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

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

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.

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

1 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 (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

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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 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: If R is an integral domain then for any a, b, cR if a0 and ab=ac, then b=c.

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

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)

Let .

Then [M22(Q);+,*] is a division ring. But it is not a field

}0,,,,|{)(2,2

dc

baQdcba

dc

baQM

Theorem 6.29: Any Field is an integral domain

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

If a,b, F-{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 1R such that 1*a =a for

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

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

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

Exercise:P367 7,8,16,17,20 1. Let Z[i] = {a + bi| a, bZ}. (1)Show that Z[i] is a commutative ring

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

field.