6.6 Rings and fields Rings Definition 21: A ring is an Abelian group [R, +] with an additional associative binary operation(denoted · such that for all a, b, c R, (1) a · (b + c) = a · b + a · c, (2) (b + c) · a = b · a + c · a. We write 0 R 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 a R (2)a*(-b)=(-a)*b=-(a*b) for a,b R (3)(-a)*(-b)=a*b for a,b R (4)(-1)*a=-a for a R (5)(-1)*(-1)=1
1:Identity of ring 0:zero of ring
[M 2,2 (Z);+, ] is an unitary ring Zero of ring (0) 2 2, Identity of ring is
2. Zero-divistors Definition 23: If a 0 is an element of a ring R for which there exists b 0 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); ,∩]
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, c R if a 0 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 a R, a=a*1=a*0=0. Hence R has only one element, In other words, If |R|>1, then 1 0.
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,c F, a*(b+c)=(a*b)+(a*c)
Let. Then [M 22 (Q);+,*] is a division ring. But it is not a field
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 1 R such that 1*a =a for all a R. (2)For each a R-{0}, need to find an element b R such that a*b = 1. Proof:(1)Let R={a 1,a 2, a n }. For c R, c 0, consider the set Rc={a 1 *c, a 2 *c, ,a n *c} R.
Exercise:P367 7,8,16,17,20 1. Let Z[i] = {a + bi| a, b Z}. (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, b Q} is a field.