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Unformatted text preview: Lenny T. Evans MATH 534 Homework 4 March 4, 2010 Lemmas Lemma 1: In a general ring R , for all x ∈ R it is true that ( e ) x = x ( e ) = x. Proof: By distributivity, ( e ) x + x = ( e + e ) x = 0 , so ( e ) x is the additive inverse of x, usually denoted x, and x ( e ) + x = x ( e + e ) = 0 , so for the same reason as before, x ( e ) = x. Thus, ( e ) x = x ( e ) = x. Lemma 2: In a general ring R , ( e ) 2 = e. Proof: Since e + ( e ) = 0 , multiplying both sides by ( e ) , ( e )( e ) + ( e ) 2 = ( e ) + ( e ) 2 = 0 . Which implies that ( e ) 2 is the additive inverse of e, which is e. 3.1.1 If p is a prime number, and R is the subset of all rational numbers m/n such that n 6 = 0 and n is not divisible by p, then R is a ring. Proof: We must show that R satisfies group properties: • R must be an Abelian group under addition R must satisfy group properties: – R must contain the additive identity Look at 0 /n = 0 ∈ R. Then, for any r ∈ R, 0+ r = r +0 = r, so by definition R has an additive identity element. – R must be closed under its binary operation Look at m 1 /n 1 ,m 2 /n 2 ∈ R, then m 1 /n 1 + m 2 /n 2 = ( m 1 n 2 + m 2 n 1 ) / ( n 1 n 2 ) . Since p 6  n 1 and p 6  n 2 , p 6  n 1 n 2 , (since p is a prime, neither prime decomposition of n 1 and n 2 contains p so their product cannot either) and since the integers are a ring, and rings are closed under multiplication and addition the numerator is in Z , so the result is a rational number whose demoninator is not dividsible by p , so R is closed under the additive binary operation. – Every element in R must have an additive inverse For any m/n ∈ R, look at m/n ∈ R. Then, m/n + ( m/n ) = ( m m ) /n = 0 and m/n + m/n = ( m + m ) /n = 0 , so by definition every element has an additive inverse. – R must be associative Look at m 1 /n 1 ,m 2 /n 2 ,m 3 /n 3 ∈ R. Then, ( m 1 /n 1 + m 2 /n 2 ) + m 3 /n 3 = ( m 1 n 2 + m 2 n 1 ) /n 1 n 2 + m 3 /n 3 = ( m 1 n 2 n 3 + m 2 n 1 n 3 + m 3 n 1 n 2 ) /n 1 n 2 n 3 . And m 1 /n 1 +( m 2 /n 2 + m 3 /n 3 ) = m 1 /n 1 + ( m 2 n 3 + m 3 n 2 ) /n 2 n 3 = ( m 1 n 2 n 3 + m 2 n 1 n 3 + m 3 n 1 n 2 ) /n 1 n 2 n 3 . So we see that ( m 1 /n 1 + m 2 /n 2 ) + m 3 /n 3 = m 1 /n 1 + ( m 2 /n 2 + m 3 /n 3 ) so R is associative in addition. – All elements in R must commute Look at m 1 /n 1 ,m 2 /n 2 ∈ R. Then, m 1 /n 1 + m 2 /n 2 = ( m 1 n 2 + m 2 n 1 ) /n 1 n 2 , and m 2 /n 2 + m 1 /n 1 = ( m 2 n 1 + m 1 n 2 ) /n 2 n 1 = ( m 1 n 2 + m 2 n 1 ) /n 1 n 2 . So, m 1 /n 1 + m 2 /n 2 = m 2 /n 2 + m 1 /n 1 , so R is commutative under addition. • R must contain the multiplicative identity Look at 1 / 1 = 1 , which is in R since 1 is not divisible by anything. Then, for m/n ∈ R, 1( m/n ) = ( m/n )1 = m/n, so by definition we see there is a multiplicative identity....
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 Spring '11
 KUMAR
 Math, Algebra, Addition, additive inverse, Abelian

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