复旦大学：《离散数学》PPT教学课件（赵一鸣）27/28

Theorem 6.29: Any Field is an integral domain ◆[Z;+,×] is an integral domain. But it is not a field

 Theorem 6.29: Any Field is an integral domain  [Z;+,] is an integral domain. But it is not a field

Theorem 6.30: A finite integral domain is a field integral domain: commutative, no zero-divisor Field: commutative, identity, inverse identity, inverse Let r; t, be a finite integral domain. ◆(1) Need to find e∈ R such that e*a=afor ala∈R ◆(2) For each a∈R-{0}, need to find an element ber such that ab=e Proof: (l)Let r=(,.. an3 ◆Forc∈R,C≠0, consider the set rc={a1tc, a2xC,…,ane∈R

 Theorem 6.30: A finite integral domain is a field. integral domain :commutative, no zero-divisor Field: commutative, identity, inverse identity, inverse  Let [R;+,*] be a finite integral domain.  (1)Need to find eR such that e*a =a for all a R.  (2)For each aR-{0}, need to find an element bR such that a*b =e.  Proof:(1)Let R={a1 ,a2 ,an }.  For cR, c 0, consider the set Rc={a1*c, a2*c, ,an*c}R

Example: Zm+, is a field iff m is a prime number a]=? If GCD(a, n)=1, then there exist k and S, S.t. ak+ns=1, where k,s∈Z ◆ns=1-ak ◆[1l=ak=a]|k ◆|k=|a ◆ Euclidean algorithn

 Example: [Zm;+,*] is a field iff m is a prime number  [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: 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, 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

O 6.6.4 Subring, Ideal and Quotient ring Subring Definition 29: A subring of a ring r is a s nonempty subset s of r which is also a ring under the same operations Example: [Q√2计为 a subring of[R;+×], where ris real number set

6.6.4 Subring, Ideal and Quotient ring 1. Subring Definition 29: A subring of a ring R is a nonempty subset S of R which is also a ring under the same operations. Example : [ Q 2;+,]is a subring of [ R;+,] , where R i s real number set

Theorem 6.34: A subset s of a ring r is a subring if and only if for a, bEs: Da+bEs (2)-a∈S 3a bES

 Theorem 6.34: A subset S of a ring R is a subring if and only if for a, bS：  (1)a+bS  (2)-aS  (3)a·bS

Example: Let R; +, be a ring. Then C={xx∈R, and ax= xa for all a∈R}isa subring of r Proof: Forex,y∈C,x+y,-X∈?C,xy?∈Ci,e VaER, a (x+y=?(x+y) a, a (-x=? KX)'a a'(x'y)=?(x'y).a

Example: Let [R;+,·] be a ring. Then C={x|xR, and a·x=x·a for all aR} is a subring of R. Proof: For  x,yC, x+y,-x?C, x·y?C i.e.  aR,a·(x+y)=?(x+y)·a,a·(-x)=?(- x)·a,a·(x·y) =?(x·y)·a