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 Posted: Sat 14:08, 25 Dec 2010 Post subject: air jordans F and F on the Ring of Infinite Equiva |
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F and F on the Ring of Infinite Matrices equivalence rings
heringQ. (See [S]) Corrollary1 (Xu [5 ~) LetFbenringwithidentity, thenF ~ n. z. c. M (F) Corrollary2, LetFbeHringwithidentityand ¡Ü lI ¡Ü IrI, thenF ~ n. z. e. M (F); WealsohaveF ~ n. z. e. M (F) 3. RemarkTheresultF ¡Ö M (F) failstobetruegenerallyfortheringM (F) (¡Ü lr1) andM (F), whereM (F) (¡Ü lr1) isthesubringofMr (F) consistingofallthematriceswithatmostnon-zeroelementsineachcolumnandM (F) isthesubringofM (F) consistingofalIthematriceswithatmostfiniten0n-zeroetementsineachcolumn. Infaet, ifthereex-istssomei ¡Ê zsuchthatK, F ¡Ù 0hereK, FisQuillenshigherKL-group: 2], thatisKtF = l (BGL (F) 4 -), (i ¡Ý 1). thenwehave64-J_If: (F) especiallyFM (F) andFM (F) iInfact, ifF ¡Ö M: (F),[link widoczny dla zalogowanych], thenKLF Zhu KLM (F) = 0ibytheresuctestablishedin [6 ~,[link widoczny dla zalogowanych], wehaveKIM (F) = 0,[link widoczny dla zalogowanych], thereforeKiF Ò» 0, acontradiction. Accord-lngtoAlgebraicK-Theory], itiswellknownthatKlD ¡Ù 0, K2Z ¡Ù 0,[link widoczny dla zalogowanych], SOZM: (z) andDM (D), hereDisquaterniondivisionring. ButtherereallyexistsaringFsuchthatF ¡Ö M: (F) iforexample, chooseF = M: (R), hereRisaringwithidenthy, thenfromM (F) A M: (M: (R)) M (R) M: (R) a FWehaveF ¡Ö M (F); thereforewehaveaquestionwhenF ¡Ö M (F)? ReferanteslFuller ? K ? R. ,[link widoczny dla zalogowanych], DensityandEquivalel ~ e. J. Time Algebra, 29 (1974). 528 - 550.2D. Qafilen, HigherK-TheoryforCategoriesfhExac! Sequcmcu. NewdevelopmentinTopology. Oxford ? June1972.3J. R. Silvester. IntrodaetiontoAlgebraicK-Theory + Chapmen8LHal1.1981.4XuYonghua, I = omorphismetweenendomorphismrhoffreemoduleat (toappear) 5XuYonghua. AnEquivalenceBetweenRingFandInfiniteMatrixSubringoverFChin. Ann. ofMath. 1¡¿ BIl (J990) 66-699XuK at tantSomeresullsonIheK-theoryofinfinitemalrlxrings, JShandongcollegeolTextileTechnology ? vo1.6 ? No. 3.1991.68-8265
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