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, sZ. 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-11 mod p

Corollary 6.3: If p is prime number, aZ, then apa 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 e. Then following results hold.(1)For aR, pa=0. And if R is an integral domain, then p is the smallest positive number such that 0=la, where a0. (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 : R→S between two rings is a

homomorphism if for all a, bR, (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) (0R) =0S and (-a) = - (a) for all aR. where 0R and 0S denote the zero elements in R and S;

If : R→S is a ring homomorphism, (1R) = 1S?NoTheorem 6.33: Let R be an integral domain, and char(R)=p. The function :RR is given by (a)=ap for all aR. Then is a homomorphism from R to R, and it is also one-to-one.

6.6.4 Subring, Ideal and Quotient ring1. SubringDefinition 29: A subring of a ring R is a nonempty subset S of R which is also a ring under the same operations.Example :

set number real is R where ],;[R];2[Q ,, ng ofis a subri

Theorem 6.34: A subset S of a ring R is a subring if and only if for a, bS：

(1)a+bS (2)-aS (3)a·bS

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

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 srS for any rR and sS.

To show that S is an ideal of R it is sufficient to check that

(a) [S; +] is a subgroup of [R; + ]; (b) if rR and sS, then rsS and srS.

Example: [R;+,*] is a commutative ring with identity element. For aR ，(a)={a*r|rR},then [(a);+,*] is an ideal of [R;+,*].

If [R;+,*] is a commutative ring, For a R, (a)={a*r+na|rR,nZ}, then [(a);+,*] is an ideal of [R;+,*].

Principle ideas Definition 31: If R is a commutative ring and

aR, then (a) ={a*r+na|rR} 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=minaD{|a| | a0,where a D}.

3. Quotient ringTheorem 6.35: Let [R; + ,*] be a ring and let S be an ideal of R. If R/S ={S+a|aR} 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

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

Definition 33: Let be a ring homomorphism from ring [R;+,*] to ring [S;+’,*’]. The kernel of is the set ker={xR|(x)=0S}.

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;+,*].

Theorem 6.37(fundamental theorem of homomorphism for rings): Let be a ring homomorphism from ring [R;+,*] to ring [S;+’,*’]. Then

[R/ker;,] [(R);+’,*’]

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