There was a moment in an interview I conducted in the 1990s with American social documentary photographer, Milton Rogovin, at which it occurred to me that I had arrived at a single degree of separation from the great Chilean poet born Neftalí Ricardo Reyes Basoalto, later to become Pablo Neruda. Or perhaps I ought to say I would have been at one degree of remove, had Neruda lived beyond my seventh birthday. As it is he died in 1973, a result of cancer, shortly after the Kissinger-backed military coup against Salvador Allende.
Milton once showed me a book given to him by the poet after their collaboration on the project Windows That Open Inward, and just as Pablo Neruda's Wikipedia entry would have us expect, the ink was green, the colour of hope. It was Rogovin who first told me about this idiosyncrasy, although I seem to recall having read Robert Bly on the matter. I have something around a half-dozen Neruda anthologies, and it is to Bly and Alastair Reid that I am partial, not as a scholar of Spanish (which I am not) but as a casual reader. Citations in this essay will come from Reid's and Bly's editions, with use also of Anthony Kerrigan's and Jodey Bateman's translations.
The smell of barbershops makes me break into hoarse sobs.
I don't want so much misery.
and leaves tracks full of warm blood leading toward the night.
there are umbrellas everywhere, and venoms, and umbilical cords.
that it can't be a round fruit on a branch.
condemned to life in a shoe.
exploring life like a blind man.
with triangular heads like a worm's.
until the whole man stopped.
and the week with the whole year.
are washed out by the waters of night.
all of us are rain under rain.
I have no idea what they are saying.
and I know it is without a name.
it rang like a bell.
which goes on all winter.
A year is four centuries.
what am I called or not called?
if I was not while I slept?
with so much of signing of papers.
I have let the poet do most of the talking, as I well should. My purpose today has been to introduce Neruda to people who have never before read his poems. If you are one of these, please do yourself the favour of borrowing or buying a copy of his work. If you have however read Neruda's work in the past, now may be a good time to do so again.
This entry was posted in Writers and Writing and tagged Milton Rogovin, Neftalí Ricardo Reyes Basoalto, Pablo Neruda, Poems, Poetry, Windows That Open Inward. Bookmark the permalink.


text:
Todd Margolis
4.47K 0
Todd Margolis is a jiu-jitsu black belt under Pedro Sauer, founder of the Potomac MMA Academy, who worked extensively with Lloyd Irvin and Mário Yamasaki. During the early 2000s, Margolis was widely regarded as one of the top grapplers in the world, a reputation earned after he beat big names of the sport such as Marc Laimon, Pablo Popovich, Rafael Lovato Jr., Kenny Florian, Ricky Migliarese, Shawn Williams, and medaled in the ADCC US Trials. Todd Margolis was also the first American athlete to reach the final of the International Brazilian Jiu-Jitsu Federation's (IBJJF) Pan American Championships at the adult black belt level, and the first US athlete to conquer a medal in the tournament's open weight division (adult BB).
Todd Margolis Jiu-Jitsu
Full Name: Todd M. Margolis
Lineage: Carlos Gracie > Helio Gracie > Rickson Gracie > Pedro Sauer > Todd Margolis
1st Place Jitstyle Colombian Nationals (2016*)
2nd Place ADCC US Trials (2002)
3rd Place IBJJF Pan Championship (2004*)
Main Achievements (Masters):
1st Place Gracie World Championship (2001 purple, 2002 brown)
Favorite Position/Technique: Knee-Cut Guard Pass / "Margolis Shrimp"
Team/Association: Potomac MMA / Lloyd Irvin
Todd Margolis Biography
Todd Margolis was born on May 2, 1976, in La Plata, Maryland, growing up in Port Tobacco, MD, a small town in Southern Maryland, about an hour south of Washington DC.
During high school Todd picked up wrestling as a school sport, an activity he started at the age of 12. Margolis quickly excelled on the mats and became Maryland State High School's Wrestling Champion in 1994, a year after he had placed 3rd (1993).
After finishing high school Todd did not pursue a sporting lifestyle, nevertheless, once he heard of jiu-jitsu from a friend of the family (Chris Doyle), he was intrigued and decided to give it a try. He was 21 years old when he entered Mario Yamasaki's gym for the first time.
Although he was not inclined to compete when he joined the classroom, after watching his first tournament, Margolis decided he wanted in on the action. He soon fought in his first tournament and quickly gained notoriety in the local scene.
Three months into training, Todd was promoted to blue belt (1998). His excellent form also earned him an invite from coach Mario for a teaching vacancy at Yamasaki's gym, which he accepted. Within a little over 2 years since the start of his training, Margolis was promoted to purple belt. It was during his time competing in this rank that Todd became the first American to beat a Gracie family member (Igor Gracie), at the IBJJF Pan Championship of 2001.
Todd continued to train under Yamasaki while keeping a close relationship with Lloyd Irvin – someone Margolis had trained with, regularly, since 1999. As a brown belt, however, Margolis and his teammate Mike Moses decided to split from Mario's club. Moses had trained previously with Pedro Sauer and re-joined with the famous Gracie student's team. Todd also joined Sauer, earning his black belt there. From white to black belt, Margolis took solely 5 and a ½ years of training.
In his first year competing in the upper echelon of the sport, Todd Margolis made it to the final of the IBJJF Pan Championship, then becoming the first American black belt to reach the final of the prestigious event (2003), he would medal again the next year, this time in the absolute division, being the first USA athlete to do so as well.
Although destined for great things in the sport, Todd injured his spine in 2005, suffering from two herniated two disks. This injury removed the Maryland native from competition for the better part of 12 years, a time during which Margolis opted to follow the path of coaching. In 2008 Margolis opened BJJ Medellin, in Medellin, Colombia. Two years later, Todd moved back to the US, leaving his student (and first black belt) Damian Martinez in charge of the team. Todd opened Potomac MMA in September of 2018.
In 2019, Margolis returned to competition, making his official return to a big stage at the Masters' division of the Pan American Championship (No-Gi).
Todd Margolis Colombian Nationals
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Theodoro Canal
Travis Stevens
Armando Wridt
İbrahim İnal
Gerard Labinski

redpajama_set_name: RedPajamaCommonCrawl


The thought of me walking the catwalk in my fabulous mesh tank top is the last thing you want to envision, but that is not the type of modeling I am talking about.
As educators, one of the things we need to constantly remember is that we need to do much more showing than we do telling.
There are many ways for us to show our students how to do something instead of just telling them. I've done it in many different ways.
For blogging, I made sure I was comfortable with my blogging skills before I had students do it. For one year, I actually completed all of the blog posts I assigned to my students so I could show them what I was looking for in a completed blog post.
When we read plays in class, I'm the first person to stand up and take the silly voice and run with it. I want my students to feel comfortable to be silly and have fun, so I make sure I can be silly and that I have fun.
If I want students to feel comfortable failing and not knowing something, I make sure to never hide my failures and admit when I'm stumped. I will never ask something of my students that I'm not willing to do my self.
Modeling is key for admins. If you want teachers to do things, model it for them. Give them an example to follow. Show them instead of just telling them.
If you want to see teachers use more technology in the classroom, use more technology in staff PD. If you want to see more student centered classes and engaging lessons, show that in meetings.
Last year, we had PD days designed that got us up and moving. We took trips downtown and to a local historical museum. We used our mobile devices to engage in questions and enhance/facilitate discussion. This was an excellent example of modeling.
Another part of these PD days was that there were teachers involved in the planning of the days. The teacher voice was present to help make it meaningful to everyone involved. How often do we give students a voice before or after a lesson?
Students watch us like a hawk. They take our lead every day. If we come into class with big smiles and high fives, they will mirror that enthusiasm. If admins walk in to a meeting with a rain cloud over their head, the teachers will prepare for the worst. We all need to model the best for our students and staff. This is the strategy that will support a culture of learning and growth as a person.
Put on your best "Blue Steel" impression and get out there.


text:
Home Dufferin Distributed File System Tutorial
Distributed file system tutorial
What is Distributed File System? Webopedia Definition
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The Distributed File System Windows 7 Tutorial
Distributed File System (DFS) on Windows Server 2016 A
Issues in the design and use of a distributed file system
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CS6601 DS Notes, Distributed Systems Lecture Notes Download link for CSE 6th SEM CS6601 Distributed Systems Lecture Notes are listed down for File System This paper discusses an independent file facility, one that is not embedded in an operating system. The distributed file system (DFS) is so named because it is
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Distributed file system (DFS) is a technology that provides easier access and availability for file shares. The two types of DFS are server-based and domain-based. The biggest challenge facing distributed file system designers has always been how to manage the tension between * distributing and replicating data (which is
1 1 Distributed File Systems Operating Systems 2 Distributed-File Systems Background Naming and Transparency Remote File Access Stateful versus Stateless Service Installing DFS. To install DFS, an administrator of a file server on the network needs to install the File System role and select the necessary DFS-related role services.
Distributed file system (DFS) is a technology that provides easier access and availability for file shares. The two types of DFS are server-based and domain-based. CS6601 DS Notes, Distributed Systems Lecture Notes Download link for CSE 6th SEM CS6601 Distributed Systems Lecture Notes are listed down for File System
To improve the reliability and availability of file shares in an enterprise network Microsoft has developed the Distributed File System (DFS). Distributed File System Replication DFSR is a replication engine that organizations can use to synchronize folders for servers on network Tutorial for Windows DFS;
Learn from Big and Hadoop Tutorial about HDFS. Get to know about its unique features, application and various reasons why HDFS works with Hadoop. Hadoop works directly with any distributed file system that can be mounted by the underlying the list of supported file systems bundled with Apache Hadoop
Hadoop works directly with any distributed file system that can be mounted by the underlying the list of supported file systems bundled with Apache Hadoop 8/30/05 C SE70481:D is trbu ed oag page 1 Distributed File Systems: Concepts and Examples Overview paper on challenges of distributed storage systems
CS6601 DS Notes Distributed Systems Lecture Notes – CSE
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Distributed File System e-Tutorials
Apache Hadoop 2.9.1 – HDFS Users Guide. Google Code University ; This tutorial covers the basics of distributed we encounter in a distributed system. For example, file servers manage disk, Introduction. Presently, our most common exposure to distributed systems that exemplify some degree of transparency is through distributed file systems..
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Distributed File System Replication DFSR is a replication engine that organizations can use to synchronize folders for servers on network Tutorial for Windows DFS; (dis-tribВґЕ«-tid fД«l sisВґt&m) (n.) A method of storing and accessing files based in a client/server architecture. In a distributed file system, one or more central
CS6601 DS Notes, Distributed Systems Lecture Notes Download link for CSE 6th SEM CS6601 Distributed Systems Lecture Notes are listed down for File System The biggest challenge facing distributed file system designers has always been how to manage the tension between * distributing and replicating data (which is
The File System(FS) shell includes various shell-like commands that directly interact with the Hadoop Distributed File System (HDFS) as well as other file systems Distributed file systems: concepts and The purpose of a distributed file system This excellent tutorial introduces the basic concepts for understanding
The File System(FS) shell includes various shell-like commands that directly interact with the Hadoop Distributed File System (HDFS) as well as other file systems Watch videoВ В· In this video, Kevin Dankwardt discusses distributed file systems. He describes OpenAFS and GlusterFS. He demonstrates the packages and utilities involved in
This is a handy solution to distributed file system access in Linux and UNIX environment. From the article: Network File System (NFS) has been part of the world of Introduction. Presently, our most common exposure to distributed systems that exemplify some degree of transparency is through distributed file systems.
The File System(FS) shell includes various shell-like commands that directly interact with the Hadoop Distributed File System (HDFS) as well as other file systems 8/30/05 C SE70481:D is trbu ed oag page 1 Distributed File Systems: Concepts and Examples Overview paper on challenges of distributed storage systems
To improve the reliability and availability of file shares in an enterprise network Microsoft has developed the Distributed File System (DFS). 1 Distributed File Systems Dr Len Hamey COMP342 Goals of a DFS Network transparency Location transparency Location independence User mobility Fault tolerance
Distributed Systems: What is a distributed system? file service, name service - application services CoDoKi, Fig. 1.2 Figure 1.2 A typical intranet Distributed File System Replication DFSR is a replication engine that organizations can use to synchronize folders for servers on network Tutorial for Windows DFS;
Distributed Systems: What is a distributed system? file service, name service - application services CoDoKi, Fig. 1.2 Figure 1.2 A typical intranet Google Code University ; This tutorial covers the basics of distributed we encounter in a distributed system. For example, file servers manage disk
8/30/05 C SE70481:D is trbu ed oag page 1 Distributed File Systems: Concepts and Examples Overview paper on challenges of distributed storage systems The biggest challenge facing distributed file system designers has always been how to manage the tension between * distributing and replicating data (which is
The Basics of the Windows Server 2008 Distributed File System (DFS) Many File Servers and File Shares It's a common scenario. Distributed Systems: Concepts and Design distributed file systems and distributed naming before moving on to important data-related EJB Tutorials. Web Services.
8/30/05 C SE70481:D is trbu ed oag page 1 Distributed File Systems: Concepts and Examples Overview paper on challenges of distributed storage systems HDFS, the Hadoop Distributed File System, is a distributed file system designed to hold very large amounts of data (terabytes or even petabytes), and provide high
10/03/2009 · Overview In this blog post, we'll cover the basics of the Distributed File System (DFS) in Windows Server 2008, which offers users simplified access to a The Hadoop Distributed File System (HDFS) was developed following the distributed file system design principles. Running on commodity hardware, HDFS is extremely
hadoop distributed file system (hdfs) architecture documentation and hadoop distributed file system in cloud computing hadoop distributed file system interview questions Software Architecture and Design Distributed Architecture A distributed system can be information is stored and retrieved from the database or file system.
Lecture 17: INTRODUCTION TO DISTRIBUTED OPERATING SYSTEMS. CHRIS STAFF Uniform File System. Identical Operating System kernels resident on each computer This is a handy solution to distributed file system access in Linux and UNIX environment. From the article: Network File System (NFS) has been part of the world of
Distributed File System (DFS) can define as a set of client and server services that allow an organization using Microsoft Windows servers to organize many (dis-tribВґЕ«-tid fД«l sisВґt&m) (n.) A method of storing and accessing files based in a client/server architecture. In a distributed file system, one or more central
Enabling Distributed File System (DFS) and creating a Namespace Distributed File System (DFS) is a technology included with Windows Server 2012 R2 that enables This is a handy solution to distributed file system access in Linux and UNIX environment. From the article: Network File System (NFS) has been part of the world of
This blog explains how to add DFS roles and how to configure Distributed File System on Windows Server 2016. Distributed Systems: What is a distributed system? file service, name service - application services CoDoKi, Fig. 1.2 Figure 1.2 A typical intranet
Hadoop Tutorial Part 2 Hadoop Distributed File System
Distributed File System DFS - University of Michigan. Learn from Big and Hadoop Tutorial about HDFS. Get to know about its unique features, application and various reasons why HDFS works with Hadoop., This blog explains how to add DFS roles and how to configure Distributed File System on Windows Server 2016..
Windows 2003 DFS (Distributed File System) Networking
Basics of the Windows Server 2008 Distributed File System. Watch videoВ В· In this video, Kevin Dankwardt discusses distributed file systems. He describes OpenAFS and GlusterFS. He demonstrates the packages and utilities involved in Distributed File System (DFS) is the Microsoft solution to the problem: a simplified way for users to access geographically dispersed files..
Enabling Distributed File System (DFS) and creating a Namespace Distributed File System (DFS) is a technology included with Windows Server 2012 R2 that enables Learn how to create a Distributed File System Namespace in Windows Server 2012 R2 in this tutorial by Russell Smith.
This blog explains how to add DFS roles and how to configure Distributed File System on Windows Server 2016. 10/03/2009 · Overview In this blog post, we'll cover the basics of the Distributed File System (DFS) in Windows Server 2008, which offers users simplified access to a
Software Architecture and Design Distributed Architecture A distributed system can be information is stored and retrieved from the database or file system. Distributed Systems: Concepts and Design distributed file systems and distributed naming before moving on to important data-related EJB Tutorials. Web Services.
Hadoop works directly with any distributed file system that can be mounted by the underlying the list of supported file systems bundled with Apache Hadoop Distributed File System (DFS) can define as a set of client and server services that allow an organization using Microsoft Windows servers to organize many
Learn about the concept and importance of distributed file systems. Distributed file system (DFS) is a technology that provides easier access and availability for file shares. The two types of DFS are server-based and domain-based.
Introduction. Presently, our most common exposure to distributed systems that exemplify some degree of transparency is through distributed file systems. 1 1 Distributed File Systems Operating Systems 2 Distributed-File Systems Background Naming and Transparency Remote File Access Stateful versus Stateless Service
1 Distributed File Systems Dr Len Hamey COMP342 Goals of a DFS Network transparency Location transparency Location independence User mobility Fault tolerance Installing DFS. To install DFS, an administrator of a file server on the network needs to install the File System role and select the necessary DFS-related role services.
1 Distributed File Systems Dr Len Hamey COMP342 Goals of a DFS Network transparency Location transparency Location independence User mobility Fault tolerance Enabling Distributed File System (DFS) and creating a Namespace Distributed File System (DFS) is a technology included with Windows Server 2012 R2 that enables
1 Distributed File Systems Dr Len Hamey COMP342 Goals of a DFS Network transparency Location transparency Location independence User mobility Fault tolerance Distributed file systems: concepts and The purpose of a distributed file system This excellent tutorial introduces the basic concepts for understanding
10/03/2009 · Overview In this blog post, we'll cover the basics of the Distributed File System (DFS) in Windows Server 2008, which offers users simplified access to a Introduction. Presently, our most common exposure to distributed systems that exemplify some degree of transparency is through distributed file systems.
Distributed File System Replication DFSR is a replication engine that organizations can use to synchronize folders for servers on network Tutorial for Windows DFS; Software Architecture and Design Distributed Architecture A distributed system can be information is stored and retrieved from the database or file system.
Watch videoВ В· In this video, Kevin Dankwardt discusses distributed file systems. He describes OpenAFS and GlusterFS. He demonstrates the packages and utilities involved in Overview. HDFS is the primary distributed storage used by Hadoop applications. A HDFS cluster primarily consists of a NameNode that manages the file system metadata
HDFS, the Hadoop Distributed File System, is a distributed file system designed to hold very large amounts of data (terabytes or even petabytes), and provide high The biggest challenge facing distributed file system designers has always been how to manage the tension between * distributing and replicating data (which is
Distributed Systems: What is a distributed system? file service, name service - application services CoDoKi, Fig. 1.2 Figure 1.2 A typical intranet The biggest challenge facing distributed file system designers has always been how to manage the tension between * distributing and replicating data (which is
Distributed File System (DFS) can define as a set of client and server services that allow an organization using Microsoft Windows servers to organize many The Basics of the Windows Server 2008 Distributed File System (DFS) Many File Servers and File Shares It's a common scenario.
hadoop distributed file system (hdfs) architecture documentation and hadoop distributed file system in cloud computing hadoop distributed file system interview questions (dis-tribВґЕ«-tid fД«l sisВґt&m) (n.) A method of storing and accessing files based in a client/server architecture. In a distributed file system, one or more central
To improve the reliability and availability of file shares in an enterprise network Microsoft has developed the Distributed File System (DFS). Working of Hadoop Distributed File system (HDFS), Namenode, Datanode and Secondary Namenode and how it prevent from Failure
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Who loves a good mystery? If you do, then this mystery box is for you! Pulled from our Yard Sale goods, we've got a Mystery box filled with 5 bottoms (either skirts, shorts or pants) for $50 (valued at $200+!). Unbox and be surprised! Shoot an unboxing vid and tag us! We'd love to join in on the surprise :) We are VERY Limited on these, so due to the exclusivity of this item, we are only able to ship one box per customer.
Please note that these mystery boxes are from our YARD SALE. They are either Overstock or Damaged goods. Mystery boxes are FINAL SALE, and cannot be returned or exchanged online or at any brick and mortar store nor at any future pop up shops. While some items may be flawless, there might be a chance where some of the items have a small rip/tear, stain, sized wrong, or some other type of defect.


text:
The Guitar Player
Davy Graham
Release Date 01 March 2019
MOVLP2316
• 180 GRAM AUDIOPHILE VINYL
The Guitar Player was Davy Graham's first full length LP, after releasing his EP ¾ A.D. two years earlier. This unique album combines his interesting and virtuoso guitar play with some excellent arrangements. With his guitar style he contributed to the musical changes in the early 1960s, where blues, rock and folk started to play a more prominent role in the music scene. The man crosses into blues, jazz, folk, Arababic music and more, often in the same song. It's an essential album and one of the greatest instrumental acoustic guitar records released in the 1960's.
British guitar virtuoso Davy Graham inspired a multitude of artists such as Paul Simon, Jimmy Page and Jimi Hendrix. Combining different music genres, Graham proved to be one of the 20th century's most versatile guitarists and songwriters.
1 Don't Stop The Carnival
2 Sermonette
3 Take Five
4 How Long, How Long Blues
5 Sunset Eyes
6 Cry Me A River
1 The Ruby And The Pearl
2 Buffalo
3 Exodus
4 Yellow Bird
5 Blues For Betty
6 Hallelujah, I Love Her So
More products from Davy Graham
The Holly Kaleidoscope: Limited Edition Kaleidoscope Vinyl RSD 2020


text:
Portret Murzynki () – obraz francuskiej malarki Marie-Guillemine Benoist z 1800 roku, o wymiarach 81 × 65 cm, namalowany techniką olejną na płótnie.

Opis
Obraz jest portretem przedstawiającym rzeczywistą, choć anonimową, osobę. Kobieta na obrazie siedzi, ukazuje widzowi prawy profil, patrzy bezpośrednio na niego. Jest półnaga, okrywa się białą tkaniną, którą podtrzymuje lewą ręką. Elementy wyposażenia widoczne na obrazie wskazują, że kobieta nie została sportretowana we własnym domu, prawdopodobnie wcześniej była niewolnicą, po czym została uwolniona na mocy dekretu o zniesienie niewolnictwa z 1794 roku.

Obraz stanowi echo licznych portretów kobiet z wyższych kręgów społeczeństwa malowanych przez nauczyciela Benoist, Jacques'a-Louisa Davida. Jest podobny do Portretu pani Récamier Davida, m.in. z powodu minimalnego użycia akcesoriów, modelowania ich jak form rzeźbiarskich i sposobu oświetlenia. Praca podkreśla cechy rasowe, m.in. wzmacniając czerń skóry przez zastosowanie białej tkaniny jako szaty, teksturę włosów i kształt oczu, nosa i ust.

Obraz można odczytywać jako prosty portret przedstawiający czarną kobietę lub jako ukazanie różnic rasowych, ukazanie kobiety na obrazie jako istoty czującej.

Obraz został zakupiony dla państwa francuskiego przez Ludwika XVIII w 1818 roku, przechowywany jest w Luwrze.

Kolaż z 2012 roku
W sierpniu 2012 roku redakcja hiszpańskiego magazynu "Fuera de Serie" umieściła na okładce jednego ze swoich numerów obrobiony komputerowo kolaż autorstwa Karine Percheron-Daniels, w którym twarz kobiety z Portretu Murzynki została zastąpiona twarzą Michelle Obamy. Okładka magazynu wywołała duże kontrowersje.

Przypisy

Obrazy Marie-Guillemine Benoist
Obrazy malarzy epoki klasycyzmu
Portrety
Obrazy w Luwrze
Wydarzenia 1800

redpajama_set_name: RedPajamaWikipedia


text:
National Right to Life Betrays Babies
⇑ back to "Issues"
1:15PM EST 1/18/2018	 Faith2Action
(Unsplash/Dakota Corbin)
Editor's Note: Friday is the annual March for Life. As we cry out for the end of abortion, prayerfully consider how the Lord can use your testimony to bring healing, restoration and justice for families devastated by the death of an unborn child.
Congress announced the pro-life bill it will pass on the 45th anniversary of Roe v. Wade is not a bill to prevent abortions, but rather one which merely prohibits an abortionist's second attempt on a child's life in the event their first attempt fails, and the child survives.
Rather than pass HR 490, the Heartbeat Bill, to protect every child whose heartbeat can be detected, Congress chose, instead, to pass the "Born Alive Abortion Survivor's Protection Act," HR 4712, which won't prevent a single abortion. Why wouldn't a pro-life Congress pass the bill with 170 co-sponsors which is closest to ending abortion? The answer: National Right to Life.
As Congressman Steve King tweeted yesterday, "National Right to Life Committee is working against 129 pro-life groups/leaders to block a vote on HR 490, my #Heartbeat Bill ending abortion."
In today's Washington Times op-ed, former Majority Leader Tom DeLay stated, "Today, National Right to Life, the only national group that doesn't support the Heartbeat Bill's substantial protection for unborn children, thinks 'success' is to kill only a million children a year." DeLay added, "We reject the establishment's approach. We also reject those who have the thoughtlessness to continue it."
DeLay said, "We have more than enough blood on our hands, and Congresshas the means to stop this loss in nearly every case."
"For six years, Ohio Right to Life fought Ohio's Heartbeat Bill harder than Planned Parenthood, and when it finally passed, they stood with the abortion industry in calling for a veto of the nation's most protective pro-life law," stated Janet Porter, President of Faith2Action, who initiated both the state and Congressional Heartbeat Bills. "And now, National Right to Life is the one obstacle standing in the way of protecting nearly a million children with beating hearts each year."
Troy Newman, president of Operation Rescue, stated, "If you do not support the bill you oppose it."
Newman added, "There has been no greater betrayal of innocent blood since Judas betrayed our Lord in the Garden of Gethsemane but that of National Right to Life selling the innocent blood of preborn children."
Rachelle Heidlebaugh, who has suffered the devastation of abortion, said, "National Right to Life is the perfect example of insanity—doing the same thing over and over again and expecting different results." She added, "Why would anyone who wants to end abortion donate another dime to this corrupt organization working against protecting babies whose heartbeats can be heard?"
Larry Cirignano of Family First added, "It's time for the failed leadership to step aside and let those who actually want to end abortion take their place."
"We call on every pro-lifer, who wants to do more than march to end abortion, to put your faith to action right now," stated Porter. "Just click the fax icon at Faith2Action.org to send seven free faxes to the leaders who decide which bills come to the floor for a Congressional vote. Better yet, click the button to send 476 faxes to Congress calling for the Heartbeat Bill stating, 'Merely protecting children who are born alive following an attempt to kill them is woefully inadequate."
ACLJ Fights UN's Threat to Redefine 'Right to Life'
Bible-Based, Transformational Event Could Completely Alter Our Spiritual Climate
Related topics: Abortion | Caucus Issues


Amazing run in the Rawah Wilderness!
What am awesomely beautiful day to go for a run in the Rawah Wilderness! So much fun and great training was had on Saturday!
This week we'll be doing some more great training and keep moving forward with that base fitness! Remember that most of you will need to be doing a little more than just the basic team training. 30 minute runs before Yoga and training on Monday and/or Sunday make this quite easy!
Remember that I have posted the Summer Pepper Hours. So instead of looking at the bottom of this page for the extra hours just go to the link above to find what you need to do The team practices will be Habanero. You can also find this on the Training Resources page.
Don't know your pepper? Ask us! Training is just like with eating hot peppers - it doesn't matter how much you WANT to do nor how much you think you should do but how much your body is actually able to handle. Thus to reach your best performance from year to year you want to be thoughtful in your approach to increasing your training load.
Weekend: TBA, we'll see how things are looking!


text:
Canada Expands Vaccines Accepted for Travel
Alonzo Phillips November 21, 2021 2 min read
With the holiday travel season nearing, Canadian officials announced several measures on Friday meant to standardize international travel and make it easier for Canadians taking short trips to re-enter the country, Home Garden USA.
Travelers who are fully vaccinated with the shots from Sinopharm, Sinovac and Covaxin will be allowed to enter Canada starting Nov. 30, opening the door wider to people from countries like India, Brazil and China, where those vaccines are more frequently administered.
Currently, Canada only accepts travelers who have received the Pfizer-BioNTech, Moderna, Oxford-AstraZeneca or Johnson & Johnson vaccines, the four shots approved in the country. The expanded list will align with the vaccines cleared for use by the World Health Organization.
Dr. Theresa Tam, Canada's chief public health officer, said that the case rates associated with each of the vaccines were about the same.
"All that is very reassuring," Dr. Tam said. She added that although Sinopharm, Sinovac and Covaxin "are not authorized in Canada, because they've gone through the W.H.O. process in terms of evaluation of safety, of efficacy and of quality, we've taken that into account as we increase the list of vaccines for Canadian border measures."
Additionally, people leaving Canada by land or by air for less than 72 hours will no longer be required to provide proof of a negative Covid-19 test to re-enter the country. The price and processing time of P.C.R. tests — which can cost more than $100 and take longer than a day for results — were widely seen as deterrents to travel.
This change applies only to Canadians, permanent residents and Indigenous people registered under the Indian Act. It also takes effect Nov. 30.
The government will take a firm stance against unvaccinated travelers entering and leaving the country beginning on that date as well, with "very few exemptions," said Omar Alghabra, Canada's transport minister. He added that only fully vaccinated travelers would be able to fly from a Canadian airport or board a Via Rail or Rocky Mountaineer train.
Starting Jan. 15, Canada will also phase out most previously granted travel exemptions for those who are unvaccinated. This will affect professional and amateur athletes, foreign students, truck drivers and other essential workers who cross the border, and adults over 18 traveling to reunite with a family member.
Other categories of unvaccinated or partly vaccinated travelers — such as refugees, marine crews and agricultural workers — will be able to enter only with limited approval.
Visit : https://homegardenusa.com/
Tags: Accepted Beach Canada Cheap Expands Holiday Holidays News resort Resorts Tips Travel travel tips Vacation Vaccines World
Previous Traveling this Thanksgiving? Here's what to know
Next Covid News: U.S.-Mexico Border Reopens to Vaccinated Tourists
Tire Tips


text:
Сюлли-сюр-Луар (замок) — французский средневековый замок на Луаре.
 Сюлли-сюр-Луар (город) — в центральной Франции, на территории департамента Луаре, на обоих берегах реки Луары.

См. также
 Сюлли (значения)


text:
Q: How to change keyboard character in Ubuntu? How to change the letter e with ę in Ubuntu xkb. my language doesn't use e,  instead we used ę.  The character e is not used at all.  the keyboard layout that I am using is us

A: Ctrl+Shift+U+0119
 will do it.
If you don't like to change your keyboard layout to your language as waltinator mentioned before, you might have a look on various websites about
UTF / UTF8 / UTF16 key codes if you often do need some which do not usually come with the US keyboard layout …
For example, I found the code for "your" ę on
http://www.utf8-chartable.de/unicode-utf8-table.pl?start=256&unicodeinhtml=dec

redpajama_set_name: RedPajamaStackExchange


text:
// Copyright 2000-2019 JetBrains s.r.o. Use of this source code is governed by the Apache 2.0 license that can be found in the LICENSE file.
package com.intellij.execution.target;

import com.intellij.ide.wizard.AbstractWizardStepEx;
import com.intellij.ide.wizard.CommitStepException;
import com.intellij.openapi.util.NlsContexts;
import org.jetbrains.annotations.Nullable;

public abstract class TargetEnvironmentWizardStep extends AbstractWizardStepEx {

  public TargetEnvironmentWizardStep(@Nullable @NlsContexts.DialogTitle String title) {
    super(title);
  }

  /**
   * This method is abstract in super class and can't be overridden in Kotlin since it exposes protected member
   * {@link AbstractWizardStepEx.CommitType}
   */
  @Override
  public final void commit(CommitType commitType) throws CommitStepException {
    doCommit(commitType);
  }

  protected abstract void doCommit(CommitType commitType) throws CommitStepException;
}

redpajama_set_name: RedPajamaGithub


text:
Britain faces difficult economic times after Brexit – David Cameron
Jun 30, 2016 | The Citizen
British Prime Minister David Cameron warns the nation of 'difficult economic times' following the exit from the European Union.
Speaking at the parliament on Wednesday, the premier predicted a troubled economy, further opposing the suspension of fiscal rules.
"There's no doubt in my mind these are going to be difficult economic times," he said. "If we do see economic difficulties, one of the ways we have to react to that is to make sure that our public finances and economy remain strong … so I don't think it would be right to suspend the fiscal rules."
His stance on the fiscal rules runs counter to the one adopted by opposition leader Jeremy Corbyn, who has called for more investment.
The situation in Britain follows a referendum held last Thursday, in which some 52 percent of the population opted for Brexit and 48 percent voted to remain in the EU, marking a narrow win for the Out campaign.
As a result, British pound fell to a record low in 31 years against the US dollar.
Brexit has also triggered tension in the country's politics with the premier promising resignation and the opposition leader under pressure to resign.
In his parliament address, Cameron also called on Corbyn to resign, arguing "It might be in my party's interest for him to sit there; it's not in the national interest."
"For heaven's sake man, go," said the prime minister.
The Labour leader's MPs overwhelmingly passed a no confidence motion against him on Tuesday and a day later he was faced with another resignation from his Shadow Cabinet .
Corbyn said the Britons' decision to leave the EU amounts to a 'rejection of the status quo, a status quo which clearly isn't delivering.'
Cameron, however, disagreed, saying 'To claim last week's result was about the state of the British economy is complete nonsense.' - Press TV.
News Sports Lifestyle Entertainment


Health costs are a concern for women and many struggle to find affordable health insurance. According to statistics from the United Health Foundation, 12.2 percent of all women in the United States ages 18 to 44 are not covered by any type of private or public health insurance. Some options are available for pregnant or soon-to-be pregnant mothers along with low-cost insurance options for children. However, these options leave many other women without a good option for health insurance. The healthcare concerns of women are unique and differ from that of their male counterparts.
When looking at the cost of women's health insurance, it is important to examine some of the most expensive types of medical costs when compared to men.
A survey from Bankrate showed 47 percent of women were faced with a bill for medical costs that was more than they expected compared to 35 percent of men. Another study by the Henry J Kaiser Family Foundation (KFF) shows 26 percent of women compared to 19 percent of men went without health care or delayed treatment because of cost.
To learn more about the disparity of health insurance costs of women compared to men, here is a look at the differences in amounts spent on health care by gender based on data from the Centers for Medicare & Medicaid Services (CMS).
Women looking for health insurance have specific concerns when looking for insurance about the coverage options for insurance that affect them most. Here are the top 10 leading causes of death among U.S. women of all ethnicities based on data from the U.S. Department of Health & Human Services Centers for Disease Control (CDC).
Preventive care is also a leading health concern for women. The proper health care can prevent many diseases and health conditions that can lead to deteriorating health or even death. Preventive health screenings can find diseases early, such as breast cancer, while they are still in the treatable stages. It can also identify women who may be at risk for developing diseases and cancers common among women. Early detection is key to treating and avoiding medical problems among women.
Women looking to find affordable health insurance are seeking plans that cover preventive care and coverage options for their top health concerns. Here are a few coverage options for women to consider.
Women can access the Health Insurance Marketplace to explore plan options based on several factors including price, benefits, and quality of coverage. When looking at available options, be aware of the coverage for essential health benefits, preventive care, and pre-existing conditions. If you also need coverage for your children, you can learn about insurance programs for single mothers including Medicaid and the Children's Health Insurance Program (CHIP). On the HealthCare.gov website, you can search for available healthcare plans in your area.
You can also see if you may qualify for a subsidy plan to reduce your cost for health insurance.
If your employer offers health insurance to its employees, you can check into what options are available with your company's employee benefits director. Often, many different levels of coverage are available if you are looking for a more affordable plan. The employer may even pay part of the premium, lowering your costs even further. One advantage to employer-based health insurance is that employee insurance premiums are often not taxable.
If you are between jobs or have just started a new job, there may be a waiting period before you become eligible to enroll in a health insurance plan. If so, you may want to consider a short-term health insurance plan. The premiums are often very affordable, and you do not have to wait for an open enrollment period to apply for coverage. Most short-term health plans allow you to renew your coverage for up to 36 months.
What is the Average Health Insurance Premium?


In early-phase clinical trials researchers want to find the largest dose of a new drug that limits the risk of side-effects to an acceptable level, to use as an upper limit on dosing in subsequent trials. In a paper published by members of the NIHR Statistics Group, an explanation is given for a type of modern trial design that finds this dose: Continual Reassessment Method (CRM).
CRM re-analyses all the data after each patient and chooses the dose for the next patient that is expected to provide some benefit whilst controlling the risk of side-effects. Old-fashioned designs often had a fixed sequence of doses and only analysed at the end of the study. The dose chosen to take forwards for future research is more often correct and more patients are treated at, or close to, the chosen dose if the CRM design is used. This prevents the failure of promising new drugs due to the incorrect dose being used, means more is known about the side effects before more patients take the drug and leads to new drugs becoming available quicker for clinical use.
The paper offers a practical 'how to' guide for trialists and statisticians using the CRM design for the first time. It presents the planning decisions that need to be made in advance and gives guidance on operational aspects.
The paper was the result of a sequence of workshops organised by the NIHR Statistics Group that brought together several leading statisticians from academia and industry. They were aware of the inefficiencies of traditional dose-finding study designs and the lack of usage of modern methods, including CRM. Another paper from an overlapping group of authors investigates from a survey the barriers to adopting modern trial designs. This paper gives guidance on how to use the CRM design.
Self-plagiarism: When is re-purposing text ethically justifiable?


text:
\section{Introduction}

Let $k$ be a field and $R$ a commutative $k$-algebra with unity. A \emph{real}
(or \emph{rank one}) \emph{valuation} on $R$ is a function $v\colon R\to \R\cup
\{+\infty\}$ such that $v(1)=0$, $v(0)=+\infty$, $v(xy)=v(x)+v(y)$ and
$v(x+y)\geq \min\{v(x),v(y)\}$ for all $x$, $y\in R$. In this paper we consider
only real ring valuations. Assume that $k$ is endowed with some valuation $\nu$
and $K$ is some $k$-algebra with a valuation $\mu$ extending $\nu$. Then every
homomorphism $f\colon R\to K$ of $k$-algebras induces a valuation $\mu\circ f$
of $R$ extending $\nu$. The first question we address in this paper is the
following: given $R$ and $\nu$, does there exist a universal valued field $K$
such that \emph{any} valuation of $R$ that extends $\nu$ is obtained via some
homomorphism $f\colon R\to K$ as described above? We give an affirmative answer
to this question in Theorem~\ref{T:ufield} of Section~\ref{S:ufield}. Note that
the universal field we are looking for is different from the maximal immediate
extension of $k$ (\cite[p. 191]{Krull}) since we do not insist that $v$ must have
the same residue field as $\nu$.

The second problem that we study in this work is lifting points in local
tropical varieties. If $R$ is a local ring with the maximal ideal $\mathfrak{m}$,
we call a valuation $v$ of $R$ \emph{local}, if $v$ is nonnegative on $R$
and positive on $\mathfrak{m}$. Note that if $R$ contains a field $k$, any
local valuation of $R$ must be trivial on $k$, i.e., $v(x)=0$ for all $x\in k$,
$x\ne 0$. Now let $R=k[[x_1,\ldots,x_n]]$ be the ring of formal power series
in variables $x_1,\ldots,x_n$, and $I\subset R$ an ideal. The \emph{local
tropicalization} $\ptrop(I)$ of $I$ is the set
$$\ptrop(I)=\{(v(x_1),\ldots,v(x_n)\,|\,v \text{ is local},\; v|_I=+\infty\}
\subseteq \Ri^n,$$
where $v$ runs over all local valuations of $R$ that take value $+\infty$ on
the ideal $I$, and $\Ri=\R\cup\{+\infty\}$. This is a particular case of
\cite[Definition~6.6]{PPS}; see also Section~\ref{S:liftpoint}. Let us introduce
two special fields. Denote by $\bar{k}$ some algebraic closure of $k$. The
field $K=\bar{k}((t^Q))=\bigcup_{N\geq 1}\bar{k}((t^{1/N}))$ of \emph{Puiseux
series} with coefficients in $\bar{k}$ is the set of all formal sums
$$\sum_{m\in\Q} a_m t^m,$$
where $a_m\in \bar{k}$, the set $\{m\,|\,a_m\ne 0\}\subset\Q$ is bounded from
below, and all its elements have bounded denominators. The field $K'=
\bar{k}((t^\R))$ of \emph{Hahn series} is the set of all formal
sums
$$a(t)=\sum_{m\in \R} a_m t^m,$$
where the set $\{m\,|\,a_m\ne 0\}$ is well-ordered, that is, any its subset has
the least element.
Addition and multiplication in $K$ and $K'$ are defined in a natural way.
These fields carry also a natural valuation $\mu$ determined by $\mu(t)=1$.
The \emph{ring of Puiseux} (respectively \emph{Hahn}) \emph{series}
$\mathcal{O}=\bar{k}[[t^\Q]]$ (resp., $\mathcal{O}'=\bar{k}[[t^\R]]$) is the
valuation ring of $\mu$, i.e., the subring of $K$ (resp. $K'$) where $\mu$ is
nonnegative. In analogy with the theory of usual non-local tropicalization
(see, e.g., \cite{JMM}), we say that a point $w=(w_1,\ldots,w_n)\in
\ptrop(I)\cap\Qi^n$, where $\Qi=\Q\cup\{+\infty\}$, \emph{lifts to} $\mathcal{O}$,
if there exists a homomorphism $f\colon R\to K$ such that $w_i=\mu(f(x_i))$ for
all $i=1,\ldots,n$. Lifting of a point $w\in\ptrop(I)$ to $\mathcal{O}'$ can
be defined similarly. In the non-local theory of tropicalization, it is known
that if characteristic of $k$ is $0$, then each rational point of the
tropicalization of a variety admits a lifting to a $K$-point of the variety.
Several proofs and generalizations are given in \cite{Dra}, \cite{JMM},
\cite{Katz}, \cite{P}. For $k=\C$ and local tropical varieties, a similar
result was announced by N. Touda in \cite{To}, however, a complete proof
did not appear.

The possibility to lift points of local tropical varieties to $\mathcal{O}$
or $\mathcal{O}'$ means that these rings play a role of a kind of universal
domains with respect to valuations on local $k$-algebras $R$, where the field
$k$ is trivially valued. This is not the strong universality introduced
in the first paragraph of this Introduction. Indeed, if we set $R=k[[x,y]]$
and $v$ to be a monomial valuation defined by $v(x)=v(y)=1$, then it is easy
to see that $v$ is not induced by any homomorphism from $R$ to $\mathcal{O}$ or
$\mathcal{O}'$. However, it is easy to define a homomorphism $f\colon R\to
\mathcal{O}$, say, by sending $x$ and $y$ to $t$, such that $\mu(f(x))=v(x)$,
$\mu(f(y))=v(y)$. Precisely, we have the following property, which we call
\emph{weak universality} of the rings of Puiseux and Hahn series.
Assume that $R$ is a complete local Noetherian ring, $k$ is a field isomorphic to
the residue field of $R$ and contained in $R$, and $x_1,\ldots,x_n\in R$ a finite
collection of elements. Then for each local valuation $v$ of $R$ there exists a
homomorphism $f\colon R\to \mathcal{O}'$ such that $v(x_i)=\mu(f(x_i))$ for
all $i=1,\ldots,n$. If, moreover, $k$ has characteristic $0$ and $v$ takes
only rational values, then there exists a homomorphism $f\colon R\to \mathcal{O}$
with the same property. Similarly, lifting points in the non-local tropical
varieties can be expressed as the weak universality of the \emph{fields} of
Puiseux and Hahn series with respect to valuations on finitely generated
$k$-algebras $R$. We prove these properties in Section~\ref{S:wufield}
and then apply them to the local tropicalization in Section~\ref{S:liftpoint}.

Our proof of the weak universality of the Puiseux and Hahn series rings is closed
to the proof of the lifting points property of tropical varieties given in
\cite{JMM}. We also use a descent by dimension, but we descent not to dimension
$0$ but to dimension $r$ equal to the rational rank of the value group of $v$;
this allows to work always over the field $\bar{k}$ and not to pass to varieties
over the field of Puiseux series.

I thank B.~Teissier and M.~Matusinski for their comments on the draft of this
work.

\section{The universal field}\label{S:ufield}

In this section we show existence of the universal valued field for
a given valued ring $k$ and a $k$-algebra $R$. The proof is based on the
following theorem about extension of valuations to a tensor product. If $A$ is
a ring, $B$ is an $A$-algebra, and $v$ is a valuation of $B$, then $v$ induces
a valuation of $A$ which we call the \emph{restriction} of $v$ and denote
$v|_A$.

\begin{theorem}\label{T:tensor}
Let $A$ be a commutative ring with unity and $B$ and $C$ $A$-algebras, also with
unity. Let $u$ be a valuation of $B$, $v$ a valuation of $C$, and assume that
$u|_A=v|_A$. Then there exists a valuation $w$ of $B\bigotimes_A C$ that extends
both $u$ and $v$: $w|_B=u$, $w|_C=v$, where $B$ and $C$ map to $B\bigotimes_A C$
as $B\ni b\mapsto b\otimes 1$, $C\ni c\mapsto 1\otimes c$.
\end{theorem}

This is stated without a proof in \cite[1.1.14f]{H}. Below we give two proofs of
Theorem~\ref{T:tensor}. The \emph{first proof} follows from recent stronger
results of I. B. Yaacov \cite{Itai} and J. Poineau \cite{Poi}. It is known that
in the conditions of Theorem~\ref{T:tensor}, the tensor product $B\bigotimes_A C$
carries a natural seminorm $\|\cdot\|$:
$$\|z\|=\inf \max_i (\exp(-u(x_i)-v(y_i))),\quad z=\sum_i x_i\otimes y_i,$$
where the infinum is taken over all representations of $z\in B\bigotimes_A C$ as
$\sum x_i\otimes y_i$, $x_i\in B$, $y_i\in C$.

\begin{theorem}[{\cite[Theorem 6]{Itai}}]\label{T:Itai}
Let $k$ be a valued algebraically closed field, and $K$ and $L$ two field
extensions of $k$. Assume that $K$ and $L$ are endowed with valuations $u$ and
$v$ that restrict to the given valuation of $k$. Then, the natural seminorm
$\|\cdot\|$ of $K\bigotimes_k L$ is multiplicative and the function
$-\log\|\cdot\|\colon K\bigotimes_k L\to \Ri$ is a valuation extending both
$u$ and $v$.
\end{theorem}

In \cite{Itai}, this theorem is proven with a help of non-standard technique
(ultrapowers) and results on quantifier elimination in some formal theories. It
can also be deduced from \cite[Section 3]{Poi}, where the technique is the
theory of affinoid algebras. To reduce Theorem~\ref{T:tensor} to
Theorem~\ref{T:Itai}, assume first that the valuation $u$ or $v$ has a nontrivial
home. Let $\mathfrak{p}=\home(u)=\{x\in B|u(x)=+\infty\}$, $\mathfrak{q}=
\home(v)$. Denote by $\mathfrak{p}\otimes 1$ and $1\otimes\mathfrak{q}$ the
extensions of $\mathfrak{p}$ and $\mathfrak{q}$ to $B\bigotimes_A C$. Then
$u$ and $v$ can be considered as valuations on the rings $B/\mathfrak{p}$ and
$C/\mathfrak{q}$ respectively, and
$$B/\mathfrak{p} \bigotimes_A C/\mathfrak{q}\simeq (B\bigotimes_A C)/
(\mathfrak{p}\otimes 1 + 1\otimes\mathfrak{q})$$
and $B/\mathfrak{p} \bigotimes_{A/\mathfrak{p}\cap A} C/\mathfrak{q}$ are
isomorphic as $A$-algebras. It follows that we can assume from the beginning
that $A$, $B$, and $C$ are domains, and $\home(u)=\home(v)=\{0\}$. Thus,
$u$ and $v$ extend canonically to the fields of fractions of $B$ and $C$.
Then, passing to the localizations (see, e.g.,
\cite[Propositions 3.5 and 3.7]{AM}), we can assume that $A=k$, $B=K$, and $C=L$
are fields, so it remains only to reduce to an algebraically closed field $k$.
Let $\bar{k}$ be an algebraic closure of $k$. Since $K'=K\bigotimes_k \bar{k}$
and $L'=L\bigotimes_k \bar{k}$ are integral extensions of $K$ and $L$
respectively, there exist a prime ideal $\mathfrak{p}\subset K'$ lying over
$\{0\}$ and a prime ideal $\mathfrak{q}\subset L'$ lying over $\{0\}$. Now,
let $\Bar{K}$ and $\Bar{L}$ be the fields of fractions of
$K'/\mathfrak{p}$ and $L'/\mathfrak{q}$ respectively. $\Bar{K}$ is an
algebraic extension of $K$, thus, by \cite[Chapter XII, \S 3]{Lang}, $u$ extends
to $\Bar{K}$, and similarly $v$ extends to $\Bar{L}$. The tensor product
$K\otimes_k L$ maps to $\Bar{K}\otimes_{\bar{k}}\Bar{L}$, and this map induces
embeddings of valued fields $K\subseteq \Bar{K}$ and $L\subseteq \Bar{L}$. Thus,
Theorem~\ref{T:tensor} indeed follows from Theorem~\ref{T:Itai}.

The \emph{second proof} of Theorem~\ref{T:tensor} is based on the results of
G. M. Bergman. This proof uses a more standard argument of commutative algebra, so
we think it is of independent interest. A \emph{pseudovaluation} on a ring $R$ is
a map $v\colon R\to\R\cup\{+\infty\}$ satisfying the same axioms as a valuation
with the exception that instead of $v(xy)=v(x)+v(y)$ we require only
$v(xy)\geq v(x)+v(y)$.

\begin{theorem}\label{T:Bergman}
Let $p$ be a pseudovaluation on a commutative ring $R$ with unity, and $S$ a
multiplicative subsemigroup in $(R,\cdot)$ such that $p|_S$ is a semigroup
homomorphism from $S$ to $\Ri$. Let $I$ be an ideal of $R$ such that there
is no $s\in S$, $f\in I$ satisfying $v(s)=v(f)< v(f-s)$. Then there exists a
valuation $v\geq p$ on $R$ such that $v|_I=+\infty$, $v|_S=p|_S$.
\end{theorem}
\begin{proof}
This theorem is a direct generalization of \cite[Corollary 1]{Berg}. Indeed, the
formula $q(x)=\sup_{f\in I} p(x+f)$ defines a pseudovaluation on $R$. On the
semigroup $S$, this pseudovaluation coincides with $p$. Then, by
\cite[Theorem 2]{Berg} there exists a valuation $v\geq q$ on $R$ that coincides
with $q$ on $S$. Since $q|_I=+\infty$, we have also $v|_I=+\infty$.
\end{proof}

We continue with an auxiliary lemma.

\begin{lemma}\label{L:fgtensor}
Let $B$ and $C$ be two finitely generated $A$-algebras over a ring $A$. Let $u$
and $v$ be valuations of $B$ and $C$ respectively that induce the same valuation
on $A$. Finally, let $x_1,\ldots,x_m\in B$ and $y_1,\ldots,y_n\in C$ be two fixed
collections of elements. Then there exists a valuation $w$ on $B\bigotimes_A C$
such that for all $i=1,\ldots,m$ $w(x_i\otimes 1)=u(x_i)$, and for all
$j=1,\ldots,n$ $w(1\otimes y_j)=v(y_j)$.
\end{lemma}
\begin{proof}
\emph{Step 1.} First, by the same argument as above we can reduce to the case
when $A=k$ is a field, $B$ and $C$ are finitely generated domains over $k$, and
$\home(u)=\{0\}$, $\home(v)=\{0\}$.
Next, adding or eliminating some elements if necessary, we can suppose that
$x_i$ generate $B$ and $y_j$ generate $C$ as rings over $k$, and none of $x_i$,
$y_j$ is $0$. Represent $B$ as a quotient of a polynomial ring
$k[X_1,\ldots,X_m]/I$ and $C$ as $k[Y_1,\ldots,Y_n]/J$ for ideals
$I\subset k[X]=k[X_1,\ldots,X_m]$ and $J\subset k[Y]=k[Y_1,\ldots,Y_n]$; here
$x_i=X_i\mod I$, $y_j=Y_j\mod J$.

\emph{Step 2.} Let us recall some terminology. We denote also by $u$ the
valuation $u$ of $B$ restricted to $k$ ($=$ $v$ of $C$ restricted to $k$). If
$w=(w_1,\ldots,w_m)\in\R^n$ is a vector and $f=\sum a_M X^M\in k[X]$ is a
polynomial in $m$ variables, the number
$$w(f)=\min_M (u(a_M)+\langle w,M\rangle),$$
where $\langle w,M\rangle=\sum w_iM_i$, is called \emph{$w$-order} of $f$. The
polynomial
$$\init_w(f)= \sum_{u(a_M)+\langle w,M\rangle=w(f)} a_M X^M$$
is called the \emph{$w$-initial form} of $f$. For an ideal $I\subset k[X]$,
the ideal $\init_w I=(\init_w(f)|f\in I)$ generated by all $w$-initial forms
of elements of $I$ is called the \emph{$w$-initial ideal} of $I$.

Now let $w^1=(u(x_1),\ldots,u(x_m))\in\R^m$, $w^2=(v(y_1),\ldots,w(y_n))\in\R^n$.
Note that since $u$ and $v$ take finite values on $x_i$ and $y_j$, the initial
ideals $\init_{w^1} I$ and $\init_{w^2} J$ are monomial free. Denote $k[X,Y]=
k[X_1,\ldots,X_m,Y_1,\ldots,Y_n]$, $w=(w^1,w^2)\in\R^{m+n}$, and let $I\otimes 1$
and $1\otimes J$ be the extensions of the ideals $I$ and $J$ to $k[X,Y]$
respectively. Then, $\init_w(I\otimes 1)=\init_{w^1}(I)\otimes 1$, and similarly
for $J$. Indeed, consider $f=\sum f_ib_i\in I\otimes 1$, where $f_i\in k[X,Y]$,
$b_i\in I$. Write $f$ as a polynomial in $Y$: $f=\sum_N b_N Y^N$, where
$b_N=b_N(X)\in I$, $b_N\ne 0$. Consider the combination
\begin{equation}\label{E:comb1}
\sum_{w^1(b_N)+\langle w^2,N\rangle\text{ is minimal}} \init_{w^1}(b_N) Y^N.
\end{equation}
If it is $0$, then this holds identically in $Y$, and thus $\init_{w^1}(b_N)=0$
for all $b_N$ involved in \eqref{E:comb1}. But this contradicts the assumption
$b_N\ne 0$. It follows that the $w$-initial form of $f$ is a $k[X,Y]$-linear
combination of initial forms of $b_N\in I$.

\emph{Step 3.} Our next claim is that the initial ideal $\init_w(I\otimes 1+
1\otimes J)$ coincides with $\init_{w^1}(I)\otimes 1+1\otimes\init_{w^2}(J)$.
The inclusion $\supseteq$ is clear, so let us prove the inclusion $\subseteq$.
Consider an element
$$h=\sum_i f_ib_i + \sum_j g_jc_j \in I\otimes 1+1\otimes J,$$
where $f_i,g_j\in k[X,Y]$, $b_i\in I$, $c_j\in J$. We can assume that $b_i$
generate the ideal $I$, $c_j$ generate the ideal $J$, the initial forms
$\init_{w^1}(b_i)$ generate the initial ideal $\init_{w^1}I$, and
$\init_{w^2}(c_j)$ generate $\init_{w^2}J$. If the initial forms do not
cancel, that is the combination
$$\sum \init_w(f_i)\init_{w^1}(b_i) + \sum \init_w(g_j)\init_{w^2}(c_j)$$
is not $0$, then it is the $w$-initial form $h_w$ of $h$, and there is nothing
to prove. If this combination is $0$, it follows that
$$\sum \init_w(f_i)\init_{w^1}(b_i),\: \sum \init_w(g_j)\init_{w^2}(c_j)\in
\init_w(I\otimes 1)\cap\init_w(1\otimes J),$$
the last intersection being equal to the product $\init_w(I\otimes 1)\cdot
\init_w(1\otimes J)$ by Corollary~\ref{C:product} below. Thus, using Step 2,
$\sum \init_w(f_i)\init_{w^1}(b_i)=-\sum \init_w(g_j)\init_{w^2}(c_j)$ can
be written as $\sum h_{ij}\init_{w_1}(b_i)\init_{w^2}(c_j)$, $h_{ij}\in k[X,Y]$.
Then, consider
$$h=\sum f_ib_i + \sum g_jc_j \pm \sum h_{ij}b_ic_j =
\sum_i (f_i-\sum_j h_{ij}c_j)b_i + \sum_j (g_j+\sum_i h_{ij}b_i)c_j.$$
The $w$-order of $h$ is fixed, while $w$-orders of expressions in
parenthesis on the right have increased comparing with the $w$-orders of
$f_i$ and $g_j$. In this way we eventually rewrite $h$ in the form where
$w$-initial forms do not cancel, thus, $w$-initial form of $h$ is a combination
of $w$-initial forms of elements of $I$ and $J$.

\emph{Step 4.} Now we show that the ideal $\init_w(I\otimes 1+1\otimes J)$
is monomial free. Suppose that a monomial $X^MY^N$ can be represented as
\begin{equation}\label{E:comb2}
X^MY^N=\sum f_i\init_{w^1}(b_i)+\sum g_j\init_{w^2}(c_j),
\end{equation}
where $f_i$, $g_j\in k[X,Y]$, $b_i\in I$, $c_j\in J$. Since $\init_{w^1}I$
and $\init_{w^2}J$ are monomial free, they have points $x^0\in(\bar{k}^*)^m$,
$y^0\in(\bar{k}^*)^n$ respectively. Substituting the point $(x^0,y^0)\in
(\bar{k}^*)^{m+n}$ to \eqref{E:comb2}, we get a contradiction.

\emph{Step 5.} Finally, we construct a valuation $w$ on $B\bigotimes_k C$. First,
consider a monomial valuation $w'$ on $k[X,Y]$ defined by $w'(X_i)=u(x_i)$,
$i=1,\ldots,m$, $w'(Y_j)=v(y_j)$, $j=1,\ldots,n$. Let $S$ be a semigroup
generated by all the monomials $aX^MY^N$, $a\in k$, $M\in\Z_{\geq 0}^{m}$,
$N\in\Z_{\geq 0}^{n}$. By Step 4 and Bergman's Theorem~\ref{T:Bergman}, $w'$
can be pushed forward to a valuation $w$ on
$$k[X,Y]/(I\otimes 1+1\otimes J)\simeq k[X]/I \bigotimes_k k[Y]/J \simeq
B\bigotimes_k C,$$
and $w$ has all the required properties.
\end{proof}

The following two results were needed for the proof of Lemma~\ref{L:fgtensor}.
We use the notation introduced in Step 2 of the proof of Lemma~\ref{L:fgtensor}
and assume some familiarity of the reader with Gr{\"o}bner bases.

\begin{lemma}\label{L:Groebner}
Let $I\subset k[X]$ be an ideal, and $\{b_1,\ldots,b_s\}$ a Gr{\"o}bner basis of
$I$ with respect to some monomial ordering $\leq$ on $k[X]$. Then $\{b_1,
\ldots,b_s\}$ is a Gr{\"o}bner basis for $I\otimes 1\subset k[X,Y]$ with respect
to any monomial ordering on $k[X,Y]$ that restricts to the ordering $\leq$
on $k[X]$.
\end{lemma}
\begin{proof}
For any $f=\sum f_i(X,Y)b_i(X)$ write $f=\sum_N(\sum a_j(X)b_j(X))Y^N$ as
a polynomial in $Y$. The leading monomial of $f$ is present only in one of the
expressions $(\sum a_jb_j)Y^N$, thus it is divisible by the leading monomial
of one of the $b_j$.
\end{proof}

\begin{corollary}\label{C:product}
Let $I\subset k[X]$ and $J\subset k[Y]$ be ideals. Then
$$(I\otimes 1)\cap(1\otimes J)=(I\otimes 1)\cdot(1\otimes J)$$
in $k[X,Y]$.
\end{corollary}
\begin{proof}
Fix a monomial ordering on $k[X,Y]$ and Gr{\"o}bner bases $\{b_i\}$ for $I$
and $\{c_j\}$ for $J$. Take $f\in(I\otimes 1)\cap(1\otimes J)$. By
Lemma~\ref{L:Groebner}, we conclude that the leading monomial $aX^MY^N$, $a\in k$,
of $f$ is divisible by the leading monomial of one of $b_i$ and of one of $c_j$.
But the leading monomial of each $b_i$ is coprime to the leading monomial of each
$c_j$, thus $aX^MY^N$ is divisible by the leading monomial of some of the
products $b_i(X)c_j(Y)$. This implies the corollary.
\end{proof}

Let us continue the proof of Theorem~\ref{T:tensor}. Now we consider the
general case of the tensor product of $A$-algebras $B$ and $C$, not necessarily
finitely generated over $A$. It is again possible to reduce to the case when
$A=k$ is a field, and this will be assumed in the sequel. Let us recall the
construction of the tensor product. Denote by $B^\bullet$ and $C^\bullet$ the
multiplicative semigroups of the rings $B$ and $C$ respectively, and consider
the direct product of semigroups $S'=B^\bullet\times C^\bullet$. Then the
tensor product $B\bigotimes_k C$ can be identified with the quotient $k[S']/T$
of the semigroup algebra $k[S']$ by the ideal $T$ generated by all relations of
the form
\begin{align}\label{E:basis}
&(ax,y)-a(x,y),\: (x,ay)-a(x,y), \\ \notag
(x'+x'',y)- &(x',y)-(x'',y),\: (x,y'+y'')-(x,y')-(x,y''),
\end{align}
where $a\in k$, $x,x',x''\in B$, $y,y',y''\in C$. Note that these expressions
generate $J$ not only as an ideal of $k[S']$ but also as a vector space
over $k$.

The valuations $u$ and $v$ of $B$ and $C$ are semigroup homomorphisms
$u\colon B^\bullet\to \Ri$, $v\colon C^\bullet\to \Ri$. The rule
$p(x,y)=u(x)+v(x)$ defines a semigroup homomorphism
$p\colon S'\to \Ri$. Furthermore, we extend $p$ to a pseudovaluation on the
semigroup algebra $k[S']$. For $f=\sum a_s s \in k[S']$, where $a_s\in k$,
$s\in S'$, we set
$$p(f)=\min_{s\in S'} (u(a_s)+p(s))$$
(we could write $v(a_s)$ instead of $u(a_s)$). Let $S$ be a subsemigroup of
$k[S']$ generated by all the monomials $a(x,y)$, $a\in k$, $a\ne 0$,
$(x,y)\in S$. It is clear that $p|_S$ is a semigroup homomorphism.
Next we are going to show that there exists a valuation $\bar{w}$ on $k[S']$
that coincides with $p$ on $S$ and can be pushed forward to a valuation $w$ of
the quotient $k[S']/T$. It suffices only to check the conditions of
Theorem~\ref{T:Bergman}, i.e., if $f\in T$ and $s\in S$ is a monomial, then
$p(f-s)\leq p(f)$. Represent $f$ as a $k$-linear combination $\sum a_i r_i$ of
expressions $r_i$ of the form \eqref{E:basis}, and $s$ as $a(x_0,y_0)$,
where $a\in k$, $x_0\in B$, $y_0\in C$. Let $x_1,\ldots,x_m$ be all the
elements of $B$ and $y_1,\ldots,y_n$ all the elements of $C$ that are present
in the monomials of $r_i$. Consider the finitely generated rings
$B'=k[x_0,x_1,\ldots,x_m]$ and $C'=k[y_0,y_1,\ldots,y_n]$ and restrictions
$u'$ and $v'$ to $B'$ and $C'$ respectively. If we represent $B'\bigotimes_k C'$
as a quotient of a semigroup algebra $k[(B')^\bullet\times(C')^\bullet]$,
then $f$ can be considered as an element of the corresponding ideal $T'\subset k[(B')^\bullet\times(C')^\bullet]$. The pseudovaluation $p$ also naturally
restricts to this semigroup algebra. On the other hand, we know by
Lemma~\ref{L:fgtensor} that $u'$ and $v'$ extend simultaneously to
$B'\bigotimes_k C'$, thus, $p$ achieves its minimum on the monomials of $f$
at least twice. This shows that Theorem~\ref{T:Bergman} applies to $p$,
$S$ and $T$. We get a valuation $w$ on
$$k[S']/T\simeq B\bigotimes_k C$$
which is a simultaneous extension of $u$ and $v$. This finishes the second proof
of Theorem~\ref{T:tensor}.

We are ready to describe the construction of universal valued fields. Now,
let $k$ be a ring with a real valuation $\nu$ and $R$ an arbitrary
commutative $k$-algebra with unity. Denote by $\V(R,k)$ the set of all
real ring valuations of $R$ that extend $\nu$.

\begin{theorem}\label{T:ufield}
Given $k$, $R$, and $\nu$, there exist a $k$-algebra $K$ and a valuation $\mu$ on
$K$ such that $\mu|_k=\nu$, and for any $v\in\V(R,k)$ there exists a morphism
$f_v\colon R\to K$ of $k$-algebras such that the valuation $v$ on $R$ is induced
by the valuation $\mu$ on $K$ via the morphism $f_v$.
\end{theorem}
\begin{proof}
Consider a $k$-algebra
$$K=\bigotimes_{v\in\V(R,k)} R,$$
the restricted tensor product over $k$ of the $k$-algebra $R$ with itself, one
copy for each $v\in\V(R,k)$, see \cite[p. 713, Proposition~A6.7b]{Eis}. Let us
show that there exists a valuation $\mu$ on $K$ that is a simultaneous
extension of all the valuations $v$ of $R$. This is a consequence of
Theorem~\ref{T:tensor} and Zorn's lemma. Indeed, for each subset $V\subseteq
\V(R,k)$ we have a natural homomorphism of $k$-algebras
$$K_V=\bigotimes_{v\in V} R \to K,$$
and, for $U\subseteq V$, a natural homomorphism $K_U\to K_V$.
Let $\V$ be the family of all pairs $(V,\mu_V)$, where $V\subseteq\V(R,k)$ and
$\mu_V$ is a valuation on $K_V$ extending all the valuations $v$ of $R$,
$v\in V$. The family $\V$ is nonempty since it contains all the pairs $(\{v\},v)$
for one-element subsets of $\V(R,k)$. It is also ordered by the following order
relation: $(U,\mu_U)\leq (V,\mu_V)$ if and only if $U\subseteq V$ and
$(\mu_V)|_{K_U}=\mu_U$. Then, the conditions of Zorn's lemma are satisfied, thus
$\V$ contains maximal elements. But by Theorem~\ref{T:tensor} such a maximal
element must coincide with $(\V(R,k),\mu)$ for some valuation $\mu$ on $K$.
Now, let $f_v$ be the natural homomorphism
$$R\to \bigotimes_{v\in\V(R,k)} R= K.$$
Then, $\mu$ induces the valuation $v$ on $R$ via $f_v$.
\end{proof}

\begin{remark}
In the conditions of Theorem~\ref{T:ufield}, the valuation $\mu$ on $K$ naturally
extends to $K/\home(\mu)$, to its field of fractions $Q(K/\home(\mu))$, and,
further, to its algebraic closure $\overline{Q(K/\home(\mu))}$. Thus, the
universal $k$-algebra $K$ can be assumed to be a domain or even an algebraically
closed field.
\end{remark}

It would be interesting to give an explicit construction of universal valuation
fields under some reasonable restrictions on the algebra $R$. It may be that such
a universal valued field depends only on the field $k$ and the valuation $\nu$
and works for all $k$-algebras of a certain class. The next proposition gives an
example of such phenomenon.

\begin{proposition}\label{P:ufforfg}
Let $k$ be an algebraically closed trivially valued field of characteristic $0$,
and $k(x)=k(x_1,\ldots,x_n)$ the field of rational functions in $n$ variables
with coefficients in $k$. Let $K=\overline{k(x)}(t^\R)$ be the Hahn series field
with coefficients in the algebraic closure of $k(x)$. Then $K$ serves as a
universal valued field for all finitely generated $k$-algebras $R$ of
dimension $d\leq n$.
\end{proposition}
\begin{proof}
Let $R$ be a finitely generated $k$-algebra of dimension $d\leq n$. If $v$ is a
valuation of $R$ over $k$, then the residue field $k(v)$ of $v$ has transcendence
degree $r\leq d$ over $k$. Thus $k(v)$ and its algebraic closure can be embedded
to $\overline{k(x)}$. On the other hand, by \cite[Theorem~6]{Kap}, the field
$\overline{k(v)}(t^\R)$ is the maximal valued field with the residue field
$\overline{k(v)}$ and the value group $\R$. It follows that the valuation $v$ is
induced by a homomorphism from $R$ to $\overline{k(v)}(t^\R)$. It remains to
note that each field of the form $\overline{k(v)}(t^\R)$ embeds to $K$.
\end{proof}

Proposition~\ref{P:ufforfg} suffers obviously from the lack of explicit description
of the algebraic closure $\overline{k(x)}$ of the field of rational functions
in several variables.

\section{Weak universality of the field of Puiseux series}\label{S:wufield}

As in the Introduction, we denote by $K=\bar{k}((t^\Q))=
\bigcup_{N\geq 1}\bar{k}((t^{1/N}))$ the field of Puiseux series with
coefficients in an algebraically closed field $\bar{k}$, and by $K'=
\bar{k}((t^\R))$ the field of Hahn series, i.e., the set of all
formal sums
$$a(t)=\sum_{m\in \R} a_m t^m,$$
where $a_m\in\bar{k}$ and $\{m\,|\,a_m\ne 0\}$ is a well-ordered subset of $\R$.
Let $\mu$ be the natural valuation of $K$ ($K'$).
The ring of Puiseux (respectively Hahn) series $\bar{k}[[t^\Q]]$
(resp. $\bar{k}[[t^\R]]$) is the subring of $K$ (resp. $K'$) consisting of the
series of nonnegative valuation. It is known that if $\bar{k}$ has
characteristic $0$, then the field $K$ of Puiseux series is algebraically closed
(\cite{Cohn}); if $\bar{k}=\C$ is the field of complex numbers, this is the
classical Newton-Puiseux Theorem. The field $K'$ of Hahn series
is always algebraically closed (\cite[Theorem~1]{MacLane}).

\begin{theorem}\label{T:wucomplete}
Let $R$ be a complete Noetherian local ring containing a field $k$ isomorphic
to the residue field of $R$. Let $v$ be a local valuation of $R$ and $x_1,\ldots,
x_n$ a finite collection of elements of the maximal ideal of $R$. Then there
exists a homomorphism $f\colon R\to \mathcal{O}'=\bar{k}[[t^\R]]$ to the ring of
Hahn series with coefficients in $\bar{k}$ such that for all
$i=1,\ldots,n$, $v(x_i)=\mu(f(x_i))$. Moreover, if the characteristic of $k$ is
$0$, the elements $x_1,\ldots,x_n$ analytically generate $R$, and $v(x_i)\in\Q$
for all $i=1,\ldots,n$, then there exists a homomorphism $f\colon R\to
\mathcal{O}=\bar{k}[[t^\Q]]$ to the ring of Puiseux series with coefficients in
$\bar{k}$ with the same property.
\end{theorem}
\begin{proof}
\emph{Step 1.} First we reduce to the case when $k=\bar{k}$ is algebraically
closed, $R$ is a domain, and do some other preliminary reductions. Consider the
tensor product $\overline{R}=R\bigotimes_k \bar{k}$. Note that $\overline{R}$ is
integral over $R$, thus, there exists a prime ideal $\mathfrak{q}\subset
\overline{R}$ such that $\mathfrak{q}\cap R=\mathfrak{p}$, where $\mathfrak{p}=
\home(v)$. The quotient $\overline{R}/\mathfrak{q}$ remains integral over
$R/\mathfrak{p}$. Hence the local valuation $v$ of $R/\mathfrak{p}$ extends to a
valuation $\bar{v}$ of $\overline{R}/\mathfrak{q}$, and, as follows from the
method of Newton polygon (see \cite[Chapitre VI, \S 4, exercise 11]{Bour}),
$\bar{v}$ is nonnegative on $\overline{R}/\mathfrak{q}$. Passing again to the
quotient by $\home(\bar{v})$, if necessary, we can assume that $\home(\bar{v})=
\{0\}$. Let $\mathfrak{n}$ be the prime ideal of $\overline{R}/\mathfrak{q}$
where $\bar{v}$ is positive. Since $\mathfrak{n}\cap R/\mathfrak{p} =
\mathfrak{m}$ is the maximal ideal of $R/\mathfrak{p}$, the ideal $\mathfrak{n}$
is also maximal. Then, consider the localization
$(\overline{R}/\mathfrak{q})_\mathfrak{n}$ and its completion $R'$ with respect
to $\mathfrak{n}$. The valuation $\bar{v}$ extends canonically to a local
valuation $v'$ of $R'$ (see, e.g., \cite[Lemma~5.16]{PPS}). Finally, passing to
the quotient of $R'$ by $\home(v')$, we can assume that $R'$ is a domain
and $\home(v')=\{0\}$.

Recall that the \emph{rational rank} of a valuation $v$ on $R$ is the dimension
over $\Q$ of the $\Q$-vector subspace of $\R$ generated by the image of $v$.
Let $r'$ be the rational rank of $v$ and $d$ the Krull dimension of $R$. It
follows from the Abhyankar inequality (\cite[Th{\'e}or{\`e}me~9.2]{Vaq}) that
$r'\leq d$. If $r$ is the dimension of the $\Q$-vector subspace of $\R$
generated by $v(x_1),\ldots,v(x_n)$, then we have $r\leq r'\leq d$. By the Cohen
structure theorem \cite[Theorem~7.7]{Eis}, the ring $R$ is a quotient of the ring
$k[[X_1,\ldots,X_N]]$ of formal power series. Enlarging the finite set
$\{x_1,\ldots,x_n\}$, if necessary, we can assume that the elements
$x_1,\ldots,x_n$ analytically generate $R$, i.e., $N=n$ and a surjection from
$k[[X_1,\ldots,X_n]]$ to $R$ can be chosen so that $X_i\mapsto x_i$, $i=1,\ldots,
n$. Clearly, there is no loss of generality if we also assume that none of
$x_i$ is $0$.

In the reduction steps described above, we performed the following actions
with the ring $R$: we took quotients, considered the tensor product with
$\bar{k}$, localized at a maximal ideal, and took a completion. These actions
could only decrease the dimension of $R$. Thus, if the elements $x_1,\ldots,x_n$
analytically generate the ring $R$ at the beginning, their images in $R'$ form a
system of parameters for $R'$. Then, a system of analytical generators for $R'$
can be obtained from $x_1,\ldots,x_n$ by adjoining only a finite number of
elements $x'_1,\ldots,x'_l\in R'$ \emph{integral} over the subring analytically
generated by $x_1,\ldots,x_n$. It again follows from the method of Newton polygon
that if $v$ takes only rational values on $x_i$, then it also takes only rational
values on $x'_j$, $j=1,\ldots,l$. This shows that the reduction steps are
applicable also in the ``moreover'' part of the theorem, i.e., we can assume
that $R$ is a local complete Noetherian domain, $k$ is an algebraically closed
subfield of $R$ of characteristic $0$ and isomorphic to the residue field
of $R$, $x_1,\ldots,x_n$ analytically generate $R$, and $v$ takes only rational
values on $x_i$.

\emph{Step 2.} Now, we reduce to the case $r=d$, where $r$ is the dimension of
the $\Q$-vector subspace of $\R$ generated by $v(x_1),\ldots,v(x_n)$, $d=\dim R$.
Suppose that $r<d$. Then, represent $R$ as a quotient $k[[X_1,\ldots,X_n]]/I$,
where $I$ is a prime ideal of $k[[X_1,\ldots,X_n]]$. The valuation $v$ induces a
valuation of $k[[X_1,\ldots,X_n]]$; by abuse of notation, we denote this induced
valuation also by $v$. Let $w$ be the vector $(w_1,\ldots,w_n)=(v(x_1),\ldots,
v(x_n))\in \R^n$. The proof of the following elementary lemma is left to the
reader.

\begin{lemma}
Let $w=(w_1,\ldots,w_n)\in \R^n$ be a vector such that the real numbers
$w_1,\ldots,w_n$ generate a $\Q$-vector subspace of $\R$ of dimension $r$ over
$\Q$. Then there exists a unique minimal rational vector subspace $L_\Q(w)$ of
$\R^n$ (\emph{rational} means that $L_\Q(w)$ is defined by linear equations with
rational coefficients) such that $w\in L_\Q(w)$. The dimension of $L_\Q(w)$
equals $r$.
\end{lemma}

By general properties of valuations, the initial ideal $\init_w I$ is monomial
free. Consider the local Gr{\"o}bner fan of the ideal $I$ (\cite{BT}).
There exists a rational polyhedral cone $\sigma\subset\R^n$ such that $w$ is
contained in the relative interior $\mathring{\sigma}$ of $\sigma$ and for all
$w'\in\mathring{\sigma}$, $\init_{w'} I=\init_w I$. The intersection
$\sigma\cap L_\Q(w)$ is also a rational polyhedral cone. Let us fix a vector
$w'\in\mathring{\sigma}\cap L_\Q(w)$ with positive \emph{integral} coordinates.
Note that the initial ideal $\init_{w'}I=\init_w I$ is generated by
$w$-homogeneous polynomials. Then, let $J=\init_{w'}I\cap k[x_1,\ldots,x_n]$ be
the corresponding ideal of the polynomial ring. We have
$$k[[x_1,\ldots,x_n]]/\init_{w'}I \simeq (k[x_1,\ldots,x_n]/J)\sphat\,,$$
where the completion is taken with respect to the maximal ideal
$(x_1,\ldots,x_n)$. Thus, by \cite[Corollary~11.19]{AM},
$$\dim k[[x_1,\ldots,x_n]]/\init_{w'}I=\dim k[x_1,\ldots,x_n]/J.$$
But by the standard flat degeneration argument (see, e.g.,
\cite[Lemma~11.10]{PPS}), $\dim k[[x_1,\ldots,x_n]]/\init_{w'}I=\dim R$, and,
since $I$ is supposed to be prime, all minimal associated primes of $\init_{w'}I$
have the same depth $d=\dim R$. We conclude that $\dim k[x_1,\ldots,x_n]/J=d$,
and, moreover, all minimal associated primes of $J$ also have depth $d$.

Since the ideal $J$ is monomial free, it has a point
$x^0\in (k^*)^n$. Choose an irreducible component $Z$ of the zero set of $J$
such that $x^0\in Z$. Let $\gamma_1,\ldots,\gamma_r$ be a basis of the
$\Q$-vector subspace of $\R$ generated by $w_1,\ldots,w_n$. Write
$$w_i=\sum_{j=1}^r w_{ij}\gamma_j,$$
where $i=1,\ldots,n$, $w_{ij}\in\Q$. Rescaling $\gamma_j$ if necessary, we can
even suppose that $w_{ij}$ are integral. By our assumptions, the rank of the
matrix $(w_{ij})_{i=1,j=1}^{n,m}$ is $r$. Since the ideal $J$ is generated by
$w$-homogeneous polynomials, together with the point $x^0=
(x_{1}^{0},\ldots,x_{n}^{0})$ the variety $Z$ contains also all the points
$$(t_{1}^{w_{11}}\cdots t_{r}^{w_{1r}}x_{1}^{0},\ldots,
t_{1}^{w_{n1}}\cdots t_{r}^{w_{nr}}x_{n}^{0}),\quad (t_1,\ldots,t_r)\in k^r.$$
Thus we can make some $r$ of the coordinates of $x^0$, say, $x_{1}^{0},\ldots,
x_{r}^{0}$, to be equal to $1$. The dimension of the linear affine space $H$:
$x_1=\cdots=x_r=1$ is $n-r$, and the dimension of $Z$ is $d>r$. Hence
$\dim H\cap Z\geq 1$, and there exists at least one point $y^0\ne x^0$,
$y^0\in H\cap Z$. Take a polynomial $\tilde{f}(x_{r+1},\ldots,x_n)$ such that
$\tilde{f}(x_{r+1}^{0},\ldots,x_{n}^{0})=0$, $\tilde{f}(y_{r+1}^{0},\ldots,
x_{n}^{0})\ne 0$. Homogenizing $\tilde{f}$ with respect to the weight vector $w'$
we get a $w'$-homogeneous polynomial $f$ such that $f(x^0)=0$, $f(y^0)\ne 0$.
By construction, $f$ is also $w$-homogeneous, not a monomial, and not contained
in any of the minimal associated primes of $\init_w I$.

\begin{lemma}
Let $I$, $w$, and $f$ be as above. Then, in the ring $k[[x_1,\ldots,x_n]]$,
$$\init_w(I+(f))=\init_w I + (f).$$
\end{lemma}
\begin{proof}
Consider an arbitrary $fg+h\in (f)+I$, $h\in I$, $g\in k[[x_1,\ldots,x_n]]$. If
$f\cdot\init_w g+\init_w h\ne 0$, then it is the $w$-initial form of $fg+h$ and
is contained in $\init_w I+(f)$. Assume that $f\cdot\init_w g+\init_w h=0$.
Since $f$ is not a zero divisor modulo $\init_w I$, we have $\init_w g\in
\init_w I$. Let $\tilde{g}\in I$ be such that $\init_w\tilde{g}=\init_w g$.
Then $fg+h$ has the same initial form as
$$fg\pm f\tilde{g}+h=fg'+h',$$
where $h'\in I$ and $g'$ has $w$-order strictly greater than $g$. Repeating
this argument we come to the situation when the initial forms of $fg'$ and $h'$
do not cancel, and thus $\init_w(fg+h)\in \init_w I+(f)$. This proves the
inclusion
$$\init_w(I+(f))\subseteq\init_w I + (f).$$
The inverse inclusion is clear.
\end{proof}

By construction, the ideal $\init_w(I+(f))=\init_w I+(f)$ is monomial free. By
Bergman's Theorem~\ref{T:Bergman}, the monomial valuation $v'$ on
$k[[x_1,\ldots,x_n]]$ defined by $v'(x_i)=w_i$, $i=1,\ldots,n$, can be
transformed to a valuation $\bar{v}$ such that $\bar{v}(x_i)=w_i$, $i=1,\ldots,n$,
and $\bar{v}|_{I+(f)}=+\infty$. Therefore we can pass to the ring $R'=R/(f)$,
$\dim R'=d-1$.

\emph{Step 3.} It remains to consider the case $d=\dim R=r$. The rational rank
$r'$ of the valuation $v$ is not greater than $d$ by Abhyankar inequality, so
in this case $r=r'$. Choose a system of parameters $y_1,\ldots,y_d$ for $R$
such that $v_1=v(y_1)$, $\ldots$, $v_d=v(y_d)$ are linearly independent over
$\Q$. The ring $R$ is integral over $k[[y_1,\ldots,y_d]]$.

Embed $k[[y_1,\ldots,y_d]]$ to the ring of Hahn series $\mathcal{O}'
\subset k((t^\R))$ by sending $y_i$ to $t^{v_i}$, $i=1,\ldots,d$. The natural
valuation $\mu$ of $k((t^\R))$ is an extension of $v$ from $k[[y_1,\ldots,y_d]]$
via this embedding. By \cite[Chapter XII, \S 3]{Lang}, every extension of $v$
from $k[[y_1,\ldots,y_d]]$ to $R$ can be induced by an embedding of $R$ to
$$\overline{Q(k[[y_1,\ldots,y_d]])_v},$$
i.e., to the algebraic closure of the completion with respect to $v$ of the field
of fractions of $k[[y_1,\ldots,y_d]]$. Since $k((t^\R))$ is algebraically closed
and complete, we conclude that there exists an embedding of $R$ to $k((t^\R))$
such that $\mu$ induces $v$. Moreover, the method of Newton polygon shows that
the image of $R$ is contained in $\mathcal{O}'$.

Finally, assume that $v$ takes only rational values on $x_i$ and the field $k$
has characteristic $0$. This implies, in particular, that $r=d=1$. This time
we embed $k[[y]]=k[[y_1]]$ to the ring of Puiseux series $\mathcal{O}$ by
sending $y$ to $t^{v_1}$. As above, we get an embedding of $R$ to $k((t^\R))$
inducing the valuation $v$, but, since the field $k((t^\Q))$ is also
algebraically closed, the image is contained in $k((t^\Q))$ and, by the method
of Newton polygon, in $\mathcal{O}$.
\end{proof}

The following theorem shows weak universality of the field of Hahn series with
respect to valuations on finitely generated algebras.

\begin{theorem}
Let $R$ be a finitely generated algebra over a field $k$. Let $v$ be a
valuation of $R$ and $x_1,\ldots,x_n$ a finite collection of elements of $R$.
Then there exists a homomorphism $f\colon R\to \bar{k}((t^\R))$ to the field of
Hahn series with coefficients in $\bar{k}$ such that for all
$i=1,\ldots,n$, $v(x_i)=\mu(f(x_i))$. Moreover, if the characteristic of $k$ is
$0$, the elements $x_1,\ldots,x_n$ generate $R$ over $k$, and $v(x_i)\in\Q$
for all $i=1,\ldots,n$, then there exists a homomorphism $f\colon R\to
\bar{k}((t^\Q))$ to the field of Puiseux series with coefficients in $\bar{k}$
with the same property.
\end{theorem}
\begin{proof}
This theorem follows from the results on lifting points in tropical varieties,
see, e.g., \cite{JMM}. Alternatively, it can be proven by an argument parallel
to the proof of Theorem~\ref{T:wucomplete}.
\end{proof}

\section{Lifting points in local tropical varieties}\label{S:liftpoint}

In this section we recall briefly the definition of local tropicalization
following \cite{PPS} and deduce the lifting points property of local tropical
varieties from Theorem~\ref{T:wucomplete}. Let $M$ be a finitely generated
free abelian group and $M_\R$ the vector space $M\bigotimes_Z \R$. For
rational polyhedral cones $\check\sigma$ in $M_\R$, we consider additive
semigroups $\Gamma$ of the form $\check\sigma\cap M$ or, more generally,
finitely generated subsemigroups (with neutral element) of $\check\sigma\cap M$
such that $\sat(\Gamma)$, the saturation of $\Gamma$ (see
\cite[Definition~2.10]{PPS}), equals $\check\sigma\cap M$. Each semigroup
homomorphism from $\Gamma$ to $\R$ lifts uniquely to a group homomorphism from $M$
to $\R$. Thus we can identify the group $\Hom_{sg}(\Gamma,\R)$ of all semigroup
homomorphisms from $\Gamma$ to $\R$ with the dual space of $M_\R$; we denote this
dual space by $N_\R$, and by $N$ the dual lattice of $M$. The semigroup
homomorphisms from $\Gamma$ to the extended real line $\Ri=\R\cup\{+\infty\}$
live in a certain partial compactification of $N_\R$ that is denoted by
$L(\sigma,N)$ in \cite[Section~4]{PPS} and called the \emph{linear variety}
corresponding to $\sigma$ and $N$; here $\sigma$ is the dual cone of
$\check\sigma$. The cone $\sigma$ corresponds to nonnegative homomorphisms from
$\Gamma$ to $\R$; the nonnegative homomorphisms from $\Gamma$ to $\Ri$ form a
certain subspace of $L(\sigma,N)$ denoted $\overline{\sigma}$. We denote by
$\overline{\sigma}^\circ$ the interior of $\overline{\sigma}^\circ$.

Now, let $R$ be a local ring with the maximal ideal $\mathfrak{m}$. Let
$\V_{\geq 0}(R)$ be the set of all nonnegative real valuations of $R$ and
$\V_{>0}(R)$ the set of all local valuations of $R$, i.e., valuations that are
nonnegative on $R$ and positive on the maximal ideal $\mathfrak{m}$. Let
$\gamma\colon\Gamma\to R$ be a homomorphism from $\Gamma$ to the multiplicative
semigroup of $R$ and assume that none of the elements of
$\gamma^{-1}(\mathfrak{m})$ is invertible in $\Gamma$. Then, for each
$v\in\V_{\geq 0}$ (resp. $v\in\V_{>0}$), the composition $v\circ\gamma$ is an
element of $\overline{\sigma}$ (resp. $\overline{\sigma}^\circ$). In this way we
get a map
$$\trop\colon \V_{\geq 0}(R)\to \overline{\sigma}\quad
(\text{resp. } \trop\colon \V_{>0}(R)\to \overline{\sigma}^\circ),$$
which we call the \emph{tropicalization map}.

\begin{definition}
The \emph{local nonnegative tropicalization} of the morphism $\gamma$, denoted
$\nntrop(\gamma)$, is the image of $\V_{\geq 0}(R)$ in $\overline{\sigma}$ under
the tropicalization map $\trop$. The \emph{local positive tropicalization},
denoted $\ptrop(\gamma)$, is the closure in $\overline{\sigma}^\circ$ of the
image of $\V_{>0}(R)$ under the tropicalization map.
\end{definition}

Next we restrict to local rings $R$ of special type. Suppose that the cone
$\check\sigma$ is \emph{strictly convex}, in particular, the only invertible
element of $\Gamma$ is $0$. Let $k$ be a field and consider all possible
formal series
$$\sum_{m\in\Gamma} a_m \chi^m,$$
where $a_m\in k$. Any two such series can be added and multiplied in a natural
way, in particular, $\chi^m\cdot\chi^{m'}=\chi^{m+m'}$. These series form a
complete Noetherian local ring $k[[\Gamma]]$ called the \emph{ring of formal
power series} over $\Gamma$, see \cite[Section~8]{PPS}. There is a natural
semigroup homomorphism $\Gamma\to k[[\Gamma]]$ defined by $\Gamma\ni
m\mapsto \chi^m$, the elements $\chi^m\in k[[\Gamma]]$ being called the
\emph{monomials}. If $I\subset k[[\Gamma]]$ is an ideal, let $\gamma$ be the
composition of the natural maps
$$\Gamma\to k[[\Gamma]]\to R=k[[\Gamma]]/I.$$
Now, we can consider the nonnegative (resp. positive) tropicalization
$\nntrop(\gamma)$ (resp. $\ptrop(\gamma)$) of $\gamma$. In the situation just
described, we denote also $\nntrop(\gamma)=\nntrop(I)$ ($\ptrop(\gamma)=\ptrop(I)$)
and call it the \emph{local nonnegative} (resp. \emph{positive})
\emph{tropicalization of the ideal} $I$.

It is proven in \cite[Theorem~11.9]{PPS} that both nonnegative $\nntrop(I)
\subseteq\overline{\sigma}$ and positive $\ptrop(I)\subseteq
\overline{\sigma}^\circ$ tropicalizations are rational PL conical subspaces of
real dimension $d$ equal to the Krull dimension of $R$. This means, in particular,
that the part of $\nntrop(I)$ contained in $N_\R$ is a support of a rational
(with respect to the lattice $N$) polyhedral fan. The nonnegative tropicalization
$\nntrop(I)$ is stratified by the positive tropicalizations $\ptrop(I_\tau)$
of certain truncations of the ring $R$, see \cite[Section~12, in particular
Lemma~12.9]{PPS} for the details.

\begin{theorem}[Lifting points lemma for local tropical varieties]
\label{T:liftpoint}
Let $I\subset k[[\Gamma]]$ be an ideal. If $u\in\ptrop(I)$, then there exists a
homomorphism $f\colon R=k[[\Gamma]]/I\to \bar{k}[[t^\R]]$ to the ring of
Hahn series with coefficients in $\bar{k}$ such that
$u=\trop(\mu\circ f)$, where $\mu$ is the natural valuation of $\bar{k}[[t^\R]]$.
Moreover, if the field $k$ has characteristic $0$ and the point $u$ is rational,
then there exists a homomorphism $f\colon R\to \bar{k}[[t^\Q]]$ to the ring of
Puiseux series with the same property.
\end{theorem}
\begin{proof}
By \cite[Theorem~11.2]{PPS}, the image of $\V_{>0}(R)$ in
$\overline{\sigma}^\circ$ is closed under the tropicalization map, hence there
exists $v\in\V_{>0}(R)$ such that $u=\trop(v)$. If $u\in \ptrop(I)\subseteq
\overline{\sigma}^\circ$ belongs to some stratum of $\overline{\sigma}$ at
infinity, then $v$ is induced by a valuation on a quotient $R/J$ of the ring
$R$, where $R/J$ also has the form $k[[\Gamma']]$ for some semigroup $\Gamma'$,
see \cite[the end of Section~8]{PPS}. Therefore it suffices to consider the case
$u\in N_\R$. Choose monomials $x_1,\ldots,x_n\in\Gamma$ which generate the
semigroup $\Gamma$. A semigroup homomorphism from $\Gamma$ to $\R$ is uniquely
determined by the images of $x_1,\ldots,x_n$. By Theorem~\ref{T:wucomplete} there
exists a homomorphism $f\colon R\to \bar{k}[[t^\R]]$ such that $\mu(f(x_i))=
u(x_i)$ for all $i=1,\ldots,n$. If the characteristic of $k$ is $0$ and $u$ is
rational, a homomorphism $f$ can be chosen so that the target ring is
$\bar{k}[[t^\Q]]$. This $f$ is the required homomorphism.
\end{proof}

\begin{corollary}
Let $I\subset k[[\Gamma]]$ be an ideal. The following three definitions of the
local positive tropicalization of $I$ are equivalent:
\begin{itemize}
\item[1)] $\ptrop(I)$ is the image of $\V_{>0}(k[[\Gamma]]/I)$ under the
tropicalization map;
\item[2)] $\ptrop(I)$ is the set of all $w\in\overline{\sigma}^\circ$ such that
the initial ideal $\init_w I$ is monomial free;
\item[3)] $\ptrop(I)$ is the set $\{\trop(\mu\circ f)\,|\,f\colon R\to
\bar{k}[[t^\R]]\}$, where $f$ runs over all local homomorphisms from
$R=k[[\Gamma]]/I$ to $\bar{k}[[t^\R]]$.
\end{itemize}
If the field $k$ has characteristic $0$, then $\ptrop(I)$ can also be described
as
\begin{itemize}
\item[$3'$)] the closure of the set $\{\trop(\mu\circ f)\,|\,f\colon R\to
\bar{k}[[t^\Q]]\}$ in $\overline{\sigma}^\circ$, where $f$ runs over all local
homomorphisms from $R$ to $\bar{k}[[t^\Q]]$.
\end{itemize}
\end{corollary}
\begin{proof}
Equivalence of 1) and 2) is proven in \cite[Theorem~11.2]{PPS}. Equivalence of
1) and 3), and, if the characteristic of $k$ is $0$, of 1) and $3'$), follows
from Theorem~\ref{T:liftpoint} and from the fact that the local positive
tropicalization defined by 1) is a \emph{rational} conical PL space,
\cite[Theorem~11.9]{PPS}.
\end{proof}

redpajama_set_name: RedPajamaArXiv