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Assign. 7 Solution (Fall 2008)

# Assign. 7 Solution (Fall 2008) - MATH 235 ALGEBRA I...

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MATH 235 ALGEBRA I SOLUTIONS TO ASSIGNMENT 7 Page 84 Exercise 5 In the following questions it is useful to remember that under our definitions, a ring homomor- phism takes 1 to 1. (a) Prove there is no ring homomorphism Z / 5 Z -→ Z . Assume there exists a ring homomorphism f : Z / 5 Z -→ Z . By the definition, f (1 Z / 5 Z ) = 1 Z But 1 Z / 5 Z has additive order 5 and 1 Z has infinite additive order. Thus 0 Z = f (0 Z / 5 Z ) = f (1 Z / 5 Z + 1 Z / 5 Z + 1 Z / 5 Z + 1 Z / 5 Z + 1 Z / 5 Z ) = f (1 Z / 5 Z ) + f (1 Z / 5 Z ) + f (1 Z / 5 Z ) + f (1 Z / 5 Z ) + f (1 Z / 5 Z ) = 1 Z + 1 Z + 1 Z + 1 Z + 1 Z = 5 Z Which is a contradiction since 0 Z = 5 Z . Thus no such f can exist. (b) Prove that there is no ring homomorphism Z / 5 Z -→ Z / 7 Z . Assume there exists a ring homomorphism f : Z / 5 Z -→ Z / 7 Z . By the definition, f (1 Z / 5 Z ) = 1 Z / 7 Z But 1 Z / 5 Z has additive order 5 and 1 Z / 7 Z has additive order 7. Thus 0 Z / 5 Z = f (0 Z / 5 Z ) = f (1 Z / 5 Z + 1 Z / 5 Z + 1 Z / 5 Z + 1 Z / 5 Z + 1 Z / 5 Z ) = f (1 Z / 5 Z ) + f (1 Z / 5 Z ) + f (1 Z / 5 Z ) + f (1 Z / 5 Z ) + f (1 Z / 5 Z ) = 1 Z / 7 Z + 1 Z / 7 Z + 1 Z / 7 Z + 1 Z / 7 Z + 1 Z / 7 Z = 5 Z / 7 Z Which is a contradiction since 0 Z / 7 Z = 5 Z / 7 Z . Thus no such f can exist. (c) Prove that the rings Z / 2 Z × Z / 2 Z and Z / 4 Z are not isomorphic. Assume there is an isomorphism f : Z / 4 Z -→ Z / 2 Z × Z / 2 Z . We know that f (0 Z / 4 Z ) = 0 Z / 2 Z × Z / 2 Z But I also claim that f (2 Z / 4 Z ) = 0 Z / 2 Z × Z / 2 Z as well which would mean that f is not injective. To show this, we know that f takes 1 to 1, and that every element in Z / 2 Z × Z / 2 Z has additive order 2 (i.e. when you add any element in this ring to itself you get i

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the additive identity 0). Thus we have f (2 Z / 4 Z ) = f (1 Z / 4 Z + 1 Z / 4 Z ) = f (1 Z / 4 Z ) + f (1 Z / 4 Z ) = 1 Z / 2 Z × Z / 2 Z + 1 Z / 2 Z × Z / 2 Z = (1 , 1) + (1 , 1) = (0 , 0) = 0 Z / 2 Z × Z / 2 Z Thus f is not injective and hence cannot be an isomorphism. (d) Is there a ring homomorphism Z / 4 Z -→ Z / 2 Z × Z / 2 Z ? Since Z / 4 Z is generated by 1, and f respects addition, any such homomorphism f is determined by its value at 1. But by our definition, f takes 1 to 1. So that f (1 Z / 4 Z ) = 1 Z / 2 Z × Z / 2 Z . Using this f is determined as follows. I will stop using the cumbersome subscripts, and the ring that each element is in is clear from the context. f (1) = (1 , 1) f (2) = f (1 + 1) = f (1) + f (1) = (1 , 1) + (1 , 1) = (0 , 0) f (3) = f (1 + 2) = f (1) + f (2) = (1 , 1) + (0 , 0) = (1 , 1) f (0) = f (1 + 3) = f (1) + f (3) = (1 , 1) + (1 , 1) = (0 , 0) Each calculation is based on the previous one, and it is in this sense that f is “determined” by its value at 1. It is important to note that we have not yet shown f is a homomorphism of rings for we have not shown that f respects the operation of multiplication defined
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Assign. 7 Solution (Fall 2008) - MATH 235 ALGEBRA I...

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