Theorem 6.30: A finite integral domain is a field. Example: [Zm;+,*] is a field iff m is a prime number Iintegral domain? [a]-1=? If GCD(a,n)=1,then there exist k and s, s.t. ak+ns=1, where k, sZ. ns=1-ak. [1]=[ak]=[a][k] [k]= [a]-1 Euclidean algorithm

Theorem 6.31(Fermat’s Little Theorem): if p is prime number, and GCD(a,p)=1, then ap-11 mod p Corollary 6.3: If p is prime number, aZ, then apa mod p

Definition 27: The characteristic of a ring R with 1 is the smallest nonzero number n such that 0 =1 + 1 + · · · + 1 (n times) if such an n exists; otherwise the characteristic is defined to be 0. We denoted by char(R). Theorem 6.32: Let p be the characteristic of a ring R with 1(e). Then following results hold. (1)For aR, pa=0. And if R is an integral domain, then p is the smallest nonzero number such that 0=la, where a0. (2)If R is an integral domain, then the characteristic is either 0 or a prime number.

6.6.3 Ring homomorphism Definition 28: An everywhere function  : R→S between two rings is a homomorphism if for all a, bR, (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 aR. where 0R and 0S denote the zero elements in R and S;

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 aR. Then  is a homomorphism from R to R, and it is also one-to-one.

Exercise:P367 16,17,20 1. Determine whether the function : Z→Z given by f(n) =2n is a ring homomorphism.