Example: [Zm; ,*] is a field iff m is a prime number

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• Example Zm, is a field iff m is a prime
  number
• a-1?
• If GCD(a,n)1,then there exist k and s, s.t.
  akns1, where k, s?Z.
• ns1-ak.
• 1akak
• k a-1
• Euclidean algorithm

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• Theorem 6.31(Fermats Little Theorem) if p is
  prime number, and GCD(a,p)1, then ap-1?1 mod p
• Corollary 6.3 If p is prime number, a?Z, then
  ap?a mod p

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• Definition 27 The characteristic of a ring R
  with 1 is the smallest nonzero number n such that
  0 1 1 1 (n times) if such an n
  exists otherwise the characteristic is defined
  to be 0. We denoted by char(R).
• Theorem 6.32 Let p be the characteristic of a
  ring R with e. Then following results hold.
• (1)For ?a?R, pa0. And if R is an integral
  domain, then p is the smallest positive number
  such that 0la, where a?0.
• (2)If R is an integral domain, then the
  characteristic is either 0 or a prime number.

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6.6.3 Ring homomorphism

• Definition 28 A function ? R?S between two
  rings is a homomorphism if for all a, b?R,
• (1) ? (a b) ? (a) ? (b),
• (2) ? (ab) ? (a) ? (b)
• An isomorphism is a bijective homomorphism. Two
  rings are isomorphic if there is an isomorphism
  between them.
• If ? R?S is a ring homomorphism, then formula
  (1) implies
• that ? is a group homomorphism between the groups
  R and S .
• Hence it follows that
• (a) ? (0R) 0S and ? (-a) -? (a) for all a?R.
• where 0R and 0S denote the zero elements in R and
  S

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If ? R?S is a ring homomorphism, ? (1R)
1S? No Theorem 6.33 Let R be an integral domain,
and char(R)p. The function ?R?R is given by
?(a)ap for all a?R. Then ? is a homomorphism
from R to R, and it is also one-to-one.
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6.6.4 Subring, Ideal and Quotient ring 1.
Subring Definition 29 A subring of a ring R is a
nonempty subset S of R which is also a ring under
the same operations. Example
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• Theorem 6.34 A subset S of a ring R is a subring
  if and only if for a, b?S
• (1)ab?S
• (2)-a?S
• (3)ab?S

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Example Let R, be a ring. Then Cxx?R,
and axxa for all a?R is a subring of
R. Proof For ? x,y?C, xy,-x??C, xy??C i.e. ?
a?R,a(xy)?(xy)a,a(-x)?(-x)a,a(xy)
?(xy)a
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• 2.Ideal(??)
• Definition 30. Let R , be a ring. A
  subring S of R is called an ideal of R if rs ?S
  and sr?S for any r?R and s?S.
• To show that S is an ideal of R it is sufficient
  to check that
• (a) S is a subgroup of R
• (b) if r?R and s?S, then rs?S and sr?S.

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• Example R, is a commutative ring with
  identity element. For a?R,(a)arr?R,then
  (a), is an ideal of R,.
• If R, is a commutative ring, For ?a ?R,
  (a)arnar?R,n?Z, then (a), is an ideal
  of R,.

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• Principle ideas
• Definition 31 If R is a commutative ring and
  a?R, then (a) arnar?R is the principle
  ideal defined generated by a.
• Example Every ideal in Z, is a principle.
• Proof Let D be an ideal of Z.
• If D0, then it holds.
• Suppose that D?0.
• Let bmina?Da a?0,where a ?D.

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3. Quotient ring Theorem 6.35 Let R , be a
ring and let S be an ideal of R. If R/S
Saa?R and the operations ? and ? on the
cosets are defined by (Sa)?(Sb)S(a b)
(Sa)?(Sb) S(ab) then R/S ? , ? is a
ring. Proof Because S is a normal subgroup
of R, R/S? is a group. Because R is a
commutative group, R/S? is also a commutative
group. Need prove R/S? is an algebraic system,
a sumigroup, distributive laws
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• Definition 32 Under the conditions of Theorem
  6.35, R/S ? , ? is a ring which is called a
  quotient ring.

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• Definition 33 Let ? be a ring homomorphism from
  ring R, to ring S,. The kernel of ?
  is the set ker?x?R?(x)0S.
• Theorem 6.36 Let ? be a ring homomorphism from
  ring R, to ring S,. Then
• (1)?(R), is a subring of S,
• (2)ker?, is an ideal of R,.

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• Theorem 6.37(fundamental theorem of homomorphism
  for rings) Let ? be a ring homomorphism from
  ring R, to ring S,. Then
• R/ker??,?? ?(R),

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• Exercise
• 1. Determine whether the function Z?Z given by
  f(n) 2n is a ring homomorphism.
• 2. Let f R?S be a ring homomorphism, with A a
  subring of R. Show that f(A) is a subring of S.
• 3. Let f R?S be a ring homomorphism, with A an
  ideal of R. Does it follow that f(A) is an ideal
  of S?
• 4.Prove Theorem 6.36