6.6.3 Ring homomorphism Definition 28: A function (p: R-s between two rings is a homomorphism if for all a, bEr (1)q(a+b)=q(a)+q(b), (2)q(ab)=g(a)(b) An isomorphism is a bijective homomorphism. Two rings a are isomorphic if there is an isomorphism between them. If o: R-S is a ring homomorphism, then formula(1) implies that op is a group homomorphism between the groups r +]andS;+’ ◆ Hence it follows that ◆q(0)=0 s and p(-a)=-g(a) for all a∈R where OR and os denote the zero elements in R and s;
6.6.3 Ring homomorphism Definition 28: A function : R→S between two rings is a homomorphism if for all a, bR, (1) (a + b) = (a) + (b), (2) (ab) = (a) (b) An isomorphism is a bijective homomorphism. Two rings are isomorphic if there is an isomorphism between them. If : R→S is a ring homomorphism, then formula (1) implies that is a group homomorphism between the groups [R; +] and [S; +’ ]. Hence it follows that (0R) =0S and (-a) = - (a) for all aR. where 0R and 0S denote the zero elements in R and S;
If o: R-S is a ring homomorphism, p (1g)=1s? No Theorem 6.33: Let r be an integral domain, and char(r=p. The function X P: RR is given by ((a)=ap for all aER. Then ( p is a homomorphism from r to R and it is also one-to-one
If : R→S is a ring homomorphism, (1R) = 1S? No Theorem 6.33: Let R be an integral domain, and char(R)=p. The function :R→R is given by (a)=ap for all aR. Then is a homomorphism from R to R, and it is also one-to-one