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Day 28- Sunday. Today we woke up and had an egg for breakfast. I got a little crazy and had mine fried instead of the usual scrambled. My parents also gave us an extra de-hydrater they had laying around (why they had two, I will never know) and Matt was pretty excited. He spent the morning naming all of the foods he wants to dehydrate once we get home.
After breakfast we went shopping and I bought a pair of heels for my friend Kasey's wedding this upcoming weekend. Then for lunch we snacked on a Brian'd bowl. For dinner my grandparents came over and my Italian mother made some very delicious sausage-mushroom marinara sauce. The family had pasta while Matt and I stuck a couple of Spaghetti Squashes in the oven for about 45 min. Dinner ended up being delicious- and I even survived watching everyone eat some garlic bread which is usually one of my favorite parts of a pasta dinner.
By the way if you haven't noticed I learned how to collage photos. Let me just say this little trick makes my life 100x easier! We spent the rest of the night driving back to Pasadena, and got home around 12 AM. The drive was super tiring, and usually after a normal pasta dinner I am asleep within 30 min of sitting in the car. I was surprised to find that I was able to stay up the WHOLE trip back to Pasadena. If you know me, this will come as a shock. I think it has to do with the fact that spaghetti squash is much lighter than regular pasta. Another point for Paleo.
This recipe is actually pretty easy but kind of messy. Here is what you do in some steps. Begin by preheating your oven to 350.
Step 1: Set up 4 'stations'- the chicken cut into tenders, a bowl with the 3 eggs whisked together, a plate with the arrowroot flour and a plate with half the almond flour.
In the almond flour I added some garlic salt and some pepper to taste.
Step 2: Get a pan and melt some coconut oil on medium heat. Then TURN off the heat and let the oil sit. I did not do this at first and tried to brown the tenders in sets of 3 or 4- but the leftover almond flour burned and it became a mess.
Step 3: Then dip a tender in egg, then the arrowroot, then the egg, then the almond flour and set it in the pan. Continue this until the pan is full. Then turn on the heat and brown the chicken on both sides (about a minute each side). Once a tender is browned place it in an oven safe dish.
Step 4: Repeat. I also switched up the dipping a bit just to experiment. I dipped some in the egg, then arrowroot, then almond and some in the arrowroot, then egg, then almond.
Once all my tenders were done I stuck them in the oven, cleaned my counter and cut the stems off the brussel sprouts. Then I quartered them, put them in an oven safe dish, mixed them in the oil and added salt, pepper and a little garlic powder. This took around 10-15 min, then I added the brussel sprouts to the oven.
I cooked the chicken for about 30-45 minutes and the sprouts cooked for about 15-20 minutes. I know, I know, I should have watched the clock better. Just wait until the sprouts brown, and until you can cut a chicken tender in half and the juiced run clear and it is white inside. We dipped them in BBQ sauce and they were awesome!! Just what we needed after a two-a-day.
Day 30- Monday. WOW. Our 30th day. Our last day of the challenge and coincidentally my last day at work. I have a lot to say but I am so exhausted I will write a nice big post tomorrow. Today was very easy. I was running late. So all I had for breakfast was a handful of nuts. Then for lunch I had leftover chicken and brussel sprouts- which were just as delicious the next day. Then for dinner we met my dad, my uncle and my cousin for dinner. We went to a steakhouse and I got a 15 ouncer that was AWESOME. I had a salad with no dressing (boring) and some onions with my steak. I ate every bit of that steak because I was starving. I also had a small glass of wine to celebrate my last day at work.
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Pyramid aka Building the Great Pyramid is a 2002 BBC Television documentary film which tells the story of the building of the Great Pyramid at Giza through the commentary of the fictional builder, Nakht.
Production
The film was produced by the BBC in co-production with the Discovery Channel and NDR.
Awards
Primetime Emmy Award 2003
Outstanding Special Visual Effects for a Miniseries, Movie or Special
International Emmy Awards 2004
Nominated: Outstanding Special Visual Effects for a Miniseries, Movie or a Special
Media information
DVD release
Released on Region 2 DVD as a bonus disc with Egypt.
Companion book
References
External links
2002 television films
2002 films
Films set in ancient Egypt
British docudrama films
Pyramids
Discovery Channel original programming
BBC television documentaries about prehistoric and ancient history
2000s British films
British drama television films
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The most effective tax plan starts with knowing the updated rules and doing a lot of small things correctly throughout year. Tax plan does not happen on yearend time and see your accountant, it is a daily fight through a whole year.
The flowing is year-round tax planning calendar which will help you to do your monthly tax planning by yourself.
1st – Record the odometer reading for your vehicle on the first day for allowable auto mobile deduction. Deducting vehicle costs can save your tax dollars greatly, but vehicle expense is an easy target for audits. To ensure the safety of your automobile cost deduction, you need to keep your km log, receipts related to your vehicle well from the first day of year.
28th (Feb 29th in a leap year)-deadline to file T4, T4A, T4 Summary and T5 and T5 Summary to government.
1st-the deadline for making your RRSP contribution for last calendar year.
15th-the deadline for the first tax installment payment if required.
31st-the deadline to file a partnership information return and send the T5013 to patterner (this due date assumes a December 31 partnership year-end).
30th– the deadline to file personal tax return and pay you tax payable if you and your spouse do not carry incorporated business. Even if you and your spouse carry on unincorporated business, the deadline for personal tax payment is also April 30.
15th – The second tax installment due.
30th-The deadline to file T5018 for construction industry business. (This due date assumes a Dec 31 as end reporting period.) The deadline for file T5018 is 6 months after fiscal year or calendar year.
Review your installment option for the upcoming tax installment.
If your tax bill over 3000, you must pay installment. You have 3 different method to calculate your installment: non-calculation option, previous year option or current year option. Which option to use depends the change of your finical situation.
If you have been charged penalty for late filing, consider use voluntary disclosure program (VDP) to waive it.
15th-the third tax installment is due.
If you want to switch you GST calculation method from Direct Method to Quick Method, apply now for the upcoming year.
Review your tax installment for the upcoming final tax installment.
15th – the final tax installment is due.
31th- record your odometer for calculate allowable vehicle deduction.
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The Knicks have selected 6'7 SG/SF Thanasis Antetokounmpo with the 51st pick in the NBA Draft. Thanasis is the brother of Milwaukee's Giannis Antetokounmpo. Thanasis, like his brother, is described as an athletic superfreak with solid upside.
This entry was posted in Articles and tagged ante knicks, antetokounmpo brother, antetokounmpo knicks, greek freak, greek freak brother, greek freak knicks, KNICKS, Knicks news, thanasis antetokounmpo on June 26, 2014 by Tommy Rothman.
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The work of the car designer blends inventive design and automotive styling abilities with elements of engineering, ergonomics and advertising. At SPD designers are in a position to face complex tasks using a multidisciplinary approach. They are known as on to ascertain progressive transportation concepts that fit a future situation the place cars are no extra the unequalled participant. Challenging police wrongdoing is tough for some: many officers cowl themselves in a narrative of heroism, sacrifice and risk every time their actions are questioned. But, just because a person signed on to do a harmful job doesn't give him or her the appropriate to maliciously injure or recklessly take the lives of the people who law enforcement officials are sworn to serve and defend. And when an officer stops serving and protecting, she or he ought to be severely punished both for the violation of that particular person's rights and the violation of the general public's trust.
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With all the net downside fixing knowledge out there to workshops, the trendy office needs to be brimming with new tools that is much more productive than up to now. This highly effective microscope named Titan 80-300 Cubed was installed on the University early in the summer, and since then it has been put via its paces to realize unprecedented resolution. Please recommendation can i accept the visa or not & which points i need to substantiate from the company.
Data and communications technology ―ICT and collaborative communication at the moment are offering new possibilities for giving Africa a voice throughout the globe‖ (Grey, 2010). Since 2009, the capability of Africa's fibre optic cable connections has expanded almost 300-fold. The SEACOM fibre optic cable system was launched to help East and Southern African countries with cheap bandwidth, thus removing the international infrastructure bottleneck.
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Indonesia: Caning of gay men an outrageous act of cruelty
24 May 2017, 10:23 UTC | Indonesia
© Private
Responding to news that two men have been caned 83 times each for having sex with each other in Indonesia's Aceh province, Amnesty International's Deputy Director for Southeast Asia and the Pacific, Josef Benedict, said:
"This sickening spectacle, carried out in front of more than a thousand jeering spectators, is an act of utmost cruelty. These two men had their privacy forcefully invaded when they were ambushed inside their own home, and their 'punishment' today was designed to humiliate as well as physically injure them.
"This sickening spectacle, carried out in front of more than a thousand jeering spectators, is an act of utmost cruelty. These two men had their privacy forcefully invaded when they were ambushed inside their own home, and their 'punishment' today was designed to humiliate as well as physically injure them."
Josef Benedict, Amnesty International's Deputy Director for Southeast Asia and the Pacific
"The authorities in Aceh and Indonesia must immediately repeal the law which imposes these punishments, which constitute cruel, inhuman and degrading treatment and may amount to torture.
"Flogging sentences and the criminalization of same sex relations are both flagrant violations of international human rights law. The international community must put pressure on Indonesia to create a safer environment for the LGBTI community before the situation deteriorates further. Nobody should be punished for consensual sex."
The two men were arrested on 28 March 2017 and charged with consensual same-sex sexual relations (liwath) under the Aceh Islamic Criminal Code. They were sentenced to 85 strokes each by the Banda Aceh Shari'a Court on 17 May, but the number was reduced because they had spent two months in detention.
Shari'a bylaws have been in force in Aceh since the enactment of the province's Special Autonomy Law in 2001, and are enforced by Islamic courts.
This is the first time gay men have been caned under Shari'a law in the province.
Consensual same-sex relations are not treated as crimes under the Indonesian Criminal Code (KUHP).
International Region:
Malaysia: Convictions of two women sentenced to caning for having sexual relations must be quashed
5 LGBTI activists to look out for this IDAHOTB 2018
Japan: End LGBTI discrimination & legalise marriage equality
Indonesia: Police must protect - not attack - transgender women living under threat in Aceh
Tanzania: Activists imprisoned without charge in LGBTI crackdown must be released
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Presented by Utah Shakespeare Festival at Engelstad Shakespeare Theatre, Cedar City UT
NEW DATES: July 21–September 3, 2020
Directed by Kent Thompson
PERICLES will be performed for only the third time in the Festival's history. Pursued by an evil king, Pericles sets sail on an odyssey of romance and spectacle. But dangers face him on every seashore as he drifts from country to country, from year to year, from intrigue to intrigue. His loves die. His friends deceive him. The gods are against him. But, in true storybook fashion, those who were lost are miraculously found, strange
PERICLES will be performed for only the third time in the Festival's history. Pursued by an evil king, Pericles sets sail on an odyssey of romance and spectacle. But dangers face him on every seashore as he drifts from country to country, from year to year, from intrigue to intrigue. His loves die. His friends deceive him. The gods are against him. But, in true storybook fashion, those who were lost are miraculously found, strange events reunite loved ones, and mysterious dreams bring family back together again, at last bringing joy and safe harbor to all.
Contact: 800-PLAYTIX
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Mon, Jun 21, 2021
8:00 pm at Engelstad Shakespeare Theatre
Thu, Jul 1, 2021
Engelstad Shakespeare Theatre
200 W College Ave, Cedar City, UT 84720
Click "What's Nearby" button below to find local Eats, Drinks, and Sleeps!
MORE FROM Utah Shakespeare Festival
Suiting the Action to the Word: Investigating Hamlet
Presented by Utah Shakespeare Festival at Online/Virtual Space
Dec 8, 2020 - May 31, 2021
The Props Treasure Trove. A Peek into the Festival's Prop Warehouse
Every Brilliant Thing - Production, Process, and Tour
The Technical Director's Approach to Scenic Design
Canon Fodder: An Actor's Reflections on Performing All of Shakespeare's Works
CONNECT WITH Utah Shakespeare Festival
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Stop build night came and went. We've now got a robot ready to compete at our two upcoming competitions in St. Louis and Peoria. We've also uploaded a reveal video to our YouTube channel, and underneath the video tab, for those interested in its capabilities. We look forward to watching the first few matches this week. See everybody at the regionals!
It's now officially the start of stop build night and as usual we have a lot of work to do. We will be enjoying a pot-luck dinner from our team parents and working on wrapping up the build season on a strong note. We are hoping to reveal our robot soon!
This past week has been a lot of work! We've been practicing with our robot and refining its shooting. The electronics team rewired the robot and we're hoping to get back to driving today. Stop build day is Tuesday so we are working hard to get as much practice as possible.
Week five is officially over and week six begins. Over the weekend we visited the competition field set up by the gracious Ratchet Rockers, team 1706. We had a great time and were able to identify and sort out some issues with our robot. We are looking forward to nailing down our autonomous in the week to come, as well as practice further with our robot!
After two years and some change we have decided to actually update our website. Today is the beginning of a new era! The robot is almost done, the code is almost done, chairman's is submitted week five is close to ending and stop build day is around the corner. You can look at our week one and two updates under our video tab. We should have a reveal video out around the stop build day (Feb 21st) where you can see in greater detail what we will be working with at competition.
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Center for Cultural Analysis (CCA)
Past Faculty Fellows
Past Postdoctoral Associates
Past Graduate Fellows
Affiliated Fellows
Past Affiliated Fellows
2020-2021 Seminar: What is Photography?
2019-2020 Seminar: The University and its Public Worlds
2019-2020 Post-doctoral Fellow Symposium
Sponsored Working Group
Americanist Seminar
The Americanist Seminar hosted two major conferences intopic "War Genres After War, or, Endless Enemies." In March, with the CCA's Pragmatism Working Group, the Seminar cohosted the half-day conference "Humboldt, Darwin and Pragmatist Worldviews." The event featured presentations by Laura Walls (Notre Dame), Jonathan Elmer (Indiana), Steven Meyer (Washington University at St. Louis), Joan Richardson (CUNY), and George Levine (Rutgers). The Seminar invited Emily Ogden (University of Virginia) to discuss her provocative new book, Credulity: A Cultural History of US Mesmerism (Chicago 2018), Leila Gómez (University of Colorado), a specialist in Latin American travel literature and empire, to lead a series of discussions on Alexander von Humboldt's legacy in Mexico and Peru, and Jameson Sweet (Rutgers, American Studies), to discuss his new work on American Indian intellectual and political history, and Jameson Sweet (Rutgers, American Studies) to discuss his new work on American Indian intellectual and political history. In the Spring Semester, with the Drama Group, Sunny Stalter Pace (Auburn, Rutgers English Ph.D. 2007) returned to Rutgers for a talk on her new project, "Imitation Modernism: Gertrude Hoffman, Pirated Ballets Russes, and Its Relation to American Popular Performance." The Seminar also cosponsored visits to Rutgers for graduate workshops by Elizabeth Maddock Dillon (Northeastern) and Gene Jarrett (NYU).
Upcoming Events for 2019-2020
Co-Sponsored with the CCA Working Group on Pragmatism:
Fall Mini Seminar—3 Meetings in 6 Weeks—"Pragmatism in the World"
Sept. 25, Discussion of selections from Thomas Alexander's The Human Eros
Oct. 8, Discussion of selections from Inventing the Modern Self and John Dewey: Modernities and the Traveling of Pragmatism in Education (2005), Democracy as Culture: Deweyan Pragmatism in a Globalizing World (2009)
Oct. 23, Seminar with Scott Stroud (Department of Communication Studies, University of Texas at Austin) Dewey in South Asia (or India). On Social Justice in India: Bhimrao Ambedkar's Brush with Dewey.
Co-Sponsored with the Seminar on Literature and Political Theory (CCA)
November 19-20
Jack Turner (University of Washington).Our meeting(s) with Turner will be concerned with democracy and kinship, and Whitman will be the main literary writer we'll be studying. (Turner is writing a book on love, death, and politics in Whitman).
New Books in Focus
Nathasha Hurley, Univ. of Alberta, Circulating Queerness: Before the Gay and Lesbian Novel.Discussion of Hurley's new book with her, Gretta LaFleur (Yale), Kyla Schuller (Rutgers), and Dana Luciano (Rutgers). Organized by Elizabeth Dean.
Professor Evans is a specialist in 19th and 20th century American literature and culture and the history of anthropology. He is the author of Before Cultures: The Ethnographic Imagination in American Literature (2005) and Ephemeral Bibelots: How an International Fad Buried American Modernism (forthcoming, 2019). He also co-produced the restoration of a silent feature film that premiered in 1914, In the Land of the Head Hunters, which was directed by the photographer Edward Curtis and starred an all-indigenous cast from the Kwakwaka'wakw community of British Columbia, Canada. The film is now listed in the National Film Registry of the Library of Congress. Each of these projects has been quite different from the other, covering a broad range of media, but they were inspired by something the anthropologist Franz Boas wrote in 1911. In a major study of American Indian languages, Boas demonstrated that race, language and culture circulate independently and at remarkably different rates. Evans's research has focused on historical episodes of uneven circulation—episodes that generated new thinking about the concepts of race and culture, the relation of art and anthropology, and the dynamics of artistic movements.
Working Group Events
Sponsored Working Groups
EMRG @ RU: Early Modern Research Group at Rutgers
Modernism and Globalization Research Group
Seminar on Literature and Political Theory
A Wider Conversation on Religion and the State
Art and Aesthetics
Asian Studies Initiative
Cooperation Across Domains
Experiencing the Salon
Marxism and Materialism
Race and the Early Modern World
Slavery + Freedom Studies Working Group
Society&Design Lab
Sound Studies / Media Studies
The Cliché
Affiliated Working Groups
The Rutgers Book Initiative
15 Seminary Place
Rutgers Academic Building
West Wing, Room 6107
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Health×
Arise for Social Justice
We're a Western MA low-income rights organization which believes we have the right to speak for ourselves. Our members are poor, homeless, at-risk, working, unemployed and people pushed...
CTCORE-Organize Now!
CTCORE- Organize Now! is committed to supporting collective action and moving in solidarity with marginalized groups.
Ex-Prisoners and Prisoners Organizing for Community Advancement (EPOCA)
We believe that social change can only be led by the people who most need the change. No one can give power to someone else. Therefore, those of...
Families for Justice as Healing
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Maine Prisoner Advocacy Coalition (MPAC)
Blue Hill, ME
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National Alliance of HUD Tenants
Founded in 1991, National Alliance of HUD Tenants (NAHT) is the first national membership organization of resident groups advocating for 2.1 million lower income families in privately-owned, HUD-assisted...
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When the gender-specific jail opened in Chicopee in 2008, co-founders Lisa Andrews and Marianne Bullock understood that incarcerated women are not offered the full spectrum of reproductive options,...
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{
"redpajama_set_name": "RedPajamaC4"
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using System;
using Microsoft.Extensions.Primitives;
using Wyam.Common.Util;
namespace Wyam.Razor
{
internal class EmptyChangeToken : IChangeToken
{
public IDisposable RegisterChangeCallback(Action<object> callback, object state) => EmptyDisposable.Instance;
public bool HasChanged => false;
public bool ActiveChangeCallbacks => false;
}
}
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{
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Tennis News: US Open, French Open, Wimbledon, Australian Open | Calgary Sun
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{
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From Long Island to Seattle and the West Coast... and Back, in 16 days!
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This video also shows all but a few photos - Across America and Back in 8 minutes!
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\section{Introduction}
The aim of this paper is to highlight some relations between completions, strongly flat modules and perfect rings in the non-commutative case. We explore some connections between some notions of Homological Algebra (cotorsion modules) and topological rings (completions in some natural topologies). These connections are well known for modules over commutative rings, thanks to Matlis, who proved that the completion in the $R$-toplogy for an integral domain $R$ is closely related to the cotorsion completion functor $\operatorname{Ext}_R^1(K,-)$. Here $Q$ is the field of fractions of $R$ and $K:=Q/R$. We investigate these connections in the non-commutative case, defining a suitable $R$-topology on any module over a not-necessarily commutative ring $R$. This leads us to the study of strongly flat modules, because the completion of $R$ in its $R$-topology turns out to be a strongly flat $R$-module (Theorem~\ref{4.9}).
We consider strongly flat modules over non-commutative rings as defined in \cite[Section~3]{submitted}.
The class of strongly flat modules lies between the class of projective modules and the class of flat modules.
In particular, we study when the class of strongly flat modules is covering, because this is related to an open problem posed by Enochs, whether ``every covering class is closed under direct limit" (see for example \cite [Open problem 5.4] {approx}).
Since flat modules are direct limits of projective modules, the class of strongly flat modules is closed under direct limits if and only if flat modules are strongly flat.
Bazzoni and Salce \cite{silsal} gave a complete answer to this question for modules over commutative domains, completely determining when the class of strongly flat modules over a commutative domain is covering.
Subsequently, Bazzoni and Positselski generalized this to arbitrary commutative rings in \cite{Leonidsilvana}. They proved that, for a commutative ring $R$,
the class $ \mathcal{SF}$ of strongly flat modules is covering if and only if
flat modules are strongly flat, if and only if
$R/aR$ is a perfect ring for every regular element $a \in R$. In our Example~\ref{5.18}, we will show that there exist non-invariant chain domains $R$ for which $\operatorname{End}(R/I)$ is perfect for every non-zero principal right or left ideal
$I$ of $R$, but the class of strongly flat left $R$-modules is not covering. Very recent papers related to these topics are the articles \cite{BP,P}
by Bazzoni and Positselski.
For a commutative ring $R$, the set of regular elements is always an Ore set, and if $Q$ denotes the classical quotient ring of $R$, the class of strongly flat modules is $^\bot \{Q^\bot \} $ \cite{ FSALCE}.
The generalization of strongly flat modules to non-commutative rings given in \cite {submitted} depends on the choice of the overring $Q$ of $R$. More precisely, if
$\varphi \colon R\to Q$ is a bimorphism in the category of rings, that is, $\varphi$ is both a monomorphism and an epimorphism, we assume that ${}_RQ$ is a flat left $R$-module. We view at $R$ as a subring of $Q$ and $\varphi \colon R\to Q$ as the inclusion.
Then a left $R$-module $_RM$ is {\em Matlis-cotorsion} if $\operatorname{Ext}^1(_RQ,{}_RM)=0$ \cite{submitted}.
Let $\mathcal {MC}$ denote the class of Matlis-cotorsion left $R$-modules. For any class of left $R$-modules $ \mathcal {A}$, set $ ^ \bot \mathcal {A}:= \{ \,B\in R\mbox{\rm -Mod} \mid \operatorname{Ext}_R ^1 (B, A) = 0$ for every $A \in \mathcal {A}\,\}$ and $\mathcal {A}^ \bot := \{ \,B\in R\mbox{\rm -Mod} \mid\operatorname{Ext}_R ^1 (A, B) = 0$ for every $A \in \mathcal {A}\,\}$.
A left $R$-module is {\em strongly flat} if it is in $ ^\bot \mathcal {MC} $.
The class of strongly flat left $R$-modules will be denoted by $\mathcal {SF}$.
By
\cite [Theorem 6.11] {approx}, the cotorsion pair $(\mathcal{SF} , \mathcal{MC})$ is
complete, that is, every left $R$-module has a special
$\mathcal{MC}$-preenvelope (or, equivalently, every left module has a special $\mathcal{SF}$-precover).
Thus, by \cite [Corollary 6.13] {approx}, the class $\mathcal{SF} $ consists of
all direct summands of
modules $N$ such that $N$ fits into an exact sequence of the form
$$ 0 \to F \to N \to G \to 0,$$
where $F$ is a free $R$-module and $G$ is $\{Q\}$-filtered. For the terminology, see \cite{submitted}.
Whenever $R$ is a right Ore domain, i.e., the class of regular elements is a right Ore set,
the class of strongly flat left $R$-modules is the class $^\bot \{Q^\bot\} $, where $Q$ is the classical right quotient ring of $R$.
Several of our results about strongly flat modules are for modules over a nearly simple chain domain. Recall that a chain domain $R$, that is, a not-necessarily commutative integral domain for which the modules $R_R$ and $_RR$ are uniserial, is {\em nearly simple} if it has exactly three two-sided ideals, necessarily $R$, its Jacobson radical $J(R)$ and $0$. The reason why we concentrate on chain domains $R$ with classical quotient ring $Q$ is due to the fact that for these rings the $R$-module $_RK:=Q/R$ is uniserial, and thus, in the study of $\operatorname{End}(_RK)$, we can take advantage of our knowledge of the endomorphism rings of uniserial modules \cite{DungAlberto, DungFacchini, Facchinitransaction, Facsal, Pavel, puninsky}. In our Example~\ref{5.18}, we also take advantage of our knowledge of the endomorphism rings of cyclically presented modules over local rings \cite{Aminis}.
If $R$ is a right chain domain and the class of strongly flat $R$-modules is
covering, then
$R$ is right invariant, that is, $aR = Ra$ for every $a \in R$. In this case, flat modules are strongly flat
(equivalently, the class $ \mathcal{SF}$ of strongly flat modules is closed under direct limit).
We began this paper in September 2017, when both of us where visiting the Department of Algebra of Charles University in Prague, and continued in March 2018 when the first named author was visiting the IPM (Institute for Research in Fundamental Sciences) in Tehran. We are very grateful to both institutions for their hospitality.
\section{The $R$-topology} \label{completions}
In Sections \ref{completions}, \ref{3} and \ref{4} of this paper, we suppose that {\em we have a ring $R$ and a multiplicatively closed subset $S$ of $R$ satisfying:} (1) {\em If $a,b\in R$ and $ab\in S$, then $a\in S$.} (2) {\em $S$ is a right Ore set in $R$.} (3) {\em The elements of $S$ are regular elements of~$R$}. (4) {\em The right ring of quotients $Q:=R[S^{-1}]$ of $R$ with respect to $S$ is a directly finite ring}. That is, our setting is that of \cite[Section~4]{submitted}.
Correspondingly, we have a Gabriel topology $\Cal G$ on $R$ consisting of all the right ideals $I$ of $R$ with $I\cap S\ne\emptyset$ (cf.~\cite[\S VI.6]{20}). In particular, the Gabriel topology $\Cal G$ consists of dense right ideals of $R$, the canonical embedding $\varphi \colon R\to Q:=R[S^{-1}]$ is an epimorphism in the category of rings, we view $R$ as a subring of $Q$ and $\varphi$ as the inclusion mapping, and ${}_RQ$ turns out to be a flat left $R$-module \cite[\S XI.3]{20}.
There is a hereditary torsion theory $(\Cal T,\Cal F)$ on $\operatorname{Mod-\!} R$ in which
the torsion submodule of any right $R$-module $M_R$ consists of all the elements $x\in M_R$ for which there exists an element $s\in S$ with $xs=0$. If we indicate the torsion submodule of $M$ by
$t(M)$, then clearly $t(M) \otimes _R Q = 0$. A right $R$-module $M_R$ is in $\Cal F$, that is, is torsion-free, if and only if right multiplication $\rho_s\colon M_R\to M_R$ by $s$ is an abelian group monomorphism for every $s\in S$. Dually, we will say that a right $R$-module $M_R$ is {\em divisible} if right multiplication $\rho_s\colon M_R\to M_R$ by $s$ is an abelian group epimorphism for every $s\in S$, that is, if $Ms=M$ for every $s\in S$. Every homomorphic image of a divisible right $R$-module is divisible. If $A$ is a submodule of a right $R$-module $B_R$ and both $A_R$ and $B/A$ are divisible, then $B_R$ is divisible. Any sum of divisible submodules is a divisible submodule, so that every right $R$-module $M_R$ contains a greatest divisible submodule, denoted by $d(M_R)$. A right $R$-module $M_R$ is {\em reduced} if $d(M_R)=0$. For every module $M_R$, $M_R/d(M_R)$ is reduced.
We have that $\Cal G=\{\,I\mid I$ is a right ideal of $R$, and $\varphi(I)Q=Q\,\}$, and $\Cal G$ has a basis consisting of the principal right ideals $sR$, $s\in S$. Let $M_R$ be any right $R$-module.
By \cite [XI, Proposition 3.4] {20},
the kernel of the canonical right $R$-module morphism $M_R\to M\otimes_R Q$ is
equal to $t(M)$. Note that if we set $K : = Q/R$, then
$_RK_R$ is an $R$-$R$-bimodule and $t(M_R)\cong\operatorname{Tor}_1^R(M_R, {}_RK)$ (see (15) and (16) in \cite[Section 3]{submitted}).
\bigskip
We now define a topology on any right $R$-module in the attempt of generalizing the $R$-topology studied by
Matlis \cite{dim} for a commutative ring $R$. Our definition is as follows. Let $R$ be any ring with identity, not
necessarily commutative, and $S$ be a subset of $R$ with the properties written at the beginning of this section.
Given any right $R$-module $M_R$, the {\em $R$-topology} on $M_R$ has a
neighborhood base of $0$ consisting, for every non-empty finite set of elements $s_1,\dots,s_n\in S$, of the
submodules $$U(s_1,\dots,s_n):=\{\,x\in M_R\mid xR\subseteq Ms_1\cap\dots\cap Ms_n\,\}$$ of $M_R$. For the
regular right module $R_R$, the
$R$-topology on $R$ has a neighborhood base of $0$ consisting, for every non-empty finite set of elements $s_1,\dots,s_n\in S$, of the right ideals $$U(s_1,\dots,s_n):=\{\,x\in R\mid xR\subseteq Rs_1\cap\dots\cap Rs_n\,\}$$ of $R$.
\begin{Lemma} On the right $R$-module $R_R$, the right ideals $U(s)$ are two-sided ideals of $R$, $U(s)$ is the annihilator of the left $R$-module $R/Rs$, and the $R$-topology is a ring topology on $R$.\end{Lemma}
\begin{proof} Clearly, $U(s)=\{\,x\in R\mid xR\subseteq Rs\,\}$ is the annihilator of the cylic left $R$-module $R/Rs$, and hence $U(s)$ is a two-sided ideal. Moreover, $R$ is a right linearly topological ring \cite[p.~144]{20}, because every filter of two-sided ideals of a ring is a fundamental system of neighborhoods of $0$ for a right and left linear topology on the ring \cite[p.~144]{20}.\end{proof}
We will use $R_{R\operatorname{-top}}$ to denote the topological ring $R$ with the $R$-topology.
\begin{Lemma}\label{12.7} Every right $R$-module, with respect to its $R$-topology, is a linearly topological module over the topological ring $R_{R\operatorname{-top}}$.\end{Lemma}
\begin{proof} It suffices to check property {\em TM}\,3 in \cite[p.~144]{20}. That is, we must prove that $(U_M(s):x)\supseteq U_R(s)$ for every $s\in S$, $x\in M_R$. Equivalently, that $xU_R(s)\subseteq U_M(s)$. Now if $r\in U_R(s)$, then $rR\subseteq Rs$, so that $xrR\subseteq xRs\subseteq Ms$, i.e., $rx\in U_M(s)$.\end{proof}
\begin{Lemma}\label{Matlis} If the ring $R$ is commutative, the linear topology on any right $R$-module $M$ defined by the submodules $U(s)$, $s\in S$, coincides with the $R$-topology defined by Matlis in \cite{dim}.\end{Lemma}
\begin{proof} $U(s)=\{\,x\in M\mid xR\subseteq Ms\,\}=Ms$.\end{proof}
In the next proposition, we consider the behavior of continuity of right $R$-module morphisms when the modules involved are endowed with the $R$-topology. Recall that a submodule $M$ of a right $R$-module $N_R$ is an {\em RD-pure submodule} if $Mr=M\cap Nr$ for every $r\in R$ (equivalently, if the natural homomorphism $M\otimes R/Rr\to N\otimes R/Rr$ is injective for every $r\in R$, or if the natural homomorphism $\operatorname{Hom}(R/rR,N)\to\operatorname{Hom}(R/rR,M)$ is surjective for every $r\in R$.) See \cite[Proposition~2]{WarfPur}.
\begin{proposition}\label{easy} {\rm (a)} Every right $R$-module morphism $f\colon M_R\to N_R$ between two right $R$-modules $M_R$ and $N_R$ endowed with their $R$-topologies is continuous.
{\rm (b)} For every right $R$-module $N_R$ and every $s\in S$, the $R$-submodule $U(s)$ of $N_R$ is the largest $R$-submodule of $N_R$ contained in $Ns$.
{\rm (c)} A submodule $M_R$ of a right $R$-module $N_R$ endowed with the $R$-topology is an open submodule of $N_R$ if and only if $M_R\supseteq U(s)$ for some $s\in S$.
{\rm (d)} A right $R$-module morphism $f\colon M_R\to N_R$ between two right $R$-modules $M_R$ and $N_R$ with their $R$-topologies is an open map if and only if $f(M_R)\supseteq U(s)$ for some $s\in S$.
{\rm (e)} Every right $R$-module epimorphism $f\colon M_R\to N_R$ between two right $R$-modules $M_R$ and $N_R$ is an open continuous map.
{\rm (f)} Every right $R$-module isomorphism $f\colon M_R\to N_R$ is a homeomorphism when the two right $R$-modules $M_R$ and $N_R$ are endowed with their $R$-topologies.
{\rm (g)} If $M_R$ is an RD-pure submodule of a right $R$-module $N_R$ and $M_R,N_R$ are endowed with their $R$-topologies, then the embedding $M_R\hookrightarrow N_R$ is a topological embedding.\end{proposition}
The proofs are easy and we omit them.
\section{The right $R$-module $\operatorname{Hom}(K_R, M\otimes_RK)$}\label{3}
In this section, the hypotheses on $R$ and $S$ are the same as in the previous section.
For any right $R$-module $M_R$, we will be interested in the right $R$-module $$\operatorname{Hom}(K_R, M\otimes_RK).$$ Here the right $R$-module structure is given by the multiplication defined, for every $f\in \operatorname{Hom}(K_R, M\otimes_RK)$ and $r\in R$, by $(fr)(k)=f(rk)$ for all $k\in K$.
For any right $R$-module $M_R$, the right $R$-module $$\operatorname{Hom}(K_R, M\otimes_RK)$$ can be endowed with the $R$-topology, defined by the submodules $U(s_1,\dots,s_n):=U(s_1)\cap\dots\cap U(s_n)$ as a
neighborhood base of $0$. But we have that:
\begin{Lemma} For the modules $\operatorname{Hom}(K_R, M\otimes_RK)$, one has that $U(s)=V(s)$, where, for every element $s\in S$, $$V(s):=\{\,f\in \operatorname{Hom}(K_R, M\otimes_RK)\mid f((Rs^{-1})/R)=0\,\}.$$\end{Lemma}
\begin{proof} $(\subseteq)$. Let $f$ be an element of $U(s)$, so that $f\in\operatorname{Hom}(K_R, M\otimes_RK)$ and $fR\subseteq \operatorname{Hom}(K_R, M\otimes_RK)s$. In order to show that $f\in V(s)$ we have to prove that $f((Rs^{-1})/R)=0$. Fix $r\in R$. Then $fr=gs$ for some $g\in \operatorname{Hom}(K_R, M\otimes_RK)$. Hence $f(rs^{-1}+R)=(fr)(s^{-1}+R)=(gs)(s^{-1}+R)=g(ss^{-1}+R)=0$. Thus $f((Rs^{-1})/R)=0$.
$(\supseteq)$. Suppose $f\in V(s)$, so that $f((Rs^{-1})/R)=0$. In order to prove that $f\in U(s)$, we must show that, for every fixed element $r\in R$, there exists $g\in \operatorname{Hom}(K_R, M\otimes_RK)$ with $fr=gs$. Define $g\colon K_R\to M\otimes_RK_R$ by $g(q+R)=f(rs^{-1}q+R)$ for all $q\in Q$. Then $g$ is a well defined right $R$-module morphism, because if $q\in R$, then $f(rs^{-1}q+R)=f(rs^{-1}+R)q\in f((Rs^{-1})/R)R=0$, and $fr=gs$.\end{proof}
\bigskip
We will denote by $V(s_1,\dots,s_n)$ the intersection $V(s_1)\cap\dots\cap V(s_n)$, but it is necessary to remark that:
\begin{Lemma} For every $s,s'\in S$, there exists $t\in S$ such that $V(s)\cap V(s')\supseteq V(t)$.\end{Lemma}
\begin{proof}
Given $s,s'\in S$, there exist $t\in S$ and $r,r'\in R$ with $t=sr=s'r'$ \cite[Lemma~4.21]{goodwar}. Then $s^{-1}=rt^{-1}$, so that $Rs^{-1}=Rrt^{-1}\subseteq Rt^{-1}$. Therefore $V(t)\subseteq V(s)$, because if $f\in \operatorname{Hom}(K_R, M\otimes_RK)$ and $f(Rt^{-1}/R)=0$, then $f(Rs^{-1}/R)=0$, that is, $f\in V(s)$. Similarly, $V(t)\subseteq V(s')$.\end{proof}
A right (or left) $R$-module $M_R$ is {\em $h$-divisible} if every homomorphism $R_R\to M_R$ extends to an $R$-module morphism $Q_R\to M_R$ \cite[Section 2]{submitted}. Any right (or left) $R$-module $M$ contains a unique largest $h$-divisible submodule $h(M)$ that contains every $h$-divisible submodule of $M$. An $R$-module $M_R$ is {\em $h$-reduced} if $h(M_R)=0$, or, equivalently, if $\operatorname{Hom}(Q_R,{}M_R)=0$ \cite{submitted}.
Obviously, $h$-divisible right $R$-modules are divisible.
\begin{proposition} \label{equal} Divisible torsion-free right $R$-modules are
$Q$-modules.
In particular, $h(M_R)=d(M_R)$ for any torsion-free right $R$-module~$M_R$.\end{proposition}
\begin{proof} Suppose $M_R$ torsion-free and divisible. Then right multiplication by $s$ is an automorphism of the abelian group $M$ for every $s\in S$. By the universal property of $Q=R[S^{-1}]$, the canonical ring antihomomorphism $R\to\operatorname{End}_\mathbb{Z}(M)$ extends to a ring antihomomorphism $Q\to\operatorname{End}_\mathbb{Z}(M)$ in a unique way. That is, there is a unique right $Q$-module structure on $M$ that extends the right $R$-module structure of $M_R$. Thus $M$ is a right $Q$-module. In particular, it is an $h$-divisible right $R$-module.
\end{proof}
\bigskip
Let $M_R$ be a right $R$-module. For every element $x\in M_R$, there is a right $R$-module morphism $R_R\to M_R$, $1\mapsto x$. Tensoring with $_RK$, we get a right $R$-module morphism $\lambda_x\colon K_R\to M\otimes_RK$, defined by $\lambda_x(k)=x\otimes k$. The canonical mapping $\lambda\colon M_R\to\operatorname{Hom}(K_R, M\otimes_RK)$, defined by $\lambda(x)=\lambda_x$ for every $x\in M_R$, is a right $R$-module morphism, as is easily checked. In the rest of this section, all $R$-modules are endowed with their $R$-topologies.
\begin{theorem}\label{righthreduced} Let $M_R$ be an $h$-reduced torsion-free right $R$-module. Then the canonical mapping $\lambda\colon M_R\to\operatorname{Hom}(K_R, M\otimes_RK)$ is an embedding of topological modules and $\operatorname{Hom}(K_R, M\otimes_RK)$ is complete.\end{theorem}
\begin{proof} The canonical mapping $\lambda\colon M_R\to\operatorname{Hom}(K_R, M\otimes_RK)$ is injective by \cite[Theorem~4.5]{submitted}. In order to show that $\lambda\colon M_R\to\operatorname{Hom}(K_R, M\otimes_RK)$ is an embedding of topological modules, it suffices to show that $\lambda^{-1}(V(s_1,\dots,s_n))=U(s_1,\dots,s_n)$ for every $s_1,\dots,s_n\in S$. Now $x\in \lambda^{-1}(V(s_1,\dots,s_n))$ if and only if $\lambda_x\in V(s_1,\dots,s_n)$, that is, if and only if $x\otimes (Rs_1^{-1}+\dots+Rs_1^{-1}/R)=0$ in $M\otimes_RK$. Equivalently, if and only if $x\otimes (rs_i^{-1}+R)=0$ in $M\otimes_RK$ for every $r\in R$ and $i=1,2,\dots,n$. By \cite[Step 3 of the proof of Theorem~4.5]{submitted}, this is equivalent to $xr\in Ms_i$ for every $r\in R$ and $i=1,2,\dots,n$, that is, if and only if $x\in U(s_1,\dots,s_n)$.
In order to prove that $\operatorname{Hom}(K_R, M\otimes_RK)$ is complete, we must show that every Cauchy net converges. Let $A$ be a directed set with order relation $\le$ and let $\{f_\alpha\}_{\alpha\in A}$ be a Cauchy net in $\operatorname{Hom}(K_R, M\otimes_RK)$. Define a morphism $f\in \operatorname{Hom}(K_R, M\otimes_RK)$ as follows. Since we are dealing with a Cauchy net, for every $s\in S$ there exists $\alpha\in A$ such that $f_\beta-f_\gamma\in V(s)$ for every $\beta,\gamma\in A$, $\beta,\gamma\ge\alpha$. Set $f(rs^{-1}+R)=f_\alpha(rs^{-1}+R)$ for every $r\in R$. We leave to the reader the easy verification that $f$ is a well defined mapping.
Let us check that $f(kr)=f(k)r$ for every $k\in K_R$ and $r\in R$.
We have that $k=as^{-1}+R$ for some $a\in R$, $s\in S$. By the right Ore condition, there exist $r'\in R$ and $t\in S$ such that $as^{-1}r=r't^{-1}$. Since $A$ is directed, there exists $\alpha$ such that
$f ( r't^{-1} + R) = f_\alpha ( r't^{-1}+ R)$ and $f( as^{-1}+R) r = f_\alpha ( as^{-1}+R) r $.
Therefore $f (kr) = f(k) r$.
It is now easily seen that $f$ is the limit of the Cauchy net.
\end{proof}
For any right $R$-module $M_R$ endowed with its $R$-topology, the {\em (Hausdorff) completion} of $M_R$ is $\displaystyle \widetilde{M_R}:= \lim_{\longleftarrow} M/U(s_1,\dots,s_n)$. Notice that the set of all the submodules $U(s_1,\dots,s_n)$ of $M_R$ is downward directed under inclusion. Here $\{s_1,\dots,s_n\}$ ranges in the set of all finite subsets of $S$. There is a canonical mapping $\eta\colon M\to \widetilde{M_R}$, whose kernel is the closure $\overline{\{0\}}$ of $0$ in the $R$-topology of $M_R$. Clearly, $\overline{\{0\}}=\bigcap_{s_1,\dots,s_n\in S} U(s_1,\dots,s_n)=
\bigcap_{s\in S} U(s)=\{\,x\in M_R\mid xR\subseteq \bigcap_{s\in S}Ms\,\}$.
From Lemma \ref{12.7}, we get that if $M_R$ is a right $R$-module, the right $R$-module $\operatorname{Hom}(K_R, M\otimes_RK)$ with the topology defined by the submodules $V(s)$ is a topological module over the topological ring $R_{R\operatorname{-top}}$.
\begin{proposition} The right $R$-submodules $V(s)$ of the ring $\operatorname{End}(K_R)$ are two-sided ideals of $\operatorname{End}(K_R)$. The topology they define on $\operatorname{End}(K_R)$ is a ring topology. If $R$ is commutative, this topology on $\operatorname{End}(K_R)$ coincides with the topology on the completion $H$ of $R$ with respect to the $R$-topology \cite[p.~15]{dim}.\end{proposition}
\begin{proof} When we consider $M = R_R$, then, by
\cite[Step 2 of the proof of Theorem~4.5]{submitted}, the elements of $K$ annihilated by right multiplications of an element $s\in S$ are those of $Rs^{-1}/R$. It follows that $Rs^{-1}/R$ is a fully invariant submodule of $K_R$. From this we get that every $V(s)$ is a two-sided ideal of the ring $\operatorname{End}(K_R)$.
Every filter of two-sided ideals of a ring is a fundamental system of neighborhoods of $0$ for a right and left linear topology on the ring \cite[p.~144]{20}. Thus the topology defined by the two-sided ideals $V(s)$ is a ring topology on $\operatorname{End}(K_R)$. Moreover, if $R$ is commutative, the submodules $V(s)$ define the $R$-topology on the right $R$-module $\operatorname{Hom}(K_R, M\otimes_RK)$ for every module $M$ (Lemma~\ref{12.7}), which coincides with the $R$-topology defined by Matlis in \cite{dim} by Lemma~\ref{Matlis}.
Finally, Matlis' $R$-topology on $\operatorname{End}(K_R)$ coincides with the topology on the completion $H$ of $R$ with respect to the $R$-topology, because the topology on the completion $H$ coincides with the $R$-topology on $H$.
\end{proof}
\section{Torsion-free modules}\label{4}
In this section, we keep the same hypotheses and notations as in the previous two sections.
As we have seen, for any right $R$-module $M_R$, there is a right $R$-module morphism $\lambda\colon M_R\to\operatorname{Hom}(K_R, M\otimes_RK)$, defined by $\lambda(x)=\lambda_x$ for every $x\in M_R$, where $\lambda_x\colon k\to x\otimes k$, and there is a canonical mapping $\eta\colon M\to \widetilde{M_R}$ of $M_R$ with its $R$-topology into its Hausdorff completion.
\begin{proposition} \label {torsionfreeright} Let $M_R$ be a torsion-free right $R$-module. Then: {\rm (a)} $\ker\lambda$ is the closure of $0$ in the $R$-topology; {\rm (b)} $\ker\lambda$ is the kernel of the canonical mapping $\eta\colon M\to \widetilde{M_R}$; and {\rm (c)} $\ker\lambda$ is equal to $h(M_R)$.\end{proposition}
\begin{proof} We have already remarked that the kernel of $\eta$ is the closure of $\overline{\{0\}}$ of $0$. Hence (a)${}\Leftrightarrow{}$(b).
The right $R$-module $\operatorname{Hom}(K_R,N_R)$ is $h$-reduced for every right $R$-module $N_R$ \cite[Theorem~2.8]{submitted}. Let $M_R$ be a torsion-free right $R$-module. Since $$\lambda\colon M_R\to \operatorname{Hom}(K_R, M\otimes_RK)$$ is a homomorphism into an $h$-reduced $R$-module, it follows that $h(M)\subseteq \ker\lambda$.
Let us prove that $\ker\lambda\subseteq \overline{\{0\}}$. Suppose $x\in \ker\lambda$. Then $x\otimes (rs^{-1}+R)$ is equal to zero in the tensor product $M\otimes K$. By \cite[Theorem~3.1(1)]{submitted}, there exists an element $y_{r,s}\in M_R$ such that $x\otimes rs^{-1}=y_{r,s}\otimes 1$ in $M\otimes_RQ$. Thus $xr\otimes 1=y_{r,s}s\otimes 1$ in $M\otimes_RQ$. Since $M_R$ is torsion-free, it follows that $xr=y_{r,s}s$ in $M_R$ by \cite[Theorem~3.1(1)]{submitted} again. This proves that $xR\subseteq \bigcap_{s\in S}Ms$, and so $\ker\lambda\subseteq \overline{\{0\}}$.
Conversely, $\overline{\{0\}}\subseteq\ker\lambda$, because if $x\in \overline{\{0\}}$, then $xR\subseteq Ms$ for every $s\in S$, that is, for every $s\in S$ and every $r\in R$ there exists $m_{r,s}\in M$ with $xr=m_{r,s}s$. Then, for every element $rs^{-1}+R\in K$, we have that $x\otimes(rs^{-1}+R)=xr\otimes (s^{-1}+R)=m_{r,s}s\otimes (s^{-1}+R)=m_{r,s}\otimes s(s^{-1}+R)=0$ in $M\otimes_RK$. Thus $x\in\ker\lambda$. This proves that $\overline{\{0\}}=\ker\lambda$. Therefore (a) and (b) hold.
We now show that $\ker\lambda$ is divisible. For every $s\in S$, $s$ is invertible in $Q$, hence $sQ=Q$, so $sK=K$. Now if $x\in \ker\lambda$ and $t\in S$, then $x\in \overline{\{0\}}$, hence $x=yt$ for some $y\in M_R$. We must prove that $y\in \ker\lambda$, that is, that $y\otimes K=0$ in $M\otimes K$. But $y\otimes K=y\otimes sK=ys\otimes K=x\otimes K=0$ in $M\otimes K$. This proves that $\ker\lambda=\overline{\{0\}}$ is divisible.
Thus $ \ker\lambda = h(M)$ by Proposition \ref{equal}.
\end{proof}
Clearly, from Proposition~\ref{torsionfreeright}, we have that:
\begin{corollary}
If $M_R$ is a torsion-free module, then $\widetilde{M_R} \cong \widetilde{M_R / h(M) }$.
\end{corollary}
\begin{lemma} \label{fourparts} Let $M$ be torsion-free right $R$-module. Then:
{\rm (a)} Every element of $M\otimes_RK$ can be written in the form $x\otimes (s^{-1}+R)$ for suitable elements $x\in M_R$ and $s\in S$.
{\rm (b)} Let $s$ be an element of $S$. The elements $y$ of $M\otimes_RK$ such that $ys=0$ are those that can be written in the form $x\otimes (s^{-1}+R)$ for a suitable $x\in M_R$.
{\rm (c)} If $x\in M_R$, $r\in R$ and $s\in S$, then $x\otimes(rs^{-1}+R)=0$ in $M\otimes_RK$ if and only if $xr\in Ms$.
{\rm (d)} The set $\{\, U(s) \mid s \in S\,\} $ is downward directed.
\end{lemma}
\begin{proof}
In the proof of Steps 1, 2 and 3 of \cite [Theorem 4.5] {submitted}, we do not use the fact that $M$ is $h$-reduced. So the proofs of (a), (b) and (c) are like those of Steps 1, 2 and 3 in \cite [Theorem 4.5] {submitted}.
(d) Assume that $s, t \in S$. Then there exist $u \in S$ and $r_1,r_2\in R$ such that $ s ^{-1}= r_1u^{-1} $ and
$ t ^{-1}= r_2 u^{-1} $. If $m \in U(u)$ and $r \in R$, then $m \otimes (rs^{-1}+R) =
m \otimes (r r_1u^{-1} +R)=0$. Part (c) implies that $m \in U(s)$, and so $U(u) \subseteq U(s)$.
Similarly, $U(u) \subseteq U(t)$.
\end{proof}
\begin{remark}\label{remarktorsionfree}
{\rm By Lemma~\ref{fourparts}(d), for
$M$ torsion-free, we have that $$\widetilde{M}= \lim_{\longleftarrow} M/U(s).$$ Notice that the kernel of
the canonical mapping $\eta \colon M\to \widetilde{M}$ is divisible by Theorem \ref{torsionfreeright}. } \end{remark}
\bigskip
Now let $M_R$ be a torsion-free right $R$-module, so that $$\lambda\colon M_R\to\operatorname{Hom}(K_R,\linebreak M\otimes_RK)$$ is continuous with respect to the $R$-topologies (Proposition~\ref{easy}(a)) and $\operatorname{Hom}(K_R,\linebreak M\otimes_RK)$ is Hausdorff. Notice that $M\otimes_RK$ and $M/h(M)\otimes_RK$ are isomorphic, so that $\operatorname{Hom}(K_R, M\otimes_RK)$ is complete (Theorem~\ref{righthreduced}). Thus $\lambda$ extends in a unique way to a continuous morphism $\widetilde{\lambda}\colon \widetilde{M} \to \operatorname{Hom}(K_R, M\otimes_RK)$. In Theorem~\ref{completion} and Example~\ref{quasismall}, we see { that $\widetilde{\lambda}$ is a continuous monomorphism, but not necessary an
isomorphism.
\begin{theorem}\label{completion} Let $M_R$ be a torsion-free right $R$-module. Then there exists a right $R$-module monomorphism $\widetilde{\lambda}\colon \widetilde{M} \to \operatorname{Hom}(K_R, M\otimes_RK)$ such
that $\lambda=\widetilde{\lambda}\eta$.
\end{theorem}
\begin{proof}
Define $\widetilde{\lambda}$ as follows. We know that $$\displaystyle\widetilde{M}= \lim_{\longleftarrow} M/U(s)\le\prod _{s\in S} M/U(s),$$ so that every element of $\widetilde{M}$ is of the form $\widetilde{m}=(m_{s}+U(s))_{s\in S}$.
Set $\widetilde{\lambda}(\widetilde{m})(rs^{-1}+R)=m_{s}\otimes (rs^{-1}+R)$ for every $r\in R$, $s\in S$.
In order to prove that $\widetilde{\lambda}(\widetilde{m})\colon K_R\to M\otimes_RK$ is a well defined mapping and is $R$-linear, note first of all that
if $s, t \in S$ are such that $U(t) \subseteq U(s)$ and $r \in R$, then $m_s - m_{t} \in U (s)$ implies that
$m_s \otimes rs^{-1}+ R = m_{t} \otimes rs^{-1}+ R$ by Lemma~\ref{fourparts}(c). From this, it is easily shown that $\widetilde{\lambda}$ is a well defined $R$-module morphism. Also notice that $\lambda=\widetilde{\lambda}\eta$.
Now we prove that $\widetilde{\lambda}$ is a monomorphism.
Suppose that $\widetilde{m}=(m_{s}+U(s))_{s\in S}$ is in $\ker{\widetilde{\lambda}}$. Then, for any $k\in K$ and any $s\in S$ with $ks=0$, we have that $m_{s}\otimes k=0$ in $M\otimes_RK$. In particular, for every $r\in R$, $s\in S$, the identity $(rs^{-1}+R)s=0$ implies that $m_{s}\otimes (rs^{-1}+R)$ in $M\otimes_RK$. By Lemma~\ref{fourparts}(c), this means that $m_{s}r\in Ms$ for every $r$ and $s$. Hence $m_{s}\in U(s)$ for every $s\in S$. This shows that
$\widetilde{\lambda}$ is injective.
\end{proof}
}
\begin{example} \label {quasismall}{\rm
Let $R$ be the nearly simple chain domain in \cite [Example 6.5] {chainringandprimeideal}.
In that example, the $R$-module $Q/R$ can be chosen to be countably generated, because the
group $G$ is countable, and so is its positive cone $P$.
If the skew field $K$ in that example is countable, then $K[P]$
is countable. In order to construct the ring $R$, the authors consider a right and left Ore subset $S$ of
$K[P]$, which is necessarily countable because $K[P]$ is countable, and
then they set $R := K[P] S^{-1}$.
Therefore if the skew field $K$ is countable, then
$R$ is countable, and so $Q/R$ is a countably generated $R$-module.
As $R_R$ is torsion-free, its completion is $\displaystyle\lim_{\longleftarrow} R/U(s)$ by
Remark~\ref{remarktorsionfree}, and, for every non-zero element $s$ of $J(R)$,
$U(s) = 0$ because $R$ is nearly simple. So $ \displaystyle R = \lim_{\longleftarrow} R/U(s)$. Let us prove that $R \ncong \operatorname{End} (K_R) $.
The module $K_R$ is a countably generated uniserial torsion locally coherent module (that is, every finitely generated submodule is coherent). By \cite[Proposition~8.1]{puninsky}, the module $K_R$ is not quasi-small. Since uniserial modules with a local endomorphism ring are quasi-small \cite{DungFacchini}, the ring $\operatorname{End}(K_R)$ cannot be isomorphic to $R$.
The same argument applies to any nearly simple chain domain $R$ with $Q/R$ countably generated.}
\end{example}
\begin{proposition}
If $R$ is a topological ring with a basis $B$ of neighborhoods of zero consisting of two-sided ideals, and
$R/I$ is a local ring for every proper ideal $I\in B$, then the Hausdorff completion of $R$ is either $0$ or
a local ring.
\end{proposition}
\begin{remark}{\rm
The case of completion of $R$ equal to zero concernes only the trivial case of $B=\{R\}$. We will not consider this case in the proof.}
\end{remark}
\begin{proof}
Let $M_I $ be the maximal ideal of $R$ such that $M_I/I$
is the maximal ideal of $R/I$ for every proper ideal $I\in B$. If $I,J\in B$, then considering the canonical projection
$ R/I\cap J\to R/I$, one sees that
$M_{(I\cap J)}=M_I$. It follows that there exists a maximal ideal $M$ of
$ R$ such that $M_I=M$ for every proper ideal $ I\in B$. The completion of $R$ is the inverse limit of the rings
$R/I$, which is a subring of the ring $\prod_{I\in B}R/I$, which has $\prod_{I\in B}M/I$ as a two-sided ideal, whose intersection $N$ with the inverse limit is a two-sided ideal of the inverse limit. Let us prove that the inverse limit is a local ring with maximal ideal $N$. It suffices to show that every element of the inverse limit not in $N$
is invertible. Let $(x_I+I)_{I\in B}$ be an element in the inverse limit, but not in $N$. Thus
$x_I\in R$ and, for $I,J\in R$ with $I\subseteq J$, we have that $ x_I-x_J\in J$, i.e., $x_I+I $
is mapped to $ x_J+J $ via the canonical projection $R/I\to R/J.$ Also, $ x_I\notin M$ for some proper ideal $I$ of
$B$. It follows that $x_I\notin M $ for every proper ideal $ I$ of $ B$. Thus $x_I+I\notin M/I,$
hence is invertible in $R/M$. Let $y_I+I$ be the inverse of $x_I+I$ in $R/I$. Now the ring morphism
$R/I\to R/J$
maps inverses to inverses. This shows that $(y_I+I)_{I\in B}$ is an element of the inverse limit, and concludes the proof.
\end{proof}
Therefore the completion of any local ring in the $R$-topology is a local ring.
\medskip
Note that, by Theorem \ref {completion} and \cite [Proposition 2.6]{submitted}, if
$M_R$ is torsion-free, then $\widetilde{M_R}$ is torsion-free.
\begin{theorem}\label{4.9} Let $R$ be a right Ore domain and $\widetilde{R_R}$ the completion of $R_R$ in the $R$-topology. Then $\widetilde{R_R}$ is a strongly flat right $R$-module.\end{theorem}
\begin{proof} We can apply the
results of \cite[Section~3]{submitted}, which are right/left symmetric, that is, hold for both right $R$-
modules and left $R$-modules. Notice that $R_R$ is $h$-reduced. We have the short exact sequence\begin{equation}
\xymatrix{
0 \ar[r] & R_R \ar[r] & \operatorname{End}(K_R) \ar[r] & \operatorname{Ext}^1_R({}_RQ_R,R_R) \ar[r] & 0.
}\end{equation} We know
that $\widetilde{R_R}$ is a submodule of $\operatorname{End}(K_R)$ that contains $R_R$. Hence $\widetilde{R_R}/R_R$ is
isomorphic to a submodule of $\operatorname{Ext}^1_R({}_RQ_R,R_R) $. In particular,
$\widetilde{R_R}/R_R$ is torsion-free, because $\operatorname{Ext}^1_R({}_RQ_R,R_R) $ is a $Q$-module, hence torsion-free. Let us
prove that $\widetilde{R_R}/R_R$ is divisible, i.e., that $(\widetilde{R_R}/R_R)r=\widetilde{R_R}/R_R$ for every non-zero $r\in R$. Equivalently, we must prove that $\widetilde{R_R}\subseteq \widetilde{R_R}r+R_R$. Now $R_R$ is
dense in $\widetilde{R_R}$,
so that, for every $\widetilde{r}\in\widetilde{R_R}$ and every
non-zero element $s$ of $R$, we have that $(\widetilde {r} + U(s)) \cap R_R \neq \emptyset. $ In particular,
$(\widetilde{r}+U(r))\cap R_R\ne\emptyset. $
Notice that $U(r)\subseteq \widetilde{R_R}r$, because, for every $x\in U(r)$, we have that $xR\subseteq \widetilde{R_R}r$, hence $x\in \widetilde{R_R}r$. It follows that $(\widetilde{r}+\widetilde{R_R}r)\cap R_R\ne\emptyset. $ Thus there exists $\widetilde{r'}\in \widetilde{R_R}$ and $r''\in R_R$ with $
\widetilde{r}+\widetilde{r'}r=r''$. Therefore $\widetilde{r}=-\widetilde{r'}r+r''\in\widetilde{R_R}r+R_R$. This proves that $\widetilde{R_R}/R_R$ is divisible and torsion-free, hence a module over the division ring $Q$. Thus $\widetilde{R_R}/R_R\cong Q^{(X)}$ for some set $X$. The short exact sequence \begin{equation*}
\xymatrix{
0 \ar[r] & R_R \ar[r] & \widetilde{R_R}\ar[r] & Q^{(X)} \ar[r] & 0.
}\end{equation*} shows that $\widetilde{R_R}$ is strongly flat.\end{proof}
\section{Strongly flat modules}
{\em In all this section, we consider two rings $R$ and $Q$, a bimorphism $\varphi \colon R\to Q$ in the category of rings, that is, $\varphi$ is both a monomorphism and an epimorphism, and we assume that ${}_RQ$ is a flat left $R$-module. For simplicity, we will view $R$ as a subring of $Q$ and $\varphi \colon R\to Q$ as the inclusion.}
Let us recall some properties of such an inclusion $\varphi \colon R\hookrightarrow Q$. It is always possible to suppose $Q \subseteq Q_{\rm max}(R) $, the maximal ring of quotients of $R$ \cite[proof of Theorem XI.4.1]{20}. The
inclusion $\varphi \colon R\to Q$ is an epimorphism in the category of rings if and only if the canonical $R$-$R$-bimodule morphism $Q \otimes _R Q \to Q$ induced by the multiplication $\cdot\colon Q\times Q\to Q$ of the ring $Q$ is an $R$-$R$-bimodule isomorphism \cite[Proposition~XI.1.2]{20}.
The family of all the subrings $Q$ of $Q_{\rm max}(R) $ with $\varphi \colon R\hookrightarrow Q$ a bimorphism and ${}_RQ$ flat is directed under inclusion \cite[Lemma XI.4.2]{20}. Its direct limit is the ``maximal flat epimorphic right ring of quotients''
$Q_{\rm tot}(R)$ of $R$ (see the paragraph after the proof of Corollary~\ref{12}).
By \cite [Theorem 4.8] {rep},
$\operatorname{Ext}^1(_RM,{} _RN) \cong \operatorname{Ext}^1(_QM,{} _QN) $ for any pair $M,N$ of left $Q$-modules, and similarly for right $Q$-modules.
\begin{lemma}\label{Divisiblestronglyflat}
Divisible strongly flat left $R$-modules are
projective $Q$-modules.
\end{lemma}
\begin{proof}
Assume that $_RD$ is a divisible strongly flat module. Since $K \otimes Q = 0$, we have $K \otimes D = 0$. Since $_RD$ is flat, we have
$D \cong Q \otimes D$. For any exact sequence
$0 \to R^{(X)} \to D \oplus T \to Q^{(Y)} \to 0 $, the corresponding exact sequence $0 \to Q^{(X)} \to D \oplus Q\otimes T \to Q^{(Y)} \to 0 $ splits. Therefore
$D$ is a projective $Q$-module.
\end{proof}
Recall that any left perfect ring is directly finite. The following result shows that
when $_R \mathcal{SF}$ is covering, then $Q$ is left perfect. Thus the results is the same as in
the commutative case, but the proof is necessarily different.
\begin{theorem} \label{Qisperfect}
If all left $Q$-modules have a strongly flat cover as left $R$-modules,
then
$Q$ is left perfect.
\end{theorem}
\begin{proof}
Assume that $_QM$ is a left $Q$-module and
$f\colon{}_RS \to{} _RM$ is a strongly flat cover of $_RM$. Then we have
an epimorphism $1\otimes f \colon Q\otimes S \to M$, $1\otimes f\colon
q\otimes s\mapsto qf(s)$.
Since $_RS$ is strongly flat, $_QQ\otimes_R S$ is a direct
summand of a direct sum of copies of $Q$, i.e.,
it is a projective left $Q$-module. Since projective left $Q$-modules
are strongly flat left $R$-modules, the left $R$-module $_RQ\otimes S$ is
strongly flat.
But $f$ is a strongly flat precover of $M$, so that there exists $g \colon Q\otimes
S \to S$ with
$f g = 1\otimes f$.
Note that $_RS$ is flat, and so $S$ can be embedded in $Q\otimes S$, that
is, there is a left $R$-module monomorphism $h\colon _RS\to _RQ\otimes_R
S$, defined by $h\colon s \mapsto 1\otimes s$.
Then $f (gh) = f $, and thus $gh $ is an automorphism of $_RS$ because
$f\colon _RS \to _RM$ is a cover. Thus $(gh)^{-1}gh=1$, so that
$e:=h(gh)^{-1}g$ is an idempotent endomorphism of the left $R$-module
$_RQ\otimes S$. Hence $e$ is an
idempotent endomorphism of the left $Q$-module $_QQ\otimes S$. This shows that
$_QQ\otimes S$ is the direct sum of the image and the kernel of $e$,
which are $Q$-modules. But the image of $e$ is the image of $h$. Hence
the splitting monomorphism $h\colon s \mapsto 1\otimes s$ induces by
corestriction a right $R$-module isomorphism of $_RS$ onto the
$Q$-module $_Qh(S)$. By \cite[Section~2(7)]{submitted},
if a left $R$-module $_RA$ is a left $Q$-module $_QA$, then its
unique left $Q$-module structure is given by the canonical isomorphism
$\operatorname{Hom}(_RQ,_RA) \to{} _RA$. Therefore $S$ has a unique left $Q$-module
structure, which extends its left $R$-module structure, and as such
$_QS$ is a projective $Q$-module. Thus $f\colon{} _QS \to {}_QM$ is a left
$Q$-module morphism.
Note that projective $Q$-modules are strongly flat, and so $f\colon{} _QS \to{}
_QM$ is a projective cover of $_QM$. Therefore
$Q$ is left perfect.
\end{proof}
The following result has a proof similar to that of \cite[Proposition 2.4 ((1) and (2))]{silsal}.
\begin{lemma}\label{B-small}
Let $A$ be a module with a strongly flat cover and let
\begin{equation} 0 \to C \to M \to A \to 0 \label{(1)}\end{equation}
be a special strongly flat precover of $A$. Then the exact sequence~{\rm (\ref{(1)})} is a strongly flat cover
if and only if $C$ is $\mathcal{MC}$-small (i.e., $C + H = M$ and $C\cap H $ Matlis-cotorsion imply
$H = M$).
\end{lemma}
\begin{theorem} \label{twosidedidealIQ}
Let
$I$ be a two-sided ideal of $R$ such that $IQ = Q$.
If all left $R/I$-modules have a strongly flat cover as left
$R$-modules, then
$R/I$ is left perfect.
\end{theorem}
\begin{proof}
It is enough to show that every left $R/I$-module has a projective
$R/I$-cover.
Let $M$ be an $R/I$-module and $f\colon{} _RA \to {}_RM$ be a strongly flat
cover of $_RM$.
Since $IM = 0$, we have that $IA \subseteq \ker (f)$.
Since ${}_RA$ is strongly flat, there exists an exact sequence
$ 0 \to R^{(X)} \to A \oplus T \to Q^{(Y)} \to 0 $, where $X$ and $Y$ are sets.
Since $IQ = Q$, we have $R/I \otimes Q = 0$. Thus we see that $A/IA$ is a projective left $R/I$-module.
So $f$ induces a map $h\colon A/IA \to M$, $h\colon a+ IA \mapsto f(a)$, and $\ker(h) = \ker(f)/IA$.
Now
$_RA$ is strongly flat and $IQ = Q$, and so $A/IA$ is a projective
left $R/I$-module.
We now show that $h$ is a cover for $M$ or, equivalently, that
$\ker(f)/IA$ is small in $A/IA$.
Assume that $T + \ker(f) = A$, where $T$ is an $R$-module of $A$ that
$IA \subseteq T $. Since $IQ = Q$, Hom$(Q, \ker(f)/ \ker(f) \cap T) = 0$. On the other hand,
since $f\colon{} _RA \to {}_RM$ is a strongly flat
cover of $_RM$, the module $\ker(f)$ is Matlis-cotorsion by Wakamatsu Lemma (see \cite [Lemma 5.13] {approx}), and thus
$ \ker(f) \cap T$ is Matlis-cotorsion. Therefore $T = A$ by Lemma \ref{B-small}.
\end{proof}
\begin{lemma} \label{3.3}
Assume that $R$ is a local ring with Jacobson radical $J$. Let
$0 \to C \to S \to M \to 0$ be an $\mathcal{SF}$-cover for $M$. Then
$C \leq JS$.
\end{lemma}
\begin{proof} Assume that $C \nleq JS$. Then $JS \neq S$. Since $R/J$ is a division ring, there exists a proper submodule $T/JS$ of $S/JS$ such that
$T/JS + (C + JS)/JS = S/JS$. Consequently $T + C = S$.
Consider the exact sequence
$0 \to T \cap C \to C \to S/ T \to 0$.
Let us show that $\operatorname{Hom}(Q, S/T) = 0$.
Note that $R_R$ is essential in $Q_R$ (because $Q$ is a subring of $Q_{\rm max}(R)$).
Thus if $x \in Q \setminus R$,
then the right ideal of $I = \{\,r \mid xr \in R\,\}$
is proper ideal of $R$, and so $I \leq J$.
By \cite [Part (b) of Theorem 3.9]{Goodearlnonsingular},
$IQ = Q$ and so $JQ = Q$.
If Hom$(Q, S/T) \neq 0$, then there exists a proper submodule
$E$ of $Q$ such that $Q/E$ is isomorphic to a submodule of $S/T$.
Thus $ Q = JQ \leq E$, which is a contradiction. Therefore
Hom$(Q, S/T) = 0$, and so $T\cap C \in Q^\bot$.
Since $C$ is $\mathcal{MC}$-small, we have $T = S$, which is a contradiction.
\end{proof}
It is known that if $R$ is commutative, $Q$ is the field of fractions of $R$ and
$_R \mathcal{SF}$ is covering, then $\pdim(_RQ) \leq 1$. We do not know what occurs in the
non-commutative case. Therefore we now study the projective dimension of~${}_RQ$.
\begin{proposition}\label{3.4'} Suppose ${}_RQ$ is a projective left $R$-module. Then ${}_RQ$ is a finitely generated left $R$-module. \end{proposition}
\begin{proof} Since ${}_RQ$ is projective, it has a dual basis \cite[Exercise 11, pp.~202-203]{andersonfuller}, that is, there are elements $x_\alpha\in Q$ and morphisms $f_\alpha\colon {}_RQ\to {}_RR$ ($\alpha\in A)$, such that, for all $x\in Q$, $f_\alpha(x)\ne0$ for only finitely many $\alpha\in A$ and $x=\sum_{\alpha\in A}f_\alpha(x)x_\alpha$. Applying the functor ${}_QQ\otimes_R-\colon R\mbox{\rm -Mod}\to Q\mbox{\rm -Mod}$, we get left $Q$-module morphisms $1\otimes f_\alpha\colon {}_QQ\otimes_RQ\to {}_QQ\otimes_RR$. Now there are left $Q$-module isomorphisms $Q\to {}_QQ\otimes_RQ$, $q\mapsto 1\otimes q$, and ${}_QQ\otimes_RR\to {}_QQ$, $q\otimes r\mapsto qr$. Composing, we get left $Q$-module endomorphisms ${}_QQ\to {}_QQ$, which are necessarily right multiplications by elements $y_\alpha\in Q$. Now, for all $x\in Q$, $f_\alpha(x)\ne0$ for only finitely many $\alpha\in A$. For $x=1$, we get that there is a finite subset $F$ of $A$ such that $f_\alpha(1)=0$ for every $\alpha\in A\setminus F$. Thus $(1\otimes f_\alpha)(1\otimes 1)=0$ for every $\alpha\in A\setminus F$. It follows that right multiplication by $y_\alpha$ maps $1$ to $0$, that is, $y_\alpha=0$ for every $\alpha\in A\setminus F$. It follows that $1\otimes f_\alpha\colon {}_QQ\otimes_RQ\to {}_QQ\otimes_RR$ is the zero mapping for every $\alpha\in A\setminus F$. Thus $(1\otimes f_\alpha)(q\otimes q')$ is the zero element of ${}_QQ\otimes_RR$ for every $q,q'\in Q$. Hence $1\otimes f_\alpha(q')$ is the zero element of ${}_QQ\otimes_RR$. It remains to show that the mapping $_RR\to {}_QQ\otimes_RR$, $r\to 1\otimes r$, is injective, which is easily seen because $\operatorname{Tor}_1^R(K,R)=0$. This proves that $f_\alpha=0$ for every $\alpha\in A\setminus F$. As a consequence, $_RQ$ is isomorphic to a direct summand of ${}_RR^F$, so that ${}_RQ$ is a finitely generated left $R$-module. \end{proof}
\begin{corollary}\label{12} Let $R$ be a ring, $S$ a multiplicatively closed subset of regular elements of $R$, and suppose that $S$ is a right denominator set, so that the right ring of fractions $Q:=R[S^{-1}]$ exists. If ${}_RQ$ is a projective left $R$-module, then $Q=R$, that is, all the elements of $S$ are invertible in $R$. \end{corollary}
\begin{proof} By Proposition \ref{3.4'}, there are finitely many elements $r_1s_1^{-1},\dots,r_ns_n^{-1}$ that generate $Q$ as a left $R$-module. Reducing to the same denominator \cite[Lemma~4.21]{goodwar}, we find elements $r'_i\in R$ and $s\in S$ such that $s_ir'_i=s$ for every~$i$. Multiplying by $s^{-1}$ on the right and by $s_i^{-1}$ on the left, we get that $r'_is^{-1}=s_i^{-1}$. Thus $Q=\sum_{i=1}^n Rr_1s_1^{-1}\subseteq Rs_1^{-1}$. This proves that $Q=Rs_1^{-1}$. In particular, $s^{-2}\in Rs_1^{-1}$, from which $1\in Rs$. Let $t\in R$ be such that $1=ts$. Then $t=s^{-1}$ in~$Q$. Thus $Q=Rs_1^{-1}=Rt\subseteq R$, hence $Q=R$.\end{proof}
On page 235 of \cite{20}, Stenstr\"om asks for necessary and sufficient conditions
for $Q_{\rm max}(R) $ to be equal to $Q_{\rm tot}(R)$. He shows that if $Q_{\rm max}(R)$ is a
right Kasch ring (i.e., a ring that contains a copy of its simple right modules), then $Q_{\rm max}(R) = Q_{\rm tot}(R)$.
If $R$ is right hereditary right noetherian \cite[Example 3, p.~235]{20} or
commutative noetherian
\cite[Example 4, p.~237]{20} or a right Goldie ring \cite[Theorem XII 2.5] {20}, then
$Q_{\rm max}(R)$ is known to be Kasch.
\begin{example}
{\rm Here is an example of a ring $R$ for which $Q_{\rm tot}(R) = Q_{\rm max}(R)$ is a projective right and left $R$-module, but $R \neq Q_{\rm max}(R)$. Let $R$ be the ring of all lower triangular $2 \times 2 $ matrices over a field $F$.
The ring $R$ is right nonsingular and $E(R_R)=S^{0}R=Q_{\rm max}(R)$ is a projective right and left $R$-module \cite [Exercise 14 on Page 78, and Corollary 2.31] {Goodearlnonsingular} (in Goodearl's notation, $S^{0}A := E(A/ Z(A_A))$, the injective envelope of $A/ Z(A_A)$ for any ring $A$).
More precisely,
$Q_{\rm max} (R) $ is the $2 \times 2 $ matrix ring over the field $F$, which is a semisimple artinian ring, hence a right and left Kasch ring, and
so $Q_{\rm max}(R) = Q_{\rm tot}(R)$ as we have seen above.}
\end{example}
We are now ready to consider the case of $\pdim(_RQ)\le 1$. Recall that a cotorsion pair $(\Cal A,\Cal B)$ is said to be {\em hereditary} if $\operatorname{Ext}_R^i(A,B)=0$ for all $i\ge 1$, $A\in\Cal A$ and $B\in\Cal B$.
Note that if $\Cal F$ is the class of flat modules and $\Cal E \Cal C$ the class
of Enochs cotorstion modules, the the cotorsion pair
$(\Cal F,\Cal E \Cal C)$ is always a hereditary cotorsion pair.
Similarly to \cite[Lemma 7.53] {approx}, we can show that:
\begin{lemma}
The following conditions are equivalent for the pair of rings $R\subseteq Q$:
{\rm (a)} $\pdim(_RQ) \leq 1$.
{\rm (b)} The cotorsion pair $(\mathcal{SF}, \mathcal{MC})$ is hereditary.
\end{lemma}
\begin{proof}
(a)${} \Rightarrow{}$(b). Assume that $\pdim(_RQ) \leq 1$. Then strongly flat modules, which are summands of
extensions of a direct sum of copies of $Q$ by a free module, are of $\pdim$ at most $1$. Thus
the cotorsion pair $(\mathcal{SF}, \mathcal{MC})$ is hereditary.
(b)${} \Rightarrow{}$ (a). By \cite [Theorem 3.5]{AS}, it is enough to show that that $\operatorname{Ext}^1(K, M)$ is $h$-reduced Matlis-cotorsion.
Using the exact sequence $0 \to M \to E(M) \to E(M)/M \to 0$, we have the exact sequence
$0 \to A \to B \to \operatorname{Ext}^1(K, M) \to 0$, where
$A = $ Hom$(K, E(M))/$ Hom $(K, M)$ and $B = $ Hom $(K, E(M)/M)$.
Note that, for every module $N$, Hom$(K, N)$ is Matlis-cotorsion and $h$-reduced by \cite [Theorem 2.8] {submitted}.
So $\operatorname{Ext}^1(K, M)$ is $h$-reduced if and only if $A \in Q^\bot$. Now $A \in Q^\bot$
follows from the fact that $(\mathcal{SF}, \mathcal{MC})$ is hereditary and the exact sequence
$0 \to $ Hom $(K, M) \to $ Hom$(K, E(M)) \to A \to 0$. As the module
$A$ is Matlis-cotorsion, from the exact sequence $0 \to A \to B \to \operatorname{Ext}^1(K, M) \to 0$ and the fact that
$(\mathcal{SF}, \mathcal{MC})$ is hereditary, we get that $\operatorname{Ext}^1(K, M)$ is Matlis-cotorsion.
\end{proof}
As a consequence, $_R\mathcal{SF} ={} _R\mathcal{F}$ implies $\pdim(_RQ) \leq 1$.
\begin{lemma}\label{stronglyflatreduced}
Let $R$ be a right Ore domain and $Q $ the classical right quotient ring of $R$. If $S$ is a strongly flat left $R$-module, then $S/h(S)$ is also strongly flat.
\end{lemma}
\begin{proof}
Assume that $R$ is not a division ring. There exists an exact
sequence $0 \to R^{(X)} \to S \oplus C \to Q^{(Y)} \to 0$.
We claim that Hom$(Q, R) = 0$. Otherwise, i.e., if $_RQ$ can be embeded in $_RR$, there exists a monomorphism $\varepsilon\colon _RQ\to _RR$. Then
$\varepsilon$ can be viewed as a monomorphism $_RQ\to{} _RQ$. This monomorphism $\varepsilon$ is right multiplication by an element $q$ of $Q$. Now $\varepsilon$ a monomorphism implies $q\ne 0$, and $R$ right Ore domain implies $Q$ division ring.
Hence $q$ is invertible in
$Q$, so that $R=Q$, which is a contradiction. This proves our claim. Now
we have the embedding Hom$(Q, S \oplus C) \to $ Hom $(Q, Q^{(Y)})$.
So we have an exact sequence $0 \to R^{(X)} \to (S\oplus C)/h(S\oplus C) \to Q^{(Y)}/ h(S\oplus C)\to 0$.
Since $h(S\oplus C)$ is a torsion-free divisible module, it is a $Q$-module. But $Q$ is division ring, so
$h(S\oplus C)$ is a direct summand of $Q^{(Y)}$. It follows that $S /h(S)$ is strongly flat.
\end{proof}
A {\em left coherent} ring is a ring over which every finitely generated left ideal is finitely presented or, equivalently,
intersection of two finitely generated left ideals is finitely generated.
\begin{theorem}\label{ideal}
Assume that $R$ is a left coherent Ore domain with classical right quotient $Q$. A left ideal ${}_RI$ of $R$ is a strongly flat left module if and only if $_RI$ is finitely generated projective.
\end{theorem}
\begin{proof}
Assume $_RI$ a non-zero strongly flat.
We have the exact sequence of $R$-$R$-bimodules $0 \to R \to Q \to Q/R = K \to 0$.
Since $_RI$ is flat, we get the exact sequence of left $R$-modules
$0 \to R \otimes I \to Q \otimes I \to K \otimes I \to 0$.
Therefore $ K \otimes I \cong (Q \otimes I) / (R \otimes I)$.
We want to show $R/I$ embeds in $K \otimes I$ as $R$-modules.
Consider the sequence of left $R$-modules
$0 \to{}_RI\to{}_RQ\to{}_RQ/I\to 0$ and apply to it the functor $Q\otimes_R-$. Since $Q_R$ is flat, we get to
an exact sequence $0 \to Q\otimes_R I \to Q\otimes_R Q \to Q\otimes_R Q/I \to 0$.
Under the natural isomorphism $f\colon Q \otimes _R Q \to Q$, the image of $Q \otimes I $ is $QI = Q$, because $I$ is non-zero,
and the image of
$R\otimes I$ is $I$, and so $ K \otimes I \cong (Q \otimes I)/( R\otimes I) \cong Q/I$ as a left $R$-module. Now $R/I \leq Q/I$ implies that $R/I$ embeds in $K \otimes I$ as $R$-module.
There exists an exact
sequence $0 \to R^{(X)} \to I \oplus T \to Q^{(Y)} \to 0$.
Since $K \otimes _R Q = 0$,
we conclude that $K \otimes I$, and so $R/I$, embed in $K^{(X)}$ as left $R$-modules.
Consequently, there exists an element $x \in{} _R K^{(X)}$
whose annihilator is equal to $I$.
But the annihilator of an element of $ K^{(X)}$
is equal to the intersection of finitely many annihilators of
elements of $K$.
If $ab^{-1} + R \in {}_RK$, then ann$(ab^{-1} + R) = R \cap Rba^{-1}$.
Note that $ R \cap Rba^{-1} \cong Ra \cap Rb$, which is a finitely generated left ideal of $R$ because $R$ is left coherent. Thus $I$ is a finitely generated left ideal of $R$ and, since it is flat, $I$ is projective \cite [Theorem 4.30] {lam2}.
\end{proof}
\begin{lemma}\label{Sflat}
Let $R$ be a right Ore ring with classical right quotient ring $Q$. Then the strongly flat cover of any $h$-reduced flat left $R$-module is $h$-reduced.
\end{lemma}
\begin{proof}
Assume that $ M $ is a flat $h$-reduced module and $0 \to C \to S \to M \to 0$ is a strongly flat cover of
$M$. Since $M$ is $h$-reduced, we can assume that $D:= h(C) = h(S)$.
So we have an exact sequence $ 0 \to C/D \to S/D \to M \to 0$. By Lemma \ref{stronglyflatreduced}, $S/D$ is strongly flat.
We can easily see that this sequence is a strongly flat precover for $M$.
Note that $C$ is torsion-free, and so $D$ is a left $Q$-module. Thus $\operatorname{Ext}_R^1(Q, D) = 0$. Since
$C$ is $\mathcal{MC}$-small in $S$, we see that $C/D$ is
$\mathcal{MC}$-small. It follows that
$0 \to C/D \to S/D \to M \to 0$ is a strongly flat cover of $M$ by Lemma~\ref {B-small}(2), and so $S \cong S/D$. Therefore $D = 0$.
\end{proof}
\begin{proposition} \label{final}
Assume that $R$ is an Ore local domain with classical quotient ring $Q$.
Suppose that $K \otimes _R S$ is
direct sum of copies of $K$ for every strongly flat module $_RS$.
If $_R \mathcal{SF}$ is a covering class, then $_R \mathcal{SF} ={} _R \mathcal{F}$.
\end{proposition}
\begin{proof}
Firstly, notice that left $Q$-modules are
injective as $R$-modules because $R$ is both a right and a left Ore domain. So, if $M$ is flat, then
$h(M)$ is
a direct summand of $M$, and therefore $M \cong h(M) \oplus M/h(M)$.
Clearly, $Q$-modules are strongly flat, and thus it is enough to show that any flat $h$-reduced module is strongly flat.
Let $M$ be an $h$-reduced flat left module and $0 \to C \to S \to M \to 0$ be a strongly flat cover of
$M$. By Lemma \ref{Sflat}, $S$ is also $h$-reduced, and thus $C$ is an $h$-reduced flat left $R$-module.
Assume that $C \neq 0$, and let $0 \to C' \to S' \to C \to 0 $ be a strongly flat cover of $C$.
Then $S'$ is Matlis-cotorsion $h$-reduced strongly flat.
Note that by the left version of \cite [Theorem 4.6] {submitted},
we have an exact sequence
$0 \to S' \to \operatorname{Hom}(K, K \otimes S') \to \operatorname{Ext}^1 (Q, S') \to 0$.
Thus $S' \cong \operatorname{Hom}(K, K \otimes S')$.
Since $S'$ is strongly flat, it is a direct summand of a direct sum of copies of $K$,
and thus
$K \otimes S'$ is isomorphic to a direct summand of a direct sum of copies of $K$, $K \otimes S' \cong K^{(Z)}$ say.
Thus $ S' \cong \operatorname{Hom}(K, K \otimes S') \cong \operatorname{Hom}(K, K^{(Z)}) \cong \operatorname{Hom} (K, K \otimes R^{(Z)})$.
By \cite [Theorem 4.6] {submitted}, we have an exact sequence
$0 \to R^{(Z)} \to \operatorname{Hom}(K, K \otimes R^{(Z)}) \to \operatorname{Ext}^1 (Q, R^{(Z)}) \to 0$.
Since $\operatorname{Ext}^1 (Q, R^{(Z)})$ is a left $Q$-module,
$R/J \otimes \operatorname{Ext}^1 (Q, R^{(Z)}) = 0$, where $J$ denotes the Jacobson radical of $R$.
Consequently, $JS' \neq S'$.
On the other hand,
by Lemma~\ref{3.3} and considering the pure exact sequence
$0 \to C \to S \to M \to 0$, we see that $JC = C$. By
Lemma \ref{3.3} and considering the exact sequence
$0 \to C' \to S' \to C \to 0 $ again, we see that $J S' = S'$, which is a contradiction. This proves that $C = 0$, so that
$M$ is strongly flat.
\end{proof}
For any left module $_RM$, let $\Add(_RM)$ denote the class of all left $R$-modules isomorphic to direct summands of direct sums of copies of $_RM$.
We will say that $\Add(_RM)$ is {\em trivial} if every direct summand of a direct sum of copies of $_RM$ is a direct sum of copies of $_RM$.
\begin{lemma}\label{Pavel}
Let $R$ be a nearly simple chain domain and let $_RK$ be the uniserial left $R$-module $Q/R$. Suppose $\Add(_RK)$ not trivial. Then
there exists a submodule $V$ of $_RK$ that is not quasismall. Moreover, all the elements of $\Add(_RK)$ are isomorphic to $R$-modules of the form $_RK^{(X)} \oplus _RV^{(Y)}$.
\end{lemma}
\begin{proof}
See \cite[Theorem 1.1(ii)] {Pavel}.
\end{proof}
In the next proposition, we describe uniserial strongly flat modules over Ore domains.
\begin{proposition}
If $R$ is
an Ore domain with classical quotient ring $Q$, then every non-zero uniserial strongly flat left module over $R$ is isomorphic to $_RQ$ or~$_RR$.
\end{proposition}
\begin{proof}
Let $_RU$ be a non-zero uniserial strongly flat left module over an Ore domain $R$.
Since $_RU$ is flat, considering the exact sequence $0 \to R \to Q$, we have an embedding $U\to Q\otimes_RU$.
Hence the annihilator of every non-zero element of $ _RU$ is zero, and so cyclic submodules of $U$ are isomorphic to $_RR$. In particular, the ring $R$ is a left chain ring.
Moreover, $U$ is the union of cyclic submodules isomorphic to
$ _RR$, that is, a direct (linearly ordered) system of copies of $_RR$, where the connecting homomorphisms are right
multiplications by non-zero elements of $R$. Applying the functor $ _RQ\otimes_R-$, since tensor product commutes with direct limits, we get that $_RQ\otimes_RU$ is a direct limit of a direct system copies of $_RQ$, in which the
connecting isomorphisms are right multiplications by non-zero elements of $ R$, that is, the connecting isomorphisms are all left $R$-module automorphisms of $_RQ$. That is, $_RQ\otimes_RU \cong _RQ$.
Hence $_RU$ embeds into $_RQ\otimes_RU\cong{}_RQ.$ If this embedding is onto, then $_RU\cong {}_RQ$. If the embedding is not onto, then $_RU$ is isomorphic to a proper submodule of $_RQ$, hence to a left ideal of $R$.
By Theorem~\ref{ideal}, $_RU$ is cyclic, and so isomorphic to $_RR$.
\end{proof}
\begin{lemma}\label{JV}
Let $R$ be a nearly simple chain domain with Jacobson radical $J$. If $\Add(K)$ is not trivial,
$V$ is as in Lemma {\rm \ref{Pavel}} and $M := \operatorname{Hom} (K, V^{(X)}),$ where $X$ is a non-empty set,
then $JM \neq M$.
That is, $M$ has maximal submodule.
\end{lemma}
\begin{proof} The module $M=\operatorname{Hom}(_RK,{}_RV)$
is a left $R$-module because $_RK_R$ is a bimodule.
Notice that $_RM$ always has a direct summand isomorphic to
$\operatorname{Hom}(K, V)$, so that we can suppose that $X$ has exactly one element.
By \cite [(ii) of Theorem 1.1] {Pavel}, $K$
has an endomorphism whose image is contained in $V$, say $\varphi\colon {}_RK\to{}_RV $, that is injective but
not surjective. Let us show that $\varphi$ is in $M$ but not in $JM$.
For every $j \in J$ and $\psi\in\operatorname{Hom}(_RK,{}_RV)$, the left $R$-module morphism $j\psi$
is not injective. In fact,
$ j\psi$ is right multiplication by $j$ viewed as a morphism $ _RK\to{} _RK $
composed with $\psi\colon{}_RK\to{}_RV. $
Thus the first morphism annihilates the element $j^{-1}+R,$
so that the kernel of $ j\psi$ is non-zero. (This proves that $ j\psi$ is not injective for $j\ne 0.$
But also when $j=0$, $j\psi$ is not injective.)
Now every element of $JM $ a finite sum of elements of the form $j\psi,$
i.e., of non-injective homomorphisms, hence is not injective because $_RK$ is
uniserial, hence uniform.
Therefore $\varphi\colon {}_RK\to{}_RV$ is not an element of $JM$.
\end{proof}
Recall that a two sided ideal $I$ of $R$ is {\em completely prime} if $xy \in I$ implies that $x \in I $ or $y \in I$ for every $x, y \in R$.
\begin{theorem}
If $R$ is a right chain domain with classical right quotient ring $Q$ such that
$_R \mathcal{SF}$ is a covering class, then
$R$ is invariant and
$_R \mathcal{SF} = {}_R \mathcal{F}$.
\end{theorem}
\begin{proof}
If $I$ is a non-zero completely prime two-sided ideal of $R$,
$R/I$ is a left perfect domain by Theorem~\ref{twosidedidealIQ}, and so it is is a division ring. Since $J(R)/I$ is an ideal of $R/I$, we conclude that
the only proper non-zero completely prime ideal of $R$ is $J(R)$.
A chain domain $R$ is said to be of rank one if $J(R)$ is its only non-zero completely prime ideal.
By \cite{chainringandprimeideal}, such a ring
is either invariant, i.e., $aR = Ra$
for all $a \in R$, or it
is nearly simple, in which case $ 0$ and $J(R)$ are the only two-sided ideals, or
$ R$
is exceptional
and there exists a non-zero prime ideal
$P$
properly contained in $J(R)$. In this last case, $\bigcap_n {P^n} = 0$ and
there are no further ideals between $P$ and $J(R)$.
In the second and the third case, $J(R)$ is not neither right nor left finitely generated and $J^2 = J$.
Now we break the proof in three steps.
\medskip
{\em Step 1: The ring $R$ cannot be exceptional. }
The Jacobson radical of $R/P$ is $J/P$, which cannot be nilpotent because $J^2 = J$.
Thus $R/P$ cannot have a
$T$-nilpotent Jacobson radical (see for example the proof of \cite [Lemma~3.33]{quasifereb}), and so $R/P$ is neither a right nor a left perfect ring, and so the class of strongly flat left modules is not covering by Theorem~\ref{twosidedidealIQ}.
\medskip
{\em Step 2: The ring $R$ cannot be nearly simple chain domain. }
Suppose $R$ a nearly simple chain domain.
For every strongly flat module $_RS$, $K \otimes S$ is direct summand of a direct sum of copies of $K$, so that $K \otimes S$ belongs to $\Add(K)$.
We have two cases:
$\Add(K)$ is trivial or not. If $\Add(K)$ is trivial, then $_R \mathcal{SF}$ covering implies $_R \mathcal{SF} = {}_R \mathcal{F}$ by Proposition \ref{final}. But every cyclic (=\,finitely generated) ideal of $R$ is flat (= projective), so $_RJ$ must be flat, hence strongly flat (see for example \cite [Theorem 39.12(2)] {wis}). Thus
$J$ must be finitely generated by Theorem~\ref{ideal}, which is a contradiction.
Now assume that $\Add(K)$ is not trivial.
By Lemma \ref{Pavel},
there exists a uniserial module $V$ which is not quasismall and every element in $\Add(K)$ is in form of $K^{(Y)} \oplus V^{(X)}$ for suitable sets $X$ and $Y$.
Let $0 \to C \to S \to J \to 0$ be a strongly flat cover of
$J$. By Lemma \ref{Sflat}, $S$ is also $h$-reduced, and so $C$ is an $h$-reduced flat left module.
Assume $C \neq 0$, and let $0 \to C' \to S' \to C \to 0 $ be a strongly flat cover of $C$.
Then $S'$ is Matlis-cotorsion $h$-reduced strongly flat.
By the left version of \cite [Theorem~4.6] {submitted},
we have an exact sequence
$0 \to S' \to \operatorname{Hom}(K, K \otimes S') \to \operatorname{Ext}^1 (Q, S') \to 0$.
So $S' \cong \operatorname{Hom}(K, K \otimes S')$.
Since $S'$ is strongly flat, $K \otimes S'$ is a
direct summand of a direct sum of copies of $K$.
Therefore there exist sets $X$ and $Y$ such that
$K \otimes S' \cong K^{(Y)} \oplus V^{(X)}$.
So $ S' \cong \operatorname{Hom}(K, K \otimes S') \cong \operatorname{Hom}(K, K^{(Y)}) \oplus \operatorname{Hom}(K, V^{(X)})$.
As we saw in the proof of Theorem \ref{final}, if $Y$ is non-empty, we can consider
the exact sequence $ 0 \to R^{Y} \to \operatorname{Hom}(K, K \otimes R^{(Y)}) \to \operatorname{Ext}^1_R ( Q, R^{(Y)}) \to 0 $,
and conclude that
$J \operatorname{Hom}(K, K^{(Y)}) \neq \operatorname{Hom}(K, K^{(Y)})$. Similarly,
by Lemma \ref {JV}, if $X$ is non-empty, $J \operatorname{Hom}(K, V^{(X)}) \neq \operatorname{Hom}(K, V^{(X)})$.
Consequently, $JS' \neq S'$.
By Lemma \ref{3.3}, considering the pure exact sequence
$0 \to C \to S \to J \to 0$, we see that $JC = C$. By Lemma \ref{3.3} again, from the exact sequence
$0 \to C' \to S' \to C \to 0 $, we get that $J S' = S'$, which is a contradiction. This proves that $C = 0$, so that
$J$ is strongly flat, which contradicts Theorem~\ref{ideal}.
\medskip
{\em Step 3: The ring $R$ is invariant and $_R \mathcal{SF} = {}_R \mathcal{F}$.}
By Steps 1 and 2, the ring $R$ must be invariant. Therefore the endomorphism ring of every uniserial module is local
(the proof is similar to the commutative case, because, like in the proof of \cite [Corollary 3] {Facsal},
every uniserial module is unshrinkable, and so the endomorphism ring of every uniserial module is local like in the proof of
\cite [Example 2.3(e)] {Facchinitransaction}). Thus
$\operatorname{End} (K _R)$ is local and every direct summand of copies of $K$ is isomorphic to a direct sum of copies of $K$ because $K_R$ is uniserial by \cite [Proposition 2.2] {DungAlberto}. Thus $_R \mathcal{SF} ={} _R \mathcal{F}$ by Proposition~\ref{final}. \end{proof}
We conclude with an example concerning right noetherian right chain domains. In a right noetherian right chain domain $R$, all right ideals are principal and two-sided \cite[Lemma~3.2]{bessen}. In particular, $J(R)=pR$ for some $p\in R$. The right noetherian right chain domain $R$ is said to be {\em of type $\omega$} \cite[p.~26 and Lemma 3.4]{bessen} if its chain of right ideals (=\,two-sided ideals) is the chain $$R=p^0R\supset J(R)=pR\supset p^2R\supset\dots\supset 0=\bigcap_{n\ge0}p^nR.$$ Thus for every non-zero right ideal $I$ of $R$, we have that $\operatorname{End} (R_R/I)\cong R/I$ is a right artinian ring, hence a perfect ring. In the next example, we show that this is also true for every non-zero principal left ideal $I$ of a right noetherian right chain domain $R$ of type $\omega$ which is not left Ore. Notice that in our example of right noetherian right chain domain of type $\omega$ which is not left Ore, the ring is not left chain (otherwise it would be left Ore) and is not left noetherian \cite[Proposition 3.7]{bessen}. The main example of such a ring can be constructed with the skew poynomial ring with coefficients in a field $F$, where $F$ has an endomorphism that is not an automorphism of $F$.
\begin{example}\label{5.18}
{\rm Let $R$ be a right noetherian right chain domain of type $\omega$ which is not left Ore.
For every non-zero principal left ideal $I$ of $R$, the endomorphism ring $\operatorname{End} (_RR/I)$ is a perfect ring. }
\end{example}
\begin{proof}
For every non-zero element $x\in R$, we have that $xR=p^nR$ for some $n\ge 0$.
Therefore $x=p^nu$ for some invertible element $u\in R$. Right multiplication by $u$
induces an isomorphism $R/Rp^n\to R/Rx$. Hence it suffices to show that
$\operatorname{End}(R/Rp^n)$ is right and left perfect for $n\ge 1$.
Notice that $ Rp^n\subseteq p^nR$. Set
$S:=\operatorname{End}(R/Rp^n)\cong E/Rp^n$, where $E: = \{\, r \in R \mid p^n r \in R p^n \,\}$ denotes the idealizer of $Rp^n$ in $R$, and set $K : = \{\, r \in R \mid p^n r \in J p^n \,\}$. By \cite [Theorem 2.1]{Aminis}, $S$ has at most two maximal ideals, the ideals $K/Rp^n$ and $(J \cap E)/ Rp^n $.
Let us show that $ K \subseteq J$.
Assume the contrary, so that $K$ contains a unit $u$ of $R$. Therefore
$p^n u = r p^n$ for some $r\in J$.
Then $ r = p^j v $ for some unit $v$ of $R$ and some $j \geq 1$.
Thus $p^n = p^j v p^n u^{-1}$.
If $j \geq n$, then $ 1 = p^{j -n } v p^n u^{-1}$, which implies $J(R)=R$, a contradiction.
If $j < n$, then $p^{n -j} = v p^n u^{-1}$. Thus $p^{n -j}$ belongs to the two-sided ideal $pR$ of $R$, which is a contradiction because $n-j<n$.
Therefore $ K \subseteq J$, so
$S$ is local with maximal ideal $(J \cap E)/ Rp^n $.
We claim that if $p^n y \in E$, then $p^ny \in Rp^n$.
To prove the claim, assume that $p^n y \in E$. Then there exists $s\in R$ such that $p^n p^n y = sp^n$. Similarly, there exists $i \geq 0$ and a unit $u$ in $R$ such that
$s = p^i u$. If $i \geq n$, then we are done, the claim is proved. Otherwise, if $i < n$, by supposing that $y = p^j v$ for some unit $v$, we get that
$ u^{-1} p^l v = p^n$, so $l > n$, which is a contradiction by \cite [Lemma 2.3] {Aminis}.
Therefore $(J\cap E)^n\subseteq J^n\cap E=p^nR\cap E\subseteq Rp^n$, to that the Jacobson radical $J(S)=(J\Cal E)/Rp^n$ of the local ring $S$ is nilpotent. It follows that $S=\operatorname{End}(R/Rp^n)$ is a right and left perfect ring.
\end{proof}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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The museum is open to the public during the day and the curators conduct regular tours. Cadets may access items from the museum's collection for research purposes, along with the museum's extensive library of history files, ship's logs, and related historical information. Cadets interested in a detailed study of Starfleet's history can apply for an internship at the museum.
The Academy's Botellin Center for the Arts hosts performances by Starfleet Academy cadets and guest artists, including plays, dramatic readings, and music concerts. Performances are open to the public, with preference given to Starfleet cadets and active personnel who wish to attend. The Arts Center also works in cooperation with the Federation Council on the Arts to sponsor displays and performances of fine arts from other member worlds. The Academy encourages students to take part in the campus artistic community to broaden their experiences and learn about other cultures through the arts.
Starfleet Academy is home to a modern athletic stadium, used for a variety of activities and events. The Academy sponsors athletic events with other institutions on Earth and throughout the Federation, including a parrises squares tournament, gymnastics, a martial arts tournament, the Academy marathon and decathlon, soccer and vrex ball games, and other events as requested by the cadets and faculty. Participation in athletic events in open to all interested cadets. Cadets interested in organizing an athletic event should speak with the Academy's Athletic Director.
Cadets can also use earned credits for a variety of holosuite athletic activities, both solo and team-oriented. The Academy's database contains programs for more than six hundred different athletic simulations, including sports, swimming, outdoor hiking and climbing, riding, and skydiving.
Maintenance staff extensively landscape the Academy grounds in an effort to preserve much of the area's natural beauty. Paths are lined with trees and grass, and flowerbeds showcase a broad variety of plant species from Earth and elsewhere.
In addition to the grounds, the Academy maintains a botanical garden featuring numerous plant species. Some grow outside, arranged in paths to provide shade and comfortable places to enjoy the natural beauty. Other plant species flourish indoors under controlled conditions, particularly those species not acclimated to the San Francisco Bay area or Earth. The biology school uses the greenhouses to study different elements of plant biology, including species not native to Earth.
Starfleet Academy attends to the individual needs of cadets, providing housing facilities in which they're expected to live during their education. Campus housing emphasizes a multicultural living environment to introduce cadets to living and working with individuals from many different species and cultures.
First- and second-year cadets share quarters to learn to coexist with others under conditions not unlike those aboard a starship or starbase. Third- and fourth-year cadets may apply for individual quarters, available based on class ranking and need. Cadets with questions or concerns about campus housing should contact the Office of Student Affairs.
Dormitory facilities consist of central living, dining, and recreational areas for cadets, along with modular private quarters, all based around a community model. Dormitories provide modern amenities, including sonic showers, computer access to multimedia entertainment, communications, and replicator facilities for meals and the production of common items like clothing.
The Academy expects all cadets to exercise appropriate behavior and follow the honor code in and out of their dormitories. Inappropriate behavior in the dormitories, or anywhere on campus, may result in demerits or other punitive action.
Special Academy programs, including intensive language and science majors, reserve certain dormitories and dorm areas. All cadets living in these dormitories follow the same major and course path, allowing them to assist each other and enhance their educational experience. The dormitory facilities set aside for the members of Red Squad follow this principle.
Starfleet Academy relies on the nearby Starfleet Medical School to provide medical services, including infirmary space for cadets requiring physical care, routine physical check-ups, and examinations for cadets. Although accidental injuries are rare, the expert staff of the infirmary can ensure they receive immediate treatment with the most advanced medical technology available to the Federation.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,818
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Q: Asp.net Mvc Model class as inherited Entity class I have extended a entity framework class (SQL server table) and added some extra properties to the child class but when I want to insert into the table which I already extended I get this exception:
Mapping and metadata information could not be found for EntityType 'Student.Models.Add.SubjectToStageModel'.
My controller:
[HttpPost]
public ActionResult SubjectToStage(SubjectToStageModel model)
{
try
{
if (ModelState.IsValid)
{
using (StudentEntities studentEntities = new StudentEntities())
{
int intCount =
studentEntities.SubjectToStageTbls.Count(
x => x.StageId == model.StageId && x.SubjectId == model.SubjectId);
if (intCount == 0)
{
studentEntities.SubjectToStageTbls.Add(model);
studentEntities.SaveChanges();
}
}
}
}
catch (Exception exception)
{
Debug.WriteLine(exception.ToString());
TempData["error"] = "An error occured";
}
return RedirectToAction("SubjectToStage");
}
My base class:
public partial class SubjectToStageTbl
{
public SubjectToStageTbl()
{
this.StudentMarkTbls = new HashSet<StudentMarkTbl>();
}
public int SubjectToStageId { get; set; }
public int SubjectId { get; set; }
public int StageId { get; set; }
public int Point { get; set; }
public virtual StageTbl StageTbl { get; set; }
public virtual ICollection<StudentMarkTbl> StudentMarkTbls { get; set; }
public virtual SubjectTbl SubjectTbl { get; set; }
}
My subclass:
public class SubjectToStageModel : SubjectToStageTbl
{
public IEnumerable<SelectListItem> StageListItem
{
get
{
List<SelectListItem> listsSelectListItems = new List<SelectListItem>();
try
{
using (StudentEntities studentEntities = new StudentEntities())
{
IQueryable<StageTbl> queryableStage = studentEntities.StageTbls;
foreach (var stage in queryableStage)
{
SelectListItem selectListItem = new SelectListItem();
selectListItem.Value = stage.StageId.ToString();
selectListItem.Text = stage.StageName;
listsSelectListItems.Add(selectListItem);
}
}
}
catch (Exception exception)
{
Debug.WriteLine(exception.ToString());
}
return listsSelectListItems;
}
}
public IEnumerable<SelectListItem> SubjectListItem
{
get
{
List<SelectListItem> listsSelectListItems = new List<SelectListItem>();
try
{
using (StudentEntities studentEntities = new StudentEntities())
{
IQueryable<SubjectTbl> queryableSubject = studentEntities.SubjectTbls;
foreach (var stage in queryableSubject)
{
SelectListItem selectListItem = new SelectListItem();
selectListItem.Value = stage.SubjectId.ToString();
selectListItem.Text = stage.SubjectName;
listsSelectListItems.Add(selectListItem);
}
}
}
catch (Exception exception)
{
Debug.WriteLine(exception.ToString());
}
return listsSelectListItems;
}
}
}
A: You didn't explicitly map the SubjectToStageModel class. If you want Entity Framework to work with derived classes you should add them to the model as well. But I don't think you intended to do that in the first place.
In fact, SubjectToStageModel is a view model. It may look convenient to derive a view model from an entity class, but I think generally it's not a good idea. View models should be tailored to the view (or use case) they're used in. A couple of reasons:
*
*It's very likely that the entity class contains more properties than you need in the view. In later maintenance it's always a pain to keep checking what you do and don't need.
*While view evolves, it may require a differently structured model than the entity.
*The view may require different validations.
*The view may be allowed to return a state that absolutely shouldn't be stored (you may require some post processing of entered data), so it's good to ensure it can't possibly be stored.
*It creates a dependency between the data layer model and the view.
Maybe these consideration don't apply in your case. Still I'd prefer to have an independent view model. However, if you're lazy (we developers prefer the word pragmatic), you may succeed by doing:
studentEntities.SubjectToStageTbls.Add((SubjectToStageTbl)model);
(I never tried though).
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 3,299
|
Home Report: Steve Nash Was Promised Pau Gasol Will Not Be Traded
Report: Steve Nash Was Promised Pau Gasol Will Not Be Traded
Kobe Bryant, Lionel Messi Star in Turkish Airlines Commercial
The Los Angeles Lakers had one of the best off-seasons in league history this summer with the additions of Steve Nash and Dwight Howard. Nash was the first one to join the Lakers, and Howard eventually teamed up with the two-time NBA MVP.
Nash and Howard joined an already talented Lakers team which included Kobe Bryant and Pau Gasol, forming one of the best starting fives in the league.
Gasol is one of the most skilled big men in the NBA, and even though he is struggling now, there is a good chance he gets his act back together. Gasol has the ability to stretch the floor with his mid-range shooting, and he can also do a great job of finding teammates for open looks. A Nash/Gasol tandem is something not many teams can compete against, and according to Ric Bucher, Nash was told that Gasol will not be traded when he signed with the team.
When Steve Nash signed with the Los Angeles Lakers, a source says he was promised that Pau Gasol would not be traded.
The Mike D'Antoni system is made to be run by an elite point guard, and Steve Nash is the perfect fit for this offense. Everyone involved with the Lakers is hoping that Gasol can improve once Nash returns to the lineup. This team is not playing to the level that it should be playing at, but there is a chance that Nash can help with the process.
Nash and Gasol are two playmaking athletes, and a lot of great can come if they are both on the same page. However, if Gasol continues to struggle even with Nash in the lineup, there's the possibility that he will end up getting dealt.
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Andrew Meshot
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The point guard position is the deepest one in the entire NBA, which makes it tough for Los Angeles Lakers rookie Lonzo Ball…
Lakers News: Russell Westbrook Maintains Focus Is On Improving & Being Consistent As Season Goes On
Russell Westbrook's first season with the Los Angeles Lakers has been a rollercoaster as the 2017 NBA MVP's acclimatization…
Lakers News: Frank Vogel Feels No Added Pressure Despite Rumors Surrounding Job Security
With over half the 2021-2022 season, the Los Angeles Lakers have clearly underwhelmed…
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 9,910
|
Q: Intel Quad NIC 82571EB gets no link in installed Ubuntu Server, but works in Live Desktop I have an Intel Quad-port Gigabit Ethernet controller in an HP ML310e Gen 8 v2 server (which also has an onboard dual-port Broadcom BCM5720).
When I boot from an Ubuntu Desktop live USB (16.04), all six ports work exactly as expected: when a cable is connected they immediately get a link, the LED on the port lights immediately and the corresponding one on my switch lights.
I installed Ubuntu Server (16.04), and none of the Intel ports get a link at all. The LEDs never light and ethtool reports "Link detected: no" for each port. Using proved-good cables. The Broadcom ports work fine. I am able to make the LEDs for each port on the Intel card blink with ethtool -p.
I found a report of an ostensibly similar problem here: https://bbs.archlinux.org/viewtopic.php?id=159454 However that was resolved by:
echo on > /sys/bus/pci/devices/0000\:09\:00.0/power/control
(for each of the correct PCI ids) - and in my case that key is already set to 'on'.
As the card works perfectly when I boot into a Live USB desktop, I believe that card and cabling is fine, but I'm unable to get it to work from an installed Ubuntu Server. This is the third from-scratch install I've tried now. In previous attempts I also tried downloading and building e1000e drivers from Intel's site, but that made no difference at all.
Would really appreciate any advice the community can give - thanks.
lshw reports:
driver=e1000e
driverversion=3.2.6-k
firmware=5.12-2
Full relevant output from lscpi, lshw and uname:
~$ lspci -nnk | grep -i eth
03:00.0 Ethernet controller [0200]: Broadcom Corporation NetXtreme BCM5720 Gigabit Ethernet PCIe [14e4:165f]
Subsystem: Hewlett-Packard Company NetXtreme BCM5720 Gigabit Ethernet PCIe [103c:2133]
03:00.1 Ethernet controller [0200]: Broadcom Corporation NetXtreme BCM5720 Gigabit Ethernet PCIe [14e4:165f]
Subsystem: Hewlett-Packard Company NetXtreme BCM5720 Gigabit Ethernet PCIe [103c:2133]
09:00.0 Ethernet controller [0200]: Intel Corporation 82571EB Gigabit Ethernet Controller (Copper) [8086:10bc] (rev 06)
09:00.1 Ethernet controller [0200]: Intel Corporation 82571EB Gigabit Ethernet Controller (Copper) [8086:10bc] (rev 06)
0a:00.0 Ethernet controller [0200]: Intel Corporation 82571EB Gigabit Ethernet Controller (Copper) [8086:10bc] (rev 06)
0a:00.1 Ethernet controller [0200]: Intel Corporation 82571EB Gigabit Ethernet Controller (Copper) [8086:10bc] (rev 06)
~$ lshw -C net
*-network:0 DISABLED
description: Ethernet interface
product: 82571EB Gigabit Ethernet Controller (Copper)
vendor: Intel Corporation
physical id: 0
bus info: pci@0000:09:00.0
logical name: enp9s0f0
version: 06
serial: 00:24:81<redacted because paranoia>
capacity: 1Gbit/s
width: 32 bits
clock: 33MHz
capabilities: pm msi pciexpress bus_master cap_list ethernet physical tp 10bt 10bt-fd 100bt 100bt-fd 1000bt-fd autonegotiation
configuration: autonegotiation=on broadcast=yes driver=e1000e driverversion=3.2.6-k firmware=5.12-2 latency=0 link=no multicast=yes port=twisted pair
resources: irq:30 memory:fbce0000-fbcfffff memory:fbc00000-fbc7ffff ioport:4000(size=32)
*-network:1 DISABLED
description: Ethernet interface
product: 82571EB Gigabit Ethernet Controller (Copper)
vendor: Intel Corporation
physical id: 0.1
bus info: pci@0000:09:00.1
logical name: enp9s0f1
version: 06
serial: 00:24:81:<redacted>
capacity: 1Gbit/s
width: 32 bits
clock: 33MHz
capabilities: pm msi pciexpress bus_master cap_list ethernet physical tp 10bt 10bt-fd 100bt 100bt-fd 1000bt-fd autonegotiation
configuration: autonegotiation=on broadcast=yes driver=e1000e driverversion=3.2.6-k firmware=5.12-2 latency=0 link=no multicast=yes port=twisted pair
resources: irq:32 memory:fbbe0000-fbbfffff memory:fbb00000-fbb7ffff ioport:4020(size=32)
*-network:0 DISABLED
description: Ethernet interface
product: 82571EB Gigabit Ethernet Controller (Copper)
vendor: Intel Corporation
physical id: 0
bus info: pci@0000:0a:00.0
logical name: enp10s0f0
version: 06
serial: 00:24:81:<redacted>
capacity: 1Gbit/s
width: 32 bits
clock: 33MHz
capabilities: pm msi pciexpress bus_master cap_list ethernet physical tp 10bt 10bt-fd 100bt 100bt-fd 1000bt-fd autonegotiation
configuration: autonegotiation=on broadcast=yes driver=e1000e driverversion=3.2.6-k firmware=5.12-2 latency=0 link=no multicast=yes port=twisted pair
resources: irq:33 memory:fbee0000-fbefffff memory:fbe00000-fbe7ffff ioport:5000(size=32)
*-network:1 DISABLED
description: Ethernet interface
product: 82571EB Gigabit Ethernet Controller (Copper)
vendor: Intel Corporation
physical id: 0.1
bus info: pci@0000:0a:00.1
logical name: enp10s0f1
version: 06
serial: 00:24:81:<redacted>
capacity: 1Gbit/s
width: 32 bits
clock: 33MHz
capabilities: pm msi pciexpress bus_master cap_list ethernet physical tp 10bt 10bt-fd 100bt 100bt-fd 1000bt-fd autonegotiation
configuration: autonegotiation=on broadcast=yes driver=e1000e driverversion=3.2.6-k firmware=5.12-2 latency=0 link=no multicast=yes port=twisted pair
resources: irq:34 memory:fbde0000-fbdfffff memory:fbd00000-fbd7ffff ioport:5020(size=32)
*-network:0
description: Ethernet interface
product: NetXtreme BCM5720 Gigabit Ethernet PCIe
vendor: Broadcom Corporation
physical id: 0
bus info: pci@0000:03:00.0
logical name: eno1
version: 00
serial: 10:60:4b:ee:39:30
size: 1Gbit/s
capacity: 1Gbit/s
width: 64 bits
clock: 33MHz
capabilities: pm vpd msi msix pciexpress bus_master cap_list rom ethernet physical tp 10bt 10bt-fd 100bt 100bt-fd 1000bt 1000bt-fd autonegotiation
configuration: autonegotiation=on broadcast=yes driver=tg3 driverversion=3.137 duplex=full firmware=5720-v1.30 NCSI v1.1.15.0 ip=10.0.1.141 latency=0 link=yes multicast=yes port=twisted pair speed=1Gbit/s
resources: irq:17 memory:fabf0000-fabfffff memory:fabe0000-fabeffff memory:fabd0000-fabdffff memory:f8000000-f801ffff
*-network:1 DISABLED
description: Ethernet interface
product: NetXtreme BCM5720 Gigabit Ethernet PCIe
vendor: Broadcom Corporation
physical id: 0.1
bus info: pci@0000:03:00.1
logical name: eno2
version: 00
serial: 10:60:4b:ee:39:31
capacity: 1Gbit/s
width: 64 bits
clock: 33MHz
capabilities: pm vpd msi msix pciexpress bus_master cap_list rom ethernet physical tp 10bt 10bt-fd 100bt 100bt-fd 1000bt 1000bt-fd autonegotiation
configuration: autonegotiation=on broadcast=yes driver=tg3 driverversion=3.137 firmware=5720-v1.30 NCSI v1.1.15.0 latency=0 link=no multicast=yes port=twisted pair
resources: irq:18 memory:fabc0000-fabcffff memory:fabb0000-fabbffff memory:faba0000-fabaffff memory:f8020000-f803ffff
$ uname -a
Linux victorious 4.4.0-31-generic #50-Ubuntu SMP Wed Jul 13 00:07:12 UTC 2016 x86_64 x86_64 x86_64 GNU/Linux
Further outputs as requested:
$ dmesg | grep 09:00
[ 0.132795] pci 0000:09:00.0: [8086:10bc] type 00 class 0x020000
[ 0.132824] pci 0000:09:00.0: reg 0x10: [mem 0xfbce0000-0xfbcfffff]
[ 0.132834] pci 0000:09:00.0: reg 0x14: [mem 0xfbc00000-0xfbc7ffff]
[ 0.132843] pci 0000:09:00.0: reg 0x18: [io 0x4000-0x401f]
[ 0.132915] pci 0000:09:00.0: PME# supported from D0 D3hot D3cold
[ 0.132964] pci 0000:09:00.1: [8086:10bc] type 00 class 0x020000
[ 0.132993] pci 0000:09:00.1: reg 0x10: [mem 0xfbbe0000-0xfbbfffff]
[ 0.133002] pci 0000:09:00.1: reg 0x14: [mem 0xfbb00000-0xfbb7ffff]
[ 0.133012] pci 0000:09:00.1: reg 0x18: [io 0x4020-0x403f]
[ 0.133083] pci 0000:09:00.1: PME# supported from D0 D3hot D3cold
[ 5.898627] e1000e 0000:09:00.0: Interrupt Throttling Rate (ints/sec) set to dynamic conservative mode
[ 6.070261] e1000e 0000:09:00.0 eth0: (PCI Express:2.5GT/s:Width x4) 00:24:81:<redacted because paranoia>
[ 6.070262] e1000e 0000:09:00.0 eth0: Intel(R) PRO/1000 Network Connection
[ 6.070341] e1000e 0000:09:00.0 eth0: MAC: 0, PHY: 4, PBA No: D98771-007
[ 6.070434] e1000e 0000:09:00.1: Interrupt Throttling Rate (ints/sec) set to dynamic conservative mode
[ 6.242251] e1000e 0000:09:00.1 eth1: (PCI Express:2.5GT/s:Width x4) 00:24:81:<redacted>
[ 6.242252] e1000e 0000:09:00.1 eth1: Intel(R) PRO/1000 Network Connection
[ 6.242331] e1000e 0000:09:00.1 eth1: MAC: 0, PHY: 4, PBA No: D98771-007
[ 7.625342] e1000e 0000:09:00.0 enp9s0f0: renamed from eth0
[ 11.802316] e1000e 0000:09:00.1 enp9s0f1: renamed from eth1
$ cat /etc/network/interfaces
# This file describes the network interfaces available on your system
# and how to activate them. For more information, see interfaces(5).
source /etc/network/interfaces.d/*
# The loopback network interface
auto lo
iface lo inet loopback
# The primary network interface
auto eno1
iface eno1 inet static
address <redacted IP in 10. range>
netmask 255.255.255.0
gateway <redacted>
$ ls -l /etc/network/interfaces.d
total 0
A: The interfaces showing as DISABLED are exactly the interfaces that are not declared in /etc/network/interfaces. Please amend the file to:
# This file describes the network interfaces available on your system
# and how to activate them. For more information, see interfaces(5).
source /etc/network/interfaces.d/*
# The loopback network interface
auto lo
iface lo inet loopback
# The primary network interface
auto eno1
iface eno1 inet static
address <redacted IP in 10. range>
netmask 255.255.255.0
gateway <redacted>
dns-nameservers 10.whatever.the.gateway 8.8.8.8
#auto enp9s0f0
iface enp9s0f0 inet dhcp
#auto enp9s0f1
iface enp9s0f1 inet dhcp
#auto enp10s0f0
iface enp10s0f0 inet dhcp
#auto enp10s0f1
iface enp10s0f1 inet dhcp
Incidentally, if the declared interface, eno1 is expected to reach the internet, DNS nameservers are required. I have taken the liberty to suggest it in my proposed edit above.
Reboot and check:
sudo lshw -C network
Are the Intel ports no longer disabled? If so, you can transfer the cable to one of them, rearrange /etc/network/interfaces and proceed.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,484
|
Technical description: Brass; angled layout; tuning-slide with wide bow; 3 valves for right hand, 4th for left hand; ligature fitting at joint. Valve type: 4 Berlin-type valves (ports of each in a plane, with straight through path when not operated; less then 90° windway deflection when valves operated).
Measurements:Length along bell section 682; length along valve section 524; bell 167.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 9,568
|
Technical Car Experts Answers everything you need: Where is Knock Sensor located on 1999 Ford F150 4.2L?
Where is Knock Sensor located on 1999 Ford F150 4.2L?
On Ford F150 engines the "Knock Sensor" is located on the right front of the motor.
Below is the Knock sensor Location diagram on Ford F150 4.2l engine.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 1,360
|
"I was introduced to Harbottle & Lewis via a number of very strong personal recommendations and they exceeded my expectations in every way. Their service was exemplary from their knowledge of the employment market to the care and attention I received from their people throughout. I cannot recommend them more highly and will most certainly be using them as my primary law firm in the future."
Creative Director of UK TV Broadcaster
Home Employment Senior executive
If you are director or a senior executive we can advise you on all aspects of your employment relationship from negotiating terms to providing pragmatic legal advice when the employment relationship encounters problems.
Our focus is your objectives and priorities and our advice will be tailored to meet your desired outcomes. We understand that offering cost effective and commercial solutions is especially important when acting for individuals.
The breakdown of an employment relationship and problems at work can be very stressful, especially for directors and senior executives who invest significant time and emotion into working life. Difficulties may arise as a result of a restructure, a change in duties and responsibilities, discrimination, bullying and harassment, disciplinaries or grievances. We can help you find amicable resolutions to workplace problems, taking into account your reputation and financial considerations including advising on 'without prejudice' or 'protected conversation' negotiations and settlement negotiations. If it is not possible to resolve matters, our employment lawyers have an excellent reputation in acting for executives in Employment Tribunal and High Court litigation.
We work with senior executives operating in a number of industry sectors, including media & entertainment, technology, fashion, retail, publishing, advertising, telecoms, investment banking, legal and accountancy.
We can provide advice on:
Negotiating your employment contract or service agreement
Employment bonus, incentive schemes, remuneration packages and tax implications
Without prejudice and/or protected conversation negotiations
Disciplinaries
Workplace disputes and grievances
Regulatory issues and directors' duties
Non-executive director appointments
Post-termination restrictions and team moves
Reputation management.
Media professional
We advised a media professional on the termination of an expatriate arrangement including ensuring that he was able to benefit from the generous tax relief applicable in such circumstances.
Chief Finance Officer, global technology company
We advised a CFO on the proposed termination of their employment following an acquisition.
Chief HR Officer, global technology company
We advised a Chief HR Officer on the proposed termination of their employment. Our client proposed the termination as a result of being unhappy with changes being made by their employer. We acted on the negotiations in relation to the settlement amount and share options.
Chief Executive Officer, advertising agency
We advised the CEO of an advertising agency on securing their exit. We put together a proposal and negotiated on the clients behalf, including in relation to restrictive covenants.
Howard Hymanson
E: howard.hymanson@harbottle.com
Marian Derham
E: marian.derham@harbottle.com
Yvonne Gallagher
E: yvonne.gallagher@harbottle.com
Poppy Lucas
E: poppy.lucas@harbottle.com
Sarah Verrecchia
E: sarah.verrecchia@harbottle.com
Harbottle & Lewis are by some margin the best law firm I have worked with in the employment market. In my career I have worked with 5 other law firms so I have a good basis for comparison. They have an edge in terms of quality of people, customer service and knowledge of the market.
Neil Jones Group Commercial Director Johnston Press PLC
I have been impressed with every contact I have made within the firm. The lawyers are all very good, easily understood and demonstrate strong knowledge. This should be highlighted as a key strength of the firm...consistency in performance is a target every business strives to achieve!
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Harbottle & Lewis were recommended to me by an ex-colleague in relation to my employment situation . The team were outstanding throughout a very difficult period and I found them to be responsive, thoroughly professional and patient. I found them to be people with a real interest in my case and determined to help me resolve the issues. Very impressive client service.
Former Tax Director of a top four Accountancy Firm
They are extremely conscientious and prompt, and the advice that they give is very pragmatic.
Over more than a decade, the employment and media partners of Harbottle & Lewis have provided me with advice that has met the test of every challenge I've ever encountered as an entrepreneur. Their highly responsive professionalism, expertise and timely execution gives me confidence in all situations.
Mike Tims, CEO and founder of MTN-I Limited
The employment team at Harbottle and Lewis were both professional and compassionate when dealing with my sudden loss of employment. The advice and service I received was done in such a way that I always felt part of the Harbottle team and we were able to act fast and efficiently .The high emotion I felt at the time was understood and the team really helped to deal with both the case and the anxiety to ensure the best outcome possible. I have and will continue to recommend them to anyone who is considering employment litigation or needing advice on exit strategy and negotiations.
Executive Group Publishing Director for leading Publishing House
On behalf of all our family I would like to take this opportunity to thank you and all your team for their efforts in bringing the claim to a successful and highly satisfactory conclusion.
Employment client
Your help is most appreciated and you have done a tremendous job, thank you.
Harbottle & Lewis has handled Wolford's Employment matters for many years. They keep us up to date, legal and safe and when we need their help and advice it is quick professional and totally reliable.
Christine Bordoni, Finance Director, Wolford
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|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
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Charming two bedroom, one full bath and one half bath home, completely renovated in 2011. Located on a prime Bella Vista/Queen Village block in the Meredith school district. Enter the home to a tile foyer, then step into the living room with ceramic tile flooring, a large window providing lots of light, a ceiling fan and a coat closet. Continue into the eat-in kitchen, which boasts maple cabinetry, granite countertops, a tile backsplash, recessed lights, stainless steel appliances, a ceiling fan and a granite-capped pass-thru to the living room. A powder room and the door out to the decked-in yard with alley to the street are also accessed from the kitchen. Classic space-saving Trinity steps, which make the home feel bigger than homes with similar square footage, lead up to the second floor. Step into an open den with a window and recessed lighting, then continue into the second floor bedroom with a ceiling fan, neutral carpet and paint and crisp white trim. The full bath with tile tub and modern vanity completes the second floor. On the third floor is a bedroom with a ceiling fan, two windows and a closet. The basement houses the mechanicals and laundry and provides storage space. This fantastic location is close to Weccacoe and Cianfrani, mere blocks to your choice of restaurants, cafes, pubs, and shops, close to Septa routes and highways, and a quick commute to Center City or University City.
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Published 04/25/2019 09:57:22 pm at 04/25/2019 09:57:22 pm in Rv Shower Door.
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Kalamazoo area physicians are invited to refer patients to an observational study being conducted through the Center for Clinical Research at the Western Michigan University Homer Stryker, M.D. School of Medicine.
Research clinical investigators are seeking 500 participants over the next three years, between the ages of 35 and 55 who will receive tests including Coronary Computed Tomography (CCTA), a Coronary Artery Calcium Score and a blood draw.
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If you had told me a few years ago that I would turn into an animal loving, tree-hugging vegan, I would have told you that you were crazy.
One side of my family is Italian and Irish, while the other side is Jewish. Our typical family dinner consisted of things like meatballs, corned beef and challah bread, all of which are laden with ingredients from animals.
Despite my heritage, here I sit as an ethical vegan who has a job dedicated to animal activism. Here is how that happened.
I don't have a big dramatic story about going vegan; I pretty much just decided to do it overnight. I had been a vegetarian for about six months before deciding to become vegan, so it wasn't a complete shock to me.
After watching documentaries like "Cowspiracy" and "If Slaughterhouses Had Glass Walls," I was hooked.
I had been living in blissful ignorance about where my food came from, but the images and statistics shown to me through the many documentaries I watched were a shock to my system.
I knew I could never go back.
Many people who go to college and begin to think for themselves without their parents' guidance become vegetarians once they are aware of the cruelty that is standard practice on factory farms in our country. Some decide to take it a step further and become vegans.
There are a lot of different labels for diets floating around, so it is probably best to start off explaining the important differences between vegetarians and vegans.
Vegetarians do not eat any meat from animals, including beef, chicken, pork and fish. Vegans, on the other hand, abstain from all animal products.
Whether it is vegetarian or vegan, there are demonstrable benefits that come with both.
Vegetarians are 12 percent less likely to die from diseases such as hypertension, diabetes and heart disease and they tend to live up to eight years longer than omnivorous eaters, according to a 2013 Loma Linda University study.
The advantages extend beyond health to the environment.
On average, meat production contributes to 50 percent of greenhouse gas emissions, according to the Worldwatch Institute. Going vegetarian or vegan lessens that problem.
These benefits of the vegetarian diet are only increased by going vegan.
Studies done by animal rights groups like PETA, Mercy For Animals and The Humane League have shown that a vegan diet gives people more energy and can increase antioxidant levels in the body, which help fight off diseases like cancer.
I could give you all of the gory details to argue for an ethical take on veganism, but I would be here all day.
I will only say that a vegan diet is crucial if you are looking at animal rights from an ethical perspective. The dairy and egg industries cover up their own atrocities that they make animals go through, it is not only the meat industry.
I know veganism can sometimes have negative connotations behind it because of the stereotypes that society puts on it.
However, if you do your research and do it for the right reasons, then I guarantee you will be making the best decision of your life. If you're considering making a change in your diet, then veganism is the right way to go.
Going forward, this column will focus on issues animal welfare, the environment and nutrition, as well as how all these topics are interconnected.
It will not be a weekly column of simply talking about veganism, although that will certainly be a recurring theme.
I look forward to exploring this topic more thoroughly.
I hope that all of you will enjoy taking part in that exploration as well.
Lippy, a sophomore nutrition and foods major from Huntersville, is a columnist.
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Interview of Andreas Kraushaar, Head of the Prepress Technology Department at Fogra
By altavia In Blog, News, Our partners December 18 2019 no comments
I'm Andreas Kraushaar, born and living in Germany. Besides my work at Fogra, in my spare time I'm a passionate football referee – the perfect combination, as it's all about neutrality and rules, both on the pitch and at Fogra.
Your background
I studied media technology at the Technical University of Ilmenau. During my studies I specialised in colour image processing and did my doctorate in colour science at Aachen University.
Your role at FOGRA
I have been departmental head for prepress technology at Fogra since 2001. My work focuses on ICC colour management, colour image processing and image quality evaluation, especially in digital printing, and 3D printing.
What is the TC130 committee and what is your role?
ISO TC 130 (Technical Committee – Graphic Technology) represents the international standardisation body for the printing industry and consists of a group of international experts working in the fields of terminology, prepress, printing, post-press, climate neutrality, materials and certification who meet about twice a year. This year, for example, we had conferences in Hong Kong in the spring and in Levi, USA, in the autumn. I'm the convenor for Working Group 3 (Process control and related metrology).
The ISO News section on the Fogra website offers all those who are interested an up-to-date overview of all the relevant developments regarding the printing industry.
What do you think of standardisation and why is it so important?
Working on ISO standards is a long-term investment on the part of Fogra, but for any individual ISO project/standard there is some sort of member interest supporting this. Contrary to many other organisations that focus on proprietary or national standards we prefer to do things the hard way, with lengthy discussions and reaching consensuses with international experts but then having this internationally agreed upon reference to base our FograCert services on.
Can you tell us a little about PSO?
This is the piano I've learnt to play since I started at Fogra. My former boss and lifelong mentor was Dr Fred Dolezalek – the father of PSO, I would say…
PSO (Process Standard Offset) is a system for creating optimal, stable and reliable print products in accordance with the international ISO 12647 series of standards. It was developed and standardised internationally by Fogra in cooperation with the German Print and Media Associations (bvdm).
It outlines an industrially oriented and standardised approach for manufacturing print products, from data collection to the final print product.
In order to cope with the strong international demand for PSO (ISO 12647) certification, Fogra runs a programme in non-German-speaking countries through which qualified partners help print shops to obtain the valuable certification. Our PSO certification partner carries out the on-site certification procedure, whilst Fogra and the bvdm evaluate the work and issue the certificate.
What is your vision of the future for print industries?
It's not going to be a walk in the park, I guess. There are great opportunities and niches but it will be a lifelong challenge to keep up the pace – stick with Fogra to keep up with the frontrunners.
About Fogra
The objective of Fogra Forschungsinstitut für Medientechnologien e.V. (Re- search Institute for Media Technologies) is to promote print engineering and its future-oriented technologies in the fields of research and development, and to enable the printing industry to utilize the results of this activity.
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10, rue Blanqui Saint-Ouen, 93400 France
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Allow me to introduce myself. My name is Wolfie.
I am half Bichon and half Shih-Tzu, also known as a Teddy Bear Dog or a Zuchon. I am sixteen and a half pounds and I barely stand a foot off the floor, but do not let my size fool you. I may be small, but I am mighty.
I've been with my family for five months now. Everyone keeps saying the Easter Bunny brought me as a surprise.
But I remember the two biggest people in the family—the mom and the dad people—came to pick me up in a conference center off the highway in a small town called Portsmouth. We drove around for hours and then snuck home because they said the kids were finally asleep. I never did see a bunny.
Pee on couch. Look adorable.
Poop on rug. Appear irresistible.
Pee on floor. Tilt head to one side with cutest expression possible.
This is me after I peed on the ottoman.
This went on for a while until the dad guy said I was on something called thin ice and I'd better get house trained soon. He had just stepped in one of my puddles wearing only his socks.
He pretends he doesn't like me, this dad. But I'm not fooled by him. When it's late at night and all the small people have gone to bed and the mom is upstairs reading, he sits on the big red couch and he calls to me in a quiet voice.
I sit next to him and we watch shows that the mom doesn't like; baseball and politics and something weird called The First 48. But I can tell by the absentminded way he rubs my foot that he's only half-listening to the television. Instead he's thinking about his patients and his children and tax returns and healthcare and insurance.
There are a lot of people in this house. Seven. Two big people and five kids. One time a man came and delivered some food in a brown paper bag that smelled delicious. When he stepped into the kitchen and saw all the kids at the counter, he asked if we were having a birthday party.
This Henry boy is the smallest, but he isn't too small. Life is very, very exciting for him, and he is very loud about it all. He fills up every room with his chatter and his laughter and his drawings, and he is squishy and delicious and curious and smart. He is so alive, you can almost see his heart beating through his favorite Batman shirt.
There are all these boys and only one little girl. When you look at her you just think about the color pink. She is sweetness and light and airy and calm, like the most delicate wafer cookie you hold on your tongue until it melts.
But she works very hard. She is the first one awake to take me out in the morning, even before I ask, and all day long she is trying to do things for other people; pack their snacks or sweep the floor or straighten the playroom.
Her mother worries and the Dad guy hopes if he tells her how beautiful and smart and sweet she is, she will believe it forever and never listen if someone accuses her of being fat or ugly or stupid or worthless.
I'm not worried, because this pink girl is so very strong.
There is a very big boy, a boy who stands taller than the mom's shoulder. He wears glasses. They call him a tween sometimes, and I don't know what that means but it seems to annoy him.
He and the mom can really get each other going laughing. They both like the same jokes. But there is a strain that I don't think was there before. It feels new.
It feels like the beginning of something and the end of something all at the same time. He is starting to cleave from them, to long for video games and something called an IPhone and movies that are PG-13. The mom, she knows this, and her heart is aching to make the most of the time she has left, before this tween will pack up his glasses and his gym shorts and his Nook and drive down the driveway to a faraway place called college.
I may be very close to the floor, but I see it all.
There's another big boy, just about as tall as the first boy. He also wears glasses. From behind they look like the same boy and sometimes people mix them up, but I never do.
From what I understand, I was supposed to help this boy. He has something called autism and he was very, very afraid of dogs, even little ones like me.
When the mom first brought me in from the garage where I was hiding and trying to stay very, very quiet, all of the kids squealed and laughed and clapped their hands. But he didn't. His face was all twisted up and his voice was very loud and angry-sounding.
I don't know anything about autism or how to help people who have it. So I just did the only thing I knew how: I waited. I waited and waited and one afternoon when no one was watching he crept over to where I was lying on the couch. With one finger he stroked my paw.
This boy gets very, very mad. One day over the summer his temper rose until it felt like the sun was shining inside the house, the rays too hot to touch. He was screaming and hitting his head over and over again.
I did not know what a para is, but the mom seemed to because she kept talking softly, telling him to take a deep breath and calm down, they would talk about it.
Then he came for her. With his fists curled into the tightest balls he charged her wordlessly. She grabbed his wrists and held them with her long fingers and said, "Enough Jack," so sharply her voice was like a knife cutting through the hot, still room.
He dropped his arms to his sides and the only sound was his whimpering, no para no para no para. I barked once, twice, my voice not as sharp as hers, more like an ice cube clattering into a smooth glass.
He fell to his knees next to me and buried his fingers into the fur around my neck, where it's longest and deepest. Through his fingertips, I understood. I knew. Somehow, because of this strange thing called a para, the boy felt different. He felt worried and alone and disappointed.
There is another boy. He looks just like the dad, with dark hair and deep brown eyes that make you think of chocolate. He is all fun, this one.
But every once in a while a shadow crosses his face and his eyes get cloudy, like the rain is coming. That's when I know he needs a little extra cuddle and I just turn on my back so he can rub my soft, white belly. He rubs it until the sun shines again.
Slowly the three of us walked back up the driveway. They looked down and started talking to me in a funny voice with funny words. "You a wittle doggy, wight?Just a wittle pup-pup." I felt confused.
Then I understood. Their babies were gone. Now I was the baby.
Last weekend we all went to a big field to play with a black and white ball. The mom and dad kicked it around with the kids, but the second boy said he only wanted to hold my leash and run with me.
So we did. We ran and ran through the fields together. And with each big step he took I could tell, for the moment, he was free. Free of the shame and rage and confusion and panic that follow him around all day like uninvited guests.
Running by my side through the rich green grass, he wasn't a child with autism or a fifth grader with a para or a brother who is not like the rest.
He was, quite simply, just a boy and his dog.
I have read a lot of your postings but wow – this one really got to me – ASD – dreaded words that way too many parents hear. What a special dog you have to be able to write such a touching story!!
Hello! If you like, you could e-mail me at carrie@carriecariello.com so I could respond more thoroughly to your beautiful comment.
Oh Carrie this is one of my favorite stories. I teared up. Thanks for the stories.
As a para and a dog lover, I thoroughly enjoyed reading your story, and immediately ordered "What Color is Monday?" I have worked as a para for 15 years, and love watching children grow, and seeing the world through the eyes of a child with autism. Jack is so fortunate to have such a loving family, but you are more fortunate to have Jack in your life! Thank you for sharing your family and educating the world!
Your words are beautiful and brilliant and heart wrenching. The way sweet little "Wolfie" described the kids sent me to tears. It made me ache for a mother who would feel and think and write such precious things about me. They must be the luckiest of small people in the world.
I read your posts and I just don't know what to write. Your posts are just awesome and are the deepest voice in my head. I have 2 autistic children, none of em as «open» as Jack so, lots of misunderstanding, lots of crisis and lots of tears from mom. Thanks for your «voice».
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