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Unformatted text preview: Problem Set #2 Solutions Chapter III. 2.2. 1. If F E K and E is stable, then G E / G = G ( K : F ) . Proof. Let G E , G , and a E . Since E is stable, ( a ) E , and so ( a ) = ( a ). Thus  1 ( a ) =  1 ( a ) = a , so  1 G E . 2. If H / G , then F H is stable. Proof. Let G = G ( K : F ) and H . Since H is a normal subgroup of G we have  1 H , and so  1 ( a ) = a whenever a F H . But then ( a ) = ( a ), so ( a ) F H (since H was abitrary), and hence F H is stable (since a F H was abitrary). 3.1. Suppose F is a field of characteristic p > . 1. ( a + b ) p = a p + b p and ( a b ) p = a p b p for all a,b F . Proof. We have ( a b ) p = p X i =0 p i a i ( b ) p i . Furthermore, p  p ! i !( p i )! = p i whenever i 6 = 0 ,p, since p is prime and 0 < i,p i < p (that is, p divides the numerator but not the denominator). But char( F ) = p so p 1 F = 0, and hence all the middle terms of ( a b ) p vanish. If p is odd, ( 1) p = 1; otherwise p = 2, in which case ( 1) p = ( 1) 2 = 1 = 1. Thus either way ( a b ) p = a p + ( b ) p = a p b p . 2. Let K = F ( t ) for an indeterminate t and let f ( x ) = x p t K [ x ] . Then f ( x ) is irreducible with just one root with multiplicity p in any splitting field....
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This note was uploaded on 08/06/2008 for the course MATH 220 taught by Professor Morrison during the Spring '08 term at UCSB.
 Spring '08
 MORRISON
 Algebra

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