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 Example: [Z m ;+,*] 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.

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Presentation on theme: " Example: [Z m ;+,*] 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."— Presentation transcript:

1  Example: [Z m ;+,*] 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

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

3 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 e. Then following results hold. (1)For  a  R, pa=0. And if R is an integral domain, then p is the smallest positive 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: 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  (a)  (0 R ) =0 S and  (-a) = -  (a) for all a  R.  where 0 R and 0 S denote the zero elements in R and S;

5 If  : R→S is a ring homomorphism,  (1 R ) = 1 S ? No Theorem 6.33: Let R be an integral domain, and char(R)=p. The function  :R  R is given by  (a)=a p for all a  R. Then  is a homomorphism from R to R, and it is also one-to-one.

6 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 :

7  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

8 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

9  2.Ideal( 理想 )  Definition 30:. Let [R; +, * ] be a ring. A subring S of R is called an ideal of R if rs  S and sr  S for any r  R and s  S.  To show that S is an ideal of R it is sufficient to check that  (a) [S; +] is a subgroup of [R; + ];  (b) if r  R and s  S, then rs  S and sr  S.

10  Example: [R;+,*] is a commutative ring with identity element. For a  R , (a)={a*r|r  R},then [(a);+,*] is an ideal of [R;+,*].  If [R;+,*] is a commutative ring, For  a  R, (a)={a*r+na|r  R,n  Z }, then [(a);+,*] is an ideal of [R;+,*].

11  Principle ideas  Definition 31: If R is a commutative ring and a  R, then (a) ={a*r+na|r  R} is the principle ideal defined generated by a.  Example: Every ideal in [Z;+,*] is a principle.  Proof: Let D be an ideal of Z.  If D={0}, then it holds.  Suppose that D  {0}.  Let b=min a  D {|a| | a  0,where a  D}.

12 3. Quotient ring Theorem 6.35: Let [R; +,*] be a ring and let S be an ideal of R. If R/S ={S+a|a  R} and the operations  and  on the cosets are defined by (S+a)  (S+b)=S+(a + b) ; (S+a)  (S+b) =S+(a*b); then [R/S; ,  ] is a ring. Proof: Because [S;+] is a normal subgroup of [R;+], [R/S;  ] is a group. Because [R;+] is a commutative group, [R/S;  ] is also a commutative group. Need prove [R/S;  ] is an algebraic system, a sumigroup, distributive laws

13  Definition 32: Under the conditions of Theorem 6.35, [R/S; ,  ] is a ring which is called a quotient ring.

14  Definition 33: Let  be a ring homomorphism from ring [R;+,*] to ring [S;+’,*’]. The kernel of  is the set ker  ={x  R|  (x)=0 S }.  Theorem 6.36: Let  be a ring homomorphism from ring [R;+,*] to ring [S;+’,*’]. Then  (1)[  (R);+’,*’] is a subring of [S;+’,*’]  (2)[ker  ;+,*] is an ideal of [R;+,*].

15  Theorem 6.37(fundamental theorem of homomorphism for rings): Let  be a ring homomorphism from ring [R;+,*] to ring [S;+’,*’]. Then  [R/ker  ; ,  ]  [  (R);+’,*’]

16  Exercise:  1. Determine whether the function : Z→Z given by f(n) =2n is a ring homomorphism.  2. Let f : R→S be a ring homomorphism, with A a subring of R. Show that f(A) is a subring of S.  3. Let f: R→S be a ring homomorphism, with A an ideal of R. Does it follow that f(A) is an ideal of S?  4.Prove Theorem 6.36


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