asf - Math 5125 Monday, December 5 Solutions to Sample...

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Unformatted text preview: Math 5125 Monday, December 5 Solutions to Sample Final Problems 1. (a) Define a ring homomorphism : F [ x ] F by ( f ) = f for f F and ( x ) = 1. Then is onto and ker = { p F [ x ] | p ( 1 ) = } . Clearly x- 1 ker , and by the factor theorem, ker ( x- 1 ) . Therefore ker = ( x- 1 ) and we can now apply the fundamental homomorphism theorem to deduce that F [ x ] / ( x- 1 ) = F as rings. Similarly F [ x ] / ( x + 1 ) = F . We conclude that F [ x ] / ( x- 1 ) = F [ x ] / ( x + 1 ) as rings. (b) Suppose : F [ x ] / ( x- 1 ) F [ x ] / ( x + 1 ) is an R-module isomorphism. Write I = ( x- 1 ) . Then 0 6 = x + 1 + I F [ x ] / I (this is where we use that the characteristic of F is not 2), yet ( x + 1 + I ) = ( x + 1 ) ( 1 + I ) = 0 and we have a contradiction. (c) Set J = ( x- 1 ) (an ideal of F [ x ] ) and define an F [ x 2 ]-map : F [ x 2 ] F [ x ] / J by ( p ) = p + J for p F [ x 2 ] . Clearly is onto and ( x 2- 1 ) ker , because x 2- 1 J . Also p ( x 2 ) ker implies p ( x 2 ) J , hence if we set x = 1 we obtain p ( 1 ) = 0. By the factor theorem, we deduce that p ( x 2- 1 ) and hence ker = ( x 2- 1 ) . The fundamental homomorphism theorem now tells us that F [ x 2 ] / ( x 2- 1 ) = F [ x ] / ( x- 1 ) as F [ x 2 ]-modules. Similarly F [ x 2 ] / ( x 2- 1 ) = F [ x ] / ( x + 1 ) and we conclude that F [ x ] / ( x- 1 ) = F [ x ] / ( x + 1 ) as F [ x 2 ]-modules, so the answer is yes. 2. Note that J ( M / IM ) = ( IM + JM ) / IM . Thus N / JN = M / IM J ( M / IM ) = M / IM ( JM + IM ) / IM , hence by third isomorphism theorem, N / JN = M / ( JM + IM ) (cancel the IM s). Also JN = ( JM + IM ) / IM = JM / ( IM JM ) ....
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This note was uploaded on 01/02/2012 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.

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asf - Math 5125 Monday, December 5 Solutions to Sample...

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