MATH535_HW9 - Serge Ballif MATH 535 Homework 9 November 2...

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Serge Ballif MATH 535 Homework 9 November 2, 2007 1. Show that if V is a vector space with minimal polynomial P ( X ) and v V has order P ( X ) , then there is a subspace U of V such that V = [ X ] v U , and the minimal polynomial of U divides P ( X ) . Hints: a) Let W be a subspace of V which is stable under the action of L , i.e. L ( W ) W . Show that if w / W , then it has order R ( X ) with positive degree in V/W and R ( X ) divides the minimal polynomial M ( X ) of L . b) Apply a) to W = [ X ] v + U , where U is maximal L -stable such that U [ X ] = 0 . If w / W , write R ( X ) w = S ( X ) v + u , u U , and conclude from multiplication by M ( X ) /R ( X ) that R ( X ) | S ( X ) . c) Show that the image in V/W of the element w 0 := w - S ( X ) R ( X ) v also has order R ( X ) and moreover R ( X ) w 0 U . d) Now we want to show that [ X ] v ( U + [ X ] w 0 ) = 0 . For that let A ( X ) w 0 + u 0 = B ( X ) v for some A ( X ) , B ( X ) F [ X ] . Show that R ( X ) | A ( X ) and from c), conclude that A ( X ) w 0 + u 0 = B ( X ) v = 0 . e) Finally deduce that V = [ X ] U , where both [ X ] v and U are L -stable. (a) and (b) were completed with the previous assignment. c) R ( x ) w 0 + W = R ( x ) w - S ( x ) v + W = 0. The element R ( x ) w - S ( x ) v is in W , because R ( x ) is the order of w in V/W and v is in W to start with. Therefore, R ( x ) w 0 = u is in W and in U . d) A ( X ) w 0 + u 0 = B ( X ) v , so A ( x ) w 0 W = [ x ] v + U . Therefore R ( x ) | A ( x ), since R ( x ) is the order of w 0 in V/W . Since U has trivial intersection with [ x ] v we know that A ( X ) w 0 + u 0 = B ( X ) v = 0. Therefore, [ x ] v ( U + [ x ] w 0 ) = 0.
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