刘绍学《近世代数基础》习题解答.pdf 50页

'§11.1.P4,Ex1OdetO=−1cosθsinθO=sinθ−cosθO±11−cosθ=01(1−cosθ)x−sinθy=0l:(1−cosθ)x−sinθy=0xxxcosθx+sinθy∈R2,O=lyyysinθx−cosθyxly|(1−cosθ)x−sinθy|√2−2cosθxOly(1−cosθ)(cosθx+sinθy)−sinθ(sinθx−cosθy)||(1−cosθ)x−sinθy||√=√2−2cosθ2−2cosθ1 xxOlyysinθx−cosθy−y·1−cosθ=−1cosθx+sinθy−xsinθxxOlφlyy1−cosθ=01sinθx−(1+cosθ)y=0l1:sinθx−(1+cosθ)y=0xxOl1yyφl11.2.P4,Ex2φ,ϕ,θ∈T(M)(φ·ϕ)·θ=φ·(ϕ·θ)∀m∈M[(φ·ϕ)·θ](m)=[φ·(ϕ·θ)](m)[(φ·ϕ)·θ](m)=(φ·ϕ)(θm)=φ[ϕ(θm)][φ·(ϕ·θ)](m)=φ[(ϕ·θ)(m)]=φ[ϕ(θm)][(φ·ϕ)·θ](m)=[φ·(ϕ·θ)](m)(φ·ϕ)·θ=φ·(ϕ·θ)1.3.P4,Ex3S(K)60◦;120◦;180◦;240◦;300◦12§22.1.P8,Ex1F0,1C√ai+bi2∈F,ai,bi∈Q,i=1,22 √√√(a1+b12)±(a2+b22)=(a1±a2)+(b1±b2)2∈F√√√(a1+b12)(a2+b22)=(a1a2+2b1b2)+(a1b2+b1a2)2∈F√1√=a1+(−b12)∈Fa1+b12a2−2b2a2−2b21111√F∀0=a=a1+b12∈F√a−1=a1+(−b12)∈Fa2−2b2a2−2b21111F2.2.P8,Ex2FQ⊂FAut(F:Q)⊂Aut(F)Aut(F)⊂Aut(F:Q)FQ∀φ∈Aut(F)φ(1)=1n,φ(n)=n,φ(−n)=−n,φ(n−1)=n−1q=m∈Q,m,n∈ZZnφ(m)=φ(m·n−1)=φ(m)·φ(n−1)=m·n−1=mnnφ∈Aut(F:Q)2.3.P8,Ex3√(1)x,y∈Qx+y2=0x=y=0x,y∈Q0z∈Qzx,zy(zx,zy)=1x,y0(x,y)=1√x+y2=0x2=2y2xx=2kk2k2=y2y(x,y)=1x=y=0√(2)x,y∈Qx+y6=0x=y=0√x,yx+y6=0x=y=0x,y0(x,y)=1√x+y6=0x2=6y2xx=2kk2k2=3y2y(x,y)=1x=y=03 √(3)x,y∈Qx+y3=0x=y=0√x,yx+y3=0x=y=0x,y0(x,y)=1√x+y3=0x2=3y2x3x=3kk3k2=y2y3(x,y)=1x=y=0√√(4)x,y,z∈Qx+y2+z3=0x=y=z=0√√x+y2+z3=0√√√x2=(y2+z3)2=2y2+2yz6+3z2√2y2+3z2−x2+2yz6=0(2)yz=0y=0z=0√√√y=0x+y2+z3=0⇔x+z3=0(3)x=z=0√√√z=0x+y2+z3=0⇔x+y2=0(1)x=y=0√√x+y2+z3=0x=y=z=0√√√(5)a,b,c,d∈Qa+b2+c3+d6=0a=b=c=d=0√√√a+b2+c3+d6=0√√√(b2+c3+d6)2=(−a)2√√2b2+3c2+6d2+4bd3+6dc2=a2(4)bd=0dc=0√√√√√d=0a+b2+c3+d6=0⇔a+b2+c3=0(4)a=b=c=0√√√√b=0c=0a+b2+c3+d6=0⇔a+d6=0(2)a=d=0√√√a+b2+c3+d6=0a=b=c=d=04 2.4.P8,Ex4(1)0,1∈Q(i)ak+bki∈Q(i),k=1,2(a1+b1i)±(a2+b2i)=(a1±a2)+(b1±b2)i∈Q(i)(a1+b1i)·(a2+b2i)=(a1a2−b1b2)+(a1b2+b1a2)i0=a1+b1i∈Q(i)(a+bi)−1=a1+(−b1)i∈Q(i)11a2+b2a2+b21111Q(i)√0,1∈Q(i,5)√√ak+bki+ck5+dk5i∈Q(i),k=1,2√√√√(a1+b1i+c15+d15i)±(a2+b2i+c25+d25i)√√√=(a1±a2)+(b1±b2)i+(c1±c2)5+(c1±c2)5i∈Q(i,5)√√√√(a1+b1i+c15+d15i)·(a2+b2i+c25+d25i)=(a1a2−b1b2+5c1c2−5d1d2)+(a1b2+b1a2+5c1d2+5d1c2)i√√√+(a1c2+c1a2−d1b2−b1d2)5+(a1d2+d1a2+c1b2+b1c2)5i∈Q(i,5)√√√0=a+bi+c5+d5i∈Q(i,5)√√(a+bi+c5+d5i)−1a(a2+5c2+b2+5d2)−5c(2ac+2bd)5d(2ac+2bd)−b(a2+5c2+b2+5d2)=(a2+5c2+b2+5d2)2−5(2ac+2bd)2+(a2+5c2+b2+5d2)2−5(2ac+2bd)2ic(a2+5c2+b2+5d2)−a(2ac+2bd)√d(a2+5c2+b2+5d2)+b(2ac+2bd)√√+(a2+5c2+b2+5d2)2−5(2ac+2bd)25+(a2+5c2+b2+5d2)2−5(2ac+2bd)25i∈Q(i,5)√Q(i,5)(2)Ex2Aut(F:Q)=Aut(F)φ∈Aut(F),a+bi∈Q(i)φ(a+bi)=a+bφ(i)φφ(i)i·i=−1φ(i·i)=φ(−1)=−1φ(i)·φ(i)=−1φ(i)=iφ(i)=−iφ(i)=iφ:Q(i)→Q(i)a+bi→a+bi5 Q(i)φ(i)=−iφ1:Q(i)→Q(i)a+bi→a−biQ(i)Aut(F)Aut(Q(i))={I,φ1}IQ(i)Ex2Aut(E:Q)=Aut(E)√√√φ∈Aut(E),a+bi+c5+d5i∈Q(i,5)φ(a+bi)=√√a+bφ(i)+cφ(5)+dφ(5)φ(i)√φφ(i)φ(5)i·i=−1φ(i·i)=φ(−1)=−1φ(i)·φ(i)=−1φ(i)=iφ(i)=−i√√√√√√5·5=5φ(5·5)=φ(5)=5φ(5)·φ(5)=5√√√√φ(5)=5φ(5)=−5√Aut(Q(i,5))4√√I:Q(i,5)→Q(i,5)√√√√a+bi+c5+d5i→a+bi+c5+d5i√√φ1:Q(i,5)→Q(i,5)√√√√a+bi+c5+d5i→a−bi+c5−d5i√√φ2:Q(i,5)→Q(i,5)√√√√a+bi+c5+d5i→a+bi−c5−d5i√√φ3:Q(i,5)→Q(i,5)√√√√a+bi+c5+d5i→a−bi−c5+d5i√I,φ1,φ2,φ3∈Aut(Q(i,5))√Aut(Q(i,5))={I,φ1,φ2,φ3}φ∈Aut(E:F)∀a∈Q,φ(a)=a,φ(i)=iAut(E:F)⊂Aut(E)Aut(E:F)I,φ2Aut(E:F)={I,φ2}6 §33.1.P11,Ex1123123Sf={,}1233213.2.P11,Ex2x3x12f(x,x,x)=x3x+x3x+x3x+x3x+x3x+x3x1231221133123323.3.P11,Ex3f(x,y)=0K(a,b)f(a,b)=0f(x,y)f(x,y)=f(y,x)f(b,a)=0(a,b)Kx−y=0(b,a)KKx−y=03.4.P11,Ex4√E±2f=x2−2EEf=x2−2√F±2E⊂F√FQ⊂F2∈FF√√a,b∈Q⊂F,2∈Fa+b2∈F√E⊂FE±2Ef=x2−27 §11.1.P17,Ex1(G,·)a∈Gx∈Gax=xa=ee(G,·)Gab=ac⇒x(ab)=x(ac)⇒(xa)b=(xa)c⇒eb=ec⇒b=cba=ca⇒(ba)x=(ca)x⇒b(ax)=c(ax)⇒be=ce⇒b=c(G,·)1.2.P17,Ex2(1)(S,·)x∈SxS=Se1∈Sxe1=xy∈GSy=Sz∈Gzx=yye1=(zx)e1=z(xe1)=zx=yx∈SSx=Se2∈Se2x=xy∈GyS=Sz∈Gxz=ye2y=e2(xz)=(e2x)z=zx=ye1=e2e1=e2=e(S,·)1 y∈SyS=Sy=Sy1,y2∈Sxy1=y2x=ey1=ey1=(y2x)y1=y2(xy1)=y2e=y2y1=y2y(S,·)(2)(S,·)a∈Sfa:S→S,x→axfaSSfa(S)=aSx,y∈Sfa(x)=fa(y)ax=ayx=yfaSSSfaaS=Sga:S→S,x→xagaSSga(S)=Sax,y∈Sga(x)=ga(y)xa=yax=ygaSSSgaSa=S(1)(S,·)(A)1.3.P17,Ex3φ−1HGx,y∈Ha,b∈Gφ(a)=x,φ(b)=y,φ(a·b)=φ(a)×φ(b)=x×yφ−1(x)=a,φ−1(y)=b,φ−1(x×y)=a·b=φ−1(x)·φ−1(y)φ−1(H,×)(G,·)1.4.P17,Ex4(1)(Z,⊕)(i)⊕2 (ii)(iii)1a∈Za⊕1=a+1−1=a=1⊕a(iv)a∈Z−a+2∈Za⊕(−a+2)=a+(−a+2)−1=1=(−a+2)⊕a(v)(Z,⊕)(2)φZZa,b∈Zφ(a+b)=a+b+1=(a+1)+(b+1)−1=(a+1)⊕(b+1)=φ(a)⊕φ(b)φ(Z,+)(Z,⊕)§22.1.P22,Ex1mnl0≤rHGH={e}=H={e}e=b∈GG=l∈Zb=alHa−l=b−1∈HM={k|ak∈H,k∈N+}M=∅m=minMH=∀h∈Hk∈Zh=akq,r∈Z,0≤r<0k=qm+rak=aqm+r=aqmar=(am)qarar=ak(am)−qHak∈H,am∈H,(am)−q∈Har∈Hmr=0h=ak=(am)qH=3.3.P27,Ex35 1)Gma∈Gam=eaGmnan=em|nnss(as)(s,n)=(an)(s,n)=e(s,n)=ek∈Z(as)k=e=askann|(sk)n|sk(n,s)=1(s,n)(s,n)(s,n)(s,n)n|kasn(s,n)(s,n)2)1)a(s,n)n=nasa(s,n)((s,n),n)(s,n)3)nl,k∈Z(s,n)(s,n)=ls+kna(s,n)=als+kn=(as)l(an)k=(as)l∈⊆=3.4.P27,Ex4GG={e,a,b,b2,ab,ba}·|eabb2abba−|−−−−−−e|eabb2abbaa|aeabbabb2b|bbab2eaabb2|b2abebbaaab|abb2baaebba|babaabb2e3.5.P27,Ex51)t−(ij),ip=P7.2.P46,Ex2HG2GHH,aHa∈HaH=H=Ha⇔a∈Hx∈GxH=H⇒x∈H⇒Hx=H⇒xH=HxxH=aH⇒x∈H⇒Hx=H⇒Hx=Ha⇒xH=HxHG7.3.P46,Ex31)[a1]=[a2],[b1]=[b2]p|(a1−a2),p|(b1−b2)a1b1−a2b2=a1b1−a2b1+a2b1−a2b2=b1(a1−a2)+a2(b1−b2)p|(a1b1−a2b2)[a1b1]=[a2b2][a],[b],[c]∈Zp{[0]}([a][b])[c]=[ab][c]=[(ab)c]=[a(bc)]=[a][bc]=[a]([b][c])[1]∈Zp{[0]}[a]∈Zp{[0]}14 [a][1]=[1][a]=[a]Zp{[0]}[b]∈Zp{[0]}b=kp(b,p)=1l,m∈Zlb+mp=1[lb+mp]=[1],[l][b]+[m][p]=[1],[l][b]=[1][l][b]Zp{[0]}p−1p−1(Zp{[0]},·)2)a∈Za=kpp|(ap−a)ap≡a(modp)a=kp[a]∈Zp{[0]}(Zp{[0]},·)p−1[a]p−1=[1][ap−1]=[1][ap−1][a]=[a],[ap]=[a]p|(ap−a),ap≡a(modp)7.4.P46,Ex4g,h∈GgSg−1=hSh−1⇔h−1gSg−1h=S=(h−1g)S(h−1g)−1⇔h−1g∈N(S)⇔gN(S)=hN(S)gSg−1=hSh−1⇔g,hGN(S)GS{gSg−1|g∈G}GN(S)|O(S)|=[G:N(S)]§88.1.P52,Ex115 1)hi∈Hi,hj∈HjHi,HjG−1−1hi∈Hi,hj∈Hj2)−1−1−1−1(hihj)(hihj)=(hihi)(hjhj)=e(h−1h−1)−1=hhijij(h−1h−1)−1=(h−1)−1(h−1)−1=hhijjijihihj=hjhi2)HiGg∈G1)gj∈Hj,j=1,2,...,ng=g1···gi−1gigi+1···gn1)g=g1···gi−1gi+1···gngihi∈Hi1)(g1···gi−1gi+1···gn)hi=hi(g1···gi−1gi+1···gn)hihiHi=Hihi=HigHi=(g1···gi−1gi+1···gngi)Hi=(g1···gi−1gi+1···gn)Hi=Hi(g1···gi−1gi+1···gn)=Hi(gig1···gi−1gi+1···gn)=Hi(g1···gi−1gi+1···gngi)=HigHiG3)GHi1,Hi2,...,Hin(3.1)G=Hi1Hi2···Hin1)i,jHiHj=HjHiG=H1H2···Hn=Hi1Hi2···Hin(3.2)h,h∈H,j=1,2,...,n1)ijijij(hh···h)(hh···h)=(hh···h)(hh···h)i1i2ini1i2in12n12n=(hh)(hh)···(hh)=(hh)(hh)···(hh)1122nni1i1i2i2inin16 (3.3)g=gg···g=gg···gg,g∈H,i1i2ini1i2inijijijj=1,2,...,n1)g=gg···g=gg···gi1i2ini1i2in=gg···g=gg···g,g,g∈H,j=1,2,...,n12n12njjjc)g=g,j=1,2,...,ng=g,j=1,2,...,njjijijGHi1,Hi2,...,Hin4)c)gigφiGHi(4.1)gi∈Hig=e···egie···e∈Gφi(g)=gii−1n−iφi(4.2)g,h∈G,g=g1g2···gn,h=h1h2···hn,gj,hj∈Hjb)gh=(g1h1)(g2h2)···(gnhn)φi(gh)=gihi=φi(g)φi(h)φiφiGHi(4.3)g=g1···gi−1gi+1···gn∈H1···Hi−1Hi+1···Hng=g1···gi−1egi+1···gn∈H1···Hi−1HiHi+1···Hnφi(g)=e,g∈Kerφig∈Kerφig=g1···gi−1gigi+1···gnφi(g)=gi=eg=g1···gi−1egi+1···gn=g1···gi−1gi+1···gn∈H1···Hi−1Hi+1···HnKerφi=H1···Hi−1Hi+1···Hn8.2.P53,Ex2h1h2···hn(h1,h2,...,hn)φGG(1)g,h∈Gg=(g1,g2,...,gn),h=(h1,h2,...,hn)gh=(g1h1,g2h2,...,gnhn)φ(gh)=(g1h1)(g2h2)···(gnhn)=(g1g2···gn)(h1h2···hn)=φ(g)φ(h)φGG(2)g∈Ggi∈Hi,i=1,2,...,ng=g1g2···gng=(g1,g2,...,gn)∈Gφ(g)=g17 φGG(3)g,h∈Gg=(g1,g2,...,gn),h=(h1,h2,...,hn)φ(g)=g1g2···gn,φ(h)=h1h2···hnφ(g)=φ(h)g1g2···gn=h1h2···hn∈GGHic)gi=hii=1,2,...,n(g1,g2,...,gn)=(h1,h2,...,hn)φGGφGG8.3.P53,Ex3G=H1H2GH1φ1:G→H1g→h1φ1GH1Kerφ1=H2H∼=G/H12H∼=G/H12H∼=H11KleinK={e,a,b,ab}ea2=b2=(ab)2=eH=,H=,H=K=HH=HH1121212H∼=HH=H11118.4.P53,Ex4nn=1n=2G=aimi,i=1,2(m1,m2)=1G|G|≤||·||=m1m218 aa∈G(aa)m1m2=(a)m1m2(a)m1m2=e(aa)12121212mm(aa)ll|mm(aa)l=12121212(a)l(a)l=ee=e·eG=(a)l(a)l=e121212(a)l=(a)l=em|l,m|llm,ml121212l=[m1,m2]m1,m2m1,m2l=[m1,m2]=m1m2a1a2Gm1m2n=kn=k+1H=···H,...,G=H·Hm1···mkmk+1(m1···mk,mk+1)=1G=<(a1···ak)·ak+1>(m1···mk·mk+1)§99.1.P58,Ex1:S2H={e,σ}σ2=eenHSα∈Sασα−1∈Hnnσ=γ1γ2···γsγ1,γ2,···,γsγitiγ1γ2···γt[t,···,t]t,...,tσ2=e1s1st1=···=ts=2σ=(i1i2)(i3i4)···(ij−1ij)n≥33−(iii)α=(iii)α−1=(iii)123123321ασα−1=(iii)(ii)(ii)···(ii)(iii)1231234j−1j321=(i1i2i3)(i1i2)(i3i4)(i3i2i1)···(ij−1ij)19 =(i1i3)(i2i4)···(ij−1ij)∈HHSn9.2.P58,Ex2Sn,An,{(1)}Snn=1,2n=3|S3|=62,33A32S32H={(1),(i1,i2)}(i1i2i3)∈S3(iii)(ii)(iii)−1=(ii)∈H1231212313HS3n≥5A5HSnSn,An,{(1)}HAnHAnHAnAnK=H∩Anα∈Sn,β∈Kβ∈H,αβα−1∈Hβ∈A,αβα−1∈Annαβα−1∈KKSnKAnK={(1)}K=AnK=AnKHAnHAn2HAnHSn(|An||Sn|)K={(1)}K={(1)}Hα,β∈Hαα=αβ=(1)H2Ex1SnSn,An,{(1)}20 §10§1111.1.P71,Ex1(1)g∈G,H∈MgHg−1∈MG(gHg−1)(gHg−1)=gHHg−1=gHg−1,(gHg−1)−1=(g−1)−1H−1g−1)H∈Me∈Ge×H=eHe−1=Hg,h∈G,H∈Mg×(h×H)=g×(hHh−1)=g(hHh−1)g−1=(gh)H(gh)−1=(gh)×H(2)HG⇔g∈GgHg−1=H⇔g∈Gg∈SH⇔SH=G11.2.P71,Ex2(1)g∈GgA∈MgAGmφ:A→gAa→gaa1,a2∈Gga1=ga2a1=a2φx∈gAa∈Ax=gaφ(a)=ga=xφA∈MeGe×A=eA=AA∈M,g,h∈Gg×(h×A)=g×(hA)=g(hA)=(gh)A=(gh)×A(2)∀s∈SA,a∈As×a=sa∈AASA−21 A∼:x∼y⇔s∈SAsx=y∼AxSA−AOxx∼A=Oxx∈Ax∈A|SA|=|Ox|SAOxφ:SA→Oxs→sxs1,s2∈SAs1x=s2x⇒s1=s2φy∈Oxs∈SAy=sxs∈SAφ(s)=sx=yφx,y∈A|SA|=|Ox|=|Oy|AOx|A||Ox||SA||m§1212.1.P71,Ex1[s,t][s,t]ss[s,t]s(1)a=(a)s∈⇒⊆⊆⊆∩x∈∩x∈kx=aksx∈lx=altx=ams|m,t|mnx=a[s,t]n=(a[s,t])n∈∩⊆∩=(2)as∈a(s,t),at∈a(s,t)⊆⊆·⊆x∈lx=a(s,t)lm,nms+nt=(s,t)22 x=a(s,t)l=(a(s,t))l=(a(ms+nt))l=(as)ml·(at)nl∈·⊆··=12.2.P71,Ex2G2a∈Ga2=e,eGa∈Ga−1=aa,b∈G(ab)−1=ab(ab)−1=b−1a−1=baab=baG12.3.P71,Ex3H⊆C(G)HGG/HG/HG/H=,a∈Gx,y∈Gx,yx,yk,lx=(a)k,y=(a)lc,c∈C(G)x=akc,y=alc1212xy=akcalc=akalcc=alakcc=alcakc=yxG1212122112.4.P72,Ex4Gp2G1,p,p2Gp2GGp2Gpa,b∈Gap=bp=e0·GH=·G,H23 pk||H|,pl||H|,|H|Hpk,pl(pk,pl)=1Hpkplakalakalpkplpk,plakal1,pk,pl,pkpl−1akal1akal=eak=alaap(aa)pk=apkapk=apk=eklkklkllaap(aa)pl=aplapl=apl=ekllklklkakalpkplH=a1a2···atp1p2···ptH=····HGHpk||H|,|H|Hpi,i=1,2,...,t|H|=p1p2···pta1a2···atp1p2···pts=pi1···pim,1≤i1≤...≤im≤t(a)s=e(aa···a)s=(a···a)s=eil12tj1jt−mK=···KGKl=pj1···pjt−m(aj1···ajt−m)Kss|la1a2···atp1p2···ptGa1a2···at12.6.P72,Ex612.7.P72,Ex7A,BGA∩Bh∈H,x∈ABh×x=hx×HABABx∼y⇔h∈Hh×x=y∼ABx∈ABOx={hx|∀h∈H}xx∼AB=Oxx∈AB24 x∈AB|Ox|=|A∩B|HOxφ:H→Oxh→hx(1)h1x=h2x⇒h1=h2φ(2)y∈Oxh∈Hhx=yφ|AB||A∩B|ABABAx∈A,h∈H⊂Ahx∈AOxAy∈BOyBai∈A,bi∈BOa1=Oa2,Ob1=Ob2h∈Hha1=−1−1a2,hb1=b2a2a1∈H,b2b1∈HOa1b1=Oa2b2h∈H−1−1−1ha1b1=a2b2⇒a2ha1=b2b1⇒b2b1∈H⇒ABAB|AB||A||B||A||B|=|AB|=|A∩B||A∩B||A∩B||A∩B|12.8.P72,Ex8H⊆K⊆G[G:H][G:K]≤[G:H],[K:K]≤[G:H][G:K]=s,[K:H]=tk1H,k2H,...,ktHK/HtklH=kmH⇔kl=kmg1K,g2K,...,gsKG/KsglK=gmK⇔gl=gmgikjHG/Hi=1,2,...,s,j=1,2,...,t−1x∈Ggi∈GgiK=xKgix∈K−1kj∈KgixH=kjHxH=gikjHG/H⊆{gikjH|i=1,2,...,s;j=1,2,...,t}25 gkH=gkH(gk)−1(gk)∈H⊆Kijlmlmijk−1g−1gk∈K⇒g−1g∈K⇒gK=gK⇒g=gmlijlilili⇒kjH=kmH⇒kj=kmG/Hst[G:H]=[G:K][K:H]12.9.P72,Ex9Ex7,8[G:A∩B]=[G:A][A:A∩B][G:A∩B]=[G:B][B:A∩B]{x(A∩B)|x∈A}{yB|y∈AB}{x(A∩B)|x∈A}{yB|y∈AB}φ:{x(A∩B)|x∈A}→{yB|y∈AB},x(A∩B)→xB−1(1)x1(A∩B)=x2(A∩B)⇒x2x1∈A∩B⊆B⇒x1B=x2B⇒φ−1−1(2)y1B=y2B,yi∈A,i=1,2⇒y2y1∈B⇒y2y1∈A∩B⇒y1(A∩B)=y2(A∩B)φ(3)yB∈{yB|y∈AB}z∈AzB=yBφ(z(A∩B))=zB=yBφ{yB|y∈AB}tt≤[G:B][G:A∩B]=[G:A][A:A∩B]=[G:A]t≤[G:A][G:B]t=[G:B]⇔G=AB⇔AB=BA=G12.10.P72,Ex10φ∈Aut(G)φ(G)=G=φφ(a)G=Ga,a−1φ(a)φ(a)=aφ(a)=a−1Aut(G)={I,φ}φ(a)=a−126 G=G=nak∈G,0na=0k−i>m01m01nibk−i=0R[x]0=a+ax+...+axn,0=b+bx+...+bxm∈R[x]01n01man=0,bm=0(a+ax+...+axn)·(b+bx+...+bxm)01n01mk=(ab)+(ab+ab)x+...+(ab)xk+...+abxn+m=0000110ik−inmi=0R[x]R[x]3.4.P98,Ex4a1b1a2b2(1)T=∅,∈T0c10c2a1b1a2b2a1+a2b1+b2+=∈T0c10c20c1+c2a1b1−a1−b1−=0c10−c1(T,+)(M2(Z),+)a1b1a2b2a1a2a1b2+b1c2·=∈T0c10c20c1c2(T,·)(T,+,·)(M2(Z),+,·)7 (2)(I,+)(M2(Z),+)ab02d∈T,∈I0c00ab02d02ad02dab02cd·=∈I,·=∈I0c0000000c00ITI(M2(Z),+)ab(3)T/I0ca1b1a2b2a1b1a2b2=⇔−∈I0c10c20c10c2⎧⎪⎪⎨a1=a2a1−a2b1−b2⇔∈I⇔c1=c20c1−c2⎪⎪⎩b1−b2∈2Zx1x2T/I=,x,y∈Z0y0y3.5.P98,Ex5(1)In∈D(R)D(R)=∅D(R)+D(R)=D(R),−D(R)=D(R),D(R)·D(R)=D(R)D(R)Mn(R)(2)RD(R)⊆C(D(R))Ei(i,i)10nEii∈D(R)A=(aij)n∈C(D(R))EiA=AEi8 ⎛⎞0···0···0⎜⎟⎛⎞⎜......⎟0···0a0···0⎜.···.···.⎟1i⎜⎟⎜...···.........···...⎟⎜⎜0···0···0⎟⎟⎜⎜⎟⎟⎜⎟⎜⎟⎜ai1···aii···ain⎟=⎜0···0aii0···0⎟⎜⎟⎜⎟⎜⎟⎜..........⎟⎜0···0···0⎟⎝.···...···.⎠⎜...⎟⎜⎝..···..···..⎟⎠0···0ani0···00···0···0⇒i=j,aij=0⇒A∈D(R)C(D(R))=D(R)§44.1.P104,Ex1(1)Rp⇒r∈Rpr=0Rpn(a+b)pn=Ckkbpn−kpnak=00