170 likes | 310 Views
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, c R , (1) a · ( b + c ) = a · b + a · c , (2) ( b + c ) · a = b · a + c · a .
E N D
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 then 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+.
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. Zero-divistors Definition 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
Next: fields, Subring, Ideal and Quotient ring ,Ring homomorphism • Exercise:P381(Sixth) OR P367(Fifth) 7,8, 16,17,20 • 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 units. • (2)Is Z[i] a field? Why? • 3.Show that Q[i] = {a + bi | a, bQ} is a field.