复旦大学：《离散数学 Discrete Mathematics（上）》英文课件（赵一鸣）28/30

Theorem 6.30: A finite integral domain is a field Example: Zm;+, is a field iff ns a prime number ◆ integral domain? 蒸◆ If GCD(a,n)=1, then there exist k and s, s.t. akins=l where k. sez ◆ns=1-ak ◆[1l=ak=[alk ◆[k]=|a1 ◆ Euclidean algorithm

 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-=l mod p o Corollary 6.3: If p is prime number, aEZ, then ap=a mod p

 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 a ring r with 1(e). Then following results hold. ()For VaER, pa=0. And if r is an integral domain, then p is the smallest nonzero number such that 0=la. where a=0. (2)IfR is an integral domain, then the characteristic is either 0 or a prime number

•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 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)=(a)φ(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: 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 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 aR. Then  is a homomorphism from R to R, and it is also one-to-one

Exercise: P367161720 1. Determine whether the function:z-Z given by f(n)=2n is a ring homomorphism

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