text:
\section{Introduction}

In \cite{Lech60}, Lech proved a simple inequality relating the Hilbert--Samuel multiplicity and the colength of an ideal. It states that if $(R,\mathfrak{m})$ is a Noetherian local ring of dimension $d$ and $I$ is any $\mathfrak{m}$-primary ideal of $R$, then we have $$e(I, R)\leq d!e(R)l(R/I),$$ where $e(I, R)$ denotes the Hilbert--Samuel multiplicity of $I$ and $e(R)=e(\mathfrak{m}, R)$. In the same paper, Lech conjectured that, for every flat local extension $(R,\mathfrak{m})\to (S,\mathfrak{n})$ of Noetherian local rings, one has $e(R)\leq e(S)$. This conjecture is wide open in general.  Using the above inequality, Lech obtained the estimate $e(R)\leq d!e(S)$ where $d=\dim R$ \cite{Lech60}. We refer to \cite{HSV17} and \cite{Han02} for some generalizations of Lech's inequality and to \cite{Ma17} for recent progress on Lech's conjecture.

If we consider the set $\left\{ \frac{l(R/I)}{e(I, R)} \right\}_{\sqrt{I}=\mathfrak{m}}$ of positive numbers, then Lech's inequality is simply saying that this set is bounded below by $\frac{1}{d!e(R)}$ (and, thus, is bounded away from $0$). The infimum of this set was investigated by Mumford in his study of local stability \cite{Mum77}. In a different direction,
in \cite{SV96} St\"{u}ckrad and Vogel studied whether $\left\{ \frac{l(R/I)}{e(I, R)} \right\}_{\sqrt{I}=\mathfrak{m}}$ is bounded from above (see also \cite{MV95}), and they conjectured the following \cite[Theorem 1 and Conjecture]{SV96}:

\begin{conjecture}[St\"{u}ckrad--Vogel]
\label{conj:Vogel}
Let $(R,\mathfrak{m})$ be a Noetherian local ring and let $M$ be a finitely generated $R$-module. Let $e(I, M)$ be the Hilbert--Samuel multiplicity\footnote{In this paper, we define the Hilbert--Samuel multiplicity of a finitely generated module $M$ with respect to $I$ to be $e(I, M)=\lim\limits_{n\to\infty}t!\frac{l_R(M/I^nM)}{n^t}$ where $t=\dim M$. This is always a positive integer even when $\dim M<\dim R$. We will simplify our notation when $I=\mathfrak{m}$ and write $e(M)$ for $e(\mathfrak{m}, M)$.} of $M$ with respect to $I$. Set $$n(M)=\sup_{\sqrt{I+\operatorname{Ann}(M)}=\mathfrak{m}}\left\{\frac{l(M/IM)}{e(I, M)}\right\}.$$ Then $n(M)<\infty$ if and only if $M$ is quasi-unmixed (i.e., $\widehat{M}$ is equidimensional).
\end{conjecture}

St\"{u}ckrad and Vogel proved the ``only if" direction in general and a graded version of the ``if" direction \cite[Theorem 1]{SV96}. Some other partial results were obtained in \cite{AH00}. In this paper we settle this conjecture in the affirmative. Furthermore, motivated by Conjecture \ref{conj:Vogel}, it is quite natural to inquire whether Lech's classical inequality can be extended to all finitely generated modules, i.e., whether there is a lower bound on the set $\left\{\frac{l(M/IM)}{e(I, M)} \right\}_{\sqrt{I}=\mathfrak{m}}$ for a finitely generated $R$-module $M$. We also answer this question in the affirmative. In sum, our main result is the following:

\begin{maintheoremA}[Theorem \ref{Thm:Vogel} and Theorem \ref{Thm:LechMod}]
Let $(R,\mathfrak{m})$ be a Noetherian local ring, and let $M$ be a finitely generated $R$-module of dimension $d$. Set
$$m(M)=\inf_{\sqrt{I+\operatorname{Ann}(M)}=\mathfrak{m}}\left\{\frac{l(M/IM)}{e(I, M)} \right\} \text{ and \hspace{0.5em}} n(M)=\sup_{\sqrt{I+\operatorname{Ann}(M)}=\mathfrak{m}}\left\{\frac{l(M/IM)}{e(I, M)} \right\}.$$ Then we have
$$m(M)\geq \frac{1}{d!e(\overline{R})}$$ where $\overline{R}=R/\operatorname{Ann}(M)$. Moreover, if $M$ is quasi-unmixed, then we also have
$$n(M)< \infty.$$
\end{maintheoremA}

As an application of Theorem A, we obtain the following result on Koszul homology:

\begin{maintheoremB}[Theorem \ref{Uniform0}]
Let $R$ be a Noetherian local ring, and let $M$ be a finitely generated quasi-unmixed $R$-module of dimension $d$.  For
every $\varepsilon>0$, there exists $t_0$ such that, for all $t \geq t_0$, all systems of parameters $\underline{x}:= x_1, \ldots, x_d$ of
$M$, and all $1 \leq i \leq d$,
\[
\frac{l(H_i(x_1^t, \ldots, x_d^t; M))}{l(M/(x_1^t, \ldots, x_d^t)M)}
<\varepsilon.
\]
In fact, there exists a constant $K$ such that for all $t \geq
1$, all systems of parameters $\underline{x}:= x_1, \ldots, x_d$ of
$M$, and all $1 \leq i \leq d$,
\[
\frac{l(H_i(x_1^t, \ldots, x_d^t; M))}{l(M/(x_1^t, \ldots, x_d^t)M)}
\le \frac K{t^i}.
\]
\end{maintheoremB}

It is well known that the ratio in Theorem B tends to $0$ for any {\it fixed} system of parameters $\underline{x}$.  What we achieve in Theorem B is a {\it uniform} convergence. We also point out that our Theorem A says that $m(M)$ is bounded below by $\frac{1}{d!e(\overline{R})}$, which is independent of $M$ (and only depends on $R/\operatorname{Ann}(M)$). One cannot expect the same for the upper bound $n(M)$:
\begin{example}
\label{nlarge}
Let $R=k[[x,y]]$ and let $M_t=\mathfrak{m}^t=(x,y)^t$.  Then the $M_t$ are all faithful $R$-modules of rank one, but clearly
$$n(M_t)\geq \frac{l(M_t/\mathfrak{m} M_t)}{e(\mathfrak{m}, M_t)}=\frac{l(\mathfrak{m}^t/\mathfrak{m}^{t+1})}{e(R)}=t+1.$$ Therefore, there cannot exist a constant $c$ such that $n(M_t)\leq c$ works for all $M_t$.
\end{example}
Nonetheless, inspired by this example, we will see in Remark \ref{RemMinGen} that $n(M)/\mu(M)$ is indeed bounded above by a constant depending only on $R/\operatorname{Ann}(M)$.

This paper is organized as follows. In Section 2 we prove Conjecture \ref{conj:Vogel}, which is the second part of the Theorem A, and we also prove some results about the behavior of the invariant $n(M)$ under base change. In Section 3 we extend the classical version of Lech's inequality and  prove the first part of Theorem A. In Section 4 we give many applications of Theorem A, prove Theorem B, and obtain an alternative proof of Conjecture \ref{conj:Vogel}. In the Appendix, we establish global versions of results in Section 4.

\vspace{1em}

\noindent\textbf{Acknowledgement}: The authors would like to thank Mel Hochster and Craig Huneke for valuable discussions, as well as Le Tuan Hoa for his comments. Partial results of this paper were done while the third author was visiting University of Utah. He would like to thank University of Utah for its hospitality and support.  The first author has been partially supported by NSF Grants \#1401384 and \#0943832.  The second author is supported in part by NSF Grant \#1836867/1600198 and NSF CAREER Grant DMS \#1252860/1501102 when preparing this paper. The third author is partially supported by a fund of Vietnam National Foundation for Science and Technology Development (NAFOSTED) under grant number 101.04-2017.10. Finally, the authors would like to thank the referee for his/her comments, which led to Remark \ref{RemMinGen}.

\section{Finiteness of $n(M)$: resolving the St\"{u}ckrad--Vogel conjecture}

To prove the St\"{u}ckrad--Vogel conjecture, we need the concept of extended degree of a finitely generated module introduced by Vasconcelos in \cite{Vas98, VasSixLec98}.

\begin{definition}\label{Def:Edeg}
Let $(R,\mathfrak{m})$ be a Noetherian local ring with infinite residue field. Let $\mathcal{M}(R)$ denote the category of finitely generated $R$-modules. An {\it extended degree} on $\mathcal{M}(R)$ with respect to an $\mathfrak{m}$-primary ideal $I$ is a numerical function
$$\operatorname{Deg}(I,\bullet)\colon \mathcal{M}(R)\to \mathbb{R}$$ satisfying the following conditions:
\begin{enumerate}
  \item $\operatorname{Deg}(I, M)=\operatorname{Deg}(I, \overline{M})+l(H_{\mathfrak{m}}^0(M))$, where $\overline{M}=M/H_{\mathfrak{m}}^0(M)$;
  \item $\operatorname{Deg}(I, M)\geq \operatorname{Deg}(I, M/xM)$ for every generic element $x\in I-\mathfrak{m} I$ of $M$;
  \item If $M$ is Cohen-Macaulay then $\operatorname{Deg}(I, M)=e(I, M)$.
\end{enumerate}
\end{definition}

The original definition in \cite{VasSixLec98} only deals with the case $I=\mathfrak{m}$. The above definition was taken from \cite[Definition 5.3]{CQ17}. The first question is whether, given a Noetherian local ring $(R,\mathfrak{m})$, an extended degree function exists. This question was settled in the affirmative by Vasconcelos (\cite{Vas98, VasSixLec98}), who showed that {\it homological degree} is an example of extended degree (when the residue field is infinite).\footnote{Here again, Vasconcelos's papers \cite{Vas98, VasSixLec98} focus on the case $I=\mathfrak{m}$, and in fact the main case Vasconcelos considered is the graded case. However the proofs in \cite{Vas98, VasSixLec98} work in the general set up, and we refer to \cite{CQ17} for more details.}

\begin{definition}
Let $(R,\mathfrak{m})$ be a homomorphic image of a Gorenstein local ring $(S,\mathfrak{n})$ of dimension $n$, and let $M$ be a finitely generated $R$-module of dimension $d$. Then the {\it homological degree}, $\operatorname{hdeg}(I, M)$, of $M$ with respect to an $\mathfrak{m}$-primary ideal $I$ is defined by the following recursive formula: $$\operatorname{hdeg}(I, M)=e(I, M)+\sum_{i=n-d+1}^{n}\binom{d-1}{i-n+d-1}\operatorname{hdeg}(I, \operatorname{Ext}_S^i(M, S)).$$
\end{definition}

We note that the above definition is recursive on dimension since $\dim \operatorname{Ext}_S^i(M, S)\leq n-i<d=\dim M$ for all $i=n-d+1,\dots, n$. For a long time, the homological degree was the only known explicit example of an extended degree. Quite recently in \cite{CQ17}, Cuong and the third author discovered another extended degree, this one defined in terms of the {\it Cohen-Macaulay deviated sequence} $\{U_i(M)\}_{i=0}^{d-1}$ of $M$. Roughly speaking, $U_i(M)$ is the unmixed component of $M/(x_{i+2},\dots,x_d)M$ for a certain carefully chosen system of parameters $x_1,\dots,x_d$ of $M$.  It is shown in \cite[Theorem 4.4]{CQ17} that this unmixed component is independent of the choice of $x_1,\dots,x_d$ as long as $x_1,\dots,x_d$ is a {\it $C$-system of parameters of $M$}, which always exists when $R$ is a homomorphic image of a Cohen-Macaulay local ring. Thus, $\{U_i(M)\}_{i=0}^{d-1}$ is a sequence of finitely generated $R$-modules depending only on $M$. Note that $U_{d-1}(M)$ is just the unmixed component of $M$.  We refer to \cite[Section 4]{CQ17} for more details.

\begin{definition}
Let $(R,\mathfrak{m})$ be a homomorphic image of a Cohen-Macaulay local ring, let $M$ be a finitely generated $R$-module of dimension $d$, and let $U_i(M)$, $0\leq i\leq d-1$, be the Cohen-Macaulay deviated sequence of $M$. We define the {\it unmixed degree} of $M$ with respect to an $\mathfrak{m}$-primary ideal $I$, denoted $\operatorname{udeg}(I, M)$, as follows: $$\operatorname{udeg}(I, M)=e(I, M)+\sum_{i=0}^{d-1}\delta_{i,\dim U_i(M)}e(I, U_i(M)).$$
\end{definition}

It is shown in \cite[Theorem 5.18]{CQ17} that $\operatorname{udeg}(I,\bullet)$ is an extended degree (when the residue field is infinite). We make an elementary but important observation that, for a fixed finitely generated module $M$, $\operatorname{hdeg}(I, M)$ (resp. $\operatorname{udeg}(I, M)$) is a {\it finite} sum $\sum_i e(I, M_i)$, where $\{M_i\}$ only depends on $M$: this is clear from the definition for $\operatorname{udeg}(I, M)$ and is easily seen by induction for $\operatorname{hdeg}(I, M)$. Therefore, by the associativity formula for multiplicities, for a fixed finitely generated $R$-module $M$, there exists a {\it finite} collection of prime ideals $\Lambda(M)=\Lambda \subseteq \mathrm{Supp(M)}$ (allowing repetition) such that
\begin{equation}
\label{eqn:Finite}
\operatorname{hdeg}(I, M)=\sum_{P\in\Lambda}e(I, R/P), \text{ and similarly for $\operatorname{udeg}(I, M)$}.
\end{equation}

Now we are ready to state and prove our main result in this section. We recall that a finitely generated $R$-module $M$ is called {\it quasi-unmixed} if $\widehat{M}$ is equidimensional. This is equivalent to the condition that $\widehat{\overline{R}}$ be equidimensional where $\overline{R}=R/\operatorname{Ann}(M)$.

\begin{theorem}
\label{Thm:Vogel}
Let $(R,\mathfrak{m})$ be a Noetherian local ring, and let $M$ be a finitely generated quasi-unmixed $R$-module. Then we have
$$n(M)=\sup_{\sqrt{I+\operatorname{Ann}(M)}=\mathfrak{m}}\left\{\frac{l(M/IM)}{e(I, M)} \right\}<\infty.$$
\end{theorem}
\begin{proof}
By passing to the $\frak m$-adic completion, we can assume that $R$ is a complete local ring and $M$ is equidimensional. We can assume also that the residue field is infinity. We now consider $\operatorname{Deg}(I, M)=\operatorname{hdeg}(I, M)$ (or $\operatorname{Deg}(I, M)=\operatorname{udeg}(I, M)$), which is an extended degree.  Thus, by Definition \ref{Def:Edeg} (2) we know that, for every generic element $x\in I-\mathfrak{m} I$ of $M$, we have $$\operatorname{Deg}(I, M)\geq \operatorname{Deg}(I, M/xM).$$ Therefore, for a generic sequence of elements $x_1,\dots,x_d$ of $M$ (we may choose $x_i$ sufficiently general such that $x_1,\dots,x_d$ is a system of parameters of $M$), we have
$$\operatorname{Deg}(I, M)\geq \operatorname{Deg}(I, M/x_1M)\geq \cdots \geq \operatorname{Deg}(I, M/(x_1,\dots,x_d)M)=l(M/(x_1,\dots,x_d))M \geq l(M/IM),$$
where the equality is because $M/(x_1,\dots,x_d)M$ is Cohen-Macaulay and, thus
$$\operatorname{Deg}(I, M/(x_1,\dots,x_d)M)=e(I, M/(x_1,\dots,x_d)M)=l(M/(x_1,\dots,x_d)M).$$
Thus, it is enough to prove that
$$\sup_{\sqrt{I}=\mathfrak{m}}\left\{\frac{\operatorname{Deg}(I, M)}{e(I, M)}\right\}<\infty.$$
At this point we invoke (\ref{eqn:Finite}): it is enough to prove that, for every $P\in\operatorname{Supp} (M)$,
\begin{equation}\label{eqn:supR/P}
\sup_{\sqrt{I}=\mathfrak{m}}\left\{\frac{e(I, R/P)}{e(I, M)}\right\}<\infty.
\end{equation}

In order to prove (\ref{eqn:supR/P}), we use induction on $\dim M$. If $\dim M=0$, (\ref{eqn:supR/P}) is obvious. In the general case, if $\dim R/P=\dim M$ then $e(I, R/P) \le e(I, M)$ by the associativity of multiplicities, so (\ref{eqn:supR/P}) is again obvious. Now we assume $\dim R/P<\dim R$.  We choose a prime ideal $P_0 \in \operatorname{Ass} M$ such that $\dim R/P_0 =\dim M$ and $P_0 \subseteq P$. We have $e(I, R/P_0) \le e(I,M)$ by the associativity of multiplicities again. Therefore, it is enough to prove that, for every $P\in\operatorname{Spec} R$,
\begin{equation}\label{eqn:supR/P}
\sup_{\sqrt{I}=\mathfrak{m}}\left\{\frac{e(I, R/P)}{e(I, R)}\right\}<\infty,
\end{equation}
where $R$ is a complete local domain and $\dim R/P < \dim R$. We pick $0\neq x\in P$ and a minimal prime $Q$ of $(x)$ such that $Q\subseteq P$. Since $R$ is a complete local domain, $R/(x)$ is equidimensional; in particular, $\dim R/(x)=\dim R/Q$, and, thus, $e(I, R/(x))\geq e(I, R/Q)$. Now we write
$$
\frac{e(I, R/P)}{e(I, R)}=\frac{e(I, R/P)}{e(I, R/Q)}\cdot\frac{e(I, R/Q)}{e(I, R/(x))}\cdot \frac{e(I, R/(x))}{e(I, R)}\leq \frac{e(I, R/P)}{e(I, R/Q)}\cdot\frac{e(I, R/(x))}{e(I, R)}.
$$

Since $\dim R/Q<\dim R$, $\sup_{\sqrt{I}=\mathfrak{m}}\left\{\frac{e(I, R/P)}{e(I, R/Q)}\right\}<\infty$ by induction, which means there exists a constant $c_1$ such that $\frac{e(I, R/P)}{e(I, R/Q)}\leq c_1$ for all $\mathfrak{m}$-primary ideals $I$. Since $x$ is a nonzerodivisor in a complete local ring $R$, by Lemma \ref{lem:Huneke} below, we know that there exists a constant $c_2$ such that $\frac{e(I, R/(x))}{e(I, R)}\leq c_2$ for all $\mathfrak{m}$-primary ideals $I$. Thus, putting $c=c_1c_2$ we see that
$$
\frac{e(I, R/P)}{e(I, R)}\leq c
$$
for all $\mathfrak{m}$-primary ideals $I$. This finishes the proof.
\end{proof}

\begin{lemma}\label{lem:Huneke}
Let $(R,\mathfrak{m})$ be a Noetherian complete local ring, and let $x$ be a nonzerodivisor on $R$. Then there exists a constant $k$ such that, for all $\mathfrak{m}$-primary ideals $I$, we have $$e(I, R/(x))\leq k\cdot e(I, R).$$
\end{lemma}
\begin{proof}
We consider the short exact sequence:
$$
0\to \frac{R}{I^n:x} \xrightarrow{\cdot x}\frac{R}{I^n}\to \frac{R}{I^n+(x)}\to 0
$$
Note that if $y\in I^n:x$, then $xy\in I^n\cap(x)$. By Huneke's uniform Artin-Rees lemma \cite[Theorem 4.12]{Hun92}, there exists a constant $k$ such that, for all $I\subseteq R$, $I^n\cap(x)\subseteq I^{n-k}x$. Thus, $xy\in I^{n-k}x$, and so $y\in I^{n-k}$ since $x$ is a nonzerodivisor. This shows that $I^n:x\subseteq I^{n-k}$ for all $\mathfrak{m}$-primary ideals $I$. By the short exact sequence above, we know that
$$
l\left(\frac{R}{I^n+(x)}\right)\leq l\left(\frac{R}{I^n}\right)-l\left(\frac{R}{I^{n-k}}\right)
$$
Now we let $n\to\infty$ and compute the corresponding Hilbert function to see that
$$e(I, R/(x))\leq k\cdot e(I, R)$$
for all $\mathfrak{m}$-primary ideals $I$.
\end{proof}

\begin{remark}\label{GlobalSV}
Assume that $R$ is a (not necessarily local) Noetherian ring such that
$R$ is a homomorphic image of a Noetherian
Gorenstein ring $S$ with $\dim(S) < \infty$. Further assume that $M$ is a finitely generated $R$-module such that $R/\operatorname{Ann}(M)$ is locally equidimensional and satisfies the uniform
Artin-Rees property (e.g., \cite[Theorem 4.12]{Hun92}) and that, for all $P \in \operatorname{Supp}(M)$, the residue field of $R_P$ is infinite. Then the proof of Theorem~\ref{Thm:Vogel} actually gives us a (global) upper bound of $n(M_P)$ for all $P \in \operatorname{Supp}(M)$. Also see Remark~\ref{SVbyInduction-g} for an alternative treatment.
\end{remark}

Given Theorem \ref{Thm:Vogel}, it is quite natural to ask whether the supremum is actually a maximum, i.e., whether $n(M)$ is attained at some $I$. We do not know the answer to this question. Below we prove a special case. Recall that a finitely generated $R$-module $M$ is called {\it generalized Cohen-Macaulay} if $H_{\mathfrak{m}}^i(M)$ has finite length for all $i<\dim (M)$.

\begin{theorem}
\label{max}
Let $(R,\mathfrak{m})$ be a Noetherian local ring with infinite residue field, and let $M$ be a finitely generated $R$-module.
If $M$ is generalized Cohen-Macaulay (e.g., $\dim(M) = 1$), then $n(M)$ is attained, i.e., $n(M) = l(M/IM)/e(I,M)$ for some
$\mathfrak{m}$-primary ideal $I$.
\end{theorem}
\begin{proof}
Since the residue field of $R$ is infinite, every $\mathfrak{m}$-primary ideal $I$ has a minimal reduction $(\underline{x})$ generated by a system of parameters of $M$. Because $\frac{l(M/IM)}{e(I, M)}\leq \frac {l(M/(\underline{x})M)}{e((\underline{x}),M)}$ for any minimal reduction $(\underline{x})$ of $I$, we have
\[
n(M) = \sup\left\{\frac {l(M/(\underline{x})M)}{e((\underline{x}),M)} \ \Big | \
\underline{x} \text{ is a system of parameter of } M\right\}.
\]
When $M$ is Cohen-Macaulay, it is easy to see that $n(M) = 1 = \frac{l(M/(\underline{x})M)}{e((\underline{x}),M)}$ for any ideal $(\underline{x})$ generated by a system of parameters of $M$. Thus, we assume that $M$ is not Cohen-Macaulay; hence, $n(M) > 1+\varepsilon$ for some $\varepsilon > 0$.

Since $M$ is generalized Cohen-Macaulay, it is well known that there exists $C > 0$ (e.g., $C = \sum_{i=0}^{d-1}\binom{d-1}{i}l(H_\mathfrak{m}^{i}(M))$ \cite[Theorem 3.18]{Schenzel98})
such that
\[
l(M/(\underline{x})M)  \le e((\underline{x}),M) + C
\qquad \text{hence} \qquad
\frac {l(M/(\underline{x})M)}{e((\underline{x}),M)} \le 1 + \frac C{e((\underline{x}),M)}
\]
for all systems of parameters $\underline{x}$ of $M$ (for example see \cite{Schenzel83} or \cite{SV78}). This shows that $n(M) \le
1 + \frac C{e(M)} < \infty$. There exists a positive
integer $N$ such that $1+ \frac Cn < 1+ \varepsilon$ for all $n > N$.
Therefore
\[
n(M) = \sup\left\{\frac {l(M/(\underline{x})M)}{e((\underline{x}),M)} \ \Big | \
\underline{x} \text{ is a system of parameters of $M$ such that }
e((\underline{x}),M) \le N \right\}.
\]
However, the set of numbers on the right side is finite, so $n(M)$ is
must be attained at some system of parameters of $M$.
\end{proof}

\subsection{The behavior of $n(M)$ under base change} In this subsection we study the behavior of $n(M)$ under localization, flat local extension, and the killing of a parameter. We begin with a result on localization.

\begin{theorem}
\label{Thm:localization}
Let $(R, \mathfrak{m})$ be a Noetherian local ring and $M$ a finitely generated $R$-module. Then for any $P \in \operatorname{Supp} (M)$, we have $n_{R_P}(M_P) \leq n_R(M)$.
\end{theorem}
\begin{proof}
We can assume $M$ is quasi-ummixed since otherwise $n(M)=\infty$ by \cite[Theorem 1]{SV78}, in which case there is nothing to prove. We can replace $R$ by $R/\operatorname{Ann}(M)$: this does not affect $n(M)$ or $n(M_P)$. Therefore, we can assume $R$ is quasi-unmixed. By \cite[Theorem 31.6]{Matsumura86}, this implies $R$ is equidimensional and catenary and that $R_P$ is also quasi-unmixed. We set $\dim (R)=d$.

By induction it is enough to consider the case of $\dim (R/P)=1$. Since the residue field of $R_P$ is infinite, it suffices to show that
\[
n(M) \geq \frac{{l}_{R_P}(M_P/IM_P)} { e(I, M_P)}\]
for all ideals $I$ generated by a system of parameters in $R_P$ (as in the proof of Theorem \ref{max}). We know $\dim R_P=\operatorname{ht} P=d-1$ since $R$ is equidimensional and catenary; thus, for any such $I$, by prime avoidance, we can find elements $x_1, \ldots, x_{d-1} \in R$
 that  form part of a system of parameters in $R$ and that have images in $R_P$ that generate $I$.
So, abusing notation a bit, we will call $I = (x_1, \ldots, x_{d-1}) \subseteq R$.

Suppose $x\in R$ is such that $(I, x)$ is $\mathfrak{m}$-primary. Since $M$ is faithful, $x_1,\dots,x_{d-1}, x$ form a system of parameters on both $R$ and $M$. We have
\[
{l}(M/(I, x)M) \geq e(x, M/IM)  = \sum_{Q \in \operatorname{Min}(M/IM)} e(x, R/Q){l}(M_Q/IM_Q),
\]
where the equality holds by the additivity property of multiplicity. By Lech's associativity formula for multiplicities for parameter ideals \cite{Lech57} (see also \cite{Nagata56}), we also have
\[
e ((I,x), M) = \sum_{Q \in \operatorname{Min}(M/IM)} e(x, R/Q)e(IR_Q, M_Q).
\]
Therefore, by definition,
\begin{align*}
n(M) &\geq \frac{{l}(M/(I,x)M)}{e((I,x), M)}
\geq \frac{\sum_{Q \in \operatorname{Min}(M/IM)} e(x, R/Q){l}_{R_Q} (M_Q/IM_Q)}{\sum_{Q \in \operatorname{Min}(M/IM)}e(x, R/Q)e(IR_Q, M_Q)}.
\end{align*}
Now we let $ \displaystyle y \in \bigcap_{\substack{Q\in \operatorname{Min}(M/IM)\\Q \neq P}} Q \setminus P$ and $\displaystyle z \in P \setminus \bigcup_{\substack{Q\in \operatorname{Min}(M/IM)\\Q \neq P}}  Q$.
Observe that, for any $t \geq 1$, we can use $x=y^t + z$ to complete $I$ to a full system of parameters. In this case
\begin{equation*}
\frac{\sum_{Q} e(x, R/Q){l}_{R_Q} (M_Q/IM_Q)}{\sum_{Q}e(x, R/Q)e(IR_Q, M_Q)}
= \frac
{e(y^t, R/P){l}_{R_P} (M_P/IM_P) + \sum_{Q \neq P} e(z, R/Q){l}_{R_Q} (M_Q/IM_Q)}
{e(y^t, R/P)e(IR_P, M_P) + \sum_{Q \neq P}e(z, R/Q)e(IR_Q, M_Q)}.
\end{equation*}
Since $e(y^t, R/P) = t e (y, R/P)$, if we pass to the limit as $t$ approaches infinity we obtain
\[
n(M) \geq \frac{{l}_{R_P} (M_P/IM_P)}{e(IR_P, M_P)}.
\qedhere
\]
\end{proof}

As a consequence, we show that the invariant $n(-)$ is non-decreasing under flat local extensions.

\begin{corollary}
\label{cor.flat}
Let $(R, \mf m)\to (S,\mf n)$ be a flat local extension of Noetherian local rings. Suppose $M$ is a finitely generated $R$-module. Then $n_R(M) \leq n_S(M \otimes_R S)$.
\end{corollary}
\begin{proof}
Let $P$ be a minimal prime of $\mf mS$.  By Theorem \ref{Thm:localization}, we have
\[n_{S_P}((M \otimes_R S)_P) \leq n_S(M \otimes_R S).\]
Thus, replacing $S$ by $S_P$, we may assume that $S$ is local and that $\mf mS$ is $\mf n$-primary.
For any $\mf m$-primary ideal $I$, its extension $IS$ is an $\mf n$-primary ideal, and
tensoring the composition series with $S$ shows that
\[
{l}_R (M/IM){l}_S (S/\mf mS) = {l}_S \left((M\otimes_R S)/I(M\otimes_R S)\right)
\]
for any finitely generated $R$-module $M$.
Thus, $e (I, M){l}_S (S/\mf mS) = e(IS, M\otimes_R S)$ and
\begin{align*}
n(M) &= \sup_{\sqrt{I+\operatorname{Ann}(M)}=\mathfrak{m}} \left\{\frac{{l}_R (M/IM)}{e(I, M)}\right\} \\
&=
\sup_{\sqrt{I+\operatorname{Ann}(M)}=\mathfrak{m}} \left\{ \frac{{l}_S \left ((M \otimes_R S)/I(M \otimes_R S)\right )}{e (IS, M\otimes_RS)} \right\} \\
   &\leq  \sup_{\sqrt{IS+\operatorname{Ann}(M)S}=\mathfrak{n}} \left\{ \frac{{l}_S \left ((M \otimes_R S)/I(M \otimes_R S)\right )}{e (IS, M\otimes_RS)} \right\}= n(M \otimes_R S). \qedhere
\end{align*}
\end{proof}

Given a flat local extension $(R,\mathfrak{m})\to (S, \mathfrak{n})$ and a finitely generated $R$-module $M$, it would also be interesting to bound $n(M\otimes_RS)$ in terms of $n(M)$ and $n(S/\mathfrak{m} S)$. We do not know how to obtain such a relation yet. Our last result in this subsection relates $n(M)$ and $n(M/xM)$ for a parameter $x$ on $M$ (i.e., $\dim (M/xM)=\dim(M)-1$ or, equivalently, $x$ is a parameter on $R/\operatorname{Ann}(M)$).

\begin{proposition}
Let $(R, \mf m)$ be a Noetherian local ring, and let $M$ be a finitely generated $R$-module of dimension $d$.
Then for any parameter $x$ of $M$, we have $n(M/xM) \leq n(M)$.
\end{proposition}
\begin{proof}
Replacing $R$ by $R/\operatorname{Ann}(M)$, we may assume $M$ is a faithful $R$-module. Hence $x$ is a parameter on $R$ as well.
Since $e(I, M/xM)\geq e(I, M)$ for every $\mathfrak{m}$-primary ideal $I$ that contains $x$, we have
\[
\frac{l((M/xM)/I(M/xM))}{e(I, M/xM)} =\frac{l(M/IM)}{e(I, M/xM)}
\le \frac{l(M/IM)}{e(I, M)} \leq n(M).
\]
This clearly implies $n(M/xM) \leq n(M)$, as desired.
\end{proof}

\begin{comment}
\begin{question}
Let $(R, \mf m) \to (S, \mf n)$ be a local flat map and $M$ a finitely
generated $R$-module. Is it true that $n(M \otimes_R S) \leq
n(M)n(S/\mf mS)$? (Here we may either assume infinite
residue fields or just define $n(-)$ with sop ideals.)
\end{question}

For example, is it true when $S$ is module-finite and free over $R$. A
concrete case could be $S = R[x]/(x^2)$. Positive answer in this case?

The above question has a positive answer if both $M$ and $S/\mf mS$ are
CM, as $1 = 1 \cdot 1$.
We also have the following

\begin{lemma}\label{lma:artinian}
Let $(R, \mf m)$ be a local ring and $(S, \mf n)$ be a faithfully flat
extension of $R$.
Suppose $M \neq 0$ is an $R$-module with $l_R(M) < \infty$.
Then $n(M \otimes_R S) \leq n(S/\mf mS) = n(M)n(S/\mf mS)$.
\end{lemma}

\begin{proof}
Say $l(M) = c$, hence $M$ has a filtration of length $c$ with each
consecutive quotient isomorphic to $R/\mf m$. Thus there is an induced
filtration of $M \otimes S$ of length $c$ with each
consecutive quotient isomorphic to $S/\mf mS$.

Let $I$ be any $\mf n$-primary ideal of $S$. Via the above filtration,
we see
\begin{align*}
\frac{l((M \otimes S)/I(M \otimes S))}{e(I,M \otimes S)}
&\leq \frac{c \cdot l((S/\mf mS)/I(S/\mf mS))}{c \cdot e(I,S/\mf mS)} \\
&=\frac{l((S/\mf mS)/I(S/\mf mS))}{e(I,S/\mf mS)}
\leq n(S/\mf mS).
\end{align*}
Therefore $n(M \otimes_R S) \leq n(S/\mf mS) = n(M)n(S/\mf mS)$.
\end{proof}

\begin{lemma}\label{lma:sop+sop }
Let $(R, \mf m)$ be a local ring and $(S, \mf n)$ be a faithfully flat extension of $R$.
Suppose $M \neq 0$ is a finitely generated $R$-module, and denote $M_S
= M \otimes_R S$.
Then for all (full) sop $x$ of $M$ and all (full) sop $y$ of $S/\mf mS$,
\[
\frac{l_S(M_S/(x,y)M_S)}{e((x,y), M_S)} \leq  n(M)n(S/\mf mS).
\]
\end{lemma}

\begin{proof}
Denote $N = M/xM$, so that $N$ has finite length over $R$. By
Lemma~\ref{lma:artinian} and associativity formula, we see
\begin{align*}
l_S(M_S/(x,y)M_S) &= l_S(N_S/(y)N_S) \\
&\leq e((y), N_S)n(S/\mf mS) \\
&= \sum_{P}e((y), S/P)l(N_{S_P}) n(S/\mf mS)
\quad \text{(associativity formula)}\\
&=\sum_{P}e((y), S/P)l(M_{S_P}/(x)M_{S_P}) n(S/\mf mS) \\
& \leq \sum_{P}e((y), S/P)e((x), M_{S_P}) n(M) n(S/\mf mS)  \\
&= e((x,y),M_S) n(M) n(S/\mf mS)
\quad \text{(associativity formula)}\\
\end{align*}
Therefore $\frac{l_S(M_S/(x,y)M_S)}{e((x,y), M_S)} \leq  n(M)n(S/\mf mS)$.
\end{proof}
\end{comment}

\section{The lower bound: a generalization of Lech's inequality}

Our goal in this section is to generalize Lech's inequality to all finitely generated $R$-modules, thus proving the first part of Theorem A in the introduction. We first prove a key lemma.

\begin{lemma}
\label{lem:CDHZ}
Let $(R,\mathfrak{m}, k)$ be a complete local domain with an algebraically closed residue field. Let $M$ be a finitely generated $R$-module with $\dim(R) = \dim(M)$, and let $J$ be an integrally closed $\mathfrak{m}$-primary ideal. Then we have $$l(M/JM)\geq l(R/J)\dim_K(M\otimes_RK),$$ where $K$ denotes the fraction field of $R$.
\end{lemma}
\begin{proof}
First of all, if we let $T(M)$ denote the torsion submodule of $M$, then we have $$0\to T(M)\to M\to M'\to 0$$ where $M'$ is torsion-free. Since $l(M/JM)\geq l(M'/JM')$ while $\dim_K(M\otimes_RK)=\dim_K(M'\otimes K)$, if the lemma holds for $M'$ then it also holds for $M$. Thus, in the rest of the proof we assume $M$ is torsion-free. In this case $\dim_K(M\otimes K)=\operatorname{rank} M$.

By \cite[Corollary 2.2]{CDHZ12}, we have
$$l(M/JM)\geq \bar{l}(R/J)\cdot\operatorname{rank} M,$$
where $\bar{l}(R/J)$ denotes the length of the longest chain of integrally closed ideals between $J$ and $R$. Therefore, it is enough to show $\bar{l}(R/J)=l(R/J)$. To prove this it is enough to find an integrally closed ideal $J'\supseteq J$ in $R$ such that $l(J'/J)=1$ because then $\bar{l}(R/J)=l(R/J)$ follows from an easy induction. Let $R\to S$ be the normalization of $R$. Since $R$ is a complete local domain, $S$ is local by \cite[Proposition 4.8.2]{HS06}, and so $S=(S,\mathfrak{n})$ is a normal local domain with $R/\mathfrak{m}=S/\mathfrak{n}=k$ since $k$ is algebraically closed. Now by \cite[Theorem 2.1]{Wat03}, there exists a chain $$\overline{JS}=J_0\subseteq J_1\subseteq J_2\subseteq\cdots \subseteq J_n=\mathfrak{n}$$ such that
\begin{enumerate}
  \item Each $J_i$ is integrally closed in $S$;
  \item $l(J_{i+1}/J_i)=1$ for every $i$.
\end{enumerate}
Since $J$ is integrally closed in $R$ and $S$ is integral over $R$, by \cite[Proposition 1.6.1]{HS06} we know
$$J_0\cap R=\overline{{JS}}\cap R=\bar{J}=J.$$
Let $t=\max\{i \mid J_i\cap R=J\}$.  Obviously $0\leq t<n$. Set $J'=J_{t+1}\cap R$.  It is easy to see that $J'\supseteq J$ is integrally closed in $R$ (one can use \cite[Proposition 1.6.1]{HS06} again). Moreover, $l(J'/J)>0$ by our choice of $t$ while $J'/J\hookrightarrow J_{t+1}/J_t$ shows that $l(J'/J)\leq l(J_{t+1}/J_t)=1$. Thus, we have $l(J'/J)=1$.
\end{proof}

We define $\operatorname{Assh}(M) = \{P \in \operatorname{Ass}(M) \mid \dim(R/P) = \dim(M)\}$ for a finitely generated $R$-module $M$. We are now ready to state and prove the following generalization of Lech's inequality.
\begin{theorem}
\label{Thm:LechMod}
Let $(R,\mathfrak{m}, k)$ be a Noetherian local ring, and let $M$ be a finitely generated $R$-module of dimension $d$. Then for every ideal $I$ of $R$ whose image in $\overline{R}=R/\operatorname{Ann}(M)$ is $\mathfrak{m}$-primary, we have $$e(I, M)\leq d!e(\overline{R})l(M/IM).$$
\end{theorem}
\begin{proof}
Replacing $R$ by $\overline{R}$ does not change either side of the inequality. Therefore, we may assume $\operatorname{Ann}(M)=0$ and so that $\dim(R)=\dim(M)=d$. We next take a flat local extension $(R,\mathfrak{m},k)\to (R', \mathfrak{m}',k')$ such that $\mathfrak{m}'=\mathfrak{m} R'$ and $k'=R'/\mathfrak{m}'$ is the algebraic closure of $R/\mathfrak{m}=k$.  (Such an $R'$ always exists: it is a suitable gonflement of $R$; see \cite[Corollaire in Appendice 2]{Bou06}). Then $R\to R'\to \widehat{R'}$ is a faithfully flat extension with $\mathfrak{m}\widehat{R'}=\mathfrak{m}_{\widehat{R'}}$, so passing from $R$ to $\widehat{R'}$ and replacing $M$ by $M\otimes_R \widehat{R'}$ do not affect either side of the inequality. Therefore, without loss of generality, we may assume $(R,\mathfrak{m}, k)$ is a complete local ring with $k=\bar{k}$ and $\operatorname{Assh}(R)=\operatorname{Assh}(M)$.

By the associativity formula of multiplicity, we have
$$e(I, M)=\sum_{P\in \operatorname{Assh}(M)}l_{R_P}(M_P)e(I, R/P)=\sum_{P\in \operatorname{Assh}(R)} l_{R_P}(M_P)e(\overline{IR/P}, R/P).$$
Using Lech's inequality \cite[Theorem 3]{Lech60} for each $R/P$, we have
\begin{equation}
\label{eqn:Associativity}
e(I, M)\leq \sum_{P\in \operatorname{Assh}(R)} d!e(R/P)l\left((R/P)/(\overline{IR/P})\right)l_{R_P}(M_P).
\end{equation}
\begin{claim}
For every minimal prime $P$ of $R$, we have
\begin{equation}\label{eqn:Claim}
l\left((R/P)/(\overline{IR/P})\right)\cdot l_{R_P}(M_P)\leq l(M/IM)\cdot l_{R_P}(R_P).
\end{equation}
\end{claim}
\begin{proof}[Proof of Claim]
Clearly we have $l_{R_P}(M_P)\leq l_{R_P}(R_P)\cdot l_{R_P}(M_P/PM_P)$ because $l_{R_P}(M_P/PM_P)$ is the minimal number of generators of $M_P$ as an $R_P$-module. Therefore,
$$l\left((R/P)/(\overline{IR/P})\right)\cdot l_{R_P}(M_P)\leq l\left((R/P)/(\overline{IR/P})\right)\cdot l_{R_P}(M_P/PM_P)\cdot l_{R_P}(R_P).$$
Now $R/P$ is a complete local domain with algebraically closed residue field $k=\bar{k}$, and $M/PM$ is a finitely generated $R/P$-module. Applying Lemma \ref{lem:CDHZ} and noting that $\dim_{\kappa(P)}(M/PM)\otimes \kappa(P)=l_{R_P}(M_P/PM_P)$, we have
$$l\left((R/P)/(\overline{IR/P})\right)\cdot l_{R_P}(M_P/PM_P) \leq l\left( \frac{M/PM}{(\overline{IR/P})(M/PM)}\right)\leq l\left(\frac{M}{(I+P)M}\right)\leq l(M/IM).$$ Putting the two inequalities above together, we get $$l\left((R/P)/(\overline{IR/P})\right)\cdot l_{R_P}(M_P)\leq l(M/IM)\cdot l_{R_P}(R_P).$$ This finishes the proof of the Claim.
\end{proof}
Finally, we plug in (\ref{eqn:Claim}) to (\ref{eqn:Associativity}) and use additivity of multiplicity to get
\begin{align*}
  e(I, M) &\leq  \sum_{P\in \operatorname{Assh}(R)} d! e(R/P) l_{R_P}(R_P)l(M/IM) \\
   &= d! l(M/IM) \left(\sum_{P\in \operatorname{Assh}(R)} l_{R_P}(R_P) e(R/P)\right)
   = d! e(R) l(M/IM).
\end{align*}
This finishes the proof.
\end{proof}

\section{Applications and an alternative approach}

In this section we give some applications of our results in Section 2 and Section 3. In the process, we obtain another way to prove Conjecture \ref{conj:Vogel} without using Vasconcelos's extended degree (see Remark~\ref{SVbyInduction}).

\begin{lemma}
\label{lem:Length}
If $(R, \mathfrak{m})$ is a Noetherian local ring and $M$ is a finitely generated quasi-unmixed $R$-module of dimension $d$, then there exists a constant $C$ such that
for every $h \in \mathbb R$
\[
\frac hC \le \inf_{\substack{\sqrt{I}= \sqrt{J} =\mathfrak{m} \\ e(I, M) \ge h e(J,M)}}
\left\{\frac{l(M/IM)}{l(M/JM)} \right\}
\quad \text{and} \quad
\sup_{\substack{\sqrt{I}= \sqrt{J} =\mathfrak{m} \\ e(I, M) \le h e(J,M)}}
\left\{\frac{l(M/IM)}{l(M/JM)} \right\} \le hC.
\]
In particular, there exists a constant $C$ such that such that for all $\frak m$-primary ideals $I$ and for all $n \ge 1$ we have
$$\frac{n^d}{C} \le \frac{l(M/I^nM)}{l(M/IM)} \le Cn^d.$$
\end{lemma}

\begin{proof}
Let $\dim(M) = \dim (\overline{R})=d$ where $\overline{R}=R/\operatorname{Ann}(M)$. We use Theorem \ref{Thm:Vogel} and Theorem \ref{Thm:LechMod} to see
\[
l(M/IM) \geq  e(I,M)/(d! e(\overline{R})) \ge h  e(J,M)/(d! e(\overline{R}))
\geq h l(M/JM) /(n(M)  d! e(\overline{R})),
\]
which proves the first inequality. By symmetry, we have
\[
l(M/IM) \leq n(M) e(I,M) \le n(M)h  e(J,M) \leq h n(M)  d! e(\overline{R})  l(M/JM),
\]
which proves the second inequality.
So we can take $C=n(M)  d! e(\overline{R})$ in both cases. Finally, taking $h=n^d$ and noting that $e(I^n,M) = n^d e(I,M)$ immediately proves the last claim.
\end{proof}

This lemma has an immediate consequence, which we will need.

\begin{corollary}
\label{cor.integralclosure}
If $(R,\mathfrak{m})$ is a Noetherian local ring and $M$ is a finitely generated quasi-unmixed $R$-module, then
$$\sup_{\substack{ \sqrt{I}=\mathfrak{m} \\ I\subseteq J\subseteq \overline{I}}}
\left\{\frac{l(M/IM)}{l(M/JM)} \right\}<\infty.$$
\end{corollary}
\begin{proof}
The condition $I\subseteq J\subseteq \overline{I}$ implies $e(I, M)=e(J, M)$. Therefore we apply Lemma \ref{lem:Length} with $h=1$ to get the desired claim.
\end{proof}

Next we prove a result that extends Lemma \ref{lem:CDHZ}
at the cost of precision in the inequality.

\begin{lemma}\label{lem:Length2}
Let $(R, \mathfrak{m})$ be a Noetherian local ring and $N$ be a
finitely generated quasi-unmixed $R$-module.
Then there exists a constant $C_N>0$ depending only on $N$ such that
\[
1/C_N \le \inf_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(M/IM)}{l(N/IN)} \right\}
\quad \text{and, equivalently,} \quad
\sup_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(N/IN)}{l(M/IM)} \right\} \le C_N
\]
for all finitely generated $R$-modules $M$ with $\operatorname{Supp}(M) \supseteq \operatorname{Supp}(N)$.
\end{lemma}

\begin{proof}
Since $N$ is quasi-unmixed, it is equidimensional. We set $c = \max_{P \in \operatorname{Min}(N)} l_{R_P}(N_P)$.
Let $M$ be any finitely generated $R$-module $M$ with $\operatorname{Supp}(M) \supseteq
\operatorname{Supp}(N)$. Denote $\overline M = M/\mathfrak{a} M$, where $\mathfrak{a} = \cap_{P \in
  \operatorname{Min}(N)}P$.
Then $\operatorname{Supp}(\overline M) = \operatorname{Supp}(N)$, and,
by the associativity formula for multiplicities,
\begin{align*}
ce(I,\overline M) =\sum_{P \in \operatorname{Min}(N)} c \cdot l_{R_P}(\overline M_P) e(I,R/P)
\ge \sum_{P \in \operatorname{Min}(N)} l_{R_P}(N_P) e(I,R/P)
= e(I,N)
\end{align*}
for all $\mathfrak{m}$-primary ideals $I$.
Now we use Theorem~\ref{Thm:Vogel} and Theorem~\ref{Thm:LechMod} to
obtain (with $d = \dim(N)$)
\[
l(N/IN) \le n(N) e(I,N) \le n(N)c e(I, \overline M)
\le n(N)c d! e(\overline{R}) l(\overline M/I\overline M)
\le n(N)c d! e(\overline{R}) l(M/IM)
\]
where $\overline{R} = R/\mathfrak{a}$ depends only on $N$.
So we can take $C_N={n(N)cd!e(\overline R)}$.
\end{proof}

Lemma~\ref{lem:Length2} allows us to establish the following general
result:

\begin{theorem}\label{thm:Length2}
Let $(R, \mathfrak{m})$ be a Noetherian local ring,  and let $M$ and $N$ denote
finitely generated $R$-modules. Then
\begin{enumerate}
\item $0 < \inf_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(M/IM)}{l(N/IN)} \right\}
\iff \sup_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(N/IN)}{l(M/IM)} \right\} < \infty
\iff \operatorname{Supp}(M) \supseteq \operatorname{Supp}(N)$.
\item There exists a constant $C>0$ depending only on $N$ such that
\[
1/C \le \inf_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(M/IM)}{l(N/IN)} \right\}
\quad \text{and, equivalently,} \quad
\sup_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(N/IN)}{l(M/IM)} \right\} \le C
\]
for all (finitely generated $R$-modules) $M$ with $\operatorname{Supp}(M) \supseteq \operatorname{Supp}(N)$.
\item $0 < \inf_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(M/IM)}{l(N/IN)} \right\}
\le \sup_{\sqrt{I} =\mathfrak{m}}\left\{ \frac{l(M/IM)}{l(N/IN)} \right\} < \infty
\iff \operatorname{Supp}(M) = \operatorname{Supp}(N)$.
\end{enumerate}
\end{theorem}

\begin{proof}
(1): Clearly, we only need to prove the second equivalence. For the
forward direction, assume
$\sup_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(N/IN)}{l(M/IM)} \right\} <
\infty$ and let $P \in \operatorname{Supp}(N)$. Denote $\overline M = M/PM$ and $\overline N =
N/PN$. As $\left\{\frac{l(N/IN)}{l(M/IM)} \mid \sqrt{I} =\mathfrak{m}\right \}
\supseteq \left\{\frac{l(\overline N/I\overline N)}{l(\overline M/I\overline M)}
\mid \sqrt{I} =\mathfrak{m}\right \}$, we get
$\sup_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(\overline N/I\overline N)}{l(\overline M/I\overline M)} \right\} <
\infty$, which implies $\dim(\overline N) \le \dim(\overline M)$ by considering
$\frac{l(\overline N/I\overline N)}{l(\overline M/I\overline M)}$ with $I = \mathfrak{m}^t$ for $t \gg 0$.
Note that $\dim(N/PN) = \dim(R/P)$, since $P \in \operatorname{Supp}(N)$. Thus,
$\dim(M/PM) = \dim(R/P)$, which forces $P \in \operatorname{Supp}(M)$.

For the backward direction, assume $\operatorname{Supp}(M) \supseteq \operatorname{Supp}(N)$. We can
further assume that $R$ is complete, which does not affect the
statement. We next take a prime cyclic filtration of $N$ of length $n$ with factors $N_i = R/P_i$ such that $P_i \in
\operatorname{Supp}(N)$ for $i = 1,\dotsc, n$ (note that the $P_i$ are not necessarily distinct). As $l(N/IN) \le
\sum_{i=1}^nl(N_i/IN_i)$ for every $\mathfrak{m}$-primary ideal $I$, it
suffices to show $\sup_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(N_i/IN_i)}{l(M/IM)} \right\} <
\infty$ for each $i = 1, \dotsc, n$. But this follows from
Lemma~\ref{lem:Length2} since each $N_i = R/P_i$ is quasi-unmixed (since
$R$ is complete) and $\operatorname{Supp}(M) \supseteq \operatorname{Supp}(N_i)$. In detail, let
$C_{N_i} > 0$ be as in Lemma~\ref{lem:Length2} for each $i = 1, \dotsc, n$. Then
\[
\sup_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(N/IN)}{l(M/IM)} \right\}
\le \sum_{i=1}^n\sup_{\sqrt{I} =\mathfrak{m}}\left\{\frac{l(N_i/IN_i)}{l(M/IM)}
\right\}
\le \sum_{i=1}^n C_{N_i} < \infty
\]
with $\sum_{i=1}^n C_{N_i} < \infty$ depending only on $N$.

(2): From the proof of (1) above (for the backward direction), we can set $C
= \sum_{i=1}^n C_{N_i}$, which depends only on (the completion of) $N$.

(3): This is clear from (1).
\end{proof}

\begin{lemma}
\label{ses}
Let $(R,\mathfrak{m})$ be a Noetherian local ring and $M$ a finitely generated $R$-module.
If $(y_1, \ldots, y_d) \subseteq (x_1, \ldots, x_k)$ are $\mathfrak{m}$-primary
ideals of $R$, then for all $0 \leq i \leq k$,
\[
l(H_i(x_1, \ldots, x_k;M))
\leq \sum_{j=0}^k {k \choose j} l(H_{i-j}(y_1, \ldots, y_d;M))
\le 2^k \max_{0 \le j \le k}  l(H_{i-j}(y_1, \ldots, y_d;M)),
\]
with the convention that $H_{<0}(y_1, \ldots, y_d; M)=0$.
\end{lemma}

\begin{proof}
If $\underline{f} = f_1, \ldots, f_s$ is any sequence of elements of
$R$ and $\underline{f}^- = f_1, \ldots, f_{s-1}$, then there is
a short exact sequence for each $0 \leq i \leq s-1$
\begin{equation*}
0 \rightarrow \dfrac{H_i(\underline{f}^-;M)}{f_s H_i(\underline{f}^- ;M)} \rightarrow H_i(\underline{f};M) \rightarrow \operatorname{Ann}_{H_{i-1}(\underline{f}^-;M)}(f_s) \rightarrow 0,
\end{equation*}
Using the short exact sequence above, we see that
\begin{align*}
l(H_i(x_1, \ldots, x_k;M))
&\leq l(H_i(x_1, \ldots, x_k;M)) + l(H_{i-1}(x_1, \ldots, x_k;M))\\
&= l(H_i(x_1, \ldots, x_k, y_1;M))
\quad \text{ (since $y_1 \in (x_1, \ldots, x_k)$)}\\
&\leq \dotsb \quad \text{ (by joining $y_2, \dotsc, y_d$ inductively)} \\
&\leq l(H_i(x_1, \ldots, x_k, y_1, \ldots, y_d;M))\\
&= l\left(\dfrac{H_i(x_1, \ldots, x_{k-1}, y_1, \ldots, y_d;M)}
{x_{k} H_i(x_1, \ldots, x_{k-1}, y_1, \ldots, y_d;M)}\right)
+ l\left(\operatorname{Ann}_{H_{i-1}(x_1, \ldots, x_{k-1}, y_1, \ldots, y_d;M)}(x_k)
\right) \\
&\leq l\left(H_i(x_1, \ldots, x_{k-1}, y_1, \ldots, y_d;M)\right)
+ l\left(H_{i-1}(x_1, \ldots, x_{k-1}, y_1, \ldots, y_d;M)\right) \\
&\leq \dotsb \quad \text{ (by removing $x_{k-1}, \dotsc, x_1$ inductively)} \\
&\leq \sum_{j=0}^k {k \choose j} l(H_{i-j}(y_1, \ldots, y_d;M))\\
&\leq 2^k \max_{0 \le j \le k} l(H_{i-j}(y_1, \ldots, y_d;M)),
\end{align*}
completing the proof.
\end{proof}

\begin{remark}[\cite{Schenzel98} or \cite{BHM18}]
\label{sseq}
Let $(R,\mathfrak{m})$ be a local ring of dimension $d$ that is a homomorphic
image of a local Gorenstein ring $S$ of dimension $n$. Then for every
finitely generated $R$-module $M$, every system of parameters $\underline{x} = x_1, \dotsc, x_d$ of
$R$, and every $i = 1, \dotsc, d$, we have
\[
l(H_i(x_1, \ldots, x_d;M)) \leq
\sum_{j = 0}^{d-i} l(H_{d-i-j}(x_1, \ldots, x_d;\operatorname{Ext}_S^{n-j}(M,S))).\footnote{This is written down in \cite{BHM18}. We point out that this also follows from \cite[Theorem 3.16]{Schenzel98} as follows: since $\underline{x}$ is a system of parameters of $R$, we can pick $\underline{y}=y_1,\dots,y_d$ with $(\underline{y})=(\underline{x})$ such that they form a strong filter regular sequence on $R$ and $M$ by prime avoidance.  Replacing $\underline{x}$ by $\underline{y}$ does not affect the Koszul homology, so we can assume $\underline{x}$ is a strong filter regular sequence on $R$ and $M$ and then note that we have a canonical isomorphism $H^j(x_1,\dots,x_d; N^\vee)\cong H_j(x_1,\dots,x_d; N)^\vee$ for all finitely generated $R$-modules $N$ by \cite[bottom of page 286]{Schenzel98}. (In particular, they have the same length.).  Therefore, the displayed formula is a restatement of \cite[Theorem 3.16]{Schenzel98}.}
\]
Note that $\dim(\operatorname{Ext}_S^{n-j}(M,S)) \le d-i$ for each $j = 0, \dotsc, d-i$, since $\operatorname{Ext}_S^{n-j}(M,S)^\vee\cong H_{\mathfrak{m}}^j(M)$ where $(-)^\vee$ stands for Matlis dual.
\end{remark}

\begin{theorem} \label{Thm:HomoBound}
Let $(R, \mathfrak{m})$ be a Noetherian local ring and $M$ a finitely generated $R$-module of dimension $d$.  Then there exists a constant $C$ depending on $M$ such
that, for every $k \geq d$, we have
\[
\sup_{\substack{\sqrt{(x_1,\ldots, x_{k}) + \operatorname{Ann}(M)}=\mathfrak{m}\\ 0 \leq i \leq k}}
\left\{\frac{l(H_i(x_1, \ldots, x_{k};M))}
{l(M/(x_1,\ldots,x_{k})M)} \right\} \le 2^kC.
\]
\end{theorem}

\begin{proof}
As in the first paragraph in the proof of Theorem \ref{Thm:LechMod}, we can replace $R$ by $R/\operatorname{Ann}(M)$, enlarge the residue field of $R$, and then complete $R$. Therefore, we can assume that $R$ is a complete local ring with infinite residue field and that $M$ is a faithful $R$-module. (The rest of the proof
only relies on the fact that $(R,\mathfrak{m})$ is a homomorphic image of a Gorenstein ring $S$ with infinite residue field.)

We proceed by induction on $d= \dim(M)$. When $\dim(M)= 0$, clearly $C = l(M)$ works.  Assume that the theorem holds for modules of dimension $< d$. Now let
$\dim(M) = d$. Let $R/P_1, \ldots, R/P_r$ be the (not necessarily distinct) factors appearing in
a prime cyclic filtration of $M$.  We note that, for each $0 \leq i \leq d$,
\[
\dfrac{l(H_i(x_1, \ldots, x_k;M))}{l(M/(x_1, \ldots, x_{k})M)} \leq \dfrac{\sum_{j=1}^r l(H_i(x_1, \ldots, x_{k};R/P_j))}{l(M/(x_1, \ldots, x_k)M)} \leq \sum_{j=1}^r \dfrac{ l(H_i(x_1, \ldots, x_k;R/P_j))}{l(M/((x_1, \ldots, x_k)+P_j)M)}.
\]
Now each $M/P_jM$ is a finitely generated faithful module over the
local domain $R/P_j$ with infinite residue field.  It then follows
from Lemma~\ref{lem:Length2} or Theorem~\ref{thm:Length2}
(applied to $M/P_jM$ and $R/P_j$) that
we may replace each term $l(M/((x_1, \ldots,
x_k)+P_j)M)$ by $l(R/((x_1, \ldots, x_k)+P_j)R)$ without affecting
the claim of the theorem.  We have now reduced to the
case of $M = R/P_j$ over the local domain $R/P_j$ of dimension $\leq d$.

Therefore, it suffices to verify the case of $M = R$ where $R$ is a
domain with $\dim(R) = d$. (We still have that
$R$ is a homomorphic image of a Gorenstein local ring $S$ with
infinite residue field.)

Let $(y_1, \dotsc, y_d)$ be a minimal reduction of $(x_1, \dotsc,
x_k)$. By Corollary \ref{cor.integralclosure} and Lemma~\ref{ses}, it suffices to
find a constant $D$ such that
\[
\frac{l(H_i(y_1, \ldots, y_{d};R))}
{l(R/(y_1,\ldots,y_{d}))} \le D
\]
for all systems of parameters $\underline{y}:= y_1, \dotsc, y_d$ of $R$ and for all
$i = 0, \dotsc, d$. Now by Remark \ref{sseq}, it suffices to show that, for any
fixed finitely generated $R$-module $L$ with $\dim(L) < d$, there exists a
constant $D_L$ such that
\[
\frac{l(H_i(y_1, \ldots, y_{d};L))}
{l(R/(y_1,\ldots,y_{d}))} \le D_L
\]
independent of $\underline{y}$ and $i$. Indeed, as $\l(L/(y_1,\ldots,y_{d})L)
\le \mu(L)l(R/(y_1,\ldots,y_{d}))$, we have
\[
\frac{l(H_i(y_1, \ldots, y_{d};L))}
{l(R/(y_1,\ldots,y_{d}))}
\le \mu(L) \frac{l(H_i(y_1, \ldots, y_{d};L))}
{l(L/(y_1,\ldots,y_{d})L)}.
\]
Since $\dim(L) < d$, the right hand side of the above inequality is
bounded above (independent of $\underline{y}$ and $i$) by the inductive
hypothesis (noting that $2^d$ is a constant as well).
\end{proof}

It is well known that (for example, see \cite{Dut83}) if $R$ is a complete local
domain of characteristic $p>0$ and dimension $d \geq 1$, then, for every system of parameters $(x_1, \ldots, x_d)$ of
$R$, $\frac{l(H_i(x_1^{p^e}, \ldots, x_d^{p^e}; R))}{l(H_0(x_1^{p^e}, \ldots, x_d^{p^e}; R))}
\xrightarrow{e \rightarrow \infty} 0$ for each $1 \leq i \leq d$.
This classical result is essentially saying that the length of higher Koszul homology modules tends to $0$ compared to the length of the $0$-th Koszul homology module when we raise any system of parameters to high Frobenius powers. Our final result is a generalization of this result to a characteristic-free version.  More importantly, the convergence to $0$ occurs in a way that is independent of the system of parameters!

\begin{theorem}\label{Uniform0}
Let $R$ be a Noetherian local ring and $M$ be a finitely generated
quasi-unmixed $R$-module with $\dim(M) = d$.  For
every $\varepsilon>0$, there exists $t_0$ such that, for all $t \geq
t_0$, all systems of parameters $\underline{x}:= x_1, \ldots, x_d$ of
$M$, and all $1 \leq i \leq d$,
\[
\frac{l(H_i(x_1^t, \ldots, x_d^t; M))}{l(M/(x_1^t, \ldots, x_d^t)M)}
<\varepsilon.
\]
In fact, there exists a constant $K$ such that for all $t \geq
1$, all systems of parameters $\underline{x}:= x_1, \ldots, x_d$ of
$M$, and all $1 \leq i \leq d$,
\[
\frac{l(H_i(x_1^t, \ldots, x_d^t; M))}{l(M/(x_1^t, \ldots, x_d^t)M)}
\le \frac K{t^i}.
\]
\end{theorem}

\begin{proof}
As usual, we replace $R$ by $\overline{R}=R/\operatorname{Ann}(M)$ and complete $R$ to assume that $R$ is complete and that $M$ is faithful over $R$. Our hypothesis then implies that both $M$ and $R$ are equidimensional. (The rest of the proof only relies on the fact that $M$ is equidimensional and that $(R,\mathfrak{m})$ is a homomorphic image of a Gorenstein local ring $S$.)

Because we consider only finitely many $i$, it is sufficient to fix
some $1 \leq i \leq d$.  By Remark \ref{sseq}, it suffices to show that, for
any fixed finitely generated $R$-module $L$ with $\dim(L) \le d-i$ and
any fixed $j = 0, \dotsc, d-i$, there
exists a constant $K$ such that, for all $t \ge 1$ and all $\underline{x}$,
\[
\dfrac{l(H_j(x_1^t, \ldots, x_d^t; L))}{l(M/(x_1^t, \ldots, x_d^t)M)}
\le \frac K{t^i}.
\]
By taking a prime cyclic filtration of $L$, it suffices to show that,
for any fixed $P \in \operatorname{Spec}(R)$ such that $\dim(R/P) = d' \le d-i$,
there exists a constant $K$ such that, for all $t \ge 1$ and all $\underline{x}$,
\[
\dfrac{l(H_j(x_1^t, \ldots, x_d^t;R/P))}{l(M/(x_1^t, \ldots, x_d^t)M)}
\le \frac K{t^i}.
\]
Denote $D := R/P$. By Theorem~\ref{Thm:HomoBound}, we fix
\[
C = \sup_{\substack{\sqrt{(x_1,\ldots, x_d)}=\mathfrak{m}}}
\left\{\frac{l(H_j(x_1, \ldots, x_d;D))}{l(D/(x_1,\ldots,x_d)D)}
\right\} <\infty.
\]
According to Theorem~\ref{Thm:Vogel}, we let
\[
B  = \sup_{\sqrt{I}=\mathfrak{m}}\left\{\frac{l(D/ID)}{e(I,D)} \right\}<\infty.
\]
Moreover, as $M$ is equidimensional, there exists $Q \in \operatorname{Min}(R) = \operatorname{Min}(M)$ such that $Q
\subsetneq P$ so that $D$ is a proper homomorphic image of $R/Q$, in which case
$\dim(R/Q) = d > d' = \dim(D)$ and $e(I, R/Q) \le e(I, M)$ for all
$\mathfrak{m}$-primary ideals $I$ (since $\dim(R/Q) = \dim(M)$). In light of Equation~\eqref{eqn:supR/P} in
the proof of Theorem~\ref{Thm:Vogel}, we set
\[
A = \sup_{\sqrt{I}=\mathfrak{m}}\left\{\frac{e(I, D)}{e(I, R/Q)}\right\}<\infty.
\]
Finally, we see
\begin{align*}
\frac{l(H_j(x_1^t, \dotsc, x_d^t;D))}{l(M/(x_1^t, \dotsc, x_d^t)M)}
&\leq \frac{l(H_j(x_1^t, \dotsc, x_d^t;D))}{e((x_1^t, \dotsc, x_d^t),M)} \\
&\leq C \frac{l(D/(x_1^t, \dotsc, x_d^t)D)}{e((x_1^t, \dotsc, x_d^t),M)} \\
&\leq BC \frac{e((x_1^t, \dotsc, x_d^t),D)}{e((x_1^t, \dotsc, x_d^t),M)} \\
& = BC \frac{t^{d'}e((x_1, \dotsc, x_d),D)}{t^de((x_1, \dotsc, x_d),M)} \\
& \leq \frac {BC}{t^{d-d'}}
\frac{e((x_1, \dotsc, x_d),D)}{e((x_1, \dotsc, x_d),R/Q)}
\leq \frac {ABC}{t^{d-d'}} \le \frac {ABC}{t^i},
\end{align*}
whose convergence to $0$, as $t \to \infty$, is independent of
systems of parameters $\underline{x}:= x_1, \ldots, x_d$.
\end{proof}

\begin{lemma}\label{koszul-expo}
Let $R$ be a (Noetherian) ring, $M$ a finitely generated
$R$-module, $\underline{x}:= x_1, \ldots, x_d$ a sequence of elements of $R$ such that
$l(M/(\underline{x})M) < \infty$, and $t_j \ge s_j \ge 1$ for $1 \le j \le
d$. Then for all $i$, we have
\[
l(H_i(x_1^{t_1}, \ldots, x_d^{t_d}; M))
\le l(H_i(x_1^{s_1}, \ldots, x_d^{s_d}; M)) \prod_{j=1}^d \frac{t_j}{s_j}
\]
\end{lemma}

\begin{proof}
By symmetry, it suffices to show
$l(H_i(x_1^{t}, x_2,\ldots, x_d; M))
\le l(H_i(x_1^{s}, x_2,\ldots, x_d; M)) \frac{t}{s}$ for all $t \ge s
\ge 1$. For each $i$, denote $H_i = H_i(x_2,\ldots, x_d; M)$. From the
exact sequence
\[
H_i \xrightarrow{x_1^t\cdot} H_i \to H_i(x_1^{t}, x_2,\ldots, x_d; M)
\to H_{i-1} \xrightarrow{x_1^t\cdot} H_{i-1}
\]
we see that
\begin{align*}
l(H_i(x_1^{t}, x_2,\ldots, x_d; M))
&= l(H_i/x_1^tH_i) + l((0:_{H_{i-1}} x_1^t)) \\
&=\sum_{j=1}^t l(x_1^{j-1}H_i/x_1^jH_i)
+ \sum_{j=1}^t l((0:_{H_{i-1}} x_1^j)/(0:_{H_{i-1}} x_1^{j-1})).
\end{align*}
Now, as we can do the above to $l(H_i(x_1^{s}, x_2,\ldots, x_d; M))$
as well,
it suffices to show the sequences $\{l(x_1^{j-1}H_i/x_1^jH_i)\}_j$ and
$\{l((0:_{H_{i-1}} x_1^j)/(0:_{H_{i-1}} x_1^{j-1}))\}_j$ are both
non-increasing. But this follows because the following maps induced by
multiplication by $x_1$:
\[
\frac{x_1^{j-1}H_i}{x_1^jH_i} \xrightarrow{x_1\cdot} \frac{x_1^{j}H_i}{x_1^{j+1}H_i}
\quad\text{and} \quad
\frac{(0:_{H_{i-1}} x_1^{j+1})}{(0:_{H_{i-1}} x_1^{j})} \xrightarrow{x_1\cdot}
\frac{(0:_{H_{i-1}} x_1^j)}{(0:_{H_{i-1}} x_1^{j-1})},
\]
are onto and 1-1 respectively.
\end{proof}

The next theorem generalizes the uniform convergence
established in Theorem~\ref{Uniform0}.

\begin{theorem}\label{Uniform0-ti}
Let $R$ be a Noetherian local ring and $M$ a finitely generated
quasi-unmixed $R$-module with $\dim(M) = d$.  For
every $\varepsilon>0$, there exists $t_0$ such that, for all $t_j \geq
t_0$ with $1 \le j \le d$, all systems of parameters $\underline{x}:= x_1,
\ldots, x_d$ of $M$, and all $1 \leq i \leq d$,
\[
\frac{l(H_i(x_1^{t_1}, \ldots, x_d^{t_d}; M))}{l(M/(x_1^{t_1}, \ldots, x_d^{t_d})M)}
<\varepsilon.
\]
In fact, there exists a constant $K$ such that for all $t_j \geq
1$ with $1 \le j \le d$, all systems of parameters $\underline{x}:= x_1, \ldots, x_d$ of
$M$, and all $1 \leq i \leq d$,
\[
\frac{l(H_i(x_1^{t_1}, \ldots, x_d^{t_d}; M))}{l(M/(x_1^{t_1}, \ldots, x_d^{t_d})M)}
\le \frac K{(\min_jt_j)^i}.
\]
\end{theorem}

\begin{proof}
With $t_j \geq 1$ for $1 \le j \le d$, we denote $t = \min_j
t_j$. Then for all systems of parameters $\underline{x}:= x_1, \ldots, x_d$ of
$M$ and all $1 \leq i \leq d$, we have
\begin{align*}
\frac{l(H_i(x_1^{t_1}, \ldots, x_d^{t_d}; M))}{l(M/(x_1^{t_1}, \ldots, x_d^{t_d})M)}
&\le \frac{l(H_i(x_1^{t}, \ldots, x_d^{t}; M))}
{l(M/(x_1^{t_1}, \ldots, x_d^{t_d})M)}
\prod_{j= 1}^d\frac{t_j}t && \text{(Lemma~\ref{koszul-expo})}\\
&\le \frac{l(H_i(x_1^{t}, \ldots, x_d^{t}; M))}
{e((x_1^{t_1}, \ldots, x_d^{t_d}),M)} {d!e(\overline R)}
\prod_{j= 1}^d\frac{t_j}t  && \text{(Theorem~\ref{Thm:LechMod})}\\
&=\frac{l(H_i(x_1^{t}, \ldots, x_d^{t}; M))}
{e((x_1^{t}, \ldots, x_d^{t}),M)} {d!e(\overline R)}\\
&\le \frac{l(H_i(x_1^{t}, \ldots, x_d^{t}; M))}
{l(M/(x_1^{t}, \ldots, x_d^{t})M)} n(M) {d!e(\overline R)}
&& \text{(Theorem~\ref{Thm:Vogel})}
\end{align*}
in which $\overline R = R/\operatorname{Ann}(M)$.
Now Theorem~\ref{Uniform0} completes the proof.
\end{proof}

\begin{remark}\label{Uniform0SV}
We would like to mention that, in Theorem~\ref{Uniform0} (hence in
Theorem~\ref{Uniform0-ti}), the
assumption that $M$ is quasi-unmixed is necessary
{(at least when $R$ has infinite residue field)}.
In fact, the
conclusion of Theorem~\ref{Uniform0} (i.e., the uniform convergence to $0$) for
$M$ implies $n(M) < \infty$ {provided that $R$ has
  infinite residue field}, which forces $M$ to be quasi-unmixed by
\cite[Theorem 1]{SV96}.
For details, the existence of $t$ such that
$\frac{\sum_{i=1}^d(-1)^{i-1}l(H_i(x_1^t, \ldots, x_d^t; M))}
{l(M/(x_1^t, \ldots, x_d^t)M)}
<\varepsilon < 1$ for all systems of parameters $\underline{x}:= x_1, \ldots,
x_d$ of $M$ implies
\begin{align*}
\dfrac{e((x_1, \ldots, x_d),M)}{l(M/(x_1, \ldots, x_d)M)}
& = \dfrac{e((x_1^t, \ldots, x_d^t), M)}{t^dl(M/(x_1, \ldots, x_d)M)}
\geq \dfrac{e((x_1^t, \ldots, x_d^t), M)}{t^dl(M/(x_1^t, \ldots, x_d^t)M)} \\
&\geq \dfrac{1}{t^d} \left(\dfrac{l(M/(x_1^t, \ldots,
    x_d^t)M)}{l(M/(x_1^t, \ldots, x_d^t)M)}
  - \dfrac{\sum_{i=1}^d(-1)^{i-1}l(H_i(x_1^t, \ldots, x_d^t; M))}
{l(M/(x_1^t, \ldots, x_d^t)M)}\right)
> \frac{1-\varepsilon}{t^d}
\end{align*}
for all systems of parameters $\underline{x}:= x_1, \ldots, x_d$ of $M$, which
implies $n(M) < \infty$ {(given that $R$ has infinite
  residue field)}.
\end{remark}

\begin{remark}\label{SVbyInduction}
Evidently the results of this section rely on
Theorem~\ref{Thm:Vogel}. However, a careful analysis of the proofs in
this section reveals an alternative proof of
Theorem~\ref{Thm:Vogel} by induction on the dimension of the quasi-unmixed
$R$-module $M$ without explicit usage of homological degree or unmixed
degree. Without loss of generality, assume that $(R,\mathfrak{m})$ is
complete with infinite residue field (thus, $R$ is a homomorphic image of a Gorenstein ring $S$ with
$\dim(S) = n$).
When $\dim(M) = 0$, it is clear that
Theorem~\ref{Uniform0},
Theorem~\ref{Thm:Vogel},
Lemma~\ref{lem:Length},
Corollary~\ref{cor.integralclosure},
Lemma~\ref{lem:Length2}, and
Theorem~\ref{Thm:HomoBound}
all hold. Now assume that \emph{all} these results hold
in dimension $<d$, and consider the case of dimension $d$. Then
Theorem~\ref{Uniform0} holds in dimension $d$ (because the proof of Theorem~\ref{Uniform0} only requires the aforementioned results in dimension $<d$), which implies
Theorem~\ref{Thm:Vogel} in dimension $d$ as we explained in
Remark~\ref{Uniform0SV}. Then we have
Lemma~\ref{lem:Length},
Corollary~\ref{cor.integralclosure},
Lemma~\ref{lem:Length2} and
Theorem~\ref{Thm:HomoBound} in dimension $d$, completing the
induction. Alternatively,
Theorem~\ref{Thm:HomoBound} and Theorem~\ref{Thm:Vogel} in dimension
$< d$ implies Theorem~\ref{Thm:Vogel} in dimension $d$ as follows:
It suffice to consider the case of $M = R$ being a domain. For an
arbitrary system of parameters $\underline{y}:= y_1, \dotsc, y_d$ of $R$, we
have
\[
\frac{l(R/(\underline{y}))}{e((\underline{y}),R)}
\le \frac{e((\underline{y}),R)+l(H_1(\underline{y};R))}{e((\underline{y}),R)}
=1 + \frac{l(H_1(\underline{y};R))}{e((\underline{y}),R)}.
\]
Similar to the reasoning in the proof of Theorem~\ref{Thm:HomoBound}
(plus taking prime cyclic filtration), it suffices to consider $R/P$
with $0 \neq P \in \operatorname{Spec}(R)$ and to find an upper bound for
$\frac{l(H_i(\underline{y};R/P))}{e((\underline{y}),R)}$
for all systems of parameters $\underline{y}$.
By Theorem~\ref{Thm:HomoBound} and Theorem~\ref{Thm:Vogel} in dimension
$< d$, it suffices to find an upper bound for
$
\frac{e(\underline{y},R/P)}{e((\underline{y}),R)}
$
for all systems of parameters $\underline{y}$. But this is Equation~\eqref{eqn:supR/P} in
the proof of Theorem~\ref{Thm:Vogel}.
\end{remark}

Even though the alternative proof sketched above does not use
extended degree explicitly, its approach is very similar to that of
homological degree: the alternative approach relies on Remark~\ref{sseq}
to reduce the dimension from $\dim(M)=d$ to $\dim(\operatorname{Ext}_S^{n-j}(M,S)) <
d$, $0 \le j < d$, while the homological degree involves
the same modules in its definition.

The following is an easy consequence of Theorem~\ref{Uniform0-ti}. It
says that for all systems of parameters $\underline{x} = x_1, \dotsc, x_{d}$ on
$M$, the rate of convergence of
$\frac{l(M/(x_1^{t_1}, \dotsc,x_{d}^{t_{d}})M)}{\prod_{j=1}^{d}t_j}$
to $e((\underline{x}),M)$ is uniformly controlled by $l(M/(\underline{x})M)$ and $\min_j t_j$ only.

\begin{corollary}\label{uniform-convergence}
Let $(R, \mathfrak{m})$ be a Noetherian local ring and $M$ be a finitely generated quasi-unmixed $R$-module.
Then for every constant $C >0$ and every $\epsilon > 0$, there exists
$t_0 \in \mathbb N$ such that, for all $t_j \ge t_0$ with $1 \le j \le
d$, all systems of parameters $\underline{x} = x_1, \dotsc, x_{d}$ on
$M$ such that $e((\underline{x}),M) \le C$, we have
\[
0 \le \frac{l(M/(x_1^{t_1}, \dotsc,x_{d}^{t_{d}})M)}{\prod_{j=1}^{d}t_j} - e((\underline{x}),M)
< \epsilon.
\]
In fact, there exists a constant $K$ such that,
for all $t_j \ge 1$ with $1 \le j \le d$ and all systems of parameters
$\underline{x} = x_1, \dotsc, x_{d}$ on $M$, we have
\[
0 \le \frac{l(M/(x_1^{t_1}, \dotsc,x_{d}^{t_{d}})M)}{\prod_{j=1}^{d}t_j} - e((\underline{x}),M)
\le e((\underline{x}),M) \frac K{\min_j t_j}
\le l(M/(\underline{x})M) \frac K{\min_j t_j} .
\]
\end{corollary}

\begin{proof}
For all all systems of parameters $\underline{x} = x_1, \dotsc, x_{d}$ on $M$,
we have (with $\underline{x}^{[t]} := x_1^{t_1}, \dotsc, x_{d}^{t_{d}}$)
\begin{align*}
0  \le \frac{l(M/(\underline{x}^{[t]})M)}{\prod_{j=1}^{d}t_j} - e((\underline{x}),M)
&= \frac{l(M/(\underline{x}^{[t]})M) - e((\underline{x}^{[t]}),M)}{\prod_{j=1}^{d}t_j} \\
&= \frac{l(M/(\underline{x}^{[t]})M)}{\prod_{j=1}^{d}t_j}
      \frac{\sum_{i=1}^{d}(-1)^{i-1} l(H_i(\underline{x}^{[t]};M))}{l(M/(\underline{x}^{[t]})M)}  \\
&\le\frac{e((\underline{x}^{[t]}),M) n(M)}{\prod_{j=1}^{d}t_j}
      \frac{\sum_{i=1}^{d}(-1)^{i-1} l(H_i(\underline{x}^{[t]};M))}{l(M/(\underline{x}^{[t]})M)} \\
&= e((\underline{x}),M) n(M) \frac{\sum_{i=1}^{d}(-1)^{i-1} l(H_i(\underline{x}^{[t]};M))}{l(M/(\underline{x}^{[t]})M)} \\
&\le e((\underline{x}),M) \left(n(M) \sum_{i=1}^{d}(-1)^{i-1}\frac{
    l(H_i(\underline{x}^{[t]};M))}{l(M/(\underline{x}^{[t]})M)} \right).
\end{align*}
Now Theorem~\ref{Uniform0-ti} completes the proof.
\end{proof}

Finally, we remark that even though $n(M)$ could tend to $\infty$ as $M$ varies (see Example \ref{nlarge}), $\frac{n(M)}{\mu(M)}$ has an upper bound that depends only on $R/\operatorname{Ann}(M)$ (and not on $M$). We would like to thank the referee for pointing out this question.
\begin{remark}
\label{RemMinGen}
Let $(R,\mathfrak{m})$ be a Noetherian local ring, and let $M$ be a finitely generated quasi-unmixed $R$-module. Set $\overline{R}=R/\operatorname{Ann}(M)$, which is quasi-unmixed by assumption and we have $\operatorname{Assh}(M)=\operatorname{Assh}(\overline{R})$. Let $c = \max_{P \in \operatorname{Assh}(\overline{R})} l_{\overline{R}_P}(\overline{R}_P)$. Clearly $l(M/IM)\leq \mu(M)l(\overline{R}/I\overline{R})$, and by the associativity formula for multiplicities $e(I, \overline{R})\leq ce(I, M)$.
Now for all $\mathfrak{m}$-primary ideals $I$ we have
$$\frac{l(M/IM)}{e(I,M)\mu(M)}= \frac{l(M/IM)}{\mu(M)l(\overline{R}/I\overline{R})}\cdot \frac{l(\overline{R}/I\overline{R})}{e(I, \overline{R})}\cdot \frac{e(I, \overline{R})}{e(I,M)}\leq cn(\overline{R}).$$ Therefore $\frac{n(M)}{\mu(M)}\leq cn(\overline{R})$, and the latter depends only on $\overline{R}=R/\operatorname{Ann}(M)$. Also note that if we take $R=k[[x,y]]$, $I_t=(x)\cap (x,y)^n$. Then $R/I_t$ is quasi-unmixed and $n(R/I_t)\geq t$: the ideal $(y)$ has multiplicity $1$ and colength $t$ in $R/I_t$. Therefore, in general the $n(\overline{R})$ (as $M$ varies) are not bounded in terms of invariants of $R$.
\end{remark}

redpajama_set_name: RedPajamaArXiv


The 8th hole of The North Course at Eagle Ridge-one of four championship courses on site.
GALENA, Ill. (Apr. 2, 2019) – Golfers including guys, gals, kids, and families alike are already driving, putting and chipping on one of the most scenic and rewarding courses in Illinois. The North Course has officially opened for the season at Eagle Ridge Resort & Spa (Eagle Ridge) in Galena, Illinois with the additional three championship golf courses, including one of the top one hundred courses in the country, The General, slated to open this month. While the traditional guys golf trip is still popular, Eagle Ridge has experienced a significant increase in family golf getaways and ladies golf.
Ladies group golf trips and women golfing in general at Eagle Ridge is also more popular than ever. Davis estimates that the resort is almost double the national average for women golfers. According to the National Golf Foundation's 2018 Golf Industry Report, women comprise 24 percent of all golfers, an increase of more than 13 percent over the prior five years.
Families and groups can take advantage of several golf packages that feature spa, dining, lodging and green fees. Package details are online at www.eagleridge.com, or golfers can call 800-892-2269 with questions, to make reservations or book a tee time.


Nordhausen is the northernmeost Landkreis (rural district) of Thuringia, Germany.
The district was created in 1815, when the Prussian province Saxony was created. There were many changes to the boundaries up to 1952, when some municipalities were brought into the district and others left.
The checkered red-silver part of the coat of arms is derived from the coat of arms of the County of Hohenstein, the historical territory the district belonged to. The tree trunk symbolizes the destruction of the city Nordhausen during World War II, with the new branch growing into the golden area symbolizing the reconstruction of the city and the bright future.


The Venezuelan state-owned PDVSA hopes to retain access to Bullen Bay terminal after its lease expires in December 2019, but the Isla refinery at Schottegat has lost strategic value for the company. This is according to sources within the Venezuelan company.
The refinery barely operated in 2018 because of a lack of feedstock, maintenance and domestic utility services. Normally the facility processed around 220,000 b/d.
A senior Venezuelan energy ministry official said PDVSA no longer has any commercial or financial interest in the refinery. "Isla has lost its strategic importance as a refining center for our company," the official said. "PDVSA can't supply the crude Isla needs, can't afford imported crude from other suppliers and doesn't have the financial resources to maintain the refinery even in nominal operating capacity."
"Bullen Bay is very important to PDVSA's export logistics, but PDVSA isn't interested in spending up to $3bn to upgrade the refinery," official added.
Bullen Bay is a critical transshipment hub for PDVSA's export operations, particularly involving shipments to China, India and close ally Cuba. PDVSA also leases storage on the other Dutch Caribbean islands of Aruba and St Eustatius and owns the 10mn bl Bopec storage facility on Bonaire.
The value of the logistical network was highlighted in May, when US independent ConocoPhillips imposed pre-judgment attachments on PDVSA's Dutch Caribbean assets to force the company to honor a $2bn arbitration award for the 2007 takeover of the US firm's Venezuelan assets. PDVSA reached a settlement with ConocoPhillips in August and has complied with an initial $500mn cash and in-kind payment, the US company confirms. PDVSA is otherwise in default on billions of dollars in bond, commercial and arbitration debt.
PDVSA's export operations would be significantly impaired if it loses access to Bullen Bay, according to a company official at the main Venezuelan oil terminal of Jose. PDVSA lacks sufficient domestic storage and terminal capacity to compensate for a potential loss of access to Curaçao, although continued access to Bopec, Aruba and NuStar's St. Eustatius terminal would soften the blow.
Venezuela's current production of just over 1mn b/d and significant oil-backed debtcommitments and barter deals leave little room to supply crude to Curaçao, the Venezuelan officials say.
A seasonal shift in refinery activity has helped turn gas into 'the unwanted hydrocarbon', an oil expert told CNBC recently, which is unlikely to last beyond winter.
Russian Energy Minister Alexander Novak.
The oil producers have been coordinating output for two years and have long discussed institutionalizing the partnership.
Novak cites burdensome bureaucracy and U.S. legislation that targets OPEC as reasons to abandon the plan.
The marriage between Russia and OPEC is off.
Russian Energy Minister Alexander Novak on Friday poured cold water on long-simmering plans to make Moscow's alliance with OPEC and other oil producers permanent. The group of roughly two dozen producers has been managing global petroleum supply for the last two years in order to rebalance the market after a prolonged and punishing oil price downturn.
The effort succeeded in shrinking global crude stockpiles and boosting prices to four-year highs — until the market suddenly crashed again in early October. The group has agreed to a fresh round of output cuts that begin Jan. 1.
For at least a year, OPEC Secretary General Mohammed Barkindo has discussed "institutionalizing" the arrangement. That would essentially form a supergroup of oil producers comprised of the 14-nation OPEC, Russia and nine other oil-exporting nations, which would be able to more quickly respond to problems in the market.
Energy ministers had been talking up progress toward the permanent arrangement as recently as their meeting in Vienna earlier this month.
But on Friday, Novak said the prospects for that plan now look dim, Reuters reported. He said it would create too much red tape and expose the non-OPEC members of the alliance to potential sanctions from the U.S. government.
"There is a consensus that there will be no such organization. That's because it requires additional bureaucratic brouhaha in relation to financing, cartel, with the U.S. side," Novak told reporters, according to Reuters.
The U.S. penalties in question are spelled out in legislation known as NOPEC, or the No Oil Producing and Exporting Cartels Act. The bill would authorize the Justice Department to sue groups like OPEC that are deemed cartels for price fixing and antitrust violations, stripping countries of sovereign immunity protections currently built into U.S. law.
The legislation was first introduced in 2007, during a time of rising crude prices and concerns that the world's oil reserves would run dry. It was revived earlier this year in both chambers of Congress by bipartisan groups of lawmakers.
The White House has historically opposed the legislation, but President Donald Trump's attacks on OPEC this year have raised concerns that he could back the measure.
However, like past presidents, Trump has shown he is willing to work with OPEC when it suits his needs.
His administration lobbied top OPEC producer Saudi Arabia to reverse the alliance's policy of capping production in order to prevent oil prices from spiking as the U.S. prepared to sanction Iran, OPEC's third-biggest producer. The alliance agreed to hike output in June, a decision that ultimately contributed to the current oil price slump.
Russia and Saudi Arabia were the world's top two oil producers until this year, when U.S. output surpassed production from both countries.
U.S. and Brent crude oil, the global benchmark, were both up more than 7%. Brent crude was up about 7.88% in late trading, while WTI crude climbed 8.58%.
Igor Sechin, head of Russia's Rosneft and a close ally of President Vladimir Putin, said Wednesday the drop in global crude oil prices was due to the Federal Reserve raising interest rates last week.
Sechin also said he saw oil prices at $50-53 per barrel next year "under a conservative scenario," according to a Reuters story.
Oil prices have fallen by more than a third this quarter. Last week, the Federal Reserve raised rates by a quarter percentage point for the fourth time this year to a range of 2.25% to 2.50%.
Crude has been caught up in wider financial market issues, such as the U.S. government shutdown, higher U.S. interest rates, and the U.S.-China trade dispute, which shook investors.
On Monday, global oil markets extended their declines, taking crude prices to the lowest levels since July of last year.
Exxon Mobil (XOM) rose 2.77% to $67.32. Chevron (CVX) climbed 4.41% to $105.44. BP (BP) rose 2.63% to $37.62, while Royal Dutch Shell dipped slightly to $25.03.
Aside from beseeching heaven, Quevedo in recent months has enacted a series of controversial measures that oil industry experts, PDVSA employees and contractors, and even everyday citizens say are pushing the once-profitable and respected company towards ruin.
In a little-noted reversal of the Socialist government's two-decade drive to nationalize the industry, the lack of expertise among military managers is leading PDVSA to hire outsiders to keep afloat even basic operations, like drilling and pumping oil. To the dismay of many familiar with Venezuela's oil industry, some of the contracts are going to small, little-known firms with no experience in the sector.
Combined, industry veterans say, the steps leave Venezuela's most important company - which accounts for over 90 percent of export revenue - with even fewer means to rebuild the nation's coffers, pay its many creditors and regain self-sufficiency as an oil producer.
Quevedo, who holds the title of oil minister as well as president of PDVSA, didn't respond to requests for comment for this story. Neither Venezuela's Information Ministry, responsible for communications for the government and senior officials, nor PDVSA's press office returned phone calls or emails from Reuters.
Nearly 20 years after the late Hugo Chavez launched his "Bolivarian revolution," much of Venezuela is in tatters. Food and medicines are scarce, hyperinflation has gutted purchasing power for increasingly desperate citizens and roughly three million Venezuelans have fled the country in search of a better life.
Now, critics say, military officials atop PDVSA have put aside any pretense of running it like a proper business, doing little to stem the fall in production or improve the company's financial, operational and staffing problems.
Demand remains healthy for Venezuelan oil. Operational problems under Quevedo, however, have caused production to drop 20 percent to 1.46 million barrels per day, according to the latest figures Caracas reported to OPEC, the oil cartel, of which it is a member.
Quevedo in January will assume OPEC's rotating presidency for one year. PDVSA's financial problems are likely to demand much of his attention.
The gross value of PDVSA's oil exports is expected to fall to $20.9 billion this year compared with $24.9 billion last year, according to a calculation provided to Reuters by the International Energy Center at IESA, a Venezuelan business school. Exports a decade ago were over four times as much, reaching $89 billion, according to PDVSA's accounts for 2008.
PDVSA didn't publish a 2017 report and hasn't released financial results in 2018.
Little has been publicly disclosed by PDVSA or Maduro's government about the military transformation within its ranks.
A Reuters examination based on confidential PDVSA documents – as well as interviews with dozens of current and former employees, shippers, traders, foreign oil executives and others who do business with the company – shows how Quevedo's National Guard is seeping into every facet of its operations. The documents include employment records, agreements with contractors and internal staff memos.
Quevedo has appointed more than 100 aides and advisors from the military and from a previous post as a government minister to senior positions, according to a person familiar with PDVSA's human resource records.
At its shabby concrete Caracas headquarters, once brimming with suited executives, military officers are now in charge of operations. Workers say offices in Quevedo's penthouse sanctum remain luxurious. But in the run-down halls below, socialist propaganda, including portraits of Fidel Castro and Ernesto "Che" Guevara, is among the scant decor left on the walls.
Allegations of corruption have been rife across the Venezuelan government in recent years; Maduro himself is the target of U.S. sanctions for graft and human rights violations, which he denies.
In 2017, the president leveled his own accusations against PDVSA, describing it as a den of "thieves." He accused many former executives of skimming from contracts and laundering money and argued that their graft worsened the country's crisis.
He ordered the arrest of dozens of top managers, including PDVSA's two previous presidents, chemist Nelson Martinez and engineer Eulogio Del Pino. Martinez died at a military hospital earlier this month, suffering a heart attack while undergoing kidney dialysis, two people familiar with the circumstances said.
Del Pino remains detained, awaiting trial. Reuters was unable to reach his lawyers for comment. A person familiar with Del Pino's defense said he has yet, after a year in jail, to have an initial court hearing.
Quevedo's Twitter profile often features a photo of the general, a stocky and balding man with heavy eyebrows, reviewing paperwork with the president or smiling happily alongside him. His feed consists almost exclusively of retweets of Maduro's posts.
Pleased with Quevedo's performance, Maduro in 2015 named him housing minister. In his two years in the post, he again became a fixture on state television, often wearing the red shirt of the Socialist movement and praising Maduro's "humane" housing policies.
Opposition leaders scoffed at what they saw as Quevedo's outsized boasts, including an unsubstantiated claim that the government constructed more than 2 million homes, despite widespread shortages of basic building materials. The housing ministry didn't respond to requests for comment.
In November 2017, intelligence agents arrested former PDVSA chief Del Pino in a predawn raid on unspecified graft charges. By then, Quevedo was Maduro's choice to lead the all-important company. The announcement prompted widespread skepticism in the industry.
Quevedo said he would need little time to get a handle on the oil businesses. "Give me 10 days," he told acquaintances, according to one person who spoke with him at the time.
From the start, Maduro made clear the challenge ahead. In a public address during "Powerhouse Venezuela 2018," a government conference meant to showcase business potential, the president ordered Quevedo to boost oil output by a whopping 1 million barrels per day – roughly a 50 percent increase at the time.
One of his first challenges, according to people within PDVSA, was to stanch the flow of workers, many of whom deserted the company and Venezuela altogether. PDVSA hasn't disclosed recent employment figures. But estimates by IPD Latin America, an oil and gas consultancy, indicate PDVSA has about 106,000 workers – 27 percent fewer than in 2016.
Because of cost-of-living increases that now top 1 million percent per year, according to Venezuela's National Assembly, PDVSA salaries have crumbled to the equivalent of a handful of dollars a month for most workers.
Crews fret a stray bullet from the soldiers' rifles could spark fires and complain that some of the crime afflicting the country is making its way on board. Although Quevedo has tasked the soldiers to help spot graft, some of the low-paid recruits ask for bribes themselves, shippers said, for signing off on paperwork or completing inspections.
"There are many risks," one captain told Reuters.
Venezuela's Defense Ministry, which oversees the National Guard, didn't respond to Reuters phone calls or emails requesting comment.
Even with soldiers as substitutes, PDVSA can't find the workers it needs to man many posts. From the processing of crude at refineries to contract negotiations with buyers, the shortage of skilled staffers is hobbling the company.
As the tankers loaded, PDVSA port employees noticed a mixup – the two crudes had blended. The mistake, the government said later, forced PDVSA to pay the buyers, because of contractual penalties, $2.7 million.
Quevedo has been creating new partnerships that are meant to shore up PDVSA. In August, for instance, the general said the company was "opening its doors" for seven private companies to pursue unspecified "service contracts" across the country.
But the companies are unfamiliar even to veterans of Venezuela's oil industry. None are recognized as having experience operating oilfields. Consorcio Rinoca Centauro Karina, one of those listed on the document, doesn't appear to have a web site. Reuters was unable to reach it or any of the others.
Critics of the arrangements, and government opponents, say the transactions aren't transparent. By keeping details from the public, they argue, the company faces little scrutiny over whom it chooses to do business with.
"PDVSA is looking to maintain its confederation of mafias, its quota of looting," said Jorge Millan, an opposition legislator who in September led a push in the National Assembly to denounce the contracts.
While Quevedo's militarization of PDVSA hasn't reversed the company's decline, the government shows few public signs of displeasure. In October, the government announced a PDVSA board shuffle. Among the changes: Jose Rojas, another National Guard general, replaced a civilian director.
Additional reporting by Mayela Armas and Vivian Sequera in Caracas and Ernest Scheyder in Vienna. Editing by Paulo Prada.
Chinese refiners are not buying more U.S. oil despite the three-month truce agreed by Presidents Trump and Xi last month, Reuters reports, citing cargo loading plans of Chinese downstream operators.
According the Reuters, Chinese demand for U.S. crude has been dampened by political uncertainty around the trade war and, more directly, by relatively high costs of transportation. This means that despite the truce and future positive developments in bilateral talks on trade, U.S. oil will have yet to become a major element of China's imported crude oil mix.
One Chinese analyst told Reuters that price was the top consideration of buyers and the price of U.S. oil simply wasn't competitive.
"Chinese companies have little incentive to buy U.S. crude due to the wide availability of crude supplies today from Iran and Russia," Seng Yick Tee from consultancy SIA Energy said. Yet trade tensions are not helping, either. With the constant threat of more tariffs, refiners are reluctant to change their buying habits.
U.S. crude oil exports hit a high of 23.95 million barrels in October 2017, data from the Energy Information Administration shows, but have since then declined, reaching 2.17 million barrels in September this year before Chinese refiners completely stopped buying U.S. crude in October.
Yet China's total oil imports in October, on the other hand, hit 40.80 million tons (9.61 million bpd), of which teapots imported 8.22 million tons. This was the highest monthly oil import amount on record, according to customs data from Beijing. The increase came despite depressed refining margins that could have motivated lower appetite for crude but apparently did not. The independent refiners drove the increase as they sought to fulfill their import quotas until year-end.
Corpus Christi Authority Port Authority has announced several milestones in the development of the crude oil export terminal on Harbor Island, Texas.
Locally-based Lone Star Ports will lead the development of the terminal, claimed to be the first US onshore export terminal able to handle fully-laden VLCCs.
Lone Star is a joint venture between The Carlyle Group and The Berry Group. Berry is the largest private employer in the Corpus Christi area through its numerous investments and operations in the oil and gas industry and its subsidiary Bay Ltd, a Corpus Christi-based infrastructure, construction, and fabrication contractor for the oil and gas sector.
Lone Star Ports is led by CEO Jeremiah "Jerry" Ashcroft III who has been responsible for the development of several large marine terminals.
The company has signed indicative agreements with Harvest Midstream and EPIC Crude Pipeline, which is backed by funds affiliated with Ares Management. These two pipelines will provide connectivity to more than one million barrels per day.
It also entered into an indicative agreement with Martin Midstream Partners to provide a single, integrated VLCC solution on Harbor Island.
These agreements and arrangements still remain subject to definitive documentation among the relevant parties, co-ordination with the Port, satisfactory completion of due diligence and final approval by each relevant party.
"Our partnership with the Carlyle Group is designed to assure global energy markets that requisite infrastructure will be in place and ready to support the growing exports of American crude oil. We are pleased with the progress The Carlyle Group has achieved thus far in reaching full project commercialisation," said Sean Strawbridge, Port of Corpus Christi CEO.
HOUSTON — Oil prices tumbled more than 7 percent on Tuesday, falling to their lowest levels in more than a year, after investors learned that Russia and the United States were pumping a lot more oil than had been expected.
The American benchmark fell below $47 a barrel for the first time in 15 months, capping a slide that has brought prices down by more than a third since early October.
Bountiful, cheap oil supplies are an unanticipated holiday bonus for American consumers. The average price of regular gasoline has fallen to $2.37 a gallon, according to the AAA motor club, 26 cents lower than a month ago. For most of the year gasoline prices were rising, but the recent decline in oil prices has driven gasoline down by about a nickel a gallon compared with a year ago.
Oil prices had stabilized earlier this month when the Organization of the Petroleum Exporting Countries and Russia agreed to slash production by 1.2 million barrels a day. But Russia announced on Monday that its output had increased to more than 11.4 million barrels a day, a record, putting in doubt its commitment to coordinate policies with Saudi Arabia and other oil producers.
On the same day, the Energy Department reported that the United States was producing 11.6 million barrels of crude oil a day, nearly a million barrels more than a year ago. The department projects that shale-oil production will climb to record levels this month, and increase by 134,000 barrels a day in January.
A shortage of pipelines has driven down oil prices from the Permian Basin of West Texas and New Mexico, the most productive American oil field. But the completion of a series of pipelines in late 2019 should benefit producers and increase exports, adding even more barrels to the global glut.
The oil industry has struggled since prices plummeted to below $30 a barrel in 2016. Hundreds of small producers and oil-services companies sought bankruptcy protection, and more than 160,000 jobs were cut. As prices increased in 2017 and earlier this year, the industry regained some of that lost ground — a recovery now under threat.
"It sure will hurt if it lasts very long," said Tom Dunlap, president of Tripledee Drilling of Ardmore, Okla. Mr. Dunlap said a price of $60 to $70 a barrel was typically needed to drill a new well profitably in his state.
Oil companies have become more efficient by relying on robotics and other technologies to replace workers, enabling them to explore and produce at lower prices. Those adjustments could come in handy as businesses are forced to cut back again in the coming months.
So far, there are few signs that companies will meaningfully reduce exploration in 2019, though that could change if prices keep falling and the economy slows.
A version of this article appears in print on Dec. 19, 2018 , on Page B2 of the New York edition with the headline: Oil Prices Plummet 7% As Investors Fear a Glut.
OPEC has its work cut out to persuade the market that its output caps will stabilize oil prices -- and U.S. producers aren't helping.
Crude settled below $50 a barrel in New York on Monday for the first time in more than a year and continued falling in after-hours trading. The slide began after data provider Genscape Inc. was said to report growing inventories at the biggest American storage hub and intensified as the U.S. Energy Department forecast higher output in the country's shale plays.
Oil prices are on track for a third straight monthly decline despite efforts by OPEC, Russia and other major exporters to halt the slide. Crude had slunk near $50 in recent weeks but always rebounded. Crossing the threshold was "significant," said Michael Loewen, a commodities strategist at Scotiabank in Toronto.
"We're probably going to see a supply slowdown in the U.S.," he said by telephone. "I do think that producers will react."
A dive for U.S. equities added to the pressure on Monday. The S&P 500 hit a 14-month low as investors anticipated a Federal Reserve interest-rate hike that could slow the economy.
West Texas Intermediate for January delivery fell $1.32 to settle at $49.88 a barrel on the New York Mercantile Exchange. Bears gained steam after the official close, with oil falling to $49.01, the lowest level since September 2017. The WTI February contract fell to $49.47.
Brent for February settlement closed down 67 cents to $59.61 on London's ICE Futures Europe exchange. The global benchmark traded at a premium of $9.41 a barrel to same-month WTI.
It typically takes about six weeks for OPEC nations to implement supply changes, and Saudi Arabia, the group's biggest producer, faces added political pressure from U.S. President Donald Trump to keep the taps open, Molchanov said.
Gasoline futures fell 1.7 percent to $1.4104 a gallon in New York trading.
A former lobbyist who's helped drive the Trump administration's policies to expand drilling on public lands is poised to take over at the U.S. Interior Department.
Kuwaiti Oil Minister Bakheet Al-Rashidi and two top executives of state-owned energy companies resigned in the latest sign of disputes within OPEC's fifth-biggest producer.
The oil producer's group has been seriously weakened by its need to bring an outsider into its heart to broker a deal on output.
If anyone doubted that OPEC is now little more than a zombie organization, the last 10 days have proved it.
The group has shown itself incapable of making it own decisions. Its smaller members have borne the brunt of an agreement to cut output that was only achieved after Russia took control of discussions from the heart of OPEC's head office. Then, even after Saudi Arabia announced it would reduce supply by nearly a million barrels a day by January, oil traders merely shrugged.
It is a sad result for an organization that once made governments tremble. Here's how it unfolded.
OPEC oil ministers gathered in the group's secretariat building on a wintry Thursday in Vienna. Weakening demand growth and soaring U.S. production prompted agreement that they needed to reduce production in order to balance the market in 2019. What they didn't see eye to eye on was how to share that burden.
Saudi Arabia insisted that all members should play an equal part, cutting by the same percentage from a new baseline set at October's production level. Naysayers argued that, given how the group's biggest producer had boosted its own output by more than a million barrels a day since May, it should therefore bear the brunt of the cuts needed to get the market back into balance. Furthermore, Iran led a contingent of countries claiming their special circumstances warranted exemptions.
The stand-off looked very similar to the one that had scuppered a deal in Doha in April 2016 – something that nobody wanted to repeat. But by the end of that Thursday, Dec. 6, officials had failed to reach an agreement and Saudi oil minister Khalid Al-Falih said he was "not confident" one was possible. The gala dinner at the Liechtenstein Palace was sparsely attended, with both the Saudi and Iranian delegations, among others, skipping the event.
Friday, the day originally scheduled for the group to meet with its partners, began little better. OPEC ministers were once again locked away in discussions. The talks went nowhere, and the OPEC+ meeting got pushed back to the afternoon.
Bilateral discussions to hammer out negotiating positions are a regular feature of the days and hours before the main OPEC gatherings. They take place in the suites of Vienna's grandest hotels, where the various delegations are holed up. In recent years, the Russians have been part of this scene.
But when Russian Energy Minister Alexander Novak arrived Friday morning, he moved in to the office of OPEC's Secretary General. If ever there was a symbol of OPEC's demise, it was this. He then summoned first Iran's oil minister, then Saudi Arabia's, for about 45 minutes each. Two hours later, the group reached a deal. And Iran, Libya and Venezuela got their exemptions, even if they didn't appear in the final communiqué.
The agreement that emerged is, on paper, fair and reasonable. OPEC will cut production by 800,000 barrels a day from a new baseline of October 2018 production, with participating members cutting by 3 percent. The group's partners will reduce their supply by 2 percent, contributing a further 400,000 barrels a day. That combined reduction is just about enough to balance supply and demand in the first half of next year.
Saudi Arabia went even further. Al-Falih said the kingdom's production would fall to 10.2 million barrels a day in January, down from 11.1 million last month.
That looks like a huge cut, but it is still 150,000 barrels a day above its target under the original deal.
And this is where the inequality of the new OPEC+ deal becomes apparent. Saudi Arabia, Russia, the United Arab Emirates and Iraq boosted their combined output by almost 1.6 million barrels a day between May and October. That not only contributes to the current glut, it also gives them much higher starting points for the latest cuts than for the previous ones. Other OPEC members all face lower starting points.
The effect has been to shift a disproportionate share of the burden of OPEC's supply management since 2016 onto the group's smaller producers.
Handing control of OPEC decision making to the Kremlin has come at a high cost for the group, and most particularly for its smaller members. Some of the latter were already feeling marginalized. This latest deal will do nothing to change their view and the divisions between the organization's "haves" and "have nots" will only widen.
Russia has done very well out of this. It agreed to cut output by 230,000 barrels a day from its October output level of 11.42 million. That would reduce its production to 11.19 million barrels a day, a figure that is just 15,000 barrels below its original 2016 baseline — a cut of just 0.1 percent.
Contrast that with OPEC member Algeria, which produces around a tenth as much oil as Russia. Its new target will be 1.023 million barrels a day. That's a cut of 66,000 barrels a day, or 6.1 percent below the 2016 baseline.
And what did this all achieve?
Oil markets have reacted with indifference. After a brief rally when the deal was revealed, Brent subsequently sank back below the level it was trading at before the meeting began. Perhaps traders don't believe the group will be able to implement the arrangement, or are waiting to see evidence that it's taken effect.
OPEC has lost a lot for very little gain.
RBC's Croft: What does an OPEC cut really look like?
NEW DELHI (Reuters) - India's monthly oil imports from Iran plunged to their lowest in a year in November with Tehran dropping two places to become only the sixth biggest supplier after New Delhi cut purchases due to the impact of U.S. sanctions, according to ship tracking data and industry sources.
Last month, the United States introduced tough sanctions aimed at crippling Iran's oil revenue-dependent economy. Washington did, though, give a six-month waiver from sanctions to eight nations, including India, and allowed them to import some Iranian oil.
India's imports from Iran in November, included some parcels that were loaded in October. In November, Iraq and Saudi Arabia continued to be the top two oil sellers to India.
"Iran do not have vessels to export oil on time ... some November loading vessels will arrive in December," said an industry source, with knowledge of the matter.
Government sources say Reuters' calculations showing India's oil imports from Iran in this fiscal year would be higher than the 452,000 bpd, or 22.6 million tonnes, it imported in the previous year, are correct.


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Our unique approach to the client side of Document Capture and Business Process Management (BPM) is reflected in the Quillix Web Client, our browser-based client for scanning, indexing and working with documents from anywhere in the world. Its uniqueness comes from its enterprise feature set, and its flexible licensing. With the emergence of cloug computing, we set out to provide all the features in our browser-based client that you would normally only find in traditional desktop capture software. Features such as database lookup, ad-hoc OCR and document classification are standard features in our web client.
For businesses with demanding time and/or volume requirements, Quillix Capture has many advanced tools in its arsenal. More than 20 processing extensions are available for everything from text and handwriting recognition to automated redaction. All of this processing intelligence is powered by our innovative Business Process Engine and Visual Process Design Tools, enabling you to easily create sophisticated, powerful and responsive business workflows.
Quillix Capture is designed to be flexible, and to provide maximum value to any organization. One of the ways we achieve this is by our flexible licensing approach. No click fees, page limits or CALS. Our licensing is straightforward and simple. Our goal is to uncouple your business from the constraints traditionally imposed by enterprise management software. License are fluid and are dynamically allocated based on need, not to specific computers or people.
Quillix Capture is a document processing powerhouse. Its server architecture was designed from the beginning to be infinitely scalable and fast. Quillix Capture employs an inherent multi-server, multi-processor architecture that takes advantage of available computing resources to provide throughput into the millions of pages per month.
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