Definition of GL(n,R) - Math Stack Exchange
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If R is a commutative ring with identity, then for an integer n≥1, GLn(R) is the set of n×n matrices in g∈Mn(R) (coefficients in R) such ... MathematicsStackExchangeisaquestionandanswersiteforpeoplestudyingmathatanylevelandprofessionalsinrelatedfields.Itonlytakesaminutetosignup. Signuptojointhiscommunity Anybodycanaskaquestion Anybodycananswer Thebestanswersarevotedupandrisetothetop Home Public Questions Tags Users Unanswered Teams StackOverflowforTeams –Startcollaboratingandsharingorganizationalknowledge. CreateafreeTeam WhyTeams? Teams CreatefreeTeam Teams Q&Aforwork Connectandshareknowledgewithinasinglelocationthatisstructuredandeasytosearch. LearnmoreaboutTeams Definitionof$\text{GL}(n,R)$ AskQuestion Asked 9years,8monthsago Modified 9years,8monthsago Viewed 1ktimes 4 $\begingroup$ Howdooneusuallydefinethegenerallineargroupoveraring$R$,denotedby$\text{GL}(n,R)$.Iwastoldinapaperthat$\text{GL}(n,R)$isagroup,andIpresumedthat$$\text{GL}(n,R)=\{A\inM_{n\timesn}(R)|\text{det}(A)~\mbox{isaunitin}~R\}.$$However,Itriedgoogleitandfound$$\text{GL}(n,R)=\{A\inM_{n\timesn}(R)|\text{det}(A)\neq0\}.$$Seeforexample,http://gmcninch.math.tufts.edu/Math215-Fall-2012/storage/HW4.pdf.As$R$isnotnecessarilyaunitalring,soitwouldhappenthat$\text{GL}(n,R)$isnotagroup.Couldanyexperttellmewhichunderstandingiscorrect?Andalso,couldyourecommendanytextbookwhichprovidesdetaileddiscussionaboutthiskindofgroup?Ineedtolearnthismore,thankyouverymuch! linear-algebragroup-theorymatricesring-theory Share Cite Follow askedFeb6,2013at12:03 EasyEasy 4,3951414silverbadges2929bronzebadges $\endgroup$ 9 1 $\begingroup$ Wewantinverses,sothefirstdefinitionisdefinitelytherightone. $\endgroup$ – ZhenLin Feb6,2013at12:05 $\begingroup$ Ifso,howdoexplainthedefinitioninthewebsite?Whyhestillcallsitagroup? $\endgroup$ – Easy Feb6,2013at12:12 $\begingroup$ Thehomeworksheetyoulinktoafter"Seeforexample"doesnotappeartocontainanydefinitionofthegenerallineargroupoveranarbitraryring.Itdefines$GL_2(\mathbbZ/n\mathbbZ)$,butitsdefinitionisthenon-unitdeterminantone. $\endgroup$ – hmakholmleftoverMonica Feb6,2013at12:18 $\begingroup$ Whatever,couldyourecommendanytextbookwhichintroducesthegenerallineargroupsoveraunitalcommutativering?Iamstrugglingtofindaproperreference.Thanks $\endgroup$ – Easy Feb6,2013at12:20 $\begingroup$ sorry,IthinkIshoulddropthe"unital"condition.. $\endgroup$ – Easy Feb6,2013at12:22 | Show4morecomments 1Answer 1 Sortedby: Resettodefault Highestscore(default) Datemodified(newestfirst) Datecreated(oldestfirst) 6 $\begingroup$ If$R$isacommutativeringwithidentity,thenforaninteger$n\geq1$,$\mathrm{GL}_n(R)$isthesetof$n\timesn$matricesin$g\in\mathrm{M}_n(R)$(coefficientsin$R$)suchthat$\mathrm{det}(g)\inR^\times$.Thisispreciselythegroupofinvertibleelementsofthering$\mathrm{M}_n(R)$ofmatrices. If$R$isafield,then$R^\times=R\setminus\{0\}$,sointhiscase,thecondition$\det(g)\inR^\times$isequivalentto$\det(g)\neq0$,butthisisnottruefor(non-zero)commutativeringswhicharen'tfields. Share Cite Follow answeredFeb6,2013at12:07 KeenanKidwellKeenanKidwell 24.9k22goldbadges5858silverbadges100100bronzebadges $\endgroup$ 2 1 $\begingroup$ why$det(g)\inR^\times$implies$g$invertible?Forexample,suppose$R=\mathbb{Z}_4$and$det(g)=2$,then$det(g^{-1})=2^{-1}$,whichdoesn'texist. $\endgroup$ – Easy Feb6,2013at12:11 5 $\begingroup$ @user60079:Here$R^\times$isthegroupofunitsinthering,so$2\notinR^\times$inyourexample.Ifthedeterminantofamatrixisaunit,thenCramer'sruleproducesaninverseforit. $\endgroup$ – hmakholmleftoverMonica Feb6,2013at12:13 Addacomment | YourAnswer ThanksforcontributingananswertoMathematicsStackExchange!Pleasebesuretoanswerthequestion.Providedetailsandshareyourresearch!Butavoid…Askingforhelp,clarification,orrespondingtootheranswers.Makingstatementsbasedonopinion;backthemupwithreferencesorpersonalexperience.UseMathJaxtoformatequations.MathJaxreference.Tolearnmore,seeourtipsonwritinggreatanswers. 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