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3.1%20Defn%20and%20Examples%20of%20Rings

# 3.1%20Defn%20and%20Examples%20of%20Rings - Math 302 Modern...

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Math 302 Modern Algebra Spring 2009 Homework Solutions 3.1 DEFINITIONS AND EXAMPLES OF RINGS 2. Let R = { 0 , e, b, c } with addition and multiplication defined by the following tables. It can be shown that R is a ring with this addition and multiplication. + 0 e b c 0 0 e b c e e 0 c b b b c 0 e c c b e 0 · 0 e b c 0 0 0 0 0 e 0 e b c b 0 b b 0 c 0 c 0 c (a) Verify axioms R4 and R5. R4. Existence of Additive Identity. We need to show there exists an element 0 R R such that for all a R , a + 0 R = a. Proof: Choose 0 R = 0. Then 0 R R . We need to show ( a R ) [ a + 0 R = a ] . To show this, we can use a proof by exhaustion. From the table, 0 + 0 = 0 , 0 + e = e = e + 0 , 0 + b = b = b + 0 , 0 + c = c = c + 0 . Therefore a + 0 = a for all a R so 0 is an additive identity for R . Thus R satisfies R4. R5. Existence of Additive Inverses. We need to show for each a R , there exists x R such that a + x = 0 R . Proof: Use a proof by exhaustion. From the addition table, we have 0 + 0 = 0 , e + e = 0 , b + b = 0 , c + c = 0 , which implies that each element is its own additive inverse. Therefore R satisfies R5. (b) Is R commutative? R9. Commutative Law of Multiplication. We need to show for all x, y R , x · y = y · x. Proof: To check this, we again use a proof by exhaustion. We need to check all possible pairs of (distinct) elements from R : 0 · e = 0 = e · 0 0 · b = 0 = b · 0 0 · c = c = c · 0 e · b = b = b · e e · c = b = c · e b · c = 0 = c · b Therefore R satisfies R9 and so is a commutative ring. 23

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3.1 Definitions and Examples of Rings Math 302 Homework Solutions page 24 2. (continued) (c) Is R a field? Since R is a commutative ring, to determine whether or not R is a field, we need to check axioms R10 and R12. R10. Existence of Multiplicative Identity. There exists an element 1 R R such that for all a R , a · 1 R = a = 1 R · a. Proof: This property is satisfied. Choose 1 R = e . Then 1 R R . We need to show ( a R ) [ a · 1 R = a 1 R · a = a ] . To show this, we can use a proof by exhaustion. From the table, 0 · e = 0 = e · 0 , e · e = e, b · e = b = e · b, c · e = c = e · c. Therefore e is a multiplicative identity for R , so R satisfies R10 and R is a ring with identity. R12. Existence of Multiplicative Inverses. For each a R , a 6 = 0 R = 0, there exists x R such that a · x = 1 R = e . Proof: This property is not satisfied. For both b and c , there does not exist an element in R such that b · x = e and c · x = e . So neither of these elements has a multiplicative inverse. Therefore R is not a field. 13. Write out the addition and multiplication tables for (a) Z 2 × Z 3 . This is the Cartesian product of Z 2 and Z 3 , so Z 2 × Z 3 = { ( a, b ) | a Z 2 b Z 3 } = { (0 , 0) , (0 , 1) , (0 , 2) , (1 , 0) , (1 , 1) , (1 , 2) } , with addition and multiplication defined as follows. For each ( a, b ) , ( c, d ) Z 2 × Z 3 , Addition: ( a, b ) + ( c, d ) = ( a + c, b + d ) Multiplication: ( a, b ) · ( c, d ) = ( ac, bd )
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