Ring Theory Problem Set 1 { Solutions be a ring with unity ...

1 ring Theory Problem Set 1 SolutionsProblem a ring with unity 1. Show that ( 1)a= afor alla :We have 1 + ( 1) = 0 by definition. Multiplying that equation on the rightbya, we obtain(1 + ( 1)) a= 0 a= 0by theorem , part i. By the distributive law, we obtain the equation1 a+ ( 1) a= 0and therefore we havea+( 1)a= 0. We also havea+( a) = 0. Thus,a+( 1)a=a+( a).The ringRunder addition is a group. The cancellation law in that group implies that a= ( 1)awhich is the result we wanted to a field and leta, b F. Assume thata6= 0, Show that thereexists an elementx Fsatisfying the equationax+b= :SinceFis a field anda6= 0, there exists an elementa 1inFsuch thataa 1= 1.

2 Letc= b. Letx=a 1c. Thenx Fsince botha 1andcare inF. We haveax+b =a(a 1c) +b= (aa 1)c+b= 1c+b=c+b= the elementxinFchosen above has the property thatax+b= all units, zero-divisors, and nilpotent elements in the ringsZ Z,Z3 Z3, andZ4 ;In general, ifR1andR2are rings with unity , then so isR1 R2. Theunity element is (1R1,1R2). An element (a1, a2) inR1 R2is a unit if and only if thereis an element (b1, b2) inR1 R2such that (a1, a2)(b1, b2) = (1R1,1R2). By definition,(a1, a2)(b1, b2) = (a1b1, a2b2). Therefore, the element (a1, a2) is a unit if and only if thereexists elementsb1 R1andb2 R2such thata1b1= 1R1anda2b2= 1R2.

3 This means that(a1, a2) is a unit inR1 R2if and only ifa1is a unit inR1anda2is a unit units inZare 1 and -1. The units inZ3are 1 and 2. The units inZ4are 1 and 3. Theunits inZ6are 1 and 5. Therefore,The units inZ Zare (1,1),(1, 1),( 1,1), and ( 1, 1).The units inZ3 Z3are (1,1),(1,2),(2,1), and (2,2).The units inZ4 Z6are (1,1),(1,5),(3,1), and (3,5).Suppose that (a1, a2) is an element ofR1 R2and thatnis a positive integer. Thenwe clearly have (a1, a2)n= (an1, an2). The additive identity inR1 R2is (0R1,0R2). Theequation (a1, a2)n= (0R1,0R2) is equivalent to the two equationsan1= 0R1andan2= , if (a1, a2) is a nilpotent element ofR1 R2, then it follows thata1is anilpotent element inR1anda2is a nilpotent element inR2.

4 The converse is true too. Tosee this, assume thata1is a nilpotent element inR1anda2is a nilpotent element , by definition, there exists positive integerseandfsuch thatae1= 0R1andaf2= Thennis a positive integer and we havean1=aef1= (ae1)f= 0fR1= 0R1andan2=af e2= (af2)e= 0eR2= 0R2 Therefore, (a1, a2)n= (0R1,0R2) and hence (a1, a2) is a nilpotent element ofR1 R2. Insummary, we have shown that (a1, a2) is a nilpotent element ofR1 R2if and only ifa1isa nilpotent element inR1anda2is a nilpotent element only nilpotent element ofZis 0. The only nilpotent element ofZ3is 0. The nilpotentelements ofZ4are 0 and 2. The only nilpotent element ofZ6is 0.

5 It follows thatThe only nilpotent element inZ Zis (0,0). The only nilpotent element inZ3 Z3is (0,0).The nilpotent elements inZ4 Z6are (0,0) and (2,0).Suppose that (a1, a2) is an element ofR1 R2. Then (a1, a2) is a zero-divisor if andonly if there exists an element (b1, b2) inR1 R2such that(b1, b2)6= (0R1,0R2)and(a1, a2)(b1, b2) = (0R1,0R2).The second equation just means thata1b1= 0R1anda2b2= 0R2. Also, (b1, b2)6= (0R1,0R2)means thatb16= 0R1orb26= 0R2. Consequently, it follows that if (a1, a2) is a zero-divisor inR1 R2, then eithera1is a zero divisor inR1ora2is a zero divisor inR2. For the converse,suppose thata1is a zero-divisor inR1.

6 Thena1b1= 0R1for some nonzero elementb1 follows that(b1,0R2)6= (0R1,0R2)and(a1, a2)(b1,0R2) = (0R1,0R2).Therefore, (a1, a2) is a zero-divisor inR1 R2. A similar argument shows that ifa2is azero-divisor inR2, then (a1, a2) is a zero-divisor inR1 R2. In summary, we have shownthat (a1, a2) is a zero-divisor inR1 R2if and only if eithera1is a zero divisor inR1ora2is a zero divisor only zero-divisor inZis 0. The only zero-divisor inZ3is 0. The zero-divisors inZ4are 0 and 2. The zero-divisors inZ6are 0, 2, 3 and 4. The above remark shows thatThe set of zero-divisors inZ Zis{(a,0) a Z} {(0, b) b Z}.The set of zero-divisors inZ3 Z3is{(a,0) a Z3} {(0, b) b Z3}.

7 The set of zero-divisors inZ4 Z6is{(a, b) a Z4, b= 0,2,3, or4} {(a, b) b Z6, a= 0or2.}. Problem , part (a)Show that the multiplicative identity in a ring with unityRis :Suppose thate Rand thatea=a=aefor alla R. Suppose also thatf Rand thatfa=a=affor alla R. Then we havef=ef=eTherefore,e=f. Thus, there can only be one element inRsatisfying the requirements forthe multiplicative identity of the , part (b)Suppose thatRis a ring with unity and thata Ris a unitofR. Show that the multiplicative inverse ofais :Suppose thatb, c Rand thatab=ba= 1 and thatac=ca= 1. Thenwe havec= 1c= (ba)c=b(ac) =b1 =b .Hence we havec=b.

8 The multiplicative inverse ofais indeed PROBLEMS:A:Prove that ifRis a division ring , then the center ofRis a :First of all, suppose thatRis any ring with identity. LetSbe the center ofR. That is,S={s R|sr=rsfor allr R}.We will show thatSis a subring fact thatSis a subgroup ofRunder addition can be seen as follows. For thispurpose, suppose thats1, s2 S. Then, for allr R, we haves1r=rs1ands2r= , using the distributive laws forR, we have(s1+s2)r=s1r+s2r=rs1+rs2=r(s1+s2)for allr R. Therefore,s1+s2 S. Furthermore, letting 0 denote the additive identity ofR, we have 0 r= 0 andr 0 = 0. Hence 0 r=r 0. Therefore, 0 , suppose thats S.

9 Lett= s, the additive inverse ofsinR. We haves+t= ,s+t S. Sincesis inSands+tis inS, it follows that, for allr R, we havesr=rsand (s+t)r=r(s+t). Therefore, we havesr+tr=rs+rt=sr+rtThus, we have the equationsr+tr=sr+rt. Applying the cancellation law for the underlyingadditive group ofRto that equation, it follows thattr=rtfor allr R. Therefore,t is, s S. This completes the verification thatSis a subgroup ofRunder theoperation of complete the proof thatSis a subring ofR, we must show that ifs1ands2are inS, then so iss1s2. So, assume thats1, s2 S. Then, for allr R, we haves1r=rs1ands2r=rs2. Considers1s2, which is an element ofR.

10 Using the associative law formultiplication inRmany times, it follows that(s1s2)r=s1(s2r) =s1(rs2) = (s1r)s2= (rs1)s2=r(s1s2)for allr R. Therefore, we indeed haves1s2 have shown thatSis a subring a ring with unity 1, then 1r=r=r1 for allr R. Therefore 1 S. HenceSisa ring with we assume thatRis a division ring . Then, by definition,Ris a ring with unity 1,16= 0, and every nonzero element ofRis a unit ofR. Suppose thatSis the center , as pointed out above, 1 Sand henceSis a ring with unity . Also, 0 is the additiveidentity ofRand is also the additive identity of the ringS. We have 16= 0. We now provethatSis a division ring .