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Unformatted text preview: 1 Math 417 Exam III Solutions: August 5, 2009 1 . Let k be a field and let f ( x ) k [ x ] be a polynomial of degree n . Prove that f ( x ) has at most n roots in k . This is Theorem 3.50 in the book. 2 . Prove that F = { a + b 2 : a, b Q } is a field. It suffices to prove that F is a subring of R (for subrings are rings in their own right) and that every nonzero element has an inverse in F . Now F is a subring if 1 F and F is closed under subtraction and multipli cation. First, 1 = 1+0 i F . Second, ( a + b 2) ( c + d 2) = ( a c )+( b d ) 2, which is in F because a c, b d Q . Third, ( a + b 2)( c + d 2) = ( ac + 2 bd ) + ( ad + bc ) 2 F. We conclude that F is a commutative ring. Finally, assume that a + b 2 6 = 0. The equation ( a + b 2)( a b 2) = a 2 2 b 2 gives ( a + b 2) 1 = a a 2 2 b 2 + b a 2 2 b 2 2 ....
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This note was uploaded on 11/15/2010 for the course MATH 417 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff
 Math, Algebra

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