1. 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

2. 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

3. Definition 27: The characteristic of a ring R with 1 is the smallest nonzero number n such that 0 = · · · + 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.

4. 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;

5. 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.

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