Let : G1→G2 be a surjection homomorphism between two groups, N be a normal subgroup of G1, and KerN. Proof: G2/(N) is a group.

Theorem 6.22: Let [H;] be a normal subgroup of the group [G;]. Then [G/H;] is a group.

(N) is a subgroup of G2. Normal subgroup? g2

-1(n)g2?(N) For g2G2, (n)(N), g1G1 s.t

(g1)=g2 ? surjection homomorphism.

Let : G1→G2 be a surjection homomorphism between two groups, N be a normal subgroup of G1, and Ker N. Proof: G1/N G≌ 2/(N).

Corollary 6.2: If is a homomorphism function from group [G;*] to group [G';], and it is onto, then [G/K;][G';]

Prove:f :G1→G2/(N) For g1G1, f (g1)=(N)(g1) 1)f is a surjection homomorphism from G1 to

G2/ (N). 2)Kerf =N

6.6 Rings and fields 6.6.1 Rings Definition 21: A ring is an Abelian group [R, +]

with an additional associative binary operation (denoted ·) such that for all a, b, cR,

(1) a · (b + c) = a · b + a · c, (2) (b + c) · a = b · a + c · a. We write 0R for the identity element of the group

[R, +]. For a R, we write -a for the additive inverse of a. Remark: Observe that the addition operation is

always commutative while the multiplication need not be.

Observe that there need not be inverses for multiplication.

Example: The sets Z,Q, with the usual operations of multiplication and addition form rings,

[Z;+,],[Q;+,] are rings Let M={(aij)nn|aij is real number}, Then

[M;+,]is a ring Example: S,[P(S);,∩]， Commutative ring

Definition 23: A ring R is a commutative ring if ab = ba for all a, bR . A ring R is an unitary ring if there is 1R such that 1a = a1 = a for all aR. Such an element is called a multiplicative identity.

Example: If R is a ring, then R[x] denotes the set of polynomials with coefficients in R. We shall not give a formal definition of this set, but it can be thought of as: R[x] = {a0 + a1x + a2x2 + …+ anxn|nZ+, aiR}.

Multiplication and addition are defined in the usual manner; if

m

i

ii

n

i

ii xbxgandxaxf

00

)()( then

},max{

0

)()()(mn

i

iii xbaxgxf

mn

k

kji

kji

xbaxgxf0

)()()(

One then has to check that these operations define a ring. The ring is called polynomial ring.

Theorem 6.26: Let R be a commutative ring. Then for all a,bR,

where nZ+.

n

i

inin bainCba0

),()(

1. Identity of ring and zero of ring Theorem 6.27: Let [R;+,*] be a ring. Then

the following results hold. (1)a*0=0*a=0 for aR (2)a*(-b)=(-a)*b=-(a*b) for a,bR (3)(-a)*(-b)=a*b for a,bR Let 1 be identity about *. Then (4)(-1)*a=-a for aR (5)(-1)*(-1)=1

1:Identity of ring 0:zero of ring

[M2,2(Z);+,] is an unitary ring

Zero of ring (0)22,

Identity of ring is

},,,|{)(2,2 Zdcbadc

baZM

1

1

22000

01

22010

00

22010

00

00

01

ring ofdivisor -zero

2. Zero-divistorsDefinition 23: If a0 is an element of a ring R for which there exists b0 such that ab=0(ba=0), then a(b) is called a left(right) zero-divistor in R.Let S={1,2}， is zero element of ring [P(S);,∩]

6.6.2 Integral domains, division rings and fields

Definition 24: A commutative ring is an integral domain if there are no zero-divisors.

[P(S);,∩] and [M;+,] are not integral domain, [Z;+,] is an integral domain

Theorem 6.28: Let R be a commutative ring. Then R is an integral domain iff. for any a, b, cR if a0 and ab=ac, then b=c.

Proof: 1)Suppose that R is an integral domain. If ab = ac, then ab - ac=0

2)R is a commutative ring, and for any a, b, cR if a0 and ab=ac, then b=c. Prove: R is an integral domain

Let [R;+;*] be a ring with identity element 1.

If 1=0, then for aR, a=a*1=a*0=0. Hence R has only one element, In other

words, If |R|>1, then 10.

Definition 25: A ring is a division ring if the non-zero elements form a group under multiplication.

If R is a division ring, then |R|2.

Ring R has identity, and any non-zero element exists inverse element under multiplication.

Definition 26: A field is a commutative division ring.

[Z;+,]is a integral domain, but it is not division ring and field

[Q;+,], [R;+,]and[C;+,] are field

Let [F;+,*] be a algebraic system, and |F| 2,

(1)[F;+]is a Abelian group (2)[F-{0};*] is a Abelian group (3)For a,b,cF, a*(b+c)=(a*b)+(a*c)

Theorem 6.29: Any Field is an integral domain

Let [F;+,*] be a field. Then F is a commutative ring.

If a,bF-{0}, s.t. a*b =0 。 [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 eR such that e*a =a for

all a R. (2)For each aR-{0}, need to find an

element bR such that a*b =e. Proof:(1)Let R={a1,a2,an}. For cR, c 0, consider the set

Rc={a1*c, a2*c, ,an*c}R.

Exercise:P367 7,8 1.Let X be any non-empty set. Show that [P(X); , ∩] is not a ring.∪ 2. Let Z[i] = {a + bi| a, bZ}. (1)Show that Z[i] is a commutative ring

and find its identity. (2)Is Z[i] a field? Why? 3.Show that Q[i] = {a + bi | a, bQ} is a

field.