RING HOMOMORPHISMS AND THE ISOMORPHISM …

1 ring HOMOMORPHISMS AND THE ISOMORPHISM THEOREMSBIANCA VIRAYWhen learning about groups it was helpful to understand how different groups relate toeach other. We would like to do so for rings, so we need some way of moving betweendifferent (R,+R, R)and(S,+S, S)be rings. A set map :R Sis a( ring )homomorphismif(1) (r1+Rr2) = (r1) +S (r2)for allr1,r2 R,(2) (r1 Rr2) = (r1) S (r2)for allr1,r2 R, and(3) (1R) = simplicity, we will often write conditions(1)and(2)as (r1+r2) = (r1) + (r2)and (r1r2) = (r1) (r2)with the particular addition and multiplication : (R,+, ) (S,+, )is a ring homomorphism then : (R,+) (S,+)isa group any ring andS Ris a subring, then the inclusioni:S Ris a that :Q Mn(Q), (a) = .. a is a ring a field and leta F. Prove that :F[x] F, (f(x)) =f(a)is a ring Zbe a positive integer. Prove that :Z Z, (a) =naisnota ring a ring and letIbe an ideal. Prove that :R R/I, (r) =r+Iis a ring if the following maps are (1) : M2(R) R, ((a bc d))=a(2) : M2(R) R, (A) = Tr(A)(3) : M2(R) R, (A) = det(A)The three definining properties of a ring homomorphism imply other important :R Sbe a ring homomorphism.

2 Then(1) (0R) = 0S,(2) ( r) = (r)for allr R,(3)ifr R then (r) S and (r 1) = (r) 1, and(4)ifR Ris a subring, then (R )is a subring (1) and (2) hold because of Remark 1. We will repeat the proofs here forthe sake of 0R+ 0R= 0R, (0R) + (0R) = (0R). Then sinceSis a ring , (0R) has an additiveinverse, which we may add to both sides. Thus we obtain (0R) = (0R) + (0R) + (0R) = (0R) + (0R) = 0S,as R. Sincer+ r= r+r= 0R, we have (r) + ( r) = ( r) + (r) = (0R) = 0S,where the last equality comes from (1). Thus ( r) = (r) as additive inverses are letr R . Then there existsr 1 Rsuch thatr r 1=r 1 r= 1R. Then since is a ring homomorphism we have (r) (r 1) = (r 1) (r) = (1R) = (r) has a multiplicative inverse and it is (r 1).Lastly, letR Rbe a subring. To show that (R ) is a subring we must show that1S (R ) and for alls1,s2 (R ),s1 s2ands1s2are also in (R ). Sinces1,s2 (R ),there existsr1,r2 R such that (r1) =s1and (r2) =s2.

3 Thuss1 s2= (r1) (r2) = (r1) + ( r2) = (r1 r2),ands1s2= (r1) (r2) = (r1r2).SinceR is a subring,r1 r2andr1r2are contained inR . Hences1 s2ands1s2are in (R ). Furthermore, 1R R so 1S= (1R) (R ). Therefore, (R ) is a subring ofS. Exercise :R Sbe a ring homomorphism. Ifr Ris a zero divisor, is (r)a zero divisor inS? If yes, then prove this statement. If no, give an example of a ringhomomorphism and a zero divisorr Rsuch that (r)is not a zero in the case of groups, HOMOMORPHISMS that are bijective are of particular rings and let :R Sbe a set map. We say that is a( ring ) isomorphismif(1) is a ( ring ) homomorphism and(2) is a bijection on say that two ringsR1andR2areisomorphicif there exists an ISOMORPHISM between rings and let :R Sbe an ISOMORPHISM . Then:2(1) 1is an ISOMORPHISM ,(2)r Ris a unit if and only if (r)is a unit ofS,(3)r Ris a zero divisor if and only if (r)is a zero divisor ofS,(4)Ris commutative if and only ifSis commutative,(5)Ris an integral domain if and only ifSis an integral domain, and(6)Ris a field if and only ifSis a Lemma thatZ[x]andR[x]are not , image, and the ISOMORPHISM theoremsA ring homomorphism :R Syields two important :R Sbe a ring homomorphism.

4 Thekernelof isker :={r R: (r) = 0} Rand theimageof isim :={s S:s= (r) for somer R} rings and let :R Sbe a homomorphism. Prove that isinjective if and only ifker ={0}. theorem 3(First ISOMORPHISM theorem ).LetRandSbe rings and let :R Sbe ahomomorphism. Then:(1)The kernel of is anidealofR,(2)The image of is asubringofS,(3)The map :R/ker im S, r+ ker 7 (r)is a well-defined image of is a subring by Lemma 1. Let us prove that ker is an ideal. ByLemma 1, (0) = 0 so 0 ker and hence the kernel is nonempty. Leta,b ker and letr R. Then since is a homomorphism we have (a+b) = (a) + (b) = 0 + 0 = 0, (ra) = (r) (a) = (r) 0 = 0, (ar) = (a) (r) = 0 (r) = +b,ra,andarare in ker and so ker is an the map . We first show that it is well-defined. Letr,r Rbe such thatr r ker , , such thatr+ ker =r + ker . Then (r) = (r + (r r )) = (r ) + (r r ) = (r ) + 0 = (r ),3so is well defined. Letr1+I,r2+I R/I. Then since is a homomorphism we have: (r1+I+r2+I) = (r1+r2+I) = (r1+r2) = (r1) + (r2)= (r1+I) + (r2+I) ((r1+I)(r2+I)) = (r1r2+I) = (r1r2) = (r1) (r2)= (r1+I) (r2+I) (1 +I) = (1) = is a us prove that is bijective.

5 Ifr+ ker ker , then (r+I) = (r) = 0 and sor ker or equivalentlyr+ ker = ker . Thus ker is trivial and so by Exercise 9, is injective. Lets im . Then there exists anr Rsuch that (r) =sor equivalentlythat (r+ ker ) =s. Thuss im and so is surjective. Hence is an ISOMORPHISM asdesired. Exercise the kernel of where is as in(1)Exercise 1,(2)Exercise 2, and(3)Exercise 4(Second ISOMORPHISM theorem ).LetRbe a ring , letS Rbe a subring, andletIbe an ideal ofR. Then:(1)S+I:={s+a:s S,a I}is a subring ofR,(2)S Iis an ideal ofS, and(3) (S+I)/Iis isomorphic toS/(S I).Proof.(1):Sis a subring andIis an ideal so 1 + 0 S+I. Lets1+a1ands2+a2beelements ofS+I. Then(s1+a1) (s2+a2) = (s1 s2) S+ (a1 a2) Iand (s1+a1)(s2+a2) =s1s2 S+s1a2+a1s2+a1a2 +Iis a subring ofR.(2): The intersectionS Iis nonempty since 0 is contained inIandS. Leta1,a2 S Iand lets S. Thena1+a2 S IsinceSandIare both closed under addition. Furthermoresa1anda1sare inS IsinceIis closed under multiplication fromR SandSis closedunder multiplication.

6 ThereforeS Iis an ideal ofS.(3): Consider the map :S (S+I)/Iwhich sends an elementstos+I. This is a ringhomomorphism by definition of addition and multiplication in quotient rings. We claim thatit is surjective with kernelS I, which would complete the proof by the first isomorphismtheorem. Consider elementss Sanda I. Thens+a+I=s+Isincea I, sos+a+I im and hence is surjective. Lets Sbe an element of ker . Thens+I=Iwhich holds if and only ifs Ior equivalently ifs S I. Thus ker =S Iand we haveour desired result. theorem 5(Third ISOMORPHISM theorem ).LetRbe a ring and letJ Ibe ideals an ideal ofR/JandR/JI/J = ideals, they are nonempty and soI/J={a+J:a I}is alsononempty. Leta1,a2 Iand letr R. By definition of addition and multiplication ofcosets, we have(a1+J) + (a2+J) = (a1+a2) +J,(r+J)(a1+J) =ra1+J,and(a1+J)(r+J) =a1r+ an ideal,a1+a2,ra1,anda1rare contained inIsoI/Jis an ideal the map :R/J R/Ithat sendsr+Jtor+I.

7 We claim that this is awell-defined surjective homomorphism with kernel equal toI/J. (See Exercise 11.) Then(R/J)/(I/J) is isomorphic toR/Iby the first ISOMORPHISM theorem . Exercise will use the notation from theorem 5. Prove that the map :R/J R/I, r+J7 r+Iis a well-defined surjective homomorphism with kernel equal thatQ( 5)is isomorphic toQ[x]/ x2 2x+ 6 .5