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Presented by Prof . Anil D. Patil

Presented by Prof . Anil D. Patil. Department of Mathematics Yashwantrao Chavan Mahavidyalaya , Tuljapur Osmanabad. Ring Theory. Class: - B. Sc. III Year Algebra. Definition of Ring.

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Presented by Prof . Anil D. Patil

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  1. Presented byProf. Anil D. Patil Department of MathematicsYashwantraoChavanMahavidyalaya,Tuljapur Osmanabad

  2. Ring Theory Class: - B. Sc. III Year Algebra

  3. Definition of Ring A non empty set R is said to be an associative ring if in R there are defined two binary operations denoted by + and . respectivly, such that for all a, b, c Є R: 1) a+bЄ R (Closure Property with respect to) 2) a+b=b+a (Commutative Law with respect to +) 3) a+(b+c)=(a+b)+c (Associative Law with respect to +) 4) There is an element 0 in R such that a+0=a=0+a for every a Є R (Existence Identity with respect to +) 5) For every a Є R there is an element (-a) in R such that a+(-a)=0=(-a)+a (Existence of inverse with respect to +) 6) a . b Є R (Closure Property with respect to . ) 7) a . (b . c) = ( a . b) . C (Associative Law with respect to . ) 8) a . (b + c) = a . b + a . c (Left distributive Law) 9) (a + b) . c = a . c + b . c (Right distributive Law)

  4. Definition • A ring R which contains the multiplicative identity (called unity or unit element) is called a ring with unity i.e. 1 Є R such that 1.a = a.1 = a for all a Є R then ring R is called a ring with unity. Remark : A ring R which does not contain multiplicative identity is called ring without unity. Example : If R is the set of integers, positive negative and 0, + is the usual addition and . is usual multiplication of integers then R is a commutative ring with unit element.

  5. LemmaIf R is a ring then for all a, b Є R • a . 0 = 0 . a = 0 • a (- b) = (- a)b = -a . B • (- a)(- b) = a.b If, in addition, has unit element 1, then iv) (- 1)a = -a v) (-1)(-1) = 1.

  6. Proof:-i) If a Є R then a.0 = a(0+0) = a.0+a.0. Therefore 0+a.0 = a.0+a.0 (since a.0 Є R and 0+a.0 = a.0) Now since R is a group under addition therefore applying rigth cancellation law for addition in R we get 0 = a.0 Similarly we have 0.a = (0+0)a+0.a+0.a. Therefore 0+0.a = 0.a+0.a. Now applying right cancellation law we get 0=0.a. Therefore we have a.0 = 0.a = 0 ii) We have a((-b)+b) = a.0 = 0 (Since – b + b = 0) a(-b) + ab + 0 (By left distributive law and property (i)) a(-b) = -ab (Since R is a ring a + b=0) a = -b Similarly we can show that (-a)b = -ab Therefore a(-b) = (-ab).

  7. iii) We have (-a)(-b) = -((-a)b) (since a(-b) = -(ab) = -(-(ab)) (since a(-b) = -(ab)) = ab (since R is a group under addition and –(-a) = a) Therefore (-a)(-b) = ab. iv) Suppose that R has unit element 1, then a+(-1)a = 1a + (-1)a = (1 + (-1)) = 0.a = 0 Therefore (-1)a = -a. v) By property (iv) we have (-1)a = -a. In particular if we take a = -1 then we have (-1)(-1) = -(-1) = 1 (-1)(-1) = 1.

  8. THANK YOU

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