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;+,*] 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 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-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 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  : 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 : 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.

6.6.4 Subring, Ideal and Quotient ring 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 :

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

Example: Let [R;+,·] be a ring 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

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.

Example: [R;+,. ] is a commutative ring with identity element 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;+,*].

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=minaD{|a| | a0,where a D}.

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

Definition 32: Under the conditions of Theorem 6 Definition 32: Under the conditions of Theorem 6.35, [R/S;  , ] is a ring which is called a quotient ring. Example:Let Z(i)={a+bi|a,bZ}, E(i)={2a+2bi|a,bZ}. [E(i);+,*] is an ideal of ring [Z(i);+,*]. Is Quotient ring [Z(i)/E(i);  , ] a field? zero-divistor!

Example: Let R be a commutative ring, and H be an ideal of R Example: Let R be a commutative ring, and H be an ideal of R. Prove that quotient ring R/H is an integral domain  For any a,bR, if abH, then aH or bH. Proof: (1)If quotient ring R/H is an integral domain, then aH or bH when abH for any a,bR. (2)R is a commutative ring, and H be an ideal of R. If aH or bH when abH for any a,bR, then quotient ring R/H is an integral domain.

Definition 33: Let  be a ring homomorphism from ring [R;+, Definition 33: Let  be a ring homomorphism from ring [R;+,*] to ring [S;+’,*’]. The kernel of  is the set ker={xR|(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 subring A of R. Show that f(A) is a subring of S. 3. Let f: R→S be a ring homomorphism, with an ideal A of R. Does it follow that f(A) is an ideal of S? 4.Prove Theorem 6.36 5.Prove Theorem 6.37 6. Let f : R→S be a ring homomorphism, and S be an ideal of f (R). Prove: (1)f -1(S) an ideal of R, where f -1(S)={xR|f (x)S} (2)R/f -1(S) f (R)/S