IBM向往的新世界-5页文档资料

世界五百强工作准则：沃森（IBM公司）职业信条IBM实实在在的企业管理宝典现代企业员工培训的经典教程职业经理MBA提升管理技巧的最佳读物如同每一位有野心的企业家一样，老托马斯·沃森也希望他的公司财源滚滚，然而，他最希望的是，借助这些“行为准则”把IBM公司培养成美国最强大的企业，把IBM的员工培养成为美国最优秀的员工。

序无论你是管理者还是普通职员，请谨记下面这段话：竭尽全力做好每一件事，尊重所有的人，穿着整洁，诚实率直，坦诚公正，永远保持乐观向上的积极态度。

此外，最为重要的是，忠心耿耿。

——托马斯·约翰·沃森（Thomas J. Watson）（IBM创始人）出版前言-------------------------------------------------------------------------------- 作者：[美]卡罗尔·汤普森IBM (Internation Business Machines Corp.) 的创始人，全球最成功的商人之一的托马斯·约翰·沃森（Thomas J. Watson）出生在纽约州北部一个普通的农民家庭。

沃森一家经济条件并不富裕，但是品德教育却异常严格。

沃森的父亲要求他的孩子们一定要尊重所有人，穿着整洁、坦率正直、努力做好每一件事，始终保持一种乐观的积极生活态度。

此外，最重要的一点是：忠诚。

在19世纪的美国，这是一种十分普遍的家庭教育方式，几乎所有的家庭都会对子女提出类似的要求，希望他们在童年时就坚守这些原则，以便在成人之后可以从中受益。

尽管大多数父亲都认为，这些家庭教育至关重要，但是，许多人并不能完全遵守这些原则，甚至慢慢地将它们抛之脑后，直至完全忘记。

然而，托马斯·沃森，这个童年时并不怎么起眼的孩子，却严格要求自己，始终一丝不苟地恪守着这些教义，他还发誓：应该不惜一切代价地捍卫这些原则，应该不断地向他人输送这些教义，应该在自己全部的职业生涯中，尽职尽责地贯彻执行。

人类五年内进入认知时代钱颖一对话IBM女掌门罗睿兰（Ginni Rometty）：【编者按】2016年10月13日，清华经管学院顾问委员会委员、IBM公司董事长、总裁及首席执行官罗睿兰（Ginni Rometty）女士在清华经管学院伟伦楼报告厅向清华学子发表了主题为“转型，奔向认知商务时代”的演讲，随后与清华经管学院院长、《清华管理评论》主编钱颖一教授展开了精彩对话。

罗睿兰女士认为认知计算和区块链技术在未来几年将给人类社会带来改变。

9钱颖一对话IBM女掌门罗睿兰（Ginni Rometty）：人类五年内进入认知时代我想跟大家谈三件事：数字转型，这是当前所处的阶段；认知计算，这是下一个发展的时代；还有我们IBM和中国的合作。

我先讲一下数字转型。

我到世界各地拜访商业领袖的时候，我问他们，你们谁想要做一个数字化的公司，或者你的公司已经是数字化的公司？这些商业领袖，包括中国企业家，无不举手。

实际上IBM也在以数字的方式再造自己，我们在云端、大数据和移动等方面做了很多工作。

但是，数字化这一理念在不断地改变，不断地创新。

我经常提醒自己和客户，当所有公司都实现了数字化转型后，公司还有什么差异化的优势呢？当我们把数字化公司加上数字智能，又会出现哪些机会？我认为，接下来我们将步入认知的时代。

我之所以用“时代”这个词，是我相信这一趋势在今后数十年的时间里会不断地深入发展，业务和技术将会有更好的结合。

大家天天都听到大数据这个词，实际上世界上90%的数据都是在最近两年才生成的，而且其中八成是非结构化的数据，也就是视频、图象、图片、推文等这样的信息。

以前，电脑可以存储这些信息，但不能理解它们。

现在有了认知计算，无需编程，电脑就能自我学习，自我理解、消化这些非结构化的信息，就像人脑一样。

你可以用自然语言和这些数据进行互动，可以培训这些系统了解医疗、卫生、金融、教育、时尚等各领域的知识。

比如我们的Watson系统，它是以IBM公司创始人的名字命名的，我们花了几年的时间开发，我们培训Watson，让它去了解肿瘤学——如何治疗癌症。

《美丽新世界》主要情节在奥尔德斯·赫胥黎的科幻小说《美丽新世界》中，描述了一个由科技与意识控制统治的未来社会。

故事发生在遥远的2540年，这个世界已经完全由高度工业化的国家统治，人类的生育方式、思维方式以及道德观念等方面都发生了极大的扭曲和变异。

本文将对《美丽新世界》的主要情节进行叙述。

第一章：遥远的未来在《美丽新世界》的开篇，读者被带入到了遥远的未来。

赫胥黎通过描写伦敦中央孵化中心的场景，展示了一个完全工业化、强调等级制度的社会。

在这个世界中，人们通过体外受精和胚胎培育技术来生育后代，所有的人都被分为不同的等级，从Alpha 到Epsilon等不同社会地位的等级。

第二章：稳定剂和麻痹剂在这个近乎完美的社会中，一切都被科技和药物控制。

稳定剂和麻痹剂是人们日常生活中不可或缺的一部分。

稳定剂用于维持人们满足的状态，保持他们的情绪稳定和幸福感。

而麻痹剂则用于消除任何痛苦和矛盾，使人们不再思考和质疑。

第三章：中心人物伯纳德故事的主要角色之一是伯纳德·马克思。

伯纳德是一个身体略小、个性内向的Alpha等级，与其他人不同，伯纳德对这个世界的不公和无趣感到不满。

他偏好孤独，并与集体感和消费主义的社会格格不入。

然而，他又与一位女子莉莉与亲近，并存有复杂的情感。

第四章：莉莉与约翰莉莉也是一个特殊的女性，她在社交方面非常受欢迎，但她与伯纳德的不同之处在于她并没有对这个社会产生怀疑之情。

莉莉带伯纳德参观了一个野蛮人社区，并在那里遇见了一个异类——约翰。

约翰是一个胚胎在野蛮人社区出生的男孩，他在这个社会中长大，对这个全新的世界产生了深深的困惑和憧憬。

第五章：约翰的挣扎随着约翰的成长，他对于这个与伦敦社会完全不同的野蛮人社区展开了深入了解。

他沉浸于诗歌和文学中，迷恋着一个他称为“伊甸园”的理想世界。

然而，当他回到伦敦时，他意识到自己的理想与这个社会格格不入，并因此陷入了沉思和痛苦之中。

第六章：社会崩溃与结局约翰的异端行为引起了伦敦社会的关注和困扰。

BRAVE NEW WORLD 美丽新世界Aldous HuxleyPlot Overview 情节一览The novel opens in the Central London Hatching and Conditioning Centre, where the Director of the Hatchery and one of his assistants, Henry Foster, are giving a tour to a group of boys. The boys learn about the Bokanovsky and Podsnap Processes that allow the Hatchery to produce thousands of nearly identical human embryos. During the gestation period the embryos travel in bottles along a conveyor belt through a factorylike building, and are conditioned to belong to one of five castes: Alpha, Beta, Gamma, Delta, or Epsilon. The Alpha embryos are destined to become the leaders and thinkers of the World State. Each of the succeeding castes is conditioned to be slightly less physically and intellectually impressive. The Epsilons, stunted and stupefied by oxygen deprivation and chemical treatments, are destined to perform menial labor. Lenina Crowne, an employee at the factory, describes to the boys how she vaccinates embryos destined for tropical climates.The Director then leads the boys to the Nursery, where they observe a group of Delta infants being reprogrammed to dislike books and flowers. The Director explains that this conditioning helps to make Deltas docile and eager consumers. He then tells the b oys about the “hypnopaedic” (sleep-teaching) methods used to teach children the morals of the World State. In a room where older children are napping, a whispering voice is heard repeating a lesson in “Elementary Class Consciousness.”Outside, the Director shows the boys hundreds of naked children engaged in sexual play and games like “Centrifugal Bumble-puppy.” Mustapha Mond, one of the ten World Controllers, introduces himself to the boys and begins to explain the history of the World State, focusing on the State’s successful efforts to remove strong emotions, desires, and human relationships from society. Meanwhile, inside the Hatchery, Lenina chats in the bathroom with Fanny Crowne about her relationship with Henry Foster. Fanny chides Lenina for going out with Henry almost exclusively for four months, and Lenina admits she is attracted to the strange, somewhat funny-looking Bernard Marx. In another part of the Hatchery, Bernard is enraged when he overhears a conversation between Henry and the Assistant Predestinator about “having” Lenina.After work, Lenina tells Bernard that she would be happy to accompany himon the trip to the Savage Reservation in New Mexico to which he had invited her. Bernard, overjoyed but embarrassed, flies a helicopter to meet a friend of his, Helmholtz Watson. He and Helmholtz discuss their dissatisfaction with the World State. Bernard is primarily disgruntled because he is too small and weak for his caste; Helmholtz is unhappy because he is too intelligent for his job writing hypnopaedic phrases. In the next few days, Bernard asks his superior, the Director, for permission to visit the Reservation. The Director launches into a story about a visit to the Reservation he had made with a woman twenty years earlier. During a storm, he tells Bernard, the woman was lost and never recovered. Finally, he gives Bernard the permit, and Bernard and Lenina depart for the Reservation, where they get another permit from the Warden. Before heading into the Reservation, Bernard calls Helmholtz a nd learns that the Director has grown weary of what he sees as Bernard’s difficult and unsocial behavior and is planning to exile Bernard to Iceland when he returns. Bernard is angry and distraught, but decides to head into the Reservation anyway.On the Reservation, Lenina and Bernard are shocked to see its aged and ill residents; no one in the World State has visible signs of aging. They witness a religious ritual in which a young man is whipped, and find it abhorrent. After the ritual they meet John, a fair-skinned young man who is isolated from the rest of the village. John tells Bernard about his childhood as the son of a woman named Linda who was rescued by the villagers some twenty years ago. Bernard realizes that Linda is almost certainly the woman mentioned by the Director. Talking to John, he learns that Linda was ostracized because of her willingness to sleep with all the men in the village, and that as a result John was raised in isolation from the rest of the village. John explains that he learned to read using a book called The Chemical and Bacteriological Conditioning of the Embryo and The Complete Works of Shakespeare, the latter given to Linda by one of her lovers, Popé. John tells Bernard that he is eager to see the “Other Place”—the “brave new world” that his mother has told him so much about. Bernard invites him to return to the World State with him. John agrees but insists that Linda be allowed to come as well.While Lenina, disgusted with the Reservation, takes enough soma to knock her out for eighteen hours, Bernard flies to Santa Fe where he calls Mustapha Mond and receives permission to bring John and Linda back to the World State. Meanwhile, John breaks into the house where Lenina is lying intoxicated and unconscious, and barely suppresses his desire to touch her. Bernard, Lenina, John, and Linda fly to the World State, where the Director is waiting to exile Bernard in front of his Alpha coworkers. But Bernard turns the tables by introducing John and Linda. The shame of being a “father”—the very word makes the onlookers laugh nervously—causes the Director to resign, leaving Bernard free to remain in London.John becomes a hit with London society because of his strange life led on the Reservation. But while touring the factories and schools of the World State, John becomes increasingly disturbed by the society that he sees. His sexual attraction to Lenina remains, but he desires more than simple lust, and he finds himself terribly confused. In the process, he also confuses Lenina, who wonders why John does not wish to have sex with her. As the discoverer and guardian of the “Savage,” Bernard also becomes popular. He quickly takes advantage of his new status, sleeping with many women and hosting dinner parties with important guests, most of whom dislike Bernard but are willing to placate him if it means they get to meet John. One night John refuses to meet the guests, including the Arch-Community Songster, and Bernard’s social standing plummets.After Bernard introduces them, John and Helmholtz quickly take to each other. John reads Helmholtz parts of Romeo and Juliet, but Helmholtz cannot keep himself from laughing at a serious passage about love, marriage, and parents—ideas that are ridiculous, almost scatological in World State culture. Fueled by his strange behavior, Lenina becomes obsessed with John, refusing Henry’s invitation to see a feely. She takes soma and visits John at Bernard’s apartment, where she hopes to seduce him. But John responds to her advances with curses, blows, and lines from Shakespeare. She retreats to the bathroom while he fields a phone call in which he learns that Linda, who has been on permanent soma-holiday since her return, is about to die. At the Hospital for the Dying he watches her die while a group of lower-caste boys receiving their “death conditioning” wonder why she is so unattractive. The boys are simply curious, but John becomes enraged. After Linda dies, John meets a group of Delta clones who are receiving their soma ration. He tries to convince them to revolt, throwing the soma out the window, and a riot results. Bernard and Helmholtz, hearing of the riot, rush to the scene and come to John’s aid. After the riot is calmed by police with soma vapor, John, Helmholtz, and Bernard are arrested and brought to the office of Mustapha Mond.John and Mond debate the value of the World State’s policies, John arguing that they dehumanize the residents of the World State and Mond arguing that stability and happiness are more important than humanity. Mond explains that social stability has required the sacrifice of art, science, and religion. John protests that, without these things, human life is not worth living. Bernard reacts wildly when Mond says that he and Helmholtz will be exiled to distant islands, and he is carried from the room. Helmholtz accepts the exile readily, thinking it will give him a chance to write, and soon follows Bernard out of theroom. John and Mond continue their conversation. They discuss religion and the use of soma to control negative emotions and social harmony.John bids Helmholtz and Bernard good-bye. Refused the option of following them to the islands by Mond, he retreats to a lighthouse in the countryside where he gardens and attempts to purify himself by self-flagellation. Curious World State citizens soon catch him in the act, and reporters descend on the lighthouse to film news reports and a feely. After the feely, hordes of people descend on the lighthouse and demand that John whip himself. Lenina comes and approaches John with her arms open. John reacts by brandishing his whip and screaming “Kill it! Kill it!” The intensity of the scene causes an orgy in which John takes part. The next morning he wakes up and, overcome with anger and sadness at his submission to World State society, hangs himself. Context 背景Aldous Huxley was born in Surrey, England, on July 26, 1894, to an illustrious family deeply rooted in England’s literary and scientific tradition. Huxley’s father, Leonard Huxley, was the son of Thomas Henry Huxley, a well-known biologist who gained t he nickname “Darwin’s bulldog” for championing Charles Darwin’s evolutionary ideas. His mother, Julia Arnold, was related to the important nineteenth-century poet and essayist Matthew Arnold.Raised in this family of scientists, writers, and teachers (his father was a writer and teacher, and his mother a schoolmistress), Huxley received an excellent education, first at home, then at Eton, providing him with access to numerous fields of knowledge. Huxley was an avid student, and during his lifetime he was renowned as a generalist, an intellectual who had mastered the use of the English language but was also informed about cutting-edge developments in science and other fields. Although much of his scientific understanding was superficial—he was easily convinced of findings that remained somewhat on the fringe of mainstream science—his education at the intersection of science and literature allowed him to integrate current scientific findings into his novels and essays in a way that few other writers of his time were able to do. Aside from his education, another major influence on Huxley’s life and writing was an eye disease contracted in his teenage years that left him almost blind. As a teenager Huxley had dreamed about becoming a doctor, but the degeneration of his eyesight prevented him from pursuing his chosen career. It also severely restricted the activities he could pursue. Because of his near blindness, he depended heavily on his first wife, Maria, to take care of him. Blindness and vision are motifs that permeate much of Huxley’s writing.After graduating from Oxford in 1916, Huxley began to make a name for himself writing satirical pieces about the British upper class. Though these writings were skillful and gained Huxley an audience and literary name, they were generally considered to offer little depth beyond their lightweight criticisms of social manners. Huxley continued to write prolifically, working as an essayist and journalist, and publishing four volumes of poetry before beginning to work on novels. Without giving up his other writing, beginning in 1921, Huxley produced a series of novels at an astonishing rate: Crome Yellow was published in 1921, followed by Antic Hay in 1923, Those Barren Leaves in 1925, and Point Counter Point in 1928. During these years, Huxley left his early satires behind and became more interested in writing about subjects with deeper philosophical and ethical significance. Much of his work deals with the conflict between the interests of the individual and society, often focusing on the problem of self-realization within the context of social responsibility. These themes reached their zenith in Huxley’s Brave New World, published in 1932. His most enduring work imagined a fictional future in which free will and individuality have been sacrificed in deference to complete social stability.Brave New World marked a step in a new direction for Huxley, combining his skill for satire with his fascination with science to create a dystopian (anti-utopian) world in which a totalitarian government controlled society by the use of science and technology. Through its exploration of the pitfalls of linking science, technology, and politics, and its argument that such a link will likely reduce human individuality, Brave New World deals with similar themes as George Orwell’s famous novel 1984. Orwell wrote his novel in 1949, after the dangers of totalitarian governments had been played out to tragic effect in World War II, and during the great struggle of the Cold War and the arms race which so powerfully underlined the role of technology in the modern world. Huxley anticipated all of these developments. Hitler came to power in Germany a year after the publication of Brave New World. World War II broke out six years after. The atomic bomb was dropped thirteen years after its publication, initiating the Cold War and what President Eisenhower referred to as a frightening buildup of the “military-industrial complex.” Huxley’s novel seems, in many ways, to prophesize the major themes and struggles that dominated life and debate in the second half of the twentieth century, and continue to dominate it in the twenty-first.After publishing Brave New World, Huxley continued to live in England, making frequent journeys to Italy. In 1937 Huxley moved to California. An ardent pacifist, he had become alarmed at the growing military buildup in Europe, and determined to remove himself from the possibility of war. Already famous as a writer of novels and essays, he tried to make a living as ascreenwriter. He had little success. Huxley never seemed to grasp the requirements of the form, and his erudite literary style did not translate well to the screen.In the late forties, Huxley started to experiment with hallucinogenic drugs such as LSD and mescaline. He also maintained an interest in occult phenomena, such as hypnotism, séances, and other activities occupying the border between science and mysticism. Huxley’s experiments with drugs led him to write several books that had profound influences on the sixties counterculture. The book he wrote about his experiences with mescaline, The Doors of Perception, influenced a young man named Jim Morrison and his friends, and they named the band they formed The Doors. (The phrase, “the doors of perception” comes from a William Blake poem called The Marriage of Heaven and Hell.) In his last major work, Island, published in 1962, Huxley describes a doomed utopia called Pala that serves as a contrast to his earlier vision of dystopia. A central aspect of Pala’s ideal culture is the use of a hallucinogenic drug called “moksha,” which provides an interesting context in which to view soma, the drug in Brave New World that serves as one tool of the totalitarian state. Huxley died on November 22, 1963, in Los Angeles. Utopias and DystopiasBrave New World belongs to the genre of utopian literature. A utopia is an imaginary society organized to create ideal conditions for human beings, eliminating hatred, pain, neglect, and all of the other evils of the world.The word ut opia comes from Sir Thomas More’s novel Utopia (1516), and it is derived from Greek roots that could be translated to mean either “good place” or “no place.” Books that include descriptions of utopian societies were written long before More’s novel, however. Plato’s Republic is a prime example. Sometimes the societies described are meant to represent the perfect society, but sometimes utopias are created to satirize existing societies, or simply to speculate about what life might be like under different conditions. In the 1920s, just before Brave New World was written, a number of bitterly satirical novels were written to describe the horrors of a planned or totalitarian society. The societies they describe are called dystopias, places where things are badly awry. Either term, utopia or dystopia, could correctly be used to describe Brave New World.。

1、不抗拒,不评断,不执着，这是真正自由和开悟的三个面相。

2、所有的事物都是稍纵即逝的,当你觉知到事物的无常之后,你对它们的执着就会减轻,同时你对他们的认知程度也会减低.一旦你看清并接纳万物的无常和不断变化的必然性之后,你可以在它们存在的时候好好享受其中的乐趣,而不会担心或焦虑将来会失去它们。

3、凡抬高自己的,必降为卑;降卑自己的,必升为高。

4、促使你把旧痛一再重演而且让你处于无意识之中的，不是痛苦之身，而是你对它的认同。

不需任何时间，当痛苦之身被触动的时候，如果能够认出你所感觉到的就是你内在的痛苦之身，光这份知晓就足以打破你对它的认同。

当与它的认同停止时，转化就发生了。

那份知晓，可以抑制旧有情绪的升起并防止它进入你的脑袋中，也不会让它接管你的内在对话、掌控你的行为和其他人的互动。

这也就是说，痛苦之身不能再利用你然后再经由你来更新自己了。

那些老旧的情绪很可能还会在你之内存活一段时间，而且不定时地还会再度升起，有时也许还会拐骗你再度与它认同，因而模糊了那份知晓，但这是暂时的，不要将旧有的情绪投射到情境上，在你之内直接面结它，你的临在完全可以包容它。

5、这个世界不会让你一直以你所自认的假相来愚弄自己,你对不同人、事、物的因应方式（尤其是面对挑战的时候），就是对你自己了解程度深浅的最佳指标。

6、赞赏生命中已经拥有的美好事物，就是所有丰盛的基础。

事实上，你认为这个世界吝于给你的，其实你吝于给予这个世界的。

7、如果你受不了自己，那么必然存在两个你，你和你受不了的那个“自己”，而且他们之中只有一个是真的。

8、当你全神贯注，并且对着一朵花，一颗水晶或一只小鸟沉思冥想，但在心智上不去定义它们的时候，它们就会成为进入无形世界的一扇窗户。

你的内在就会有个开启（即使很小），让你因而进入心灵的领域。

9、所谓觉醒，很重要的一部分就是去辨识出那个未觉醒的自己——也就是小我，在小我思考、说话和行动的时候，辨识到它，并且辨认出那个集体受到制约的心智运作过程（它在未觉醒状态中持久不衰）。

ibm英文广告语篇一：著名英语广告语著名的英语广告语1．Goodtothelastdrop．滴滴香浓，意犹未尽。

（麦斯威尔咖啡）2．obeyyourthirst．服从你的渴望。

（雪碧）3．Poetryinmotion，dancingclosetome．动态的诗，向我舞近。

（丰田汽车）4.Justdoit.只管去做。

（耐克运动鞋）5.Feelthenewspace.感受新境界。

（三星电子）6.intelligenceeverywhere.智慧演绎，无处不在。

（摩托罗拉手机）7.Thechoiceofanewgeneration.新一代的选择。

（百事可乐）8.weintegrate,youcommunicate.我们集大成，您联络世界。

（三菱电工）9.TakeToSHiBa,taketheworld.拥有东芝，拥有世界。

（东芝电子）10.nobusinesstoosmall,noproblemtoobig.没有不做的小生意，没有解决不了的大问题。

（iBm公司）11.JoinＷorldenglish，EnjoyＥnglishworld.(新世界国际英语) 国际英语,欢乐无限。

12.impossibleisnothing(adidas)一切皆有可能！13.nothingcancomeofnothing．(莎士比亚)物有其本，事有其源。

14.Buyaustralia,Buyyouajob.(澳大利亚)买澳大利亚货，给你买份工作．15.oneworld,onedream.(北京奥运会)同一个世界,同一个梦想.16.welead．otherscopy．我们领先，他人仿效。

（理光复印机）17.impossiblemadepossible．使不可能变为可能。

（佳能）18.Taketimetoindulge．尽情享受吧！（雀巢）19.Therelentlesspursuitofperfection．不懈追求完美。

（凌志轿车）etowheretheflavouris．marlborocountry．光临风韵之境——万宝路世界。

另一种方法是，我们把一个人放到完全陌生的环境里工作。

比如：我自己在香港长大，我去过总部，也去过澳洲。

澳洲的文化跟中国文化很不一样，我在那里要管几千个澳洲人，把那里的服务业务开拓起来，这是一种磨炼。

再比如说，我们现在的CEO 钱大群先生，我们也曾经把他派到美国工作，送他到亚太总部把比如新加坡的业务带起来。

这样就能让一些领导人了解这个世界，能跟很多不同的人沟通，一起工作，并让他们了解做事情的方法，什么时候听，什么时候说。

《中外管理》：您刚才讲到，运用差异化来产生竞争力，这对于一个全球整合企业来说是不是尤为重要？周伟焜：关于怎样更融合地把这些人放到一起，发挥差异化将会是整个公司的一个挑战。

我们也没有答案。

管理的差异是很大的，我自己的体会是有两个层次的。

第一个层次：希望找很多都是跟你同一个声音的人。

最后达到的效果是“我不需要讲，我闭一闭眼睛就知道他们想什么了”。

我认为这是一个阶段，第二个层次：要愿意接受这样的情况——这个团队中的人有可能想法跟你完全不一样，而你不是想办法把他压住，而是启发他与大家相互协商碰撞，碰撞产生火花，把事情做得更好。

特别是当你去到更重要的职位，并非什么都懂的时候，这样的领导方式可能会更好。

《中外管理》：在领导力方面，IBM是如何为全球整合做准备的？如何判断需要什么样的领导力？周伟焜：我们会做一些调研和调查。

比如说，我当CEO的时候，大约每两三年就会做一次全面的调查，来看一段时期里我们的人跟全世界人才数据库在能力上的差距。

比如：两年前，差距最小的是我们对公司未来的热情，差距最大的是团队的互动能力。

那我们就知道要在什么方面努力，引进怎样的训练方法等等。

有时候我自己也会跟过去做比较，进步在哪里？还有哪些不足？上一次比较我发现：我的不足是怎样跟不同的人一起工作，而能够很有效地把自己的意见传递给大家，并让大家接受。

如果满分是4分，我五年前是3分，去年我的评价是3.9。

可以看出，这种方法是很有效的。

大灣高億分之公尺的美麗新世界中Ｋ-10億分之1公尺的美麗新世界淺談奈米科技12奈米－淺談奈米科技種子學校課程大綱壹什麼是奈米？大灣•壹、什麼是奈米？•貳、奈米事件簿高中Ｋ-•參、自然界的奈米現象12奈米•肆、奈米尺度下的物質性質•伍、伍、奈米科技奈米科技陸奈米科技的應用米種子•陸、奈米科技的應用•柒、柒、奈米科技產品奈米科技產品學校柒奈米科技產品•捌、結語大灣高中Ｋ-12奈米壹、什麼是奈米？米種子學校一個笑話大灣•奈米是什麼米？–蓬萊米？池上米？新品種的米高中Ｋ-–新品種的米？–農民四處打聽哪裡可以買到奈米的種子，12奈米準備種一種試試看，趁著現在挺熱門的，好好大撈一筆！米種子學校奈米是什麼？大灣•奈米是一種長度單位。

–奈米是英文nanometer的譯名。

奈米中的「米是「公尺的意思而高中Ｋ-–奈米中的「米」是「公尺」的意思，而「奈」這個字首，源自拉丁文NANO，意思是「矮小12奈米思是「矮小」。

•1奈米相當於10億分之1公尺，是原子米種子大小的尺度。

學校–相當於4個原子的直徑。

–大約是10個氫原子排在一起的長度。

101億分之公尺有多大？•從手背往下看，一直看到10億分之1公尺，大灣就是1奈米（10-9m ）。

高中Ｋ-手背12奈米 1 毫米10 公分100 微米1 公分白血米種子10100血球學校1 微米10 微米DNA10奈米100奈米1 奈米人高大灣針頭高中Ｋ-紅血球分子及DNA氫原子12奈米千100萬奈米米種子 1 奈米01奈米1千奈米學校0.1 奈米20億奈米101億分之公尺有多大？大灣•以地球的大小來衡量，地球高中Ｋ-直徑約為12萬7千公里千公里，，而十12奈米億分之一約是12.7厘米厘米，，就米種子大概是一顆彈珠的大小了。

學校珠的大小了大灣高中Ｋ-12奈米米種子學校大灣高中Ｋ-12奈米貳、奈米事件簿米種子學校大灣•1959年12月29日，美國物理學家理查‧費曼（Richard P F高中Ｋ-P. Feynman）在物理學會的演講中，發表了世人公認奈米技術的預言12奈米米技術的預言：–操縱與控制微小物體是可能的，因為當物體米種子操縱與控制微小物體是可能的因為當物體被縮小後，在微小的世界仍有許多空間。

Solutions to Atiyah and MacDonald’s Introduction toCommutative AlgebraAthanasios PapaioannouAugust5,20042Chapter1Rings and Ideals1.1We see that x∈R implies x∈J(the Jacobson radical),hence1+xA⊂A×.In particular,1+x is a unit. We can now easily deduce that the sum of a nilpotent element and a unit is a unit itself.1.2We have the following:(i)If f(x)=a0+a1x+...+a n x n is a unit in A[x],let g(x)=b0+b1x+...+b m x m be its inverse. We deduce that a0is a unit.We use induction on n to prove that the coefficients are nilpotent.The casen=0is a tautology.If the proposition holds for n−1,then we see that a r+1n b m−r=0(we just write downexplicitly the relations that ensue from fg=1and then multiply each of them by increasing powers of a n).In particular,this implies that a m+1n b0=0and,since b0is a unit,we deduce that a m+1n=0.Hence a n isnilpotent and we may apply the inductive hypothesis.The converse follows from exercise1and exercise2,(ii).(ii)If f(x)is nilpotent,then we can apply induction to n to show that all its coefficients are nilpotent. The case n=0is a tautology.In the general case,it’s apparent that the leading coefficient will be a m n for suitable m∈N hence a n is nilpotent.Now the inductive hypothesis applies.Conversely,if all the coefficients of f(x)are nilpotent and d∈N is such that a d i=0,0≤i≤n(e.g.let d be the sum of the orders of all the orders of the coefficients),then we see that f(x)d=0.(iii)If f is a zero divisor,then let g be a polynomial of minimal order,such that fg=0.If g= b0+b1x+...+b m x m is not of degree0,then a n b m=0,hence a n g is annihilates f but is of degree less than m,contradiction.Therefore,g is of degree0;there is a∈A,such that af=0.The converse is trivial.(iv)If fg is primitive,then so are f and g,too.The converse is just Gauss’s Lemma,in a more general context(the elementary argument still carries though).1.3The generalization follows easily by induction.1.4If J denotes the Jacobson radical and R denotes the nilpotent radical,then J⊃R,since R is the intersection of all prime ideals,while J is the intersection of all prime and maximal ideals.Therefore,we only need to show J⊂R in A[x].Indeed,if f(x)∈J,then1−f(x)g(x)∈A×,for all g(x)∈A[x].In particular1−f(x)x is a unit,hence if f(x)=a0+a1x+...+a n x n,then a0,a1,...a n are all nilpotent,hence by exercise2,(ii) f(x)∈R.This completes the proof.34CHAPTER 1.RINGS AND IDEALS1.5We have the following:(i)If f = ∞n =0a n x n ∈A [[x ]]is invertible,then obviously a 0is a unit.Conversely,if a 0is a unit,then we may let b 0be such that a 0b 0=1and then we may define b n ,n ∈N recursively by the explicit relations they have to satisfy.(ii)If f ∈A [[x ]]is nilpotent,then so is a 0∈A ,which,raised to a power,is the constant term when f is raised to a power.Therefore,f −a 0=xg (x ),g (x )∈A [[x ]],will also be nilpotent,hence g (x )will also be nilpotent.But the constant term of g (x )is a 1,which must be nilpotent,too.By this process,we show that all the coefficients will have to be nilpotent.The inverse is not true;a sufficient condition for it to be true would be the ring to be Noetherian.(iii)We easily see that 1−f (x )g (x )is a unit for all g (x )∈A [[x ]]if and only if 1−a 0b 0is a unit for all b 0∈A ,hence if and only if a 0belongs to the Jacobson radical of A .(iv)The extension mapping sends any ideal a of A to the ideal a e which consists of f (x )= a n x n ,a n ∈a .Conversely,given any ideal b of A [[x ]],b c consists of all coefficients in any element of b .It’s now clear that the contraction of a maximal ideal of A [[x ]]is maximal too and that m c =(m ,x ).(v)This also follows immediately from the above.1.6It clearly suffices to show that every prime ideal in A is maximal.Let p be a prime ideal in A and let x be a non-zero element of A −p .Then the ideal (x )={ax/a ∈A }will contain an idempotent element e =0,say a 0x .This implies that a 0x (a 0x −1)=0∈p ,hence a 0x (a 0x −1)=0in A/p ,too.However,A/p is an integral domain,therefore e =a 0x =0implies a 0x =1,or that x is a unit.Hence A/p is a field and this means that p is maximal.1.7Let p be a prime ideal of A .Form A/p ,which will be an integral domain.Given any non-zero x ∈A/p ,there will be a suitable n ∈N −{1},such that x n =x or equivalently x (x n −1−1)=0.This implies that x n −1=1,hence that x is invertible.Therefore,A/p is a field and thus p is a maximal ideal.1.8Every descending chain of prime ideals has a lower bound (the intersection of them all),hence by Zorn’s lemma the set Spec(A )has at least one minimal element (in order to apply the lemma,we order the set by ⊇rather than ⊆).1.9Since r (a )is an intersection of prime ideals (those that contain a ),we see that r (a )=a implies that a is an intersection of prime ideals.Conversely,if a is an intersection of prime ideals,then this intersection is contained in the intersection of prime ideals of r (a ),hence r (a )⊂a ,which shows that r (a )=a (the other direction is obvious by definition).1.10We have the following:(i)⇒(ii)Any maximal ideal in A (there is at least one),will be prime,hence it will coincide with the unique prime ideal a of A .Hence A is a local ring and R =a .If we consider A/R =A/a (which is a field),we deduce that every element of A is nilpotent or a unit.1.11.5(ii)⇒(iii)This direction is obvious by the definition of a field.(iii)⇒(i)The nilpotent radical is maximal (and thus prime)if A/R is a field.However,R =p ∈Spec(A )p ,hence R is included in the intersection on the right,hence every prime ideal contains R .But this implies that every prime ideal coincides with R (since R is maximal)and so there is only one prime ideal in A .1.11We have the following:(i)We just apply the given condition to x +x ,to obtain x +x =(x +x )2=x 2+x 2+2x =x +x +2x ,hence 2x =0.(ii)Every prime ideal is maximal follows from exercise 7,while the second corollary follows from (i).(iii)It suffices to show that any ideal generated by two elements of A is in fact principal.Indeed,given a,y ∈A ,we claim that (x,y )=(x +y +xy ).The direction (x,y )⊃(x +y +xy )is trivial.For the other inclusion,note that any element of (x,y )is of the form x m y n but given the conditions of idempotency,we see that the only elements that remain after the reductions in the sum will belong to (x +y +xy ),hence the other direction.1.12If m is the unique maximal ideal of the ring A ,then m contains all the non-units of A .If e ∈A were idempotent,then e (e −1)=0,hence if e or e −1were units,then e would be 0or 1.Otherwise,e ∈m and e −1∈m ,which imply 1∈m ,contradiction,since m is by definition a proper ideal of A .Construction of an algebraic closure of a field (E.Artin)1.13We will first show that a =({f (x f )}f ∈Σ)is not the unit ideal.Otherwise,given any polynomial p ∈A it would be presentable as a finite sum in the formp =f ∈Σy f f (x f ),where y f ∈A .But 1clearly cannot be represented in such a form,hence α=(1).If we now let m be the maximal ideal of A containing a ,we observe that K 1=A/m is a field extension of K in which every f ∈Σhas a root.Repeating the construction,we obtain K 2in which every f ∈Σhas two roots (if possible),and similarly we obtain K n for al n ∈N .We deduce that L = ∞i =1K i is a field extension which contains all the roots of every f ∈Σ;its algebraic elements form an algebraic closure K for K .1.14The fact that Σhas a maximal element is a trivial application of Zorn’s lemma;we just need to show that every ascending chain of ideals has a maximal element.Now,assume that m is a maximal ideal of Σand let xy ∈m ,pxy =0,with p =0.We claim that x ∈m or y ∈m .Assume the contrary.Then,m ⊂(m ,x )and (m ,x )would still be an ideal of Σ,since its elements are clearly zero divisors.This furnishes a contradiction to the maximality of m .Therefore,every maximal ideal of Σis prime.The prime spectrum of a ring6CHAPTER1.RINGS AND IDEALS 1.15We have the following:(i)The relations V(E)=V(a)=V(r(a))are obvious.(ii)Similarly,the relations V(0)=X=Spec(A)and V(1)=∅are obvious.(iii)Again,we havei∈I V(E i)=V(i∈IE i)(iv)Similarly trivial are the relations V(a∩b)=V(ab)=V(a)∪V(b).These results show that the space Spec(A)of all prime ideals of A can be endowed with a topology-the Zariski topology-if we define the V(E)to be its closed sets.1.16We immediately see the following:Spec(Z)={(p):p∈Z is prime}.Spec(R)=∅.Spec(C[x])={(p):p∈C[x]is of degree1}.Spec(R[x])={(p):p∈R[x]is irreducible}.Spec(Z[x])={(p):p∈Z[x]is irreducible}.1.17Given f∈A,we define X f={p∈Spec(A)/f/∈p}.It’s obvious that X=X1,∅=X0and O=X−V(E)=f∈E X f,hence the set{X f,f∈A}is a basis for the Zariski topology on Spec(A).We now have:(i)X f∩X g=X fg(obviously)(ii)X f=∅⇔f∈R(obviously)(iii)X f=X⇔f∈A×(obviously)(iv)X f=X g⇔r(f)=r(g)(obviously)(v)Note that for f,g∈A,X f=X g if and only if(f)=(g).In particular,X f=X=X1if and only if f∈A×.We also easily deduce(by de Morgan’s formula and exercise15)that:i∈I X fi=X({fi}i∈I)Therefore,if{X fi }i∈I is an open cover of X(and it’s only those covers of X that we need to consider,by astandard proposition in point-set topology),thenX({fi }i∈I)=X1,which implies that the{f i}i∈I generate the unit ideal.Therefore,there is afinite subset of indices J such that1=j∈Jg j f j,where g j∈A.Now,obviously the{f j}j∈J generate the unit ideal,hence the{X fj }j∈J are afinite subcoverof X.(vi)This follows by exactly the same argument as above,but considering covers of the form{X fi }i∈I,where X fi ⊂X f.(vii)If an open subspace Y of X is quasi-compact,then considering a standard cover by sets of the form X f,f∈A we see that Y must be afinite union of sets X f.Conversely,if an open subspace Y of X is a union of afinite number of sets X f,then any open cover{X fi }i∈I of Y induces an open cover for each of the X f(namely{X f∩X fi}i∈I).By(vii),each of those willhave afinite subcover and these subcovers yield afinite subcover of X.1.18.7 1.18We have the following:(i)By the definition of the Zariski topology,{x}is closed in Spec(A)if and only if{x}=V(E)for some subset E of A,hence if and only if p x is the only prime ideal that contains E,hence if and only if E=p x and p x is maximal(attaching any elements of A−E would generate the unit ideal).(ii)The relation{x}=V(p x)is obvious by our remarks above.(iii)y∈{x}=V(p x)if and only if p y⊃p x(iv)If x and y are distinct points of Spec(A),then either p x p y or p y p x;assume without loss of generality the latter.This is equivalent by our previous observations to x/∈{y},which implies that X−{y} is an open set that contains x but not y.1.19We claim that Spec(A)is irreducible if and only X f∩X g=∅for f and g non-nilpotent.Indeed,since {X f}f∈A are a basis for the Zariski topology on Spec(A),we see that any two non-empty sets will intersect if and only if any two non-empty basis elements intersect.This is equivalent to X fg=X f∩X g=∅if X f,X g=∅.The latter condition is fulfilled if and only if f and g are not nilpotent(by exercise17)hence the previous condition is equivalent to the following:there is a prime ideal p such that fg/∈p if f,g are not nilpotent hence fg/∈R if f/∈R,g/∈R.Therefore,X is irreducible if and only if the nilradical is prime.1.20We have the following:(i)If Y is an irreducible subspace if a topological space X,then Y is also irreducible,since by definition any neighborhood of a boundary point will intersect Y(hence any two open sets in Y continue to intersect in Y).(ii)We consider the setΣof all irreducible subspaces of X;it’s not empty,since x∈Σfor all x∈X. Then,by an application of Zorn’s lemma in the usual fashion(any ascending chain of irreducible subspaces will be bounded by the union of all its elements which is irreducible itself)we guarantee a maximal element forΣ.(iii)The maximal irreducible components of X are obviously closed(otherwise their irreducible closures would strictly contain them,contradiction)and they cover X(we see that any point x of X is contained in a maximal irreducible subspace by applying Zorn’s lemma to the spaceΣx which comprises the irreducible subspaces of X that contain x).In a Hausdorffspace each point is a maximal irreducible component.(iv)In the case of Spec(A)we note that the closed sets V(p),where p is a minimal prime ideal are irreducible(any two open sets will be of the form V(p)−V(E)and hence they will intersect)and that any two points x∈V(p1),y∈V(p2)can be separated by disjoint open sets.Therefore,the maximal irreducible components of Spec(A)are V(p),where p∈Spec(A)is minimal.1.21We have the following:(i)The following equivalences:q∈φ∗−1(X f)⇐⇒φ∗(q)∈X f⇐⇒f/∈φ∗(q)=φ−1(q)⇐⇒q∈Yφ(f)yield thatφ∗−1(X f)=Yφ(Xf )Nowφ∗is continuous,since the X f form a basis for the Zariski topology.(ii)The following equivalences:q∈φ∗−1(V(a))⇐⇒φ∗(q)⊇a⇐⇒q⊇a e⇐⇒q∈V(φ(a)e) yield thatφ∗−1(V(a))=V(a e),as desired.8CHAPTER1.RINGS AND IDEALS(iii)The statement thatφ∗(V(b))=V(b c)follows in the same fashion as the previous one.(iv)By proposition1.1,we know thatφ∗(Y)=V(ker(φ))andφ∗induces a bijective(and continuous, by the previous question)map from Y to V(ker(φ)).Thus we merely need to show thatφ∗−1is continuous. Let Y =V(b)be any closed subset of Y and let a=φ−1(b).Then,the following equivalences:p∈φ∗(Y )=φ∗(V(b))⇐⇒p=φ∗(q)⊇b⇐⇒p=φ−1(q)⊇b c⇐⇒p∈V(b c)imply thatφ∗(Y )=V(b c)and in particular thatφ∗(Y )is closed when Y is and therefore thatφ∗is a homeomorphism.In particular,the natural surjective projection map from A to A/R induces a homeomorphism between Spec(A)and Spec(A/R).(v)By the previous statement,we have∗=∗=V(0c)=V(ker(φ))thusφ∗(Y)is dense in X⇐⇒φ∗(Y)=V(ker(φ))=X←→ker(φ)⊆p for all prime ideals p⇐⇒ker(φ)⊆R.(vi)The desired result follows immediately by definition.(vii)By assumption,the two only prime ideals of A are p and0,which implies that p is a maximal ideal of A and thus A/p is afield.This yields that the ring B=(A/p)×K will also have only two ideals,namely q1={(x,0):x∈A}and q2={(0,k):k∈K}.The ring homomorphismφ:A−→B defined byφ(x,x)is bijective(φ∗(q1)=0andφ∗(q2)=p)and continuous.However,φ∗is not a homeomorphism.In the topological space Spec(B)={q1,q2},we have{q1}=V(q1) is closed as q1 q2,butφ∗(q1)=0is not closed in Spec(A),since0is not a maximal ideal of A(by exercise 18).1.22A decomposition of A in the formA A1×A2×...×A nyields a decompositionSpec(A) Spec(A1)×Spec(A2)×...Spec(A n)If we embed every space Spec(A i)as X i=(0,0,...,Spec(A i),0,...,0)in Spec(A),it’s a standard argument that the existence of the latter decomposition is equivalent to the decomposition of Spec(A)as a disjoint union of the X i.Given a ring A,we have the following:(i)⇒(ii)This direction follows from our previous observation and the definition of connectedness.(ii)⇒(iii)If a decomposition of the form A A1×A2(where A1,A2are non-trivial)existed,then a non-trivial idempotent element of A would be the pull-back of(1,0).(iii)⇒(ii)If e∈A is a non-trivial idempotent,then X=Spec(A)decomposes as X1 X2,whereX1={p∈X/e∈p}andX2={p∈X/e−1∈p}.It’s a trivial observation that X1∩X2=∅(as in our proof that a local ring possesses no non-trivial idempotents)and similarly trivial is the verification that X=X1∪X2.This decomposition implies that X is disconnected.1.23.91.23We have the following:(i)Each f ∈A is idempotent hence X f induces a disjoint decomposition as in the previous exercise.It’s now obvious that the sets X 1and X 2with the notation of exercise 22are simultaneously closed and open.(ii)By the formula of exercise 17,and by the fact that a Boolean ring is always a Principal Ideal Domain,we deduce that there is f ∈A such thatX (f 1,f 2,...,f n )=X f 1∪X f 2∪...∪X f n =X f(iii)The hint in the book is a full proof;let Y ⊂X be both open and closed.Since Y is open,it is a union of sets X f .Since Y is a closed subspace of a quasi-compact space,it is quasi-compact too hence it is a finite union of sets X f ,say X f 1,X f 2,...,X f n .Now,(ii)finishes the proof.(iv)X is (obviously)compact and Hausdorff.1.24There is nothing to be provided other than a tedious verification of the axioms.1.25Stone’s Theorem that every Boolean lattice is isomorphic to the lattice of open and closed sets of some compact Hausdorffspace follows immediately from exercises 23and 24.1.26We will just repeat the construction of the book,which shows that X Max(C (X )),by the map µ:X →Max(C (X ))given by x →m x ={f ∈C (X )/f (x )=0}.Note that m x is always a maximal ideal,as the kernel of the surjective map that sends f to f (x )(whence C (X )/m x R is a field).(i)Let m be any maximal ideal in X .Then,let V (m )be the set of common zeroes of functions in m ,namely V (m )={x ∈X/f (x )=0for all f ∈m }.We claim that V (m )=∅.Indeed,otherwise,for every x ∈X there is f x ∈m ,such that f x (x )=0.Since f x is continuous,there is a neighborhood U x of x on which f x does not vanish.By the compactness of X ,a finite number of these neighborhoods cover X ,say{U x i }i =1,2,...n .Then,f = n i =1f 2x i ∈m ,but f does not vanish on any point of X ,hence it’s a unit,hence m =(1),contradiction.Therefore,V (m )=∅.Let x ∈V (m ).Then,⊆m x ,which implies m =m x by the maximalityof m .Hence m ∈Im µand µis surjective.(ii)By Urysohn’s lemma,the continuous functions separate the points of C (X )and this implies that m x =m y if x =y .Hence µis injective.(iii)For f ∈C (X ),letU f ={x ∈X/f (x )=0}andU f ={m ∈X/f ∈m }.We obviously have µ(U f )=U f .Since the open sets U f (resp.U f )form bases of the topologies on X and X we deduce that µis also continuous (as is µ−1).Therefore X is homeomorphic to Max(C (X )).Affine algebraic varieties 1.27There is nothing to be proved in this exercise if we invoke the Nullstellensatz for the surjectivity of µ.10CHAPTER1.RINGS AND IDEALS 1.28We have the following situation:Ψ:[φ:X→Y,regular]↔Hom k(P(Y),P(X)),whereΨis defined byφ−→Ψ(φ):(η→η◦φ).We see thatΨis injective becauseη(φ1)=η(φ2)for allηimpliesφ1=φ2(obviously;just letηbe the natural projections).It’s also surjective;ifΨis any k-algebra homomorphism P(Y)→P(X),thenΨ=φ◦η, whereηis the polynomial transformation that sends the values ofφon P(X)to the values ofΨ.Chapter2Modules2.1Since m and n are coprime,there are integers a and b such that am+by=1.Therefore,given x⊗y∈(Z/m Z)⊗Z(Z/n Z),we see that x⊗y=1(x⊗y)=am(x⊗y)+nb(x⊗y)=a(mx⊗y)+b(x⊗ny)= a(0⊗y)+b(x⊗0)=0,and since every generator is identically0,so will the whole tensor product be.2.2If we tensor the exact sequencea incl−→Aπ−→A/a−→0,we obtain−→(A/a)⊗A M−→0a⊗A M incl⊗1−→A⊗A Mπ⊗1and this induces an isomorphism between(A/a)⊗A M and the cokernel of incl⊗1,which is(A⊗A M)/(a⊗A M) M/a M,since given any ideal a of A a trivial argument shows that a⊗A M a M(we need M to be flat for this to be true).Hence,(A/a)⊗A M M/a M,as desired.The above proposition is true even if M is not aflat A-module.A proof in this general case would proceed as follows:consider the mapφ:(A/a)⊗A M−→M/a M,defined by(a,x)→ax mod a M.This is clearly a bilinear homomorphism,which induces a linear homomorphism M−→(A/a)⊗A M whose inverse is x→1⊗A x(where1is the image of1in A/a).It is clear that a M is contained in the kernel of this last linear map,and hence the construction yields an isomorphism between M/a M and(A/a)⊗A M,as desired (Lang,Algebra,612).2.3Let m be the maximal ideal of A,k=A/m A its residuefield.We also let M k=k⊗A M M/m M,by exercise2.The condition M⊗A N=0implies(M⊗A N)k=0,hence M k⊗A N k=(M⊗A N)k⊗A k=0. But M k and N k are vectorfields over thefield k,hence dim k(M k⊗A N k)=dim k(M k)dim k(N k),hence M k⊗A N k=0implies M k=0or N k=0.Let without loss of generality the former be true.Then,since m is the unique maximal ideal,it will coincide with the Jacobson radical of A,so M/m M=M k=0implies M=m M and by Nakayama’s lemma this yields M=0.2.4If M,N,P are A-modules,we know that(M⊕N)⊗P=(M⊗P)⊕(N⊗P),1112CHAPTER2.MODULEShence we have the following equivalence:M=i∈IM i isflat if and only if an exact sequence0−→M −→M−→M −→0 remains exact after tensoring by M:0−→M ⊗(i∈I M i)−→M⊗(i∈IM i)−→M ⊗(i∈IM i)−→0.The above can be written as0−→i∈I (M ⊗M i)−→i∈I(M⊗M i)−→i∈I(M ⊗M i)−→0,and this sequence is exact if and only if each component0−→(M ⊗M i)−→(M⊗M i)−→(M ⊗M i)−→0 is exact,hence if and only if each M i isflat.2.5We observe thatA[x]=∞m=0(x m),hence A[x]is aflat A-algebra if and only if each component(x m)is aflat A-algebra(by the previous exercise). By Lang’s lemma(Lang,Algebra,618),it suffices to show that the natural mapφ:a⊗(x m)−→a(x m)is an isomorphism for any ideal a of A.Indeed,surjectivity is obvious and we may write any arbitrary generator of a⊗(x m)as a⊗x m(transferring the constants to thefirst slot;the product will be in a,because a is an ideal).Any two such generators will map to the same element in a(x m)is and only if their respectivefirst slots coincide and thus if and only if they coincide.Hence the natural map is an isomorphism,as desired.2.6First note that M[x]={m0+m1x+...+m r x r/m i∈M,r∈N}is a module over A[x],if we define the product of two polynomials in the obvious fashion.We see that since M=AM={am/a∈A,m∈M}, M[x]will coincide with(AM)[x].We also see that the homomorphismφ:A[x]⊗A M−→(AM)[x]defined as a(x)⊗A m→a(x)m has an obvious inverseψ:a(x)m→a(x)⊗A m and thus it induces an isomorphism of tensor products between A[x]⊗A M and(AM)[x]=M[x],as desired.2.7Note that if A is a ring and a is any ideal in it,thenA[x]/a[x] (A/a)[x].For the proof,just note that a[x]is the kernel of the natural projection map from A[x]to(A/a)[x].Now let p be a prime ideal of A.Then,A/p is an integral domain,and hence so is(A/p)[x]by the Hilbert Basis Theorem.But,by the above,A[x]/p[x]will be an integral domain too,thus p[x]sill be a prime ideal in A[x],as desired.In the case p is maximal in A,it doesn’t necessarily follow that p[x]is maximal in A[x];if F is afield,it doesn’t necessarily follow that F[x]is afield too.2.8.13 2.8We have the following:(i)If M and N areflat A-modules,then so is M⊗A N,since the tensor functor is associative(tensoring an exact sequencefirst by M and then by N is equivalent to tensoring by M⊗A N).(ii)Notefirst that if−→B⊗A M −→0−→B⊗A M1⊗g0−→M 1⊗fis an exact sequence and B isflat,then0−→M f−→M g−→M −→0is also exact.By proposition2.19,this boils down to the statement that if the homomorphism1⊗A f is injective,then so is the homomorphism f.For assume that1⊗A f is injective,but f isn’t.Then,there are distinct x1,x2such that f(x1)=f(x2),and for every suitable y we would have y⊗A f(x1)=y⊗A f(x2) hence y⊗A x1=y⊗A x2(by the injectivity of1⊗A f).But then,(1)⊗A(x)=0(where x=x1−x2)and since both modules arefinitely generated,we deduce(by exercise3)that x=0,contradiction.Now if0−→M f−→M g−→M −→0is exact,then by the assumptions of the exercise it will remain exact when tensoringfirst by B(after which we may regard the sequence as a sequence of B-modules)and then by N.The above lemma implies that we may remove the tensor by theflat A-module B without penalty and this will leave us with an exact sequence;but that’s merely the initial sequence tensored by N.Hence N isflat as an A-module,as desired.2.9We shall only use the assumption that0−→M f−→M g−→M −→0is exact and M isfinitely generated. We see that if x1,x2,...,x n are generators for M ,then x1,x2,...,x n in Coker(g)=M /f(M )will generate Coker(g).But by the exactness of the sequence,Coker(g) M,hence M will befinitely generated.2.10We shallfirst show the following embedding:(N/u(M))/(a(N/u(M))) →(N/a N)/u(M/a M)The mapping is the natural one:given n∈N/u(M)we send itfirst to˜n∈N/a N and then toˆ˜n∈(N/a N)/u(M/a M).It’s a trivial verification that the kernel of this A-module homomorphism is included in a(N/u(M)),hence the embedding.However,we notice that(N/a N)/u(M/a M)=0since the induced homomorphism is surjective.Therefore,we will have(N/u(M))/(a(N/u(M)))=0,too and by Nakayama’s lemma N/u(M)=0,hence N=u(M),hence u is surjective,as desired.2.11We have the following:(i)Letφ:A n−→A m be an isomorphism and let m be a maximal ideal of A.Then,m annihilates the module(A/m)⊗A A m,hence we may regard(A/m)⊗A A m(as well as(A/m)⊗A A n,of course)as a vector space over thefield A/m.But thenφinduces an isomorphism1⊗φ:(A/m)⊗A A m−→(A/m)⊗A A nbetween two vector spaces of dimensions n and m and this clearly implies m=n.(ii)Ifφ:A m−→A n is surjective,then A n A m/N,where N=kerφ.But,if x1,x2,...x m generate A m,then obviously x1,x2,...,x m generate A n,hence m≥n,as desired.14CHAPTER 2.MODULES (iii)Thanks to Nick Rozenblyum for informing me that the statement is in fact correct,but it’s probably one of the most difficult problems in the book!If φ:A m −→A n is injective,then it necessarily follows that m ≤n .Indeed,let {e 1,e 2,...,e m }be the standard canonical basis of A m and let φ(e i )=(a i 1,a i 2,...,a in )∈A n for 1≤i ≤m .Then,let D be the n ×n matrix (a ij ).Without loss of generality,we may consider the cases m,n >0and D =0(the omitted cases are trivial);also,by possibly rearranging the orders of the basis elements of A m and A n ,we may assume that the non-zero r ×r minor of D is at the upper left corner.Suppose,contrary to the desired conclusion,that m >n .Then,m ≥r +1and we denote the (r +1)×r block matrix at the upper left corner of D by D .For each j =1,2,...,n the quantity r +1i =1a ij b i can be realized as the determinant of a (r +1)×(r +1)matrix.If 1≤j ≤r ,then the matrix has two identical columns up to ±1.If r +1≤j ≤n ,then the determinant of the matrix is an (r +1)×(r +1)minor (again,up to ±1)of D .Therefore,we have r +1i =1a ij b i =0for all 1≤j ≤n .But this means thatφ(b 1,...,b r +1,0,...,0)=(r +1i =1a i 1b i ,...,r +1 i =1a in b i )=(0,...,0)∈A nwhich is a contradiction to the injectivity of φsince (b 1,...,b r +1,...,0,...,0)=(0,0,...,0)∈A m .This completes the proof that m ≤n .2.12Let e 1,e 2,...,e n be a basis of A n .Choose u 1,u 2,...,u n such that φ(u i )=e i .We now claim that M =ker φ⊕(u 1,u 2,...,u n ).Indeed,if x ∈M ,then φ(x )= n i =1y i e i =φ( n i =1y i u i )for some y i ∈A ,hence there is a unique n ∈ker φsuch that x =n + n i =1y i u i ,which clearly implies M =ker φ⊕(u 1,u 2,...,u n ).Since there is a finite number of generators x 1,x 2,...,x m of M ,we see that at most a finite number of them generate (u 1,u 2,...,u n )(we may add the u i to the x i if necessary)and hence the complement of (u 1,u 2,...,u n )in M ,namely ker φ,will be generated by the rest of the x i .In particular,it will be finitely generated.2.13In order to show the injectivity of g ,we will repeat a familiar argument:let y map to 0under g .Then,1⊗y =0,thus (1)⊗(y )=0,hence (y )=0by exercise 3,since (1)and (y )are both finitely generated.This yields y =0,as desired.Then,let p :N B −→N be defined by sending b ⊗y to by .We now claim that ker p and g (N )are direct summands of N B and moreover N B =g (N )⊕ker p .Indeed,we obviously have g (N )∩ker p =0,and also N B =g (N )+ker p ,since any generator b ⊗n of N B can be written as b (1⊗n )+0⊗n .This completes the proof.Direct limits2.14We will just repeat the construction of the book;there is nothing else to be proved.Let A be a ring,I a directed set and let (M i )i ∈I be a family of A -modules indexed by I .For each pair i,j ∈I such that i ≤j ,let µij :M i →M j be a homomorphism and suppose that the following axioms are satisfied:(i)µii is the identity mapping on M i .(ii)µij =µkj µik ,for elements i ≤j ≤k of I .Then the modules M i and homomorphisms µij are said to form a direct system M =(M i ,µij )over the directed set I .We shall construct an A -module M called the direct limit of the direct system M .Let C be the direct sum of the M i ;identify each module M i with its embedding in C .Then let D denote the submodule generated。

第五章所有权的错觉“拥有”某物，这到底是什么意思呢？将一些东西变成我所拥有的（我的），又是什么意思呢？如果你站在纽约的街头，指着一座摩天大楼说：“那栋楼是我的，我拥有它。

”你不是非常有钱，就是你有妄想症，要不就是骗子。

无论如何，你是在述说一个故事，在这个故事中，“我”这个念相和“大楼”这个念相合而为一了。

这就是所有权的心理概念运作的方式。

如果大家认同你的故事，你会有一个签署的文件来证明他们的认可：你是很有钱的。

如果没有人同意你说的故事，他们会送你去看精神科医师：你不是有妄想症，就是有强迫说谎的倾向。

在这里，很重要的一件事就是，无论人们同意与否，你要辨识出：这个故事和组成这个故事的念相与你是谁完全无关。

即使人们同意这个故事，最终它还是一个幻相。

很多人一直到了死亡迫在眉睫、外在的事物开始瓦解时，才了解到：没有任何事物和“他们是谁”的本质有关。

当死亡临近时，这整个“所有权”的概念终究显得完全没有意义了。

在生命的最后时刻，他们也了解到，他们终其一生都在寻找一个更完整的自我感，但是他们真正在寻找的本体，其实一直都在那里，只是大部分时间都因为他们对事物的认同（其实最终就是他们对心智的认同）而被掩盖了。

“灵里贫穷的人有福了，”耶稣说，“天国将是他们的。

”“灵里贫穷”是什么意思？没有内在的负累，没有认同。

不认同于任何事物，也不认同于任何让他们有自我感的心理概念。

“天国”又是什么呢？就是当你放下认同而成为“灵里贫穷”的人时，你会有的那个简单但是深远的本体的喜悦。

这就是为什么在西方和东方，弃绝所有世俗的财产，一直都是一个古老的灵修传统。

然而，弃绝财产并不能自然而然地将你从小我中解脱出来。

小我会试图借由认同于其他事物，而维持它的生存。

比方说，它可能会认同于这样的一个心理形象：我超越了对物质世界所有的兴趣，所以我比其他人更为优越，而且更有灵性。

有些人虽然弃绝了所有俗世的财产，但他的小我却比一些百万富翁还大。

如果你拿走了一种认同，小我很快地会找到另外一种。