend to say is, "I'm a mess, and it's great to see that you're a mess too. Let's get a drink!" Hopefully, this makes everyone feel a little less shame and a little more human.
Solidarity is a beautiful thing... even if hitting reply instead of forward isn't.


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Heat Nation Exclusive: Q&A With Miami Heat Guard Kendrick Nunn - Heat Nation
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Home / Editorials / Heat Nation Exclusive: Q&A With Miami Heat Guard Kendrick Nunn
Heat Nation Exclusive: Q&A With Miami Heat Guard Kendrick Nunn
By Justin Benjamin
Steve Mitchell / USA TODAY Sports
As the Miami Heat went through multiple changes this past offseason, there was one player that flew under the radar.
On the last day of the 2018-19 NBA regular season, the Heat signed guard Kendrick Nunn. However, huge moves over the summer, including the draft selection of Tyler Herro and signing of Jimmy Butler, headlined the Heat's acquisitions.
Meanwhile, Nunn had a phenomenal summer. The 6-foot-3 combo guard was named to the NBA Summer League First Team back in July after putting up 22.0 points, 5.2 assists and 4.3 rebounds on 50.5 percent shooting from the field and 38.5 percent shooting from downtown.
While he didn't get invited to any of Butler's early morning workouts before training camp, he made sure he was ready for the preseason. For those who weren't convinced by his Summer League production, Nunn put the league on notice during the preseason.
In the Heat's preseason finale against James Harden, Russell Westbrook and the Houston Rockets, the 24-year-old erupted for 40 points on 6-of-10 shooting from distance. Furthermore, the explosive performance was the highest-scoring preseason game by a Heat player in the last 20 years, surpassing a 38-point contest by four-time MVP LeBron James.
We got the chance to exclusively talk with Nunn before he begins his big rookie season on Wednesday.
How was your summer?
Nunn: My summer was great. Just here in Miami, training, working out and playing in the summer league. That's how my summer went really. I had a great summer.
Talk me through that Summer League experience.
Nunn: It was great. Just having that opportunity to sign with them (Heat) on the last day of the season. Last year was great, perfect timing. I got a chance to work with them during the summer going into Summer League. It just made it even better. I was more comfortable with everything and the development over the summer was great. The results showed. I had a great Summer League.
Is there an area of your game that enhanced over the summer?
Nunn: Every summer I believe my game has been going to another level. But particularly this summer I just worked on everything. My ball-handling, my passing, playing the combo position. More than training but just playing at the 1 and 2 position, being more of a ball-handler gives you more development I believe. That's where my game went to another level.
Were you surprised how much the Heat's roster transformed over the summer?
Nunn: Not really. This was my first summer actually being on an NBA team. I just let it happen and then I watch whoever ends up on the team after the trades. I'm glad Jimmy [Butler] came and plays for the Heat.
Do you have any goals this upcoming season? Expectations?
Nunn: For this season, I just want to contribute in any way I can, be in the rotation this upcoming year, just contribute to some wins and go on a playoff run.
What stands out about the Heat's culture?
Nunn: Just the mentality that goes into everything, hard work. There's great people in the organization. It's a lot of love. The coaches are great. We work hard, do things the right way. I think we're one of the best organizations in the league.
Do you have any personal career goals?
Nunn: I definitely want to be an All-Star.
How good is Tyler Herro going to be?
Nunn: Man, I can't put no cap on him. He's a dawg. He's nice.
Is there anybody you emulate your game after?
Nunn: I study the elite guards in the NBA. I have good footwork like C.J. McCollum. He's real crafty. The elite guards, point guards, that's where my game will go to another level, just studying that.
Which veterans are you surrounding yourself with and taking advice from?
Nunn: All the vets that have more experience than me. I'm all open ears. Typically like U.D. (Udonis Haslem) because my locker is next to his and J.J. (James Johnson). I'm in between them. They're always giving me tips and advice and stuff like that. They've been in the league so long so they have advantages. So whatever it is, helping me during practice and stuff like that, talking to me, has helped me out.
What did it mean to you to see Dwyane [Wade] encourage you on Twitter?
From what I've seen i like that kid a lot https://t.co/VV4CbnF5Tq
— DWade (@DwyaneWade) August 29, 2019
Nunn: It meant a lot. Just that he respects me and the way that I handle myself. It means a lot coming from Dwyane Wade, someone I look up to.
What do Heat fans need to know about you?
Nunn: The nickname "K9" comes from being a dog on the court. I'm a nice guy to talk to. I'm really down to earth.
c.j. mccollumFeaturedhouston rocketsJames Hardenjames johnsonjimmy butlerKendrick Nunnlebron jamesmiami heatNews Tickerrussell westbrooktyler herroUdonis Haslem
← Previous Story Why Tyler Herro Is Poised to Win This Season's Rookie of the Year Award
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About Justin Benjamin
Born and raised in South Florida, Justin has always been a passionate Miami Heat fan. An avid supporter from the time Miami got its first championship in 2006 to having a league-worst 15-67 record in 2008 to the whole LeBron James era until now, Justin has seen and stuck through it all. His all-around analysis and heart for the game has made him a premier NBA writer. He writes for a variety of sites but his commitment to the Heat is always top priority.
Former All-Star Victor Oladipo has a new NBA home after being traded to the Houston Rockets, but it reportedly...
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Septic tank cleaning in York PA and anywhere else, really, is a job best left to the experts. Septic tanks are miracle workers of sorts. Yours is out there buried in the soil in York, PA, and you hardly ever think of it. Why should you when it has served you well over the years?
Well, while they are easy to bury and forget, neglect may cause some issues that will set you back tens of thousands of dollars.
The septic system has been in existence for over 100 years and the design has hardly changed. Being an integral part of any real estate – whether residential or commercial – septic tanks hold gray water that is released from every part of a house or building. This water meets in the same place regardless of the source, and over time it begins to separate with the heavier material sinking to the bottom. Here, liquid scum floats and leaves the tank through built-in drainage into the absorption field where it blends into the soil, fertilizing it ideally. No worries, the design prevents it from reaching the water table.
Just like a car, your septic tank will serve you better and for longer if you maintain it properly and have it cleaned regularly. The heavy sludge that settles at the bottom needs regular separation so that it doesn't cost you tons of money after ten years of neglect.
The frequency of cleaning will depend on size of the tank and material that does in there – among other variables. Current designs differ from their older models and so even the method of breaking down sludge has changed. Effective cleaning will require the services of a professional servicing company with the right personnel and equipment.
When you diagnose a problem, you want it acted on fast before it escalates. We typically advise our customers to schedule regular inspections so they know the state of their septic tanks, but fret not if you forgot to get that done. Calling our offices triggers a few reactions depending on the urgency of your matter, and a team is dispatched on the same day or the next.
We not only have state-of-the-art equipment for cleaning and maintaining septic tanks, but we also have a team of experts ready to act on your issue. They will find out what your system needs and recommend the appropriate action. We offer several solutions for septic tanks, from inspections and cleaning to repairs and installation – for both commercial and residential property.
Your family stays healthy. When untreated seeps through the drainage to the water table, it contaminates water that your family uses, exposing you to all manner of diseases.
It is cheaper – Cleaning your system regularly keeps it in great shape and will cost less than replacing it in its entirety.
Property Value – a good drainage system makes a lot of difference during property valuation.
Some problems need follow-up to solve completely and we offer this appropriately. Smith's Septic has been in operation since 1959 and during this time, we have served several residents of York PA diligently. We have built a community not only through offering great services at reasonable cost, but also nurturing those relationships with regular communication.
Failure to clean your septic tank regularly could cause a big problem.
Do not wait until the stench in your backyard forces you to call us. Schedule a septic tank cleaning in York PA today for peace of mind and to save you the trouble of repairing your septic tank.


In our continuous effort to help the NGOs in raising funds for their various projects, we are pleased to announce our affiliation with popular shopping sites amazon, flipkart and snapdeal. For every purchase made using the above links at these sites, we will donate upto 1% to our Premium Member NGOs. At the end of every month, we will tabulate 1% of total sales during that period and distribute it among our Premium Member NGOs. Shoppers do not need to pay anything extra.
Every purchase you make from the above links will contribute to the social cause and will help the poor and needy. So, please be part of this campaign and spread this message among your friends and family members.


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Frederick L. Stiffler
Frederick Lamar Stiffler, age 67, passed away Thursday, December 20, 2018 at the Riverside Methodist Hospital in Columbus, Ohio. He was born March 29, 1951 in Gallipolis, Ohio to the late Delmar F. and Helen (Miller) Stiffler. In addition to his parents, Fred was preceded in death by his wife, Sharon Stiffler; his brother, Jerry M. Stiffler; his brother-in-law, Robert Hammerstein; his sister-in-law, Carol Stiffler and his beloved canine companion, Ziggy. Fred grew up on a farm in the Fourmile Community and attended grade school at Franklin Local School and was a 1969 graduate of the Jackson High School. He was a U.S. Navy veteran serving as hospital corpsman and later as a Navy Recruiter. After leaving the Navy, Fred graduated from Rio Grande University and the Cincinnati College of Mortuary Science. He was employed by Marathon Oil Co. and retired in 2012 after 17 years of service. Fred was employed as a funeral director and embalmer at the Mayhew-Brown Funeral Home in Jackson, Ohio. He is survived by his son, Christopher Stiffler of Alpharetta, Georgia; his brother, Delmar H. Stiffler of Columbus, Ohio; two very devoted sisters, Christine (Roger) Hill of Mt. Vernon, Ohio and Donna Hammerstein of Beaver, Ohio; his stepdaughters, Tonya Stumbo of Dayton, Ohio and Carla (Sean) Taylor of Jackson, Ohio; step-grandchildren: Jessica, Nathan and Ali; one step great granddaughter, Aaliyah and several nieces and nephews. No calling hours were observed. A graveside service with military hours was held at the convenience of the family. Arrangements were under the direction of the Mayhew-Brown Funeral Home in Jackson. Condolences may be sent to: www.mayhew-brownfuneralhome.com.
Frederick Lamar Stiffler, age 67, passed away Thursday, December 20, 2018 at the Riverside Methodist Hospital in Columbus, Ohio. He was born March 29, 1951 in Gallipolis, Ohio to the late Delmar F. and Helen (Miller) Stiffler. In addition to... View Obituary & Service Information
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Home Idaho State Prison
Idaho Maximum Security Institution (IMSI) Information
13400 South Pleasant Valley Road, Kuna, ID, 83634
Idaho Maximum Security Institution (IMSI) is a medium security state prison located in city of Kuna, Ada County, ID. It houses adult male inmates (above 18 years of age) who are convicted for crimes which come under Idaho state law. Most of the inmate's serving time in this prison are sentenced for the period of over a year and are sentenced for crimes which are serious in nature. The prison is operated and maintained by Idaho Department of Corrections (IDOC)
Below we have given information about the Idaho Maximum Security Institution (IMSI) including inmate search, phone number, visitation rules/hours and contact information.
Idaho Maximum Security Institution (IMSI) Prison Information
Idaho Maximum Security Institution (IMSI)
Ada County
//www.idoc.idaho.gov/content/locations/prisons/idaho_maximum_security_institution
Idaho Maximum Security Institution (IMSI) Inmate Search
If you are looking for information of any inmate currently housed at Idaho Maximum Security Institution (IMSI), you can visit the the inmate search/roster link here. Alternatively you can call on to enquire about the inmate.
If you are still not able to find the whereabouts of the inmate, then you can try searching on Idaho statewide inmate search page.
If you are visiting someone in Idaho Maximum Security Institution (IMSI) make sure you are in approved visitor list. It's good idea to confirm with the prison authorities before coming for a visit.
Idaho Correctional Center (ICC) - CCA
14601 South Pleasant Valley Road, Kuna, ID, 83634 208-331-2760
Idaho State Correctional Institution (ISCI)
South Idaho Correctional Institution (SICI)
Idaho Capp Facility - MTC
Correctional Alternative Placement Program (CAPP)
South Boise Women's Correctional Center (SBWCC)
Horseshoe Bend Jail
112 Ada StreetPO Box 246, Horseshoe Bend, ID, 83629 208-793-2219, 208-793-2255
Idaho City Police Jail
511 Main Street, Idaho City, ID, 83631 208-392-4596
McCall Police Jail
216 East Park Street, McCall, ID, 83638 208-634-7144
Boise County Jail
3851 Highway 21PO Box 189, Idaho City, ID, 83631 208-392-4411
Valley County Juvenile Detention Center
550 Deinhard Lane, McCall, ID, 83638 208-634-4131
Police Department Jail 20