it is clear that σ−1 exists, and so σ is a bijection Thus, σ is a ring isomorphism Theorem 16 2 Let f : R → S be a homomorphism of rings Then (i) f(0R)=0S
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[PDF] Chapter 4 Homomorphisms and Isomorphisms of Groups
In each case, φ is a homomorphism since ak+l = akal and φ is bijective by Theorem 2 3 since for a ∈ R∗ we have det(diag(a,1,1,···,1)) = a 4 8 Example: The map φ : R → R+ given by φ(x) = ex is a group isomorphism since it is bijective and φ(x + y) = ex+y = exey = φ(x)φ(y)
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it is clear that σ−1 exists, and so σ is a bijection Thus, σ is a ring isomorphism Theorem 16 2 Let f : R → S be a homomorphism of rings Then (i) f(0R)=0S
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LECTURE 16
Homomorphisms and Isomorphisms of Rings
Having now seen a number of diverse examples of rings, it is appropriate at this point to see how two
dierent sets might be endowed with essentially the same ring structure. Consider a setRconsisting of two elementsa;bwith the following addition and multiplication tables +a b aa b bb a a b aa a ba b It is easy to verify thatRhas a structure of a commutative ring with identity if 0R=aand 1R=b.On the other hand,Z2=f[0];[1]gis another commutative ring with identity consisting of only two elements.
If we write down the addition and multiplication tables forZ2 +[0] [1] [0][0] [1] [1][1] [0] [0] [1] [0][0] [0] [1][0] [1] we see thatZ2has about the same structure as that ofRonce we recognize the correspondencesa$[0],b$[1]. In such a case, when two setsRandShave a virtually identical ring structure, we shall say that
RandSareisomorphic. Below we formalize this concept a little more precisely. Recall that a mapffrom a setRto a setSisinjectiveif f(r) =f(r0))r=r0: f:R!Sissurjectiveif every element inScan be expressed ass=f(r) for somerinR. Finally, f:R!Sisbijectiveiffis both injective and surjective. Finally, we recall that iffis bijective, thenf has a inverse; i.e., there exists a (unique) functionf1:S!Rsuch that ff1(x)=x=f1(f(x))8x2R : Definition16.1.A mapf:R!Sbetween two rings is a called ahomomorphismif: (i)f(r+r0) =f(r) +f(r0);8r;r02R; (ii)f(rr0) =f(r)f(r0);8r;r02R : fis said to be anisomorphismif it is also bijective. Example 1.Consider the map:C!Cwhere(x+iy) =xiy(i.e.,is complex conjugation inC).Then ifz1=x1+iy1andz2=x2+iy2,
(16.1) (z1+z2) =(x1+x2+i(y1+y2)) =x1+x2i(y1+y2) =x1iy1+x2iy2 =(z1) +(z2) 6016. HOMOMORPHISMS AND ISOMORPHISMS OF RINGS 61
(16.2) (z1z2) =(x1x2y1y2+i(x1y2+y1x2)) =x1x2y1y2i(x1y2+y1x2) = (x1iy1)(x2iy2) =(z1)(z2)Thus,is a homomorphism of rings. Since
2==Identity map
it is clear that1exists, and sois a bijection. Thus,is a ring isomorphism. Theorem16.2.Letf:R!Sbe a homomorphism of rings. Then (i)f(0R) = 0S. (ii)f(r) =f(r)for everyr2R. Moreover, ifRandShave identities andfis an surjective homomorphism, then (iii)f(1R) = 1S. (iv)Whenevera2Ris a unit ofR, thenf(a)is a unit ofSandf(a)=1=fa1.Proof.
(i) Sincefis a homomorphism and 0R+ 0R= 0RinR, f(0R) +f(0R) =f(0R+ 0R) =f(0R): Addingf(0R)2Sto both sides of this equation yields f(0R) = 0S: (ii) f(r) +f(r) =f(r+ (r)) =f(0R) = 0Sby (i)
Hence f(-r) is the additive inverse of f(r) inS; i.e.,f(r) =f(r). (iii) Sincefis surjective, 1S=f(r) for somer2R. Therefore, f(1R) =f(1R)1S=f(1R)f(r) =f(1Rr) =f(r)1S: (iv) Supposeais a unit inRwith multiplicative inversea1. Then by (iii) 1S=f(1R) =fa1a=fa1f(a)
and sof(a) is a unit inSwith multiplicative inversefa1. Corollary16.3.Letf:R!Sbe a ring homomorphism. Then the image offinS image(f) =fs2Sjs=f(r)for somer2Rg is a subring ofS. Proof.From the fact thatfis a ring homomorphism it follows thatimage(f) is closed under addition and multiplication: s;s02image(f)) 9r;r02R s:t: s=f(r); s0=f(r0)
)s+s0=f(r) +f(r0) =f(r+r0)2image(f)16. HOMOMORPHISMS AND ISOMORPHISMS OF RINGS 62
s;s02image(f)) 9r;r02R s:t: s=f(r); s0=f(r0)
)ss0=f(r)f(r0) =f(rr0)2image(f)From part (ii) of the preceding theorem we have
s=f(r) for somer2R) s=f(r)2image(f)Sinceimage(f) is closed under addition, multiplication, and taking additive inverses, we have by Theorem