Math 371 Lecture #9 x3.3: Isomorphisms and Homomorphisms, Part I When are two different looking rings really the \same" ring? This is the question of \isomorphism" which we explore informally before defining it precisely. Example. The subset S = f0; 2; 4g is a subring of Z6. Here are the addition and multiplication tables for S. + 0 2 4 · 0 2 4 0 0 2 4 0 0 0 0 2 2 4 0 2 0 4 2 4 4 0 2 4 0 2 4 Is S really the same as Z3? Here are the addition and multiplication tables for Z3 = f[0]; [1]; [2]g. + [0] [2] [1] · [0] [2] [1] [0] [0] [2] [1] [0] [0] [0] [0] [2] [2] [1] [0] [2] [0] [1] [2] [1] [1] [0] [2] [1] [0] [2] [1] Notice that the elements of Z3 are written in an unusual order in the tables. Why? Because then the element 0 2 S acts likes [0] 2 Z3, the element 2 2 S acts like [2] 2 Z3, and the element 4 2 S acts like [1] 2 Z3. These tables suggest forming the bijective (i.e., injective and surjective) function f : S ! Z3 by f(0) = [0]; f(2) = [2]; f(4) = [1]: This bijection satisfies the properties (1) f(a + b) = f(a) + f(b) for all a; b 2 S, and (2) f(ab) = f(a)f(b) for all a; b 2 S, which can be verified by checking all of the possible cases for pairs a; b 2 S. Keep in mind here that the addition and multiplication inside f is that of S, while the addition and multiplication outside of f is that of Z3. So one of the cases for checking is a = 4 and b = 2 in S for which f(4 + 2) = f(0) = [0]; f(4) + f(2) = [1] + [2] = [0]; f(4 · 2) = f(2) = [2]; f(4)f(2) = [1][2] = [2]: Although the labels for the elements of S and Z3 are different looking, both rings have the same addition and multiplication tables, i.e., the same ring structure. The bijection f : S ! Z3 gives us the connection between the elements of S and the elements of Z3. This example illustrates the algebraic concept of an isomorphism from one ring to an- other. Definition A ring R is isomorphic to a ring S (written R ∼= S) if there exists a function f : R ! S such that (i) f is injective, (ii) f is surjective, and (iii) f satisfies f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b) for all a; b 2 R. Example. Is there an isomorphism from the ring Z to the ring a 0 S = : a 2 ? 0 0 Z A candidate for an isomorphism is the function f : Z ! S defined by a 0 f(a) = : 0 0 This function is injective and surjective. For a; b 2 Z we have a + b 0 f(a + b) = 0 0 and a 0 b 0 a + b 0 f(a) + f(b) = + = : 0 0 0 0 0 0 Thus f(a + b) = f(a) + f(b) for all a; b 2 Z. Again for a; b 2 Z we have ab 0 f(ab) = ; 0 0 and a 0 b 0 ab 0 f(a)f(b) = = : 0 0 0 0 0 0 Thus f(ab) = f(a)f(b) for all a; b 2 Z. Therefore f : Z ! S is an isomorphism. The multiplicative identity of Z is 1Z = 1. Does S have a multiplicative identity? Yes, it is 1 0 1 = f(1 ) = ; S Z 0 0 which is not the usual 2 × 2 identity matrix I in M(Z). Example. For any ring R the identity map iR : R ! R defined by iR(r) = r is an isomorphism. It is obviously bijective, and for a; b 2 R we have iR(a + b) = a + b = iR(a) + iR(b); iR(ab) = ab = iR(a)iR(b): p p Example. For R = Q( 5) = fa + b 5 : a; b 2 Qg, the map f : R ! R defined by p p f(a + b 5) = a − b 5 p p p is a bijection that, for a1 + b1 5; a2 + b2 5 2 Q( 5), satisfies p p p p f(a1 + b1 5 + a2 + b2 5) = f((a1 + a2) + (b1 + b2) 5) = (a1 + a2) − (b1 + b2) 5; p p p p p f(a1 + b1 5) + f(a2 + b2 5) = a1 − b1 5 + a2 − b2 5 = (a1 + a2) − (b1 + b2) 5; p p p f((a1 + b1 5)(a2 + b2 5)) = f(a1a2 + 5b1b2 + (a1b2 + a2b1) 5) p = a1a2 + 5b1b2 − (a1b2 + a2b1) 5; p p p p f(a1 + b1 5)f(a2 + b2 5) = (a1 − b1 5)(a2 − b2 5) p = a1a2 + 5b1b2 − (a1b2 + a2b1) 5: Thus f is an isomorphism. What is f ◦ f? It is iR. p Is the function g : R ! Q defined by g(a + b 5) = a an isomorphism? No, it is not injective, nor does it satisfy the multiplicative part of the third condition of an isomorphism. p How about the function h : Q ! R defined by h(a) = a+0 5? It injective, not surjective, but does satisfy the third condition of an isomorphism. Definition. For rings R and S, a function f : R ! S is called a homomorphism if f(a + b) = f(a) + f(b); f(ab) = f(a)f(b); for all a; b 2 R. Examples. (a) For any rings R and S, the zero map z : R ! S defined by z(r) = 0S is a homomorphism trivially. When both R and S contains nonzero elements, the zero map is neither injective nor surjective. (b) For the rings R = Z12 and S = Z4, the map f : R ! S defined by f([a]12) = [a]4 is well-defined because 4 j 12, is surjective because [0]4; [1]4; [2]4; [3]4 are the images of [0]12; [1]12; [2]12; [3]12, is not injective because f([7]12) = [3]4 = f([3]12), and is a homo- morphism because for all [a]12; [b]12 2 R we have f([a]12 + [b]12) = f([a + b]12) = [a + b]4; f([a]12) + f([b]12) = [a]4 + [b]4 = [a + b]4; f([a]12[b]12) = f([ab]12) = [ab]4; f([a]12)f([b]12) = [a]4[b]4 = [ab]4:.