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HW11Soln - MATH 310 HOMEWORK 11 SOLUTIONS 1 If S is a...

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MATH 310, HOMEWORK 11 SOLUTIONS 1. If S is a commutative ring with identity, let S 2 = { ( a, b ) | a, b S } with the following “addition” and “multiplication”: ( a, b ) + ( c, d ) = ( a + c, b + d ) ( a, b )( c, d ) = ( ac - bd, ad + bc ) where a, b, c, d S . Recall that S 2 is a commutative ring with identity. (a) Prove that Z 2 3 is a field with 9 elements. (b) If p is a prime and p = 1 in Z 4 , prove that Z 2 p is not a field. Hint: First prove that it is not an integral domain. Hint for the Hint: Use Homework 9, 3(b). Proof. (a) We know that Z 2 p is a commutative ring with identity (and we know that the identity is (1 , 0)), so we only need to check that every non-zero element is invertible: (1 , 0)(1 , 0) = (1 , 0) (2 , 0)(2 , 0) = (1 , 0) (0 , 1)(0 , 2) = (1 , 0) (1 , 1)(2 , 1) = (1 , 0) (2 , 2)(1 , 2) = (1 , 0) Proof. (b) From Homework 9, 3(b), we know that: p - 1 2 ! 2 = ( - 1) p +1 2 for any odd prime p . If p = 1 in Z 4 , then p = 4 k +1 for some integer k , so p +1 2 = 2 k +1, and the right hand side becomes: ( - 1) p +1 2 = ( - 1) 2 k +1 = - 1 In other words, a = p - 1 2 ! Z p satisfies a 2 = - 1. Therefore: (1 , a )(1 , - a ) = (1 + a 2 , a - a ) = (0 , 0) So Z 2 p is not an integeral domain. Since every field is an integral domain, Z 2 p is not a field if p = 1 in Z 4 .
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2. Consider the ring Z 2 [ x ] of polynomials in x with coefficients in Z 2 .
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