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Ring Theory & Definitions in Algebra - B.Sc. III Year

Presented by Prof. Anil D. Patil of Mathematics Department at Yashwantrao Chavan Mahavidyalaya, Tuljapur, Osmanabad. This class covers the definition and properties of associative rings, unity in rings, and relevant lemmas and proofs in algebra.

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Ring Theory & Definitions in Algebra - B.Sc. III Year

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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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