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RING HOMOMORPHISMS AND THE ISOMORPHISM THEOREMS

BIANCA VIRAY

When learning about groups it was helpful to understand how dierent groups relate to each other. We would like to do so for rings, so we need some way of moving between dierent rings. Denition 1.LetR= (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;r22R, (2)(r1Rr2) =(r1)S(r2)for allr1;r22R, and (3)(1R) = 1S: For simplicity, we will often write conditions(1)and(2)as(r1+r2) =(r1) +(r2)and (r1r2) =(r1)(r2)with the particular addition and multiplication implicit. Remark 1.If: (R;+;)!(S;+;)is a ring homomorphism then: (R;+)!(S;+)is a group homomorphism. Example 1.IfRis any ring andSRis a subring, then the inclusioni:S ,!Ris a ring homomorphism.

Exercise 1.Prove that

':Q!Mn(Q); '(a) =0 B

B@a0:::0

0a :::0............

0 0::: a1

C CA is a ring homomorphism.

Exercise 2.LetFbe a eld and leta2F. Prove that

':F[x]!F; '(f(x)) =f(a) is a ring homomorphism. Exercise 3.Letn2Zbe a positive integer. Prove that ':Z!Z; '(a) =na isnota ring homomorphism. Exercise 4.LetRbe a ring and letIbe an ideal. Prove that ':R!R=I; '(r) =r+I is a ring homomorphism. Exercise 5.Determine if the following maps are homomorphisms. 1 (1): M2(R)!R; a b c d =a (2): M2(R)!R; (A) = Tr(A) (3): M2(R)!R; (A) = det(A) The three denining properties of a ring homomorphism imply other important properties.

Lemma 1.Let:R!Sbe a ring homomorphism. Then

(1)(0R) = 0S, (2)(r) =(r)for allr2R, (3)ifr2Rthen(r)2Sand(r1) =(r)1, and (4)ifR0Ris a subring, then(R0)is a subring ofS. Proof.Statements (1) and (2) hold because of Remark 1. We will repeat the proofs here for the sake of completeness.

Since 0

R+0R= 0R,(0R)+(0R) =(0R). Then sinceSis a ring,(0R) has an additive inverse, which we may add to both sides. Thus we obtain (0R) =(0R) +(0R) +(0R) =(0R) +(0R) = 0S; as desired.

Letr2R. 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 unique. Now letr2R. Then there existsr12Rsuch thatrr1=r1r= 1R. Then since is a ring homomorphism we have (r)(r1) =(r1)(r) =(1R) = 1S: Thus(r) has a multiplicative inverse and it is(r1). Lastly, letR0Rbe a subring. To show that(R0) is a subring we must show that 1 S2(R0) and for alls1;s22(R0),s1s2ands1s2are also in(R0). Sinces1;s22(R0), there existsr1;r22R0such that(r1) =s1and(r2) =s2. Thus s

1s2=(r1)(r2) =(r1)+(r2) =(r1r2);ands1s2=(r1)(r2) =(r1r2):

SinceR0is a subring,r1r2andr1r2are contained inR0. Hences1s2ands1s2are in (R0). Furthermore, 1R2R0so 1S=(1R)2(R0). Therefore,(R0) is a subring ofS. Exercise 6.Let:R!Sbe a ring homomorphism. Ifr2Ris a zero divisor, is(r) a zero divisor inS? If yes, then prove this statement. If no, give an example of a ring homomorphismand a zero divisorr2Rsuch that(r)is not a zero divisor. As in the case of groups, homomorphisms that are bijective are of particular importance. Denition 2.LetRandSbe rings and let:R!Sbe a set map. We say thatis a (ring) isomorphismif (1)is a (ring) homomorphism and (2)is a bijection on sets. We say that two ringsR1andR2areisomorphicif there exists an isomorphism between them. Lemma 2.LetRandSbe rings and let:R!Sbe an isomorphism. Then: 2 (1)1is an isomorphism, (2)r2Ris a unit if and only if(r)is a unit ofS, (3)r2Ris 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 eld if and only ifSis a eld.

Exercise 7.Prove Lemma 2.

Exercise 8.Prove thatZ[x]andR[x]are not isomorphic.

1.Kernel, image, and the isomorphism theorems

A ring homomorphism':R!Syields two important sets. Denition 3.Let:R!Sbe a ring homomorphism. Thekernelofis ker:=fr2R:(r) = 0g R and theimageofis im:=fs2S:s=(r) for somer2Rg S: Exercise 9.LetRandSbe rings and let:R!Sbe a homomorphism. Prove thatis injective if and only ifker=f0g. Theorem 3(First isomorphism theorem).LetRandSbe rings and let:R!Sbe a homomorphism. Then: (1)The kernel ofis anidealofR, (2)The image ofis asubringofS, (3)The map ':R=ker!imS; r+ ker7!(r) is a well-dened isomorphism. Proof.The image ofis a subring by Lemma 1. Let us prove that keris an ideal. By Lemma 1,(0) = 0 so 02kerand hence the kernel is nonempty. Leta;b2kerand let r2R. Then sinceis a homomorphism we have (a+b) =(a) +(b) = 0 + 0 = 0; (ra) =(r)(a) =(r)0 = 0; (ar) =(a)(r) = 0(r) = 0:

Thusa+b;ra;andarare in kerand so keris an ideal.

Consider the map'. We rst show that it is well-dened. Letr;r02Rbe such that rr02ker, i.e., such thatr+ ker=r0+ ker. Then (r) =(r0+ (rr0)) =(r0) +(rr0) =(r0) + 0 =(r0); 3 so'is well dened. Letr1+I;r2+I2R=I. Then sinceis 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) = 1:

Therefore'is a homomorphism.

Let us prove that'is bijective. Ifr+ ker2ker', then'(r+I) =(r) = 0 and so r2keror equivalentlyr+ ker= ker. Thus ker'is trivial and so by Exercise 9,' is injective. Lets2im. Then there exists anr2Rsuch that(r) =sor equivalently that'(r+ ker) =s. Thuss2im'and so'is surjective. Hence'is an isomorphism as desired. Exercise 10.Compute the kernel of'where'is as in(1)Exercise 1,(2)Exercise 2, and (3)Exercise 4. Theorem 4(Second isomorphism theorem).LetRbe a ring, letSRbe a subring, and letIbe an ideal ofR. Then: (1)S+I:=fs+a:s2S;a2Igis 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 + 02S+I. Lets1+a1ands2+a2be elements ofS+I. Then (s1+a1)(s2+a2) = (s1s2)|{z}

2S+(a1a2)|{z}

2Iand (s1+a1)(s2+a2) =s1s2|{z}

2S+s1a2+a1s2+a1a2|{z}

2I:

HenceS+Iis a subring ofR.

(2): The intersectionS\Iis nonempty since 0 is contained inIandS. Leta1;a22S\I and lets2S. Thena1+a22S\IsinceSandIare both closed under addition. Furthermore sa

1anda1sare inS\IsinceIis closed under multiplication fromRSandSis closed

under multiplication. ThereforeS\Iis an ideal ofS. (3): Consider the map:S!(S+I)=Iwhich sends an elementstos+I. This is a ring homomorphism by denition of addition and multiplication in quotient rings. We claim that it is surjective with kernelS\I, which would complete the proof by the rst isomorphism theorem. Consider elementss2Sanda2I. Thens+a+I=s+Isincea2I, so s+a+I2imand henceis surjective. Lets2Sbe an element of ker. Thens+I=I which holds if and only ifs2Ior equivalently ifs2S\I. Thus ker=S\Iand we have our desired result. Theorem 5(Third isomorphism theorem).LetRbe a ring and letJIbe ideals ofR.

ThenI=Jis an ideal ofR=Jand

R=JI=J

=R=I: 4 Proof.SinceIandJare ideals, they are nonempty and soI=J=fa+J:a2Igis also nonempty. Leta1;a22Iand letr2R. By denition of addition and multiplication of cosets, we have (a1+J) + (a2+J) = (a1+a2) +J; (r+J)(a1+J) =ra1+J;and (a1+J)(r+J) =a1r+J: SinceIis an ideal,a1+a2;ra1;anda1rare contained inIsoI=Jis an ideal ofR=J. Consider the map:R=J!R=Ithat sendsr+Jtor+I. We claim that this is a well-dened surjective homomorphism with kernel equal toI=J. (See Exercise 11.) Then (R=J)=(I=J) is isomorphic toR=Iby the rst isomorphism theorem. Exercise 11.We will use the notation from Theorem 5. Prove that the map:R=J! R=I; r+J7!r+Iis a well-dened surjective homomorphism with kernel equal toI=J. Exercise 12.Prove thatQ(p5)is isomorphic toQ[x]=hx22x+ 6i. 5quotesdbs_dbs12.pdfusesText_18

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