1 / 19

6.6 Rings and fields

Let  : G 1 →G 2 be a surjection homomorphism between two groups, N be a normal subgroup of G 1 , and Ker   N. Proof: G 2 /  (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 G 2 . Normal subgroup?

jade-nunez
Download Presentation

6.6 Rings and fields

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Let : G1→G2 be a surjection homomorphism between two groups, N be a normal subgroup of G1, and KerN. 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 g2G2, (n)(N), g1G1 s.t (g1)=g2 ? • surjection homomorphism.

  2. Let : G1→G2 be a surjection homomorphism between two groups, N be a normal subgroup of G1, and KerN. Proof: G1/N≌G2/(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 g1G1, f (g1)=(N)(g1) • 1)f is a surjection homomorphism from G1 to G2/ (N). • 2)Kerf =N

  3. 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. • 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.

  4. Example:The sets Z,Q, with the usual operations of multiplication and addition form rings, • [Z;+,],[Q;+,]are rings • Let M={(aij)nn|aij is real number}, Then [M;+,]is a ring • Example: S,[P(S);,∩]， • Commutative ring

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

  6. 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|nZ+, aiR}. • 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.

  7. Theorem 6.26: Let R be a commutative ring. Then for all a,bR, • where nZ+.

  8. 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 aR • (2)a*(-b)=(-a)*b=-(a*b) for a,bR • (3)(-a)*(-b)=a*b for a,bR • Let 1 be identity about *. Then • (4)(-1)*a=-a for aR • (5)(-1)*(-1)=1

  9. 1:Identity of ring • 0:zero of ring

  10. [M2,2(Z);+,] is an unitary ring • Zero of ring(0)22, • Identity of ring is

  11. 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(b) is called a left(right) zero-divistor in R. Let S={1,2}， is zero element of ring [P(S);,∩]

  12. 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, cR if a0 and ab=ac, then b=c. • Proof:1)Suppose that R is an integral domain. If ab = ac, then ab - ac=0

  13. 2)R is a commutative ring,and for any a, b, cR if a0 and ab=ac, then b=c. Prove: R is an integral domain

  14. 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.

  15. 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

  16. 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)

  17. 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

  18. 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.

  19. 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, bZ}. • (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, bQ} is a field.

More Related